Mostly, a thought experiment:
Cantor claims: "whichever list of infinite (binary) strings you can
come up with, I will (anti)diagonalise it."
It is (AFAIK) easy enough to come up with a production of _all_ the
infinite binary strings. The problem remains how to make the list
bulletproof to diagonalisation. All we can do is falsify the
diagonalisation procedure itself, by following the claim to the
letter:
We start by producing our list, entry by entry; in parallel, our
opponent starts building the antidiagonal (I will abstract from the
technical details, unnecessary here):
List Diagonal Antidiagonal
.(0) 0 1
.(0)1 0 1
... ... ...
At this point, one migth wander: is there? is not there?
=== Cantor's diagonal is a nonhalting machine. ===
Under this perspective, accepting the diagonal argument "simply"
implies a specific choice about infinity (and infinities; and what can
or cannot be proved; and a fundamental paradox: incongruent by
construction).
Now, what about Turing?
LV


0




Reply

julio (505)

5/7/2009 12:04:15 AM 

See related articles to this posting
> We start by producing our list, entry by entry; in parallel, our
> opponent starts building the antidiagonal (I will abstract from the
> technical details, unnecessary here):
>
> =A0 =A0 List =A0 Diagonal Antidiagonal
> =A0 =A0 .(0) =A0 =A0 =A0 =A0 0 =A0 =A0 =A0 =A0 =A0 =A0 1
> =A0 =A0 .(0)1 =A0 =A0 =A0 =A00 =A0 =A0 =A0 =A0 =A0 =A0 1
> =A0 =A0 =A0... =A0 =A0 =A0 =A0 ... =A0 =A0 =A0 =A0 =A0 ...
>
> At this point, one migth wander: is there? is not there?
>
> =A0 =A0 =A0 =A0 =3D=3D=3D Cantor's diagonal is a nonhalting machine. =3D=
=3D=3D
>
> Under this perspective,
And that is not the perspective in which the proof is given.
The diagonal is not a machine, nor Turning machine, nor abstract
machine of any kind, halting or nonhalting.
Rather, the diagonal is a sequence. And it is not defined "entry by
entry" as some list is given "entry by entry".
A denumerable list of denumerable binary strings is a sequence, so a
FUNCTION.
And a each denumerable binary string is a sequence, so a FUNCTION.
And the digaonal is a sequence, so a FUNCTION.
GIVEN ANY denumerable list of denumerable binary strings, which is a
FUNCTION whose values are FUNCTIONS, there exists the diagonal of that
list, and that diagonal is a FUNCTION and it is not one of the values
of the function that is the said list of denumerable binary strings.
That's IT. There are no machines, or "entry by entry" processes, or
other junk involved.
If you change the whole matter to something about "entry by entry"
process or machines or whatever then that is a DIFFERENT matter from
the diagonal proof that there is no list of the entire set of
denumerable binary strings.
MoeBlee


0




Reply

jazzmobe (307)

5/7/2009 12:29:57 AM


On Wed, 6 May 2009, MoeBlee wrote:
>> We start by producing our list, entry by entry; in parallel, our
>> opponent starts building the antidiagonal (I will abstract from the
>> technical details, unnecessary here):
>>
>> At this point, one migth wander: is there? is not there?
>>
>> === Cantor's diagonal is a nonhalting machine. ===
>>
>> Under this perspective,
>
> And that is not the perspective in which the proof is given.
>
> The diagonal is not a machine, nor Turning machine, nor abstract
> machine of any kind, halting or nonhalting.
>
Clearly OP is has been duped into joining the currently popular religous
clut of computerism. The first tenante of computerism is that computers
know it all. Not only has no computer ever successfully passed a SAT
exam, Godel showed that there's somethings that computer will never know.
Anyway those computerism nuts are a NP hard case to get them to even admit
that there's not enough time for computers to know anything but the
simplest.
Indeed, is any computer capable of earning a master or even a graduate,
degree in computer science?
> Rather, the diagonal is a sequence. And it is not defined "entry by
> entry" as some list is given "entry by entry".
>
> A denumerable list of denumerable binary strings is a sequence, so a
> FUNCTION.
>
> And a each denumerable binary string is a sequence, so a FUNCTION.
>
> And the digaonal is a sequence, so a FUNCTION.
>
> GIVEN ANY denumerable list of denumerable binary strings, which is a
> FUNCTION whose values are FUNCTIONS, there exists the diagonal of that
> list, and that diagonal is a FUNCTION and it is not one of the values
> of the function that is the said list of denumerable binary strings.
>
> That's IT. There are no machines, or "entry by entry" processes, or
> other junk involved.
>
> If you change the whole matter to something about "entry by entry"
> process or machines or whatever then that is a DIFFERENT matter from
> the diagonal proof that there is no list of the entire set of
> denumerable binary strings.
>
> MoeBlee
>


0




Reply

marsh6245 (32)

5/7/2009 5:43:06 AM


In article <20090506222934.N59609@agora.rdrop.com>,
William Elliot <marsh@rdrop.remove.com> wrote:
> On Wed, 6 May 2009, MoeBlee wrote:
>
> >> We start by producing our list, entry by entry; in parallel, our
> >> opponent starts building the antidiagonal (I will abstract from the
> >> technical details, unnecessary here):
> >>
> >> At this point, one migth wander: is there? is not there?
> >>
> >> === Cantor's diagonal is a nonhalting machine. ===
> >>
> >> Under this perspective,
> >
> > And that is not the perspective in which the proof is given.
> >
> > The diagonal is not a machine, nor Turning machine, nor abstract
> > machine of any kind, halting or nonhalting.
> >
> Clearly OP is has been duped into joining the currently popular
> religous [cult] of computerism.
It's actually even worse than that. By framing the Cantor diagonal
construction in a computerfriendly way (e.g. for every i and j, the
"enumeration oracle" produces the j'th digit of the i'th listentry in
finite time), one can readily construct a *computable* diagonal which
provably differs from every entry in the enumeration.
So it's less a case of computerworship than selfworship (i.e.
believing that one's naive intuitions MUST be right, even when proven
wrong).
[snip]
> > Rather, the diagonal is a sequence. And it is not defined "entry by
> > entry" as some list is given "entry by entry".
> >
> > A denumerable list of denumerable binary strings is a sequence, so a
> > FUNCTION.
> >
> > And a each denumerable binary string is a sequence, so a FUNCTION.
> >
> > And the digaonal is a sequence, so a FUNCTION.
> >
> > GIVEN ANY denumerable list of denumerable binary strings, which is a
> > FUNCTION whose values are FUNCTIONS, there exists the diagonal of that
> > list, and that diagonal is a FUNCTION and it is not one of the values
> > of the function that is the said list of denumerable binary strings.
> >
> > That's IT. There are no machines, or "entry by entry" processes, or
> > other junk involved.
> >
> > If you change the whole matter to something about "entry by entry"
> > process or machines or whatever then that is a DIFFERENT matter from
> > the diagonal proof that there is no list of the entire set of
> > denumerable binary strings.
> >
> > MoeBlee
> >


 BBB b \ Barbara at LivingHistory stop co stop uk
 B B aa rrr b 
 BBB a a r bbb  Quidquid latine dictum sit,
 B B a a r b b  altum viditur.
 BBB aa a r bbb 



0




Reply

see80 (286)

5/7/2009 6:31:03 AM


On 7 Mag, 02:04, LudovicoVan <ju...@diegidio.name> wrote:
> Mostly, a thought experiment:
>
> Cantor claims: "whichever list of infinite (binary) strings you can
> come up with, I will (anti)diagonalise it."
>
> It is (AFAIK) easy enough to come up with a production of _all_ the
> infinite binary strings. The problem remains how to make the list
> bulletproof to diagonalisation. All we can do is falsify the
> diagonalisation procedure itself, by following the claim to the
> letter:
>
> We start by producing our list, entry by entry; in parallel, our
> opponent starts building the antidiagonal (I will abstract from the
> technical details, unnecessary here):
>
> =A0 =A0 List =A0 Diagonal Antidiagonal
> =A0 =A0 .(0) =A0 =A0 =A0 =A0 0 =A0 =A0 =A0 =A0 =A0 =A0 1
> =A0 =A0 .(0)1 =A0 =A0 =A0 =A00 =A0 =A0 =A0 =A0 =A0 =A0 1
> =A0 =A0 =A0... =A0 =A0 =A0 =A0 ... =A0 =A0 =A0 =A0 =A0 ...
>
> At this point, one migth wander: is there? is not there?
>
> =A0 =A0 =A0 =A0 =3D=3D=3D Cantor's diagonal is a nonhalting machine. =3D=
=3D=3D
>
> Under this perspective, accepting the diagonal argument "simply"
> implies a specific choice about infinity (and infinities; and what can
> or cannot be proved; and a fundamental paradox: incongruent by
> construction).
>
> Now, what about Turing?
>
> LV
The infinite union of infinite numerable sets is numarable?
Clearly no!
So Reals are not walkable!


0




Reply

fransisf (9)

5/7/2009 7:57:32 AM


In article
<c61e543e18cd40bf8c28b4b7e31a98be@o27g2000vbd.googlegroups.com>,
jesko <fransisf@gmail.com> wrote:
> On 7 Mag, 02:04, LudovicoVan <ju...@diegidio.name> wrote:
> > Mostly, a thought experiment:
> >
> > Cantor claims: "whichever list of infinite (binary) strings you can
> > come up with, I will (anti)diagonalise it."
> >
> > It is (AFAIK) easy enough to come up with a production of _all_ the
> > infinite binary strings. The problem remains how to make the list
> > bulletproof to diagonalisation. All we can do is falsify the
> > diagonalisation procedure itself, by following the claim to the
> > letter:
> >
> > We start by producing our list, entry by entry; in parallel, our
> > opponent starts building the antidiagonal (I will abstract from the
> > technical details, unnecessary here):
> >
> > � � List � Diagonal Antidiagonal
> > � � .(0) � � � � 0 � � � � � � 1
> > � � .(0)1 � � � �0 � � � � � � 1
> > � � �... � � � � ... � � � � � ...
> >
> > At this point, one migth wander: is there? is not there?
> >
> > � � � � === Cantor's diagonal is a nonhalting machine. ===
> >
> > Under this perspective, accepting the diagonal argument "simply"
> > implies a specific choice about infinity (and infinities; and what can
> > or cannot be proved; and a fundamental paradox: incongruent by
> > construction).
> >
> > Now, what about Turing?
> >
> > LV
>
> The infinite union of infinite numerable sets is numarable?
Yes: the infinite union of a denumerably infinite set of denumerably
infinite sets is denumerable. Let i:j indicate the j'th element of the
i'th set. One way to enumerate the infinite union is:
{1:1, 1:2, 2:1, 1:3, 2:2, 3:1, 1:4, 2:3, 3:2, 4:1, ...}
Clearly, this covers all i:j.
> Clearly no!
Clearly wrong!
> So Reals are not walkable!
A correct conclusion from an incorrect premise. (By "walkable" I assume
you mean denumerable.)


 BBB b \ Barbara at LivingHistory stop co stop uk
 B B aa rrr b 
 BBB a a r bbb  Quidquid latine dictum sit,
 B B a a r b b  altum viditur.
 BBB aa a r bbb 



0




Reply

see80 (286)

5/7/2009 8:32:50 AM


On 7 Mag, 10:32, Barb Knox <s...@sig.below> wrote:
> In article
> <c61e543e18cd40bf8c28b4b7e31a9...@o27g2000vbd.googlegroups.com>,
>
>
>
>
>
> =A0jesko <frans...@gmail.com> wrote:
> > On 7 Mag, 02:04, LudovicoVan <ju...@diegidio.name> wrote:
> > > Mostly, a thought experiment:
>
> > > Cantor claims: "whichever list of infinite (binary) strings you can
> > > come up with, I will (anti)diagonalise it."
>
> > > It is (AFAIK) easy enough to come up with a production of _all_ the
> > > infinite binary strings. The problem remains how to make the list
> > > bulletproof to diagonalisation. All we can do is falsify the
> > > diagonalisation procedure itself, by following the claim to the
> > > letter:
>
> > > We start by producing our list, entry by entry; in parallel, our
> > > opponent starts building the antidiagonal (I will abstract from the
> > > technical details, unnecessary here):
>
> > > =A0 =A0 List =A0 Diagonal Antidiagonal
> > > =A0 =A0 .(0) =A0 =A0 =A0 =A0 0 =A0 =A0 =A0 =A0 =A0 =A0 1
> > > =A0 =A0 .(0)1 =A0 =A0 =A0 =A00 =A0 =A0 =A0 =A0 =A0 =A0 1
> > > =A0 =A0 =A0... =A0 =A0 =A0 =A0 ... =A0 =A0 =A0 =A0 =A0 ...
>
> > > At this point, one migth wander: is there? is not there?
>
> > > =A0 =A0 =A0 =A0 =3D=3D=3D Cantor's diagonal is a nonhalting machine.=
=3D=3D=3D
>
> > > Under this perspective, accepting the diagonal argument "simply"
> > > implies a specific choice about infinity (and infinities; and what ca=
n
> > > or cannot be proved; and a fundamental paradox: incongruent by
> > > construction).
>
> > > Now, what about Turing?
>
> > > LV
>
> > The infinite union of infinite numerable sets is numarable?
>
> Yes: the infinite union of a denumerably infinite set of denumerably
> infinite sets is denumerable. =A0Let i:j indicate the j'th element of the
> i'th set. =A0One way to enumerate the infinite union is:
> =A0 =A0 {1:1, =A01:2, 2:1, =A01:3, 2:2, 3:1, =A01:4, 2:3, 3:2, 4:1, ...}
>
> Clearly, this covers all i:j.
>
> > Clearly no!
>
> Clearly wrong!
>
> > So Reals are not walkable!
>
> A correct conclusion from an incorrect premise. =A0(By "walkable" I assum=
e
> you mean denumerable.)
>
> 
> 
>  =A0BBB =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0b =A0 =A0\ =A0 =A0 Barbara at Liv=
ingHistory stop co stop uk
>  =A0B =A0B =A0 aa =A0 =A0 rrr =A0b =A0 =A0 
>  =A0BBB =A0 a =A0a =A0 r =A0 =A0 bbb =A0  =A0 =A0Quidquid latine dictum=
sit,
>  =A0B =A0B =A0a =A0a =A0 r =A0 =A0 b =A0b =A0 =A0 =A0altum viditur.
>  =A0BBB =A0 =A0aa a =A0r =A0 =A0 bbb =A0  =A0
>  Nascondi testo citato
>
>  Mostra testo citato 
Well, you're rigtht!
A correct conclusion from an incorrect premise
I say A wrong conclusion from a correct premise.


0




Reply

fransisf (9)

5/7/2009 9:15:26 AM


On 7 Mag, 10:32, Barb Knox <s...@sig.below> wrote:
> In article
> <c61e543e18cd40bf8c28b4b7e31a9...@o27g2000vbd.googlegroups.com>,
>
>
>
>
>
> =A0jesko <frans...@gmail.com> wrote:
> > On 7 Mag, 02:04, LudovicoVan <ju...@diegidio.name> wrote:
> > > Mostly, a thought experiment:
>
> > > Cantor claims: "whichever list of infinite (binary) strings you can
> > > come up with, I will (anti)diagonalise it."
>
> > > It is (AFAIK) easy enough to come up with a production of _all_ the
> > > infinite binary strings. The problem remains how to make the list
> > > bulletproof to diagonalisation. All we can do is falsify the
> > > diagonalisation procedure itself, by following the claim to the
> > > letter:
>
> > > We start by producing our list, entry by entry; in parallel, our
> > > opponent starts building the antidiagonal (I will abstract from the
> > > technical details, unnecessary here):
>
> > > =A0 =A0 List =A0 Diagonal Antidiagonal
> > > =A0 =A0 .(0) =A0 =A0 =A0 =A0 0 =A0 =A0 =A0 =A0 =A0 =A0 1
> > > =A0 =A0 .(0)1 =A0 =A0 =A0 =A00 =A0 =A0 =A0 =A0 =A0 =A0 1
> > > =A0 =A0 =A0... =A0 =A0 =A0 =A0 ... =A0 =A0 =A0 =A0 =A0 ...
>
> > > At this point, one migth wander: is there? is not there?
>
> > > =A0 =A0 =A0 =A0 =3D=3D=3D Cantor's diagonal is a nonhalting machine.=
=3D=3D=3D
>
> > > Under this perspective, accepting the diagonal argument "simply"
> > > implies a specific choice about infinity (and infinities; and what ca=
n
> > > or cannot be proved; and a fundamental paradox: incongruent by
> > > construction).
>
> > > Now, what about Turing?
>
> > > LV
>
> > The infinite union of infinite numerable sets is numarable?
>
> Yes: the infinite union of a denumerably infinite set of denumerably
> infinite sets is denumerable. =A0Let i:j indicate the j'th element of the
> i'th set. =A0One way to enumerate the infinite union is:
> =A0 =A0 {1:1, =A01:2, 2:1, =A01:3, 2:2, 3:1, =A01:4, 2:3, 3:2, 4:1, ...}
>
> Clearly, this covers all i:j.
>
> > Clearly no!
>
> Clearly wrong!
>
> > So Reals are not walkable!
>
> A correct conclusion from an incorrect premise. =A0(By "walkable" I assum=
e
> you mean denumerable.)
>
> 
> 
>  =A0BBB =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0b =A0 =A0\ =A0 =A0 Barbara at Liv=
ingHistory stop co stop uk
>  =A0B =A0B =A0 aa =A0 =A0 rrr =A0b =A0 =A0 
>  =A0BBB =A0 a =A0a =A0 r =A0 =A0 bbb =A0  =A0 =A0Quidquid latine dictum=
sit,
>  =A0B =A0B =A0a =A0a =A0 r =A0 =A0 b =A0b =A0 =A0 =A0altum viditur.
>  =A0BBB =A0 =A0aa a =A0r =A0 =A0 bbb =A0  =A0
>  Nascondi testo citato
>
>  Mostra testo citato 
If you map any real to N (Naturals) then you may conceive Reals as
the infinite uinion of denumerable
sets since N is always denumarable. But this is union must be not
walkable or not denumarable.
So a contradiction!
Thanks a lot!


0




Reply

fransisf (9)

5/7/2009 9:23:28 AM


On 7 Mag, 02:04, LudovicoVan <ju...@diegidio.name> wrote:
> Mostly, a thought experiment:
>
> Cantor claims: "whichever list of infinite (binary) strings you can
> come up with, I will (anti)diagonalise it."
>
> It is (AFAIK) easy enough to come up with a production of _all_ the
> infinite binary strings. The problem remains how to make the list
> bulletproof to diagonalisation. All we can do is falsify the
> diagonalisation procedure itself, by following the claim to the
> letter:
>
> We start by producing our list, entry by entry; in parallel, our
> opponent starts building the antidiagonal (I will abstract from the
> technical details, unnecessary here):
>
> =A0 =A0 List =A0 Diagonal Antidiagonal
> =A0 =A0 .(0) =A0 =A0 =A0 =A0 0 =A0 =A0 =A0 =A0 =A0 =A0 1
> =A0 =A0 .(0)1 =A0 =A0 =A0 =A00 =A0 =A0 =A0 =A0 =A0 =A0 1
> =A0 =A0 =A0... =A0 =A0 =A0 =A0 ... =A0 =A0 =A0 =A0 =A0 ...
>
> At this point, one migth wander: is there? is not there?
>
> =A0 =A0 =A0 =A0 =3D=3D=3D Cantor's diagonal is a nonhalting machine. =3D=
=3D=3D
>
> Under this perspective, accepting the diagonal argument "simply"
> implies a specific choice about infinity (and infinities; and what can
> or cannot be proved; and a fundamental paradox: incongruent by
> construction).
>
> Now, what about Turing?
>
> LV
Yes , I agree with your argument.
But consider this function.
F is defined so:
R is a real number. Every decimal digits of R can be mapped to a
subset of N
es.
5,892479847892798278974894279874289
0 246810..................................................
Since any properties over N is suitable for this mapping, the set of
Reals can be mapped to the set of all properties over Naturals.
Clearly since every property is denumerable , their union must be
denumerable or Naturals will be not walkable.
So ...........


0




Reply

fransisf (9)

5/7/2009 10:14:59 AM


In article
<c7ccbbf2af51448eb9da773e185ecf13@r36g2000vbr.googlegroups.com>,
jesko <fransisf@gmail.com> wrote:
> On 7 Mag, 10:32, Barb Knox <s...@sig.below> wrote:
> > In article
> > <c61e543e18cd40bf8c28b4b7e31a9...@o27g2000vbd.googlegroups.com>,
> >
> >
> >
> >
> >
> > �jesko <frans...@gmail.com> wrote:
> > > On 7 Mag, 02:04, LudovicoVan <ju...@diegidio.name> wrote:
> > > > Mostly, a thought experiment:
> >
> > > > Cantor claims: "whichever list of infinite (binary) strings you can
> > > > come up with, I will (anti)diagonalise it."
> >
> > > > It is (AFAIK) easy enough to come up with a production of _all_ the
> > > > infinite binary strings. The problem remains how to make the list
> > > > bulletproof to diagonalisation. All we can do is falsify the
> > > > diagonalisation procedure itself, by following the claim to the
> > > > letter:
> >
> > > > We start by producing our list, entry by entry; in parallel, our
> > > > opponent starts building the antidiagonal (I will abstract from the
> > > > technical details, unnecessary here):
> >
> > > > � � List � Diagonal Antidiagonal
> > > > � � .(0) � � � � 0 � � � � � � 1
> > > > � � .(0)1 � � � �0 � � � � � � 1
> > > > � � �... � � � � ... � � � � � ...
> >
> > > > At this point, one migth wander: is there? is not there?
> >
> > > > � � � � === Cantor's diagonal is a nonhalting machine. ===
> >
> > > > Under this perspective, accepting the diagonal argument "simply"
> > > > implies a specific choice about infinity (and infinities; and what can
> > > > or cannot be proved; and a fundamental paradox: incongruent by
> > > > construction).
> >
> > > > Now, what about Turing?
> >
> > > > LV
> >
> > > The infinite union of infinite numerable sets is numarable?
> >
> > Yes: the infinite union of a denumerably infinite set of denumerably
> > infinite sets is denumerable. �Let i:j indicate the j'th element of the
> > i'th set. �One way to enumerate the infinite union is:
> > � � {1:1, �1:2, 2:1, �1:3, 2:2, 3:1, �1:4, 2:3, 3:2, 4:1, ...}
> >
> > Clearly, this covers all i:j.
> >
> > > Clearly no!
> >
> > Clearly wrong!
> >
> > > So Reals are not walkable!
> >
> > A correct conclusion from an incorrect premise. �(By "walkable" I assume
> > you mean denumerable.)
> >
> > 
> > 
> >  �BBB � � � � � � � �b � �\ � � Barbara at LivingHistory stop co stop uk
> >  �B �B � aa � � rrr �b � � 
> >  �BBB � a �a � r � � bbb �  � �Quidquid latine dictum sit,
> >  �B �B �a �a � r � � b �b � � �altum viditur.
> >  �BBB � �aa a �r � � bbb �  �
> >  Nascondi testo citato
> >
> >  Mostra testo citato 
> If you map any real to N (Naturals) then you may conceive Reals as
> the infinite uinion of denumerable sets
It is indeed the case that each real in [0,1) can be mapped to a unique
(possibly infinite) set of naturals, e.g. by representing the fraction
in binary (not ending in 111...) and having your set of naturals be the
place positions which have a 1.
> since N is always denumarable.
A true statement, but how is it relevant here?
> But this is union must be not walkable or not denumarable.
You still haven't defined what you mean by "walkable". Tsk tsk.
As for denumerable, OF COURSE a union (even a nondenumerable union) of
subsets of N will be a subset of N. And so what?
> So a contradiction!
Nope.
> Thanks a lot!
(It doesn't seem to be helping much, n'estce pas?)
Face facts: Cantor's diagonal proof that R > N is SIMPLE. It has
even been CHECKED BY COMPUTERS. Therefore the only way that R = N
is if the various formulations of axiomatic set theory are internally
INCONSISTENT (which is emphatically NOT the same as being inconsistent
with someone's naive intuitions). Proving (say) ZFC to be inconsistent
would be worth at least a Fields Medal and a centrefold picture in the
Journal of the AMS. If it were easy (or even less than extremely
difficult) to do then it would already have been done. The chance of
you doing it is vanishingly small.
Plus, there are good reasons for believing that ZFC etc. ARE internally
consistent. For example, there is an intuitively notunreasonable model
for ZFC etc. (Goedel's "constructible universe").


 BBB b \ Barbara at LivingHistory stop co stop uk
 B B aa rrr b 
 BBB a a r bbb  Quidquid latine dictum sit,
 B B a a r b b  altum viditur.
 BBB aa a r bbb 



0




Reply

see80 (286)

5/7/2009 10:56:53 AM


On 7 Mai, 08:31, Barb Knox <s...@sig.below> wrote:
> In article <20090506222934.N59...@agora.rdrop.com>,
> =A0William Elliot <ma...@rdrop.remove.com> wrote:
>
>
>
>
>
> > On Wed, 6 May 2009, MoeBlee wrote:
>
> > >> We start by producing our list, entry by entry; in parallel, our
> > >> opponent starts building the antidiagonal (I will abstract from the
> > >> technical details, unnecessary here):
>
> > >> At this point, one migth wander: is there? is not there?
>
> > >> =3D=3D=3D Cantor's diagonal is a nonhalting machine. =3D=3D=3D
>
> > >> Under this perspective,
>
> > > And that is not the perspective in which the proof is given.
>
> > > The diagonal is not a machine, nor Turning machine, nor abstract
> > > machine of any kind, halting or nonhalting.
>
> > Clearly OP is has been duped into joining the currently popular
> > religous [cult] of computerism.
>
> It's actually even worse than that. =A0By framing the Cantor diagonal
> construction in a computerfriendly way (e.g. for every i and j, the
> "enumeration oracle" produces the j'th digit of the i'th listentry in
> finite time), one can readily construct a *computable* diagonal which
> provably differs from every entry in the enumeration.
The computed number differs from every entry checked up to the nth
line.
It does not differ from all lines, given that there are more than n +
1 for every n.
It does not differ from all lines unless the list can be checked
completely.
It does not differ from all the paths of the infinite binary tree.
Regards, WM


0




Reply

mueckenh (275)

5/7/2009 11:10:28 AM


On 7 Mai, 12:56, Barb Knox <s...@sig.below> wrote:
> Face facts: Cantor's diagonal proof that R > N is SIMPLE.
It presupposes actual infinity, i.e. a list that can be checked
completely.
If it is possible to go through a countable set (checking it) then it
is possible to construct every element. Then it is possible to
constrct the complete binarey tree (all its nodes) by means of a
contable number of steps using a countable number of paths.
< =A0It has
> even been CHECKED BY COMPUTERS.
which do what their programmers want.
>=A0Therefore the only way that R =3D N
> is if the various formulations of axiomatic set theory are internally
> INCONSISTENT
or that the assumption of actual infinity leads to a contradiction.
> (which is emphatically NOT the same as being inconsistent
> with someone's naive intuitions). =A0
That means, all counter aruments will be qualifed as naive intuition?
Here is an agument that counts and counters:
In a (not necessarily finite) sequence of finite linear sets A c B c
C ... we
have never two elements a and b of the sets such that
(a e A & b !e A) & (b e B & a !e B).
This proves that there is a largest natural number if there are all
natural numbers.
Regards, WM


0




Reply

mueckenh (275)

5/7/2009 11:21:45 AM


On 7 Mag, 12:56, Barb Knox <s...@sig.below> wrote:
> In article
> <c7ccbbf2af51448eb9da773e185ec...@r36g2000vbr.googlegroups.com>,
>
>
>
>
>
> =A0jesko <frans...@gmail.com> wrote:
> > On 7 Mag, 10:32, Barb Knox <s...@sig.below> wrote:
> > > In article
> > > <c61e543e18cd40bf8c28b4b7e31a9...@o27g2000vbd.googlegroups.com>,
>
> > > =A0jesko <frans...@gmail.com> wrote:
> > > > On 7 Mag, 02:04, LudovicoVan <ju...@diegidio.name> wrote:
> > > > > Mostly, a thought experiment:
>
> > > > > Cantor claims: "whichever list of infinite (binary) strings you c=
an
> > > > > come up with, I will (anti)diagonalise it."
>
> > > > > It is (AFAIK) easy enough to come up with a production of _all_ t=
he
> > > > > infinite binary strings. The problem remains how to make the list
> > > > > bulletproof to diagonalisation. All we can do is falsify the
> > > > > diagonalisation procedure itself, by following the claim to the
> > > > > letter:
>
> > > > > We start by producing our list, entry by entry; in parallel, our
> > > > > opponent starts building the antidiagonal (I will abstract from =
the
> > > > > technical details, unnecessary here):
>
> > > > > =A0 =A0 List =A0 Diagonal Antidiagonal
> > > > > =A0 =A0 .(0) =A0 =A0 =A0 =A0 0 =A0 =A0 =A0 =A0 =A0 =A0 1
> > > > > =A0 =A0 .(0)1 =A0 =A0 =A0 =A00 =A0 =A0 =A0 =A0 =A0 =A0 1
> > > > > =A0 =A0 =A0... =A0 =A0 =A0 =A0 ... =A0 =A0 =A0 =A0 =A0 ...
>
> > > > > At this point, one migth wander: is there? is not there?
>
> > > > > =A0 =A0 =A0 =A0 =3D=3D=3D Cantor's diagonal is a nonhalting mach=
ine. =3D=3D=3D
>
> > > > > Under this perspective, accepting the diagonal argument "simply"
> > > > > implies a specific choice about infinity (and infinities; and wha=
t can
> > > > > or cannot be proved; and a fundamental paradox: incongruent by
> > > > > construction).
>
> > > > > Now, what about Turing?
>
> > > > > LV
>
> > > > The infinite union of infinite numerable sets is numarable?
>
> > > Yes: the infinite union of a denumerably infinite set of denumerably
> > > infinite sets is denumerable. =A0Let i:j indicate the j'th element of=
the
> > > i'th set. =A0One way to enumerate the infinite union is:
> > > =A0 =A0 {1:1, =A01:2, 2:1, =A01:3, 2:2, 3:1, =A01:4, 2:3, 3:2, 4:1, .=
...}
>
> > > Clearly, this covers all i:j.
>
> > > > Clearly no!
>
> > > Clearly wrong!
>
> > > > So Reals are not walkable!
>
> > > A correct conclusion from an incorrect premise. =A0(By "walkable" I a=
ssume
> > > you mean denumerable.)
>
> > > 
> > > 
> > >  =A0BBB =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0b =A0 =A0\ =A0 =A0 Barbara at=
LivingHistory stop co stop uk
> > >  =A0B =A0B =A0 aa =A0 =A0 rrr =A0b =A0 =A0 
> > >  =A0BBB =A0 a =A0a =A0 r =A0 =A0 bbb =A0  =A0 =A0Quidquid latine di=
ctum sit,
> > >  =A0B =A0B =A0a =A0a =A0 r =A0 =A0 b =A0b =A0 =A0 =A0altum viditur.
> > >  =A0BBB =A0 =A0aa a =A0r =A0 =A0 bbb =A0  =A0
> > >  Nascondi testo citato
>
> > >  Mostra testo citato 
> > If you map any real to N (Naturals) =A0then you may conceive Reals as
> > the infinite uinion of denumerable sets
>
> It is indeed the case that each real in [0,1) can be mapped to a unique
> (possibly infinite) set of naturals, e.g. by representing the fraction
> in binary (not ending in 111...) and having your set of naturals be the
> place positions which have a 1.
>
> > since N is always denumarable.
>
> A true statement, but how is it relevant here?
>
> > But this is union must be not walkable or not denumarable.
>
> You still haven't defined what you mean by "walkable". =A0Tsk tsk.
>
> As for denumerable, OF COURSE a union (even a nondenumerable union) of
> subsets of N will be a subset of N. =A0And so what?
If Reals are defined as the set of all properties over N, then Reals
are numerable
since if not numearble then N would be so , that it is absurd.
>
> > So a contradiction!
>
> Nope.
>
> > Thanks a lot!
>
> (It doesn't seem to be helping much, n'estce pas?)
>
> Face facts: Cantor's diagonal proof that R > N is SIMPLE. =A0It has
> even been CHECKED BY COMPUTERS. =A0Therefore the only way that R =3D N=

> is if the various formulations of axiomatic set theory are internally
> INCONSISTENT (which is emphatically NOT the same as being inconsistent
> with someone's naive intuitions). =A0Proving (say) ZFC to be inconsistent
> would be worth at least a Fields Medal and a centrefold picture in the
> Journal of the AMS. =A0If it were easy (or even less than extremely
> difficult) to do then it would already have been done. =A0The chance of
> you doing it is vanishingly small.
>
> Plus, there are good reasons for believing that ZFC etc. ARE internally
> consistent. =A0For example, there is an intuitively notunreasonable mode=
l
> for ZFC etc. (Goedel's "constructible universe").
>
> 
> 
>  =A0BBB =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0b =A0 =A0\ =A0 =A0 Barbara at Liv=
ingHistory stop co stop uk
>  =A0B =A0B =A0 aa =A0 =A0 rrr =A0b =A0 =A0 
>  =A0BBB =A0 a =A0a =A0 r =A0 =A0 bbb =A0  =A0 =A0Quidquid latine dictum=
sit,
>  =A0B =A0B =A0a =A0a =A0 r =A0 =A0 b =A0b =A0 =A0 =A0altum viditur.
>  =A0BBB =A0 =A0aa a =A0r =A0 =A0 bbb =A0  =A0
>  Nascondi testo citato
>
>  Mostra testo citato 


0




Reply

fransisf (9)

5/7/2009 11:36:07 AM


On 7 Mag, 12:56, Barb Knox <s...@sig.below> wrote:
> In article
> <c7ccbbf2af51448eb9da773e185ec...@r36g2000vbr.googlegroups.com>,
>
>
>
>
>
> =A0jesko <frans...@gmail.com> wrote:
> > On 7 Mag, 10:32, Barb Knox <s...@sig.below> wrote:
> > > In article
> > > <c61e543e18cd40bf8c28b4b7e31a9...@o27g2000vbd.googlegroups.com>,
>
> > > =A0jesko <frans...@gmail.com> wrote:
> > > > On 7 Mag, 02:04, LudovicoVan <ju...@diegidio.name> wrote:
> > > > > Mostly, a thought experiment:
>
> > > > > Cantor claims: "whichever list of infinite (binary) strings you c=
an
> > > > > come up with, I will (anti)diagonalise it."
>
> > > > > It is (AFAIK) easy enough to come up with a production of _all_ t=
he
> > > > > infinite binary strings. The problem remains how to make the list
> > > > > bulletproof to diagonalisation. All we can do is falsify the
> > > > > diagonalisation procedure itself, by following the claim to the
> > > > > letter:
>
> > > > > We start by producing our list, entry by entry; in parallel, our
> > > > > opponent starts building the antidiagonal (I will abstract from =
the
> > > > > technical details, unnecessary here):
>
> > > > > =A0 =A0 List =A0 Diagonal Antidiagonal
> > > > > =A0 =A0 .(0) =A0 =A0 =A0 =A0 0 =A0 =A0 =A0 =A0 =A0 =A0 1
> > > > > =A0 =A0 .(0)1 =A0 =A0 =A0 =A00 =A0 =A0 =A0 =A0 =A0 =A0 1
> > > > > =A0 =A0 =A0... =A0 =A0 =A0 =A0 ... =A0 =A0 =A0 =A0 =A0 ...
>
> > > > > At this point, one migth wander: is there? is not there?
>
> > > > > =A0 =A0 =A0 =A0 =3D=3D=3D Cantor's diagonal is a nonhalting mach=
ine. =3D=3D=3D
>
> > > > > Under this perspective, accepting the diagonal argument "simply"
> > > > > implies a specific choice about infinity (and infinities; and wha=
t can
> > > > > or cannot be proved; and a fundamental paradox: incongruent by
> > > > > construction).
>
> > > > > Now, what about Turing?
>
> > > > > LV
>
> > > > The infinite union of infinite numerable sets is numarable?
>
> > > Yes: the infinite union of a denumerably infinite set of denumerably
> > > infinite sets is denumerable. =A0Let i:j indicate the j'th element of=
the
> > > i'th set. =A0One way to enumerate the infinite union is:
> > > =A0 =A0 {1:1, =A01:2, 2:1, =A01:3, 2:2, 3:1, =A01:4, 2:3, 3:2, 4:1, .=
...}
>
> > > Clearly, this covers all i:j.
>
> > > > Clearly no!
>
> > > Clearly wrong!
>
> > > > So Reals are not walkable!
>
> > > A correct conclusion from an incorrect premise. =A0(By "walkable" I a=
ssume
> > > you mean denumerable.)
>
> > > 
> > > 
> > >  =A0BBB =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0b =A0 =A0\ =A0 =A0 Barbara at=
LivingHistory stop co stop uk
> > >  =A0B =A0B =A0 aa =A0 =A0 rrr =A0b =A0 =A0 
> > >  =A0BBB =A0 a =A0a =A0 r =A0 =A0 bbb =A0  =A0 =A0Quidquid latine di=
ctum sit,
> > >  =A0B =A0B =A0a =A0a =A0 r =A0 =A0 b =A0b =A0 =A0 =A0altum viditur.
> > >  =A0BBB =A0 =A0aa a =A0r =A0 =A0 bbb =A0  =A0
> > >  Nascondi testo citato
>
> > >  Mostra testo citato 
> > If you map any real to N (Naturals) =A0then you may conceive Reals as
> > the infinite uinion of denumerable sets
>
> It is indeed the case that each real in [0,1) can be mapped to a unique
> (possibly infinite) set of naturals, e.g. by representing the fraction
> in binary (not ending in 111...) and having your set of naturals be the
> place positions which have a 1.
>
> > since N is always denumarable.
>
> A true statement, but how is it relevant here?
>
> > But this is union must be not walkable or not denumarable.
>
> You still haven't defined what you mean by "walkable". =A0Tsk tsk.
>
> As for denumerable, OF COURSE a union (even a nondenumerable union) of
> subsets of N will be a subset of N. =A0And so what?
>
> > So a contradiction!
>
> Nope.
>
> > Thanks a lot!
>
> (It doesn't seem to be helping much, n'estce pas?)
>
> Face facts: Cantor's diagonal proof that R > N is SIMPLE. =A0It has
> even been CHECKED BY COMPUTERS. =A0Therefore the only way that R =3D N=

> is if the various formulations of axiomatic set theory are internally
> INCONSISTENT (which is emphatically NOT the same as being inconsistent
> with someone's naive intuitions). =A0Proving (say) ZFC to be inconsistent
> would be worth at least a Fields Medal and a centrefold picture in the
> Journal of the AMS. =A0If it were easy (or even less than extremely
> difficult) to do then it would already have been done. =A0The chance of
> you doing it is vanishingly small.
>
> Plus, there are good reasons for believing that ZFC etc. ARE internally
> consistent. =A0For example, there is an intuitively notunreasonable mode=
l
> for ZFC etc. (Goedel's "constructible universe").
>
> 
> 
>  =A0BBB =A0 =A0 =A0 =A0 =A0 =A0 =A0 =A0b =A0 =A0\ =A0 =A0 Barbara at Liv=
ingHistory stop co stop uk
>  =A0B =A0B =A0 aa =A0 =A0 rrr =A0b =A0 =A0 
>  =A0BBB =A0 a =A0a =A0 r =A0 =A0 bbb =A0  =A0 =A0Quidquid latine dictum=
sit,
>  =A0B =A0B =A0a =A0a =A0 r =A0 =A0 b =A0b =A0 =A0 =A0altum viditur.
>  =A0BBB =A0 =A0aa a =A0r =A0 =A0 bbb =A0  =A0
>  Nascondi testo citato
>
>  Mostra testo citato 
The main paradox is:
" Counting infinite object is not coutable but I'm still counting "
This paradox arises when you want to count objects that are infinite
objects as reals.
Properties over N are infinite object as "To be a powers" , but no
one can admit that they don't constitute
an enumerable set. But clearly one can always construct a new property
from the preceding counted properties.
But for the same infinity notion one can always add a new natural to
Naturals.
Tha's all
Thanks.


0




Reply

fransisf (9)

5/7/2009 11:51:28 AM


On 7 Mag, 02:04, LudovicoVan <ju...@diegidio.name> wrote:
> Mostly, a thought experiment:
>
> Cantor claims: "whichever list of infinite (binary) strings you can
> come up with, I will (anti)diagonalise it."
>
> It is (AFAIK) easy enough to come up with a production of _all_ the
> infinite binary strings. The problem remains how to make the list
> bulletproof to diagonalisation. All we can do is falsify the
> diagonalisation procedure itself, by following the claim to the
> letter:
>
> We start by producing our list, entry by entry; in parallel, our
> opponent starts building the antidiagonal (I will abstract from the
> technical details, unnecessary here):
>
> =A0 =A0 List =A0 Diagonal Antidiagonal
> =A0 =A0 .(0) =A0 =A0 =A0 =A0 0 =A0 =A0 =A0 =A0 =A0 =A0 1
> =A0 =A0 .(0)1 =A0 =A0 =A0 =A00 =A0 =A0 =A0 =A0 =A0 =A0 1
> =A0 =A0 =A0... =A0 =A0 =A0 =A0 ... =A0 =A0 =A0 =A0 =A0 ...
>
> At this point, one migth wander: is there? is not there?
>
> =A0 =A0 =A0 =A0 =3D=3D=3D Cantor's diagonal is a nonhalting machine. =3D=
=3D=3D
>
> Under this perspective, accepting the diagonal argument "simply"
> implies a specific choice about infinity (and infinities; and what can
> or cannot be proved; and a fundamental paradox: incongruent by
> construction).
>
> Now, what about Turing?
>
> LV
Further one can easily prove that countability is never a property of
infinity.
So no classification of infinity can be based upon countability.
Thanks in advance
Thanks
Simply thanks.


0




Reply

fransisf (9)

5/7/2009 12:00:03 PM


On Wed, 6 May 2009 17:04:15 0700 (PDT), LudovicoVan
<julio@diegidio.name> wrote:
>Mostly, a thought experiment:
>
>Cantor claims: "whichever list of infinite (binary) strings you can
>come up with, I will (anti)diagonalise it."
>
>It is (AFAIK) easy enough to come up with a production of _all_ the
>infinite binary strings.
Exactly what does "come up with a production of" mean here?
>The problem remains how to make the list
>bulletproof to diagonalisation. All we can do is falsify the
>diagonalisation procedure itself, by following the claim to the
>letter:
>
>We start by producing our list, entry by entry; in parallel, our
>opponent starts building the antidiagonal (I will abstract from the
>technical details, unnecessary here):
>
> List Diagonal Antidiagonal
> .(0) 0 1
> .(0)1 0 1
> ... ... ...
>
>At this point, one migth wander: is there? is not there?
Is there _what_?
> === Cantor's diagonal is a nonhalting machine. ===
It's not a machine at all.
You're talking about infinite strings of digits. Can you
give an example of a _halting_ machine that produces
such a thing? No. This has nothing to do with the
diagonal argument  the question of halting machines
is simply irrelevant.
>Under this perspective, accepting the diagonal argument "simply"
>implies a specific choice about infinity (and infinities; and what can
>or cannot be proved; and a fundamental paradox: incongruent by
>construction).
You're aware that this entire post is incoherent, right?
>Now, what about Turing?
What about Turing?
>LV
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the postgrads."
in sci.logic.)


0




Reply

dullrich (221)

5/7/2009 12:13:52 PM


On 7 Mag, 02:04, LudovicoVan <ju...@diegidio.name> wrote:
> Mostly, a thought experiment:
>
> Cantor claims: "whichever list of infinite (binary) strings you can
> come up with, I will (anti)diagonalise it."
>
> It is (AFAIK) easy enough to come up with a production of _all_ the
> infinite binary strings. The problem remains how to make the list
> bulletproof to diagonalisation. All we can do is falsify the
> diagonalisation procedure itself, by following the claim to the
> letter:
>
> We start by producing our list, entry by entry; in parallel, our
> opponent starts building the antidiagonal (I will abstract from the
> technical details, unnecessary here):
>
> =A0 =A0 List =A0 Diagonal Antidiagonal
> =A0 =A0 .(0) =A0 =A0 =A0 =A0 0 =A0 =A0 =A0 =A0 =A0 =A0 1
> =A0 =A0 .(0)1 =A0 =A0 =A0 =A00 =A0 =A0 =A0 =A0 =A0 =A0 1
> =A0 =A0 =A0... =A0 =A0 =A0 =A0 ... =A0 =A0 =A0 =A0 =A0 ...
>
> At this point, one migth wander: is there? is not there?
>
> =A0 =A0 =A0 =A0 =3D=3D=3D Cantor's diagonal is a nonhalting machine. =3D=
=3D=3D
>
> Under this perspective, accepting the diagonal argument "simply"
> implies a specific choice about infinity (and infinities; and what can
> or cannot be proved; and a fundamental paradox: incongruent by
> construction).
>
> Now, what about Turing?
>
> LV
Authority wrote:
"We may begin with a dialectical argument and show as follows that
there is no such thing. If =91bounded by a surface=92 is the definition of
body there cannot be an infinite body either intelligible or sensible.
Nor can number taken in abstraction be infinite, for number or that
which has number is numerable. If then the numerable can be numbered,
it would also be possible to go through the infinite."
Aristotle : Physics III,5
From this it doesn't follow that Naturals are numbered while Reals
not!


0




Reply

fransisf (9)

5/7/2009 12:25:03 PM


On May 7, 3:56=A0am, Barb Knox <s...@sig.below> wrote:
>
> Face facts: Cantor's diagonal proof that R > N is SIMPLE. =A0It has
> even been CHECKED BY COMPUTERS. =A0Therefore the only way that R =3D N=

> is if the various formulations of axiomatic set theory are internally
> INCONSISTENT (which is emphatically NOT the same as being inconsistent
> with someone's naive intuitions).
Pardon me, but wouldn't it be more accurate to say: the only
way that R =3D N would be provable in [some] axiomatic
set theory would be if that theory was inconsistent? And even if
that were so, it would still be the case that R > N in a
consistent theory. Yes?
Marshall


0




Reply

marshall.spight (580)

5/7/2009 1:22:08 PM


WM wrote:
> The computed number differs from every entry
> checked up to the nth line.
> It does not differ from all lines, given that
> there are more than n + 1 for every n.
> It does not differ from all lines unless the
> list can be checked completely.
Another consequence of your rule is
One (1) does not differ from all even integers unless the
list (of even integers) can be checked completely.
Since there are an infinite number of even integers, it
must follow, as night does the day, that one (1) is
an even integer.
Jim Burns
> It does not differ from all the paths of the
> infinite binary tree.


0




Reply

burns.87 (39)

5/7/2009 2:10:33 PM


On May 7, 1:04=A0am, LudovicoVan <ju...@diegidio.name> wrote:
> Mostly, a thought experiment:
>
> Cantor claims: "whichever list of infinite (binary) strings you can
> come up with, I will (anti)diagonalise it."
>
> It is (AFAIK) easy enough to come up with a production of _all_ the
> infinite binary strings. The problem remains how to make the list
> bulletproof to diagonalisation. All we can do is falsify the
> diagonalisation procedure itself, by following the claim to the
> letter:
>
> We start by producing our list, entry by entry; in parallel, our
> opponent starts building the antidiagonal (I will abstract from the
> technical details, unnecessary here):
>
> =A0 =A0 List =A0 Diagonal Antidiagonal
> =A0 =A0 .(0) =A0 =A0 =A0 =A0 0 =A0 =A0 =A0 =A0 =A0 =A0 1
> =A0 =A0 .(0)1 =A0 =A0 =A0 =A00 =A0 =A0 =A0 =A0 =A0 =A0 1
> =A0 =A0 =A0... =A0 =A0 =A0 =A0 ... =A0 =A0 =A0 =A0 =A0 ...
>
> At this point, one migth wander: is there? is not there?
>
> =A0 =A0 =A0 =A0 =3D=3D=3D Cantor's diagonal is a nonhalting machine. =3D=
=3D=3D
>
> Under this perspective, accepting the diagonal argument "simply"
> implies a specific choice about infinity (and infinities; and what can
> or cannot be proved; and a fundamental paradox: incongruent by
> construction).
>
> Now, what about Turing?
>
> LV
I think what is going on is that you and several of the other
respondents are (perhaps without knowing it) mathematical
constructivists.
Classical mathematics rests on ZFC set theory. In that theory,
Cantor's argument is valid.
However, you and the other respondents have an intuition that there is
something wrong with the argument. This suggests to me that you would
prefer intuitionist mathematics (as developed by Brouwer and others),
or some other constructivist variant.
Intuitionistic mathematics is perfectly respectable, studied by
logicians, etc. Many of the results of classical mathematics can also
be proved in intuitionistic mathematics  but not all.
So I suppose the next question is, does Cantor's argument work in
intuitionist mathematics? I'm afraid I don't know the answer to that
one.


0




Reply

polyomino (2)

5/7/2009 2:57:17 PM


On May 7, 1:04=A0am, LudovicoVan <ju...@diegidio.name> wrote:
> Mostly, a thought experiment:
>
> Cantor claims: "whichever list of infinite (binary) strings you can
> come up with, I will (anti)diagonalise it."
>
> It is (AFAIK) easy enough to come up with a production of _all_ the
> infinite binary strings. The problem remains how to make the list
> bulletproof to diagonalisation. All we can do is falsify the
> diagonalisation procedure itself, by following the claim to the
> letter:
>
> We start by producing our list, entry by entry; in parallel, our
> opponent starts building the antidiagonal (I will abstract from the
> technical details, unnecessary here):
>
> =A0 =A0 List =A0 Diagonal Antidiagonal
> =A0 =A0 .(0) =A0 =A0 =A0 =A0 0 =A0 =A0 =A0 =A0 =A0 =A0 1
> =A0 =A0 .(0)1 =A0 =A0 =A0 =A00 =A0 =A0 =A0 =A0 =A0 =A0 1
> =A0 =A0 =A0... =A0 =A0 =A0 =A0 ... =A0 =A0 =A0 =A0 =A0 ...
>
> At this point, one migth wander: is there? is not there?
>
> =A0 =A0 =A0 =A0 =3D=3D=3D Cantor's diagonal is a nonhalting machine. =3D=
=3D=3D
>
> Under this perspective, accepting the diagonal argument "simply"
> implies a specific choice about infinity (and infinities; and what can
> or cannot be proved; and a fundamental paradox: incongruent by
> construction).
>
> Now, what about Turing?
>
> LV
http://en.wikipedia.org/wiki/Constructivism_(mathematics)
 does discuss Cantor's argument


0




Reply

polyomino (2)

5/7/2009 3:05:34 PM


DavidA <polyomino@f2s.com> writes:
> On May 7, 1:04�am, LudovicoVan <ju...@diegidio.name> wrote:
> > Mostly, a thought experiment:
> >
> > Cantor claims: "whichever list of infinite (binary) strings you can
> > come up with, I will (anti)diagonalise it."
> >
> > It is (AFAIK) easy enough to come up with a production of _all_ the
> > infinite binary strings. The problem remains how to make the list
> > bulletproof to diagonalisation. All we can do is falsify the
> > diagonalisation procedure itself, by following the claim to the
> > letter:
> >
> > We start by producing our list, entry by entry; in parallel, our
> > opponent starts building the antidiagonal (I will abstract from the
> > technical details, unnecessary here):
> >
> > � � List � Diagonal Antidiagonal
> > � � .(0) � � � � 0 � � � � � � 1
> > � � .(0)1 � � � �0 � � � � � � 1
> > � � �... � � � � ... � � � � � ...
> >
> > At this point, one migth wander: is there? is not there?
> >
> > � � � � === Cantor's diagonal is a nonhalting machine. ===
> >
> > Under this perspective, accepting the diagonal argument "simply"
> > implies a specific choice about infinity (and infinities; and what can
> > or cannot be proved; and a fundamental paradox: incongruent by
> > construction).
> >
> > Now, what about Turing?
> >
> > LV
>
> I think what is going on is that you and several of the other
> respondents are (perhaps without knowing it) mathematical
> constructivists.
>
> Classical mathematics rests on ZFC set theory. In that theory,
> Cantor's argument is valid.
>
> However, you and the other respondents have an intuition that there is
> something wrong with the argument. This suggests to me that you would
> prefer intuitionist mathematics (as developed by Brouwer and others),
> or some other constructivist variant.
>
> Intuitionistic mathematics is perfectly respectable, studied by
> logicians, etc. Many of the results of classical mathematics can also
> be proved in intuitionistic mathematics  but not all.
>
> So I suppose the next question is, does Cantor's argument work in
> intuitionist mathematics? I'm afraid I don't know the answer to that
> one.
As others have noted, if you are serious about constructivism, so that
not only the decimal expansions of the reals is taken to be
constructive, but also an enumeration of such reals is itself given
constructively, then lo and behold it is possible to construct effectively
a real not in the given list.
This argument is given eg by Bishop in his book "Constructive Analysis".
It might help such as LV to note that Bishop does not state the claim in
the form that one set is larger than another, but in the form that says
that for any effectively given enumeration, a missing real can be constructed.

Alan Smaill


0




Reply

smaill1 (89)

5/7/2009 4:18:54 PM


On 7 May, 13:13, David C. Ullrich <dullr...@sprynet.com> wrote:
> On Wed, 6 May 2009 17:04:15 0700 (PDT), LudovicoVan
> <ju...@diegidio.name> wrote:
> >Mostly, a thought experiment:
>
> >Cantor claims: "whichever list of infinite (binary) strings you can
> >come up with, I will (anti)diagonalise it."
>
> >It is (AFAIK) easy enough to come up with a production of _all_ the
> >infinite binary strings.
>
> Exactly what does "come up with a production of" mean here?
I mean "production" in the constructive sense: at least predicative.
Though I'm not putting this burden on Cantor's diagonal argument
itself, it's just the reference point for my counterargument. For
instance, one could give an inductive definition, then implement and
run it in, say, Prolog. The program is not more "nonhalting" than a
program to output the naturals.
> >The problem remains how to make the list
> >bulletproof to diagonalisation. All we can do is falsify the
> >diagonalisation procedure itself, by following the claim to the
> >letter:
>
> >We start by producing our list, entry by entry; in parallel, our
> >opponent starts building the antidiagonal (I will abstract from the
> >technical details, unnecessary here):
>
> > =A0 =A0List =A0 Diagonal Antidiagonal
> > =A0 =A0.(0) =A0 =A0 =A0 =A0 0 =A0 =A0 =A0 =A0 =A0 =A0 1
> > =A0 =A0.(0)1 =A0 =A0 =A0 =A00 =A0 =A0 =A0 =A0 =A0 =A0 1
> > =A0 =A0 ... =A0 =A0 =A0 =A0 ... =A0 =A0 =A0 =A0 =A0 ...
>
> >At this point, one migth wander: is there? is not there?
>
> Is there _what_?
The antidiagonal: is it in the list? Is it not? The proper question
is: will the program ever output it's antidiagonal?
> > =A0 =A0 =A0 =A0=3D=3D=3D Cantor's diagonal is a nonhalting machine. =
=3D=3D=3D
>
> It's not a machine at all.
Turing machines can be an interpretation tool here. Anti
diagonalisation is a process, not yet a total function, because
decisions about infinities are a consequence in our case, not a
premise (they are yet to be taken).
> You're talking about infinite strings of digits. Can you
> give an example of a _halting_ machine that produces
> such a thing? No.
As said, asking for a machine that produces all infinite binary
strings and halts is just as sensible as asking for a machine that
produces all naturals and halts.
> This has nothing to do with the
> diagonal argument  the question of halting machines
> is simply irrelevant.
> >Under this perspective, accepting the diagonal argument "simply"
> >implies a specific choice about infinity (and infinities; and what can
> >or cannot be proved; and a fundamental paradox: incongruent by
> >construction).
>
> You're aware that this entire post is incoherent, right?
It's all but incoherent. Maybe it's unclear, maybe it's incorrect in
many ways, maybe there is something you are still missing, maybe a bit
of all of them.
In synthesis, the incongruency I am talking about could be seen in
this fact: we are denied the existence of a complete list, in terms of
a construct that is claimed to be complete.
> >Now, what about Turing?
>
> What about Turing?
Turing leverages the diagonal argument to show the insolubility of the
halting problem (consequently failing Hilbert's program, by the way).
In any case, I find Turing and his machines interesting here because,
as far as I can see, Turing reasons at the same "primitive" (in the
theoretical sense) level as Cantor.
LV


0




Reply

julio (505)

5/7/2009 4:24:40 PM


On 7 May, 15:57, DavidA <polyom...@f2s.com> wrote:
> I suppose the next question is, does Cantor's argument work in
> intuitionist mathematics? I'm afraid I don't know the answer to that
> one.
The diagonal argument, as well as all arguments about the
uncountability of the reals, are not predicative.
LV


0




Reply

julio (505)

5/7/2009 4:30:30 PM


On 7 May, 06:43, William Elliot <ma...@rdrop.remove.com> wrote:
> Clearly OP is has been duped into joining the currently popular religous
> clut of computerism.
You are a simple man.
LV


0




Reply

julio (505)

5/7/2009 4:34:34 PM


On 7 May, 07:31, Barb Knox <s...@sig.below> wrote:
> In article <20090506222934.N59...@agora.rdrop.com>,
> =A0William Elliot <ma...@rdrop.remove.com> wrote:
> > On Wed, 6 May 2009, MoeBlee wrote:
>
> > >> We start by producing our list, entry by entry; in parallel, our
> > >> opponent starts building the antidiagonal (I will abstract from the
> > >> technical details, unnecessary here):
>
> > >> At this point, one migth wander: is there? is not there?
>
> > >> =3D=3D=3D Cantor's diagonal is a nonhalting machine. =3D=3D=3D
>
> > >> Under this perspective,
>
> > > And that is not the perspective in which the proof is given.
>
> > > The diagonal is not a machine, nor Turning machine, nor abstract
> > > machine of any kind, halting or nonhalting.
>
> > Clearly OP is has been duped into joining the currently popular
> > religous [cult] of computerism.
>
> It's actually even worse than that. =A0By framing the Cantor diagonal
> construction in a computerfriendly way (e.g. for every i and j, the
> "enumeration oracle" produces the j'th digit of the i'th listentry in
> finite time), one can readily construct a *computable* diagonal which
> provably differs from every entry in the enumeration.
Provably how?
> So it's less a case of computerworship than selfworship (i.e.
> believing that one's naive intuitions MUST be right, even when proven
> wrong).
Simple lady.
LV


0




Reply

julio (505)

5/7/2009 4:36:52 PM


On 7 May, 11:56, Barb Knox <s...@sig.below> wrote:
> Face facts: Cantor's diagonal proof that R > N is SIMPLE. =A0It has
> even been CHECKED BY COMPUTERS.
Formal proofs are irrelevant to the diagonal argument: they have it
embedded, as an informing principle, into their axioms and
definitions.
> Therefore the only way that R =3D N
> is if the various formulations of axiomatic set theory are internally
> INCONSISTENT
Incorrect (and a myopic). They can be formally consistent with unsound
or otherwise invalid premises. Indeed, the diagonal argument here
under investigation, is an informing principle to standard axiomatic
set theories.
> Proving (say) ZFC to be inconsistent
> would be worth at least a Fields Medal and a centrefold picture in the
> Journal of the AMS.
That tells a lot about the current state of mathematics and not only.
LV


0




Reply

julio (505)

5/7/2009 4:45:53 PM


On 7 May, 14:22, Marshall <marshall.spi...@gmail.com> wrote:
> On May 7, 3:56=A0am, Barb Knox <s...@sig.below> wrote:
>
> > Face facts: Cantor's diagonal proof that R > N is SIMPLE. =A0It has
> > even been CHECKED BY COMPUTERS. =A0Therefore the only way that R =3D =
N
> > is if the various formulations of axiomatic set theory are internally
> > INCONSISTENT (which is emphatically NOT the same as being inconsistent
> > with someone's naive intuitions).
>
> Pardon me, but wouldn't it be more accurate to say: the only
> way that R =3D N would be provable in [some] axiomatic
> set theory would be if that theory was inconsistent?
It would be incorrect and, strictly speaking, false.
LV


0




Reply

julio (505)

5/7/2009 4:46:55 PM


LudovicoVan <julio@diegidio.name> writes:
> On 7 May, 13:13, David C. Ullrich <dullr...@sprynet.com> wrote:
> > On Wed, 6 May 2009 17:04:15 0700 (PDT), LudovicoVan
> > <ju...@diegidio.name> wrote:
> > >Mostly, a thought experiment:
> >
> > >Cantor claims: "whichever list of infinite (binary) strings you can
> > >come up with, I will (anti)diagonalise it."
> >
> > >It is (AFAIK) easy enough to come up with a production of _all_ the
> > >infinite binary strings.
> >
> > Exactly what does "come up with a production of" mean here?
>
> I mean "production" in the constructive sense: at least predicative.
> Though I'm not putting this burden on Cantor's diagonal argument
> itself, it's just the reference point for my counterargument. For
> instance, one could give an inductive definition, then implement and
> run it in, say, Prolog. The program is not more "nonhalting" than a
> program to output the naturals.
>
> > >The problem remains how to make the list
> > >bulletproof to diagonalisation. All we can do is falsify the
> > >diagonalisation procedure itself, by following the claim to the
> > >letter:
> >
> > >We start by producing our list, entry by entry; in parallel, our
> > >opponent starts building the antidiagonal (I will abstract from the
> > >technical details, unnecessary here):
> >
> > > � �List � Diagonal Antidiagonal
> > > � �.(0) � � � � 0 � � � � � � 1
> > > � �.(0)1 � � � �0 � � � � � � 1
> > > � � ... � � � � ... � � � � � ...
> >
> > >At this point, one migth wander: is there? is not there?
> >
> > Is there _what_?
>
> The antidiagonal: is it in the list? Is it not? The proper question
> is: will the program ever output it's antidiagonal?
As you well know, a program will never produce infinite output.
So your choice of question is seriously loaded  your program will
never output a binary expansion of even one real number.
Now, maybe this is a good way to think of the potential infinite
and constructivism  but as you expressed it here, you need to
be more evenhanded about what it means to construct a number (and a list
of numbers).
> > > � � � �=== Cantor's diagonal is a nonhalting machine. ===
> >
> > It's not a machine at all.
>
> Turing machines can be an interpretation tool here. Anti
> diagonalisation is a process, not yet a total function, because
> decisions about infinities are a consequence in our case, not a
> premise (they are yet to be taken).
That's fine, it is as good a tool as it needs to be.
> > You're talking about infinite strings of digits. Can you
> > give an example of a _halting_ machine that produces
> > such a thing? No.
>
> As said, asking for a machine that produces all infinite binary
> strings and halts is just as sensible as asking for a machine that
> produces all naturals and halts.
So what are *you* asking for when you ask whether the program can *output*
its antidiagonal?
> > This has nothing to do with the
> > diagonal argument  the question of halting machines
> > is simply irrelevant.
>
> > >Under this perspective, accepting the diagonal argument "simply"
> > >implies a specific choice about infinity (and infinities; and what can
> > >or cannot be proved; and a fundamental paradox: incongruent by
> > >construction).
> >
> > You're aware that this entire post is incoherent, right?
>
> It's all but incoherent. Maybe it's unclear, maybe it's incorrect in
> many ways, maybe there is something you are still missing, maybe a bit
> of all of them.
>
> In synthesis, the incongruency I am talking about could be seen in
> this fact: we are denied the existence of a complete list, in terms of
> a construct that is claimed to be complete.
Complete in what sense?
 that there is a completed infinite totality
 that every real appears somewhere in the list
 that every constructive real appears somewhere in the list
 that the nth digit of the mth real can be computed on demand
 something else??
>
> LV

Alan Smaill


0




Reply

smaill1 (89)

5/7/2009 5:03:57 PM


LudovicoVan <julio@diegidio.name> writes:
> On 7 May, 11:56, Barb Knox <s...@sig.below> wrote:
>
> > Face facts: Cantor's diagonal proof that R > N is SIMPLE. �It has
> > even been CHECKED BY COMPUTERS.
>
> Formal proofs are irrelevant to the diagonal argument: they have it
> embedded, as an informing principle, into their axioms and
> definitions.
Why should we believe this assertion of yours?
You have made it several times, and not yet given the slightest
reason for anyone to believe it.
>
> LV

Alan Smaill


0




Reply

smaill1 (89)

5/7/2009 5:06:03 PM


On 7 May, 17:18, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> As others have noted, if you are serious about constructivism, so that
> not only the decimal expansions of the reals is taken to be
> constructive, but also an enumeration of such reals is itself given
> constructively, then lo and behold it is possible to construct effectively
> a real not in the given list.
> This argument is given eg by Bishop in his book "Constructive Analysis".
Interesting: could you maybe give an idea how this argument goes? How
to "construct effectively a real not in the given list"?
> It might help such as LV to note that Bishop does not state the claim in
> the form that one set is larger than another
Usually I don't either, so no help.
LV


0




Reply

julio (505)

5/7/2009 5:06:42 PM


On May 7, 4:21=A0am, WM <mueck...@rz.fhaugsburg.de> wrote:
> On 7 Mai, 12:56, Barb Knox <s...@sig.below> wrote:
>
> > Face facts: Cantor's diagonal proof that R > N is SIMPLE.
>
> It presupposes actual infinity,
If there is no set whose members are all and only the natural numbers,
then the very question of comparison between N and R reduces to what
'N' and 'R' would stand for; and this is not a very interesting
question. But we may put the matter in the hypothetical: IF there is a
set whose members are all and only the natural numbers (and building
from that set we get a set whose members are all and only the real
numbers) then the set of natural numbers is strictly dominated by the
set of real numbers.
In other words, yes, if there are no set of natural numbers, then we
hardly care about relative cardinality. But if there is a set of
natural numbers, then it is strictly dominated by the set of real
numbers.
You're not saying anything that is not completely understood already
by mathematicians. It is already well understood that we need to adopt
some axiom (such as the axiom of infinity) to prove that there exists
a set of all natural numbers.
> >=A0Therefore the only way that R =3D N
> > is if the various formulations of axiomatic set theory are internally
> > INCONSISTENT
>
> or that the assumption of actual infinity leads to a contradiction.
That is silly. Your 'or' is not practically an alternative since the
alternative that set theoy is inconsistent was ALREADY MENTIONED.
MoeBlee


0




Reply

jazzmobe (307)

5/7/2009 5:11:27 PM


On May 7, 4:36=A0am, jesko <frans...@gmail.com> wrote:
> If Reals are defined as the set of all properties over N
Well, they're not defined that way.
MoeBlee


0




Reply

jazzmobe (307)

5/7/2009 5:12:42 PM


LudovicoVan <julio@diegidio.name> writes:
> On 7 May, 17:18, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>
> > As others have noted, if you are serious about constructivism, so that
> > not only the decimal expansions of the reals is taken to be
> > constructive, but also an enumeration of such reals is itself given
> > constructively, then lo and behold it is possible to construct effectively
> > a real not in the given list.
> > This argument is given eg by Bishop in his book "Constructive Analysis".
>
> Interesting: could you maybe give an idea how this argument goes? How
> to "construct effectively a real not in the given list"?
>
> > It might help such as LV to note that Bishop does not state the claim in
> > the form that one set is larger than another
Let's look at total functions taking two natural numbers and returning
either 0 or 1, eg f(n,m). If we leave out the question of
multiple representations, let's suppose you have implemented
such a total function.
We can ask successive questions about the digits f(0,0), f(0,1), f(0,2), ...
for example, and take these as the digits of a constructive real.
Same story for f(1,0), f(1,1), f(1,2), ...
But the diagonal number f(0,0), f(1,1), f(2,2), ... is just as
computable as any of the individual reals. Likewise then the antidiagonal,
which differs from every real in the given enumeration at at least
one position.
So, the antidiagonal is itself something given algorithmically,
and you can easily write the procedure to return its nth digit,
provided that the given listing of reals itself is a total
function  this does not mean that all the digits involved must
be computed, of course!
> Usually I don't either, so no help.
To put it another way, he does not say that the reals are "uncountable".
> LV

Alan Smaill


0




Reply

smaill1 (89)

5/7/2009 5:30:35 PM


On 7 Mai, 16:10, James Burns <burns...@osu.edu> wrote:
> WM wrote:
> > The computed number differs from every entry
> > checked up to the nth line.
> > It does not differ from all lines, given that
> > there are more than n + 1 for every n.
> > It does not differ from all lines unless the
> > list can be checked completely.
>
> Another consequence of your rule is
>
> One (1) does not differ from all even integers unless the
> list (of even integers) can be checked completely.
No. If you know that an element exists only once in the world, you
need not search the whole universe in order to find a second one if
you have got the first.
But, yes, in order to be sure that 1 (or any other number) does not
appear in a sequence you must know the whole sequence.
> Since there are an infinite number of even integers, it
> must follow, as night does the day, that one (1) is
> an even integer.
Since there are an infinite number of even integers you cannot exclude
that there is a term a_2n = 1 unless you have searched them all.
Regards, WM


0




Reply

mueckenh (275)

5/7/2009 5:44:14 PM


On 7 Mai, 16:57, DavidA <polyom...@f2s.com> wrote:
> Classical mathematics rests on ZFC set theory. In that theory,
> Cantor's argument is valid.
In that theory N is given without construction. In that theory the
complete binary tree is given without construction. And its paths can
be shown to belong to a countable set.
>
> However, you and the other respondents have an intuition that there is
> something wrong with the argument. This suggests to me that you would
> prefer intuitionist mathematics (as developed by Brouwer and others),
> or some other constructivist variant.
>
> Intuitionistic mathematics is perfectly respectable, studied by
> logicians, etc. Many of the results of classical mathematics can also
> be proved in intuitionistic mathematics  but not all.
Only those which are correct.
>
> So I suppose the next question is, does Cantor's argument work in
> intuitionist mathematics? I'm afraid I don't know the answer to that
> one
It is asserted, erroneously, that it does.
But it does not, because there is no diagonal of the list of all
finite words
0
1
00
01
10
11
....
obviously.
Regards, WM


0




Reply

mueckenh (275)

5/7/2009 5:46:44 PM


On 7 Mai, 18:18, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> DavidA <polyom...@f2s.com> writes:
> > On May 7, 1:04=A0am, LudovicoVan <ju...@diegidio.name> wrote:
> > > Mostly, a thought experiment:
>
> > > Cantor claims: "whichever list of infinite (binary) strings you can
> > > come up with, I will (anti)diagonalise it."
>
> > > It is (AFAIK) easy enough to come up with a production of _all_ the
> > > infinite binary strings. The problem remains how to make the list
> > > bulletproof to diagonalisation. All we can do is falsify the
> > > diagonalisation procedure itself, by following the claim to the
> > > letter:
>
> > > We start by producing our list, entry by entry; in parallel, our
> > > opponent starts building the antidiagonal (I will abstract from the
> > > technical details, unnecessary here):
>
> > > =A0 =A0 List =A0 Diagonal Antidiagonal
> > > =A0 =A0 .(0) =A0 =A0 =A0 =A0 0 =A0 =A0 =A0 =A0 =A0 =A0 1
> > > =A0 =A0 .(0)1 =A0 =A0 =A0 =A00 =A0 =A0 =A0 =A0 =A0 =A0 1
> > > =A0 =A0 =A0... =A0 =A0 =A0 =A0 ... =A0 =A0 =A0 =A0 =A0 ...
>
> > > At this point, one migth wander: is there? is not there?
>
> > > =A0 =A0 =A0 =A0 =3D=3D=3D Cantor's diagonal is a nonhalting machine.=
=3D=3D=3D
>
> > > Under this perspective, accepting the diagonal argument "simply"
> > > implies a specific choice about infinity (and infinities; and what ca=
n
> > > or cannot be proved; and a fundamental paradox: incongruent by
> > > construction).
>
> > > Now, what about Turing?
>
> > > LV
>
> > I think what is going on is that you and several of the other
> > respondents are (perhaps without knowing it) mathematical
> > constructivists.
>
> > Classical mathematics rests on ZFC set theory. In that theory,
> > Cantor's argument is valid.
>
> > However, you and the other respondents have an intuition that there is
> > something wrong with the argument. This suggests to me that you would
> > prefer intuitionist mathematics (as developed by Brouwer and others),
> > or some other constructivist variant.
>
> > Intuitionistic mathematics is perfectly respectable, studied by
> > logicians, etc. Many of the results of classical mathematics can also
> > be proved in intuitionistic mathematics  but not all.
>
> > So I suppose the next question is, does Cantor's argument work in
> > intuitionist mathematics? I'm afraid I don't know the answer to that
> > one.
>
> As others have noted, if you are serious about constructivism, so that
> not only the decimal expansions of the reals is taken to be
> constructive, but also an enumeration of such reals is itself given
> constructively, then lo and behold it is possible to construct effectivel=
y
> a real not in the given list.
>
> This argument is given eg by Bishop in his book "Constructive Analysis".
> It might help such as LV to note that Bishop does not state the claim in
> the form that one set is larger than another, but in the form that says
> that for any effectively given enumeration, a missing real can be constru=
cted.
That is only true if the reals are given as infinite sequences of
digits (which is in principle impossible). If they are given in the
form of finite words, the argument fails because the lis of all finite
words does not allow a diagonal:
o
1
00
01
10
11
....
Regards, WM


0




Reply

mueckenh (275)

5/7/2009 5:49:09 PM


DavidA a �crit :
> On May 7, 1:04 am, LudovicoVan <ju...@diegidio.name> wrote:
>> Mostly, a thought experiment:
>>
>> Cantor claims: "whichever list of infinite (binary) strings you can
>> come up with, I will (anti)diagonalise it."
>>
>> It is (AFAIK) easy enough to come up with a production of _all_ the
>> infinite binary strings. The problem remains how to make the list
>> bulletproof to diagonalisation. All we can do is falsify the
>> diagonalisation procedure itself, by following the claim to the
>> letter:
>>
>> We start by producing our list, entry by entry; in parallel, our
>> opponent starts building the antidiagonal (I will abstract from the
>> technical details, unnecessary here):
>>
>> List Diagonal Antidiagonal
>> .(0) 0 1
>> .(0)1 0 1
>> ... ... ...
>>
>> At this point, one migth wander: is there? is not there?
>>
>> === Cantor's diagonal is a nonhalting machine. ===
>>
>> Under this perspective, accepting the diagonal argument "simply"
>> implies a specific choice about infinity (and infinities; and what can
>> or cannot be proved; and a fundamental paradox: incongruent by
>> construction).
>>
>> Now, what about Turing?
>>
>> LV
>
> I think what is going on is that you and several of the other
> respondents are (perhaps without knowing it) mathematical
> constructivists.
>
> Classical mathematics rests on ZFC set theory. In that theory,
> Cantor's argument is valid.
>
> However, you and the other respondents have an intuition that there is
> something wrong with the argument. This suggests to me that you would
> prefer intuitionist mathematics (as developed by Brouwer and others),
> or some other constructivist variant.
>
> Intuitionistic mathematics is perfectly respectable, studied by
> logicians, etc. Many of the results of classical mathematics can also
> be proved in intuitionistic mathematics  but not all.
>
> So I suppose the next question is, does Cantor's argument work in
> intuitionist mathematics? I'm afraid I don't know the answer to that
> one.
I am afraid you dont know much about intuitionist mathematics too


0




Reply

denis.feldmann.sansspam1 (11)

5/7/2009 5:53:28 PM


On May 7, 10:46=A0am, WM <mueck...@rz.fhaugsburg.de> wrote:
> On 7 Mai, 16:57, DavidA <polyom...@f2s.com> wrote:
>
> > Classical mathematics rests on ZFC set theory. In that theory,
> > Cantor's argument is valid.
>
> In that theory N is given without construction. In that theory the
> complete binary tree is given without construction. And its paths can
> be shown to belong to a countable set.
Of course the set of paths belongs to a countable set. The set of
paths is a member of the singleton set whose only member is the set of
paths. I don't see why you bother to remark on that fact.
But perhaps what you actually mean is that the set of paths is
countable. You've never proven that in ZFC
> > So I suppose the next question is, does Cantor's argument work in
> > intuitionist mathematics? I'm afraid I don't know the answer to that
> > one
>
> It is asserted, erroneously, that it does.
> But it does not, because there is no diagonal of the list of all
> finite words
> 0
> 1
> 00
> 01
> 10
> 11
> ...
> obviously.
Such silliness you write.
Whether an intuitionist accepts the axioms used in the diagonal
argument is one matter. But ASIDE from the axioms, the LOGIC used in
the diagonal argument IS intuitionistic.
And you've not shown any relevance of your list with the diagonal
argument.
MoeBlee


0




Reply

jazzmobe (307)

5/7/2009 6:54:43 PM


On May 7, 10:49=A0am, WM <mueck...@rz.fhaugsburg.de> wrote:
> On 7 Mai, 18:18, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > This argument is given eg by Bishop in his book "Constructive Analysis"=
..
> > It might help such as LV to note that Bishop does not state the claim i=
n
> > the form that one set is larger than another, but in the form that says
> > that for any effectively given enumeration, a missing real can be const=
ructed.
>
> That is only true if the reals are given as infinite sequences of
> digits (which is in principle impossible). If they are given in the
> form of finite words,
You're silly. Of course if you declare that reals are not represented
by denumerable sequences, then there is no diagonal argument to make.
But there is no basis at all to think that ALL the reals can be
represented by finite sequences only.
MoeBlee


0




Reply

jazzmobe (307)

5/7/2009 6:57:20 PM


In article
<c7ccbbf2af51448eb9da773e185ecf13@r36g2000vbr.googlegroups.com>,
jesko <fransisf@gmail.com> wrote:
> If you map any real to N (Naturals) then you may conceive Reals as
> the infinite uinion of denumerable
> sets since N is always denumarable. But this is union must be not
> walkable or not denumarable.
How do you propose to map R to N ?


0




Reply

vmhjr22 (27)

5/7/2009 8:06:41 PM


In article
<fc0b0e9bc3e242498e98ab95d3d2af20@g20g2000vba.googlegroups.com>,
jesko <fransisf@gmail.com> wrote:
> On 7 Mag, 02:04, LudovicoVan <ju...@diegidio.name> wrote:
> > Mostly, a thought experiment:
> >
> > Cantor claims: "whichever list of infinite (binary) strings you can
> > come up with, I will (anti)diagonalise it."
> >
> > It is (AFAIK) easy enough to come up with a production of _all_ the
> > infinite binary strings. The problem remains how to make the list
> > bulletproof to diagonalisation. All we can do is falsify the
> > diagonalisation procedure itself, by following the claim to the
> > letter:
> >
> > We start by producing our list, entry by entry; in parallel, our
> > opponent starts building the antidiagonal (I will abstract from the
> > technical details, unnecessary here):
> >
> > � � List � Diagonal Antidiagonal
> > � � .(0) � � � � 0 � � � � � � 1
> > � � .(0)1 � � � �0 � � � � � � 1
> > � � �... � � � � ... � � � � � ...
> >
> > At this point, one migth wander: is there? is not there?
> >
> > � � � � === Cantor's diagonal is a nonhalting machine. ===
> >
> > Under this perspective, accepting the diagonal argument "simply"
> > implies a specific choice about infinity (and infinities; and what can
> > or cannot be proved; and a fundamental paradox: incongruent by
> > construction).
> >
> > Now, what about Turing?
> >
> > LV
>
> Yes , I agree with your argument.
> But consider this function.
>
> F is defined so:
>
> R is a real number. Every decimal digits of R can be mapped to a
> subset of N
>
> es.
>
>
> 5,892479847892798278974894279874289
> 0 246810..................................................
>
> Since any properties over N is suitable for this mapping, the set of
> Reals can be mapped to the set of all properties over Naturals.
> Clearly since every property is denumerable
I am not quite sure of what you mean by "properties", but the
denumerability of each property, whatever it is, in no way guarantees
the denumerability of the set of properties.
> , their union must be
> denumerable or Naturals will be not walkable.
> So ...........


0




Reply

vmhjr22 (27)

5/7/2009 8:10:24 PM


In article
<bca5b803b94d41d581dd1fd80dbb8051@t11g2000vbc.googlegroups.com>,
WM <mueckenh@rz.fhaugsburg.de> wrote:
> The computed number differs from every entry checked up to the nth
> line.
> It does not differ from all lines, given that there are more than n +
> 1 for every n.
Which line does it not differ from?
The nth line has an nth digit different from the nth digit of the
antidiagonal. What more is needed?
> It does not differ from all lines unless the list can be checked
> completely.
The construction rule assures the diagonal is different from every line
in at least one digit position, What more is needed?
> It does not differ from all the paths of the infinite binary tree.
True, but that set of all paths is already uncountable.


0




Reply

vmhjr22 (27)

5/7/2009 8:17:15 PM


In article
<aeec55446d64408eb1a879d6923f072b@j12g2000vbl.googlegroups.com>,
WM <mueckenh@rz.fhaugsburg.de> wrote:
> On 7 Mai, 12:56, Barb Knox <s...@sig.below> wrote:
>
> > Face facts: Cantor's diagonal proof that R > N is SIMPLE.
>
> It presupposes actual infinity, i.e. a list that can be checked
> completely.
And all of WM's charms are not enough to banish actual infinity from
mainstream mathematics.
>
> >�Therefore the only way that R = N
> > is if the various formulations of axiomatic set theory are internally
> > INCONSISTENT
>
> or that the assumption of actual infinity leads to a contradiction.
Which neither WM nor anyone else has demonstrated to the satisfaction of
anyone but themselves.
>
> > (which is emphatically NOT the same as being inconsistent
> > with someone's naive intuitions). �
>
> That means, all counter aruments will be qualifed as naive intuition?
> Here is an agument that counts and counters:
> In a (not necessarily finite) sequence of finite linear sets A c B c
> C ... we
> have never two elements a and b of the sets such that
> (a e A & b !e A) & (b e B & a !e B).
>
> This proves that there is a largest natural number if there are all
> natural numbers.
Not to anyone other than WM himself.
Wm has repeated this argument too many times now, but has still found no
one who accepts it as valid.
He should retire it.
>
> Regards, WM


0




Reply

vmhjr22 (27)

5/7/2009 8:23:10 PM


In article
<ef9fad85dbf04fdb800074590d35e8ff@e20g2000vbc.googlegroups.com>,
jesko <fransisf@gmail.com> wrote:
> If Reals are defined as the set of all properties over N, then Reals
> are numerable
> since if not numearble then N would be so , that it is absurd.
There are quite a few definitions of the reals, none of which are
anything like that, so why suggest it?
And until you can develop the reals as a complete Archimedean ordered
field strictly from that definition alone, no one else will accept it.


0




Reply

vmhjr22 (27)

5/7/2009 8:27:06 PM


In article
<4509f4d7fa184e05ae0261e28767c4f5@t11g2000vbc.googlegroups.com>,
LudovicoVan <julio@diegidio.name> wrote:
> On 7 May, 07:31, Barb Knox <s...@sig.below> wrote:
> > In article <20090506222934.N59...@agora.rdrop.com>,
> > �William Elliot <ma...@rdrop.remove.com> wrote:
> > > On Wed, 6 May 2009, MoeBlee wrote:
> >
> > > >> We start by producing our list, entry by entry; in parallel, our
> > > >> opponent starts building the antidiagonal (I will abstract from the
> > > >> technical details, unnecessary here):
> >
> > > >> At this point, one migth wander: is there? is not there?
> >
> > > >> === Cantor's diagonal is a nonhalting machine. ===
> >
> > > >> Under this perspective,
> >
> > > > And that is not the perspective in which the proof is given.
> >
> > > > The diagonal is not a machine, nor Turning machine, nor abstract
> > > > machine of any kind, halting or nonhalting.
> >
> > > Clearly OP is has been duped into joining the currently popular
> > > religous [cult] of computerism.
> >
> > It's actually even worse than that. �By framing the Cantor diagonal
> > construction in a computerfriendly way (e.g. for every i and j, the
> > "enumeration oracle" produces the j'th digit of the i'th listentry in
> > finite time), one can readily construct a *computable* diagonal which
> > provably differs from every entry in the enumeration.
>
> Provably how?
Look it up!
>
> > So it's less a case of computerworship than selfworship (i.e.
> > believing that one's naive intuitions MUST be right, even when proven
> > wrong).
>
> Simple lady.
There are varieties of simplicity, some good, some not.
Hers is considerably better than yours.
>
> LV


0




Reply

vmhjr22 (27)

5/7/2009 8:38:05 PM


In article
<f4166d965f734a8693ebe3c63924beb4@t10g2000vbg.googlegroups.com>,
LudovicoVan <julio@diegidio.name> wrote:
> On 7 May, 11:56, Barb Knox <s...@sig.below> wrote:
>
> > Face facts: Cantor's diagonal proof that R > N is SIMPLE. �It has
> > even been CHECKED BY COMPUTERS.
>
> Formal proofs are irrelevant to the diagonal argument: they have it
> embedded, as an informing principle, into their axioms and
> definitions.
>
> > Therefore the only way that R = N
> > is if the various formulations of axiomatic set theory are internally
> > INCONSISTENT
>
> Incorrect (and a myopic). They can be formally consistent with unsound
> or otherwise invalid premises.
How does one test individual premises or sets of premises for
unsoundness or invalidity? Unless one has some absolute standards and
can make everyone agree with them, one can never be absolutely sure.
> Indeed, the diagonal argument here
> under investigation, is an informing principle to standard axiomatic
> set theories.
But until YOUR standards of consistency and soundness are universally
accepted, such questions are moot.


0




Reply

vmhjr22 (27)

5/7/2009 8:44:41 PM


In article
<881bf5de515843818db95541ca973881@g20g2000vba.googlegroups.com>,
LudovicoVan <julio@diegidio.name> wrote:
> On 7 May, 14:22, Marshall <marshall.spi...@gmail.com> wrote:
> > On May 7, 3:56�am, Barb Knox <s...@sig.below> wrote:
> >
> > > Face facts: Cantor's diagonal proof that R > N is SIMPLE. �It has
> > > even been CHECKED BY COMPUTERS. �Therefore the only way that R = N
> > > is if the various formulations of axiomatic set theory are internally
> > > INCONSISTENT (which is emphatically NOT the same as being inconsistent
> > > with someone's naive intuitions).
> >
> > Pardon me, but wouldn't it be more accurate to say: the only
> > way that R = N would be provable in [some] axiomatic
> > set theory would be if that theory was inconsistent?
>
> It would be incorrect and, strictly speaking, false.
Can you justify that claim of falsehood?


0




Reply

vmhjr22 (27)

5/7/2009 8:45:48 PM


In article
<3ab27d4989cc4aec9644e62d883f6439@21g2000vbk.googlegroups.com>,
WM <mueckenh@rz.fhaugsburg.de> wrote:
> On 7 Mai, 16:10, James Burns <burns...@osu.edu> wrote:
> > WM wrote:
> > > The computed number differs from every entry
> > > checked up to the nth line.
> > > It does not differ from all lines, given that
> > > there are more than n + 1 for every n.
> > > It does not differ from all lines unless the
> > > list can be checked completely.
> >
> > Another consequence of your rule is
> >
> > One (1) does not differ from all even integers unless the
> > list (of even integers) can be checked completely.
>
> No. If you know that an element exists only once in the world, you
> need not search the whole universe in order to find a second one if
> you have got the first.
> But, yes, in order to be sure that 1 (or any other number) does not
> appear in a sequence you must know the whole sequence.
Then WM is incapable of proving that 1 is not an even number.
Others might note that there is a property which distinguishes members
of the set of all even naturals from nonmembers, and testing for that
property test for membership in that set.
>
> > Since there are an infinite number of even integers, it
> > must follow, as night does the day, that one (1) is
> > an even integer.
>
> Since there are an infinite number of even integers you cannot exclude
> that there is a term a_2n = 1 unless you have searched them all.
You can in ZFC, where its membership in a set can be determined by
whether the object has, or does not have, some property.
Perhaps WM needs to go back to his Gymnasium for a refresher.


0




Reply

vmhjr22 (27)

5/7/2009 8:52:42 PM


On 7 Mai, 19:53, Denis Feldmann <denis.feldmann.sanss...@neuf.fr>
wrote:
> DavidA a =E9crit :
>
>
>
> > On May 7, 1:04 am, LudovicoVan <ju...@diegidio.name> wrote:
> >> Mostly, a thought experiment:
>
> >> Cantor claims: "whichever list of infinite (binary) strings you can
> >> come up with, I will (anti)diagonalise it."
>
> >> It is (AFAIK) easy enough to come up with a production of _all_ the
> >> infinite binary strings. The problem remains how to make the list
> >> bulletproof to diagonalisation. All we can do is falsify the
> >> diagonalisation procedure itself, by following the claim to the
> >> letter:
>
> >> We start by producing our list, entry by entry; in parallel, our
> >> opponent starts building the antidiagonal (I will abstract from the
> >> technical details, unnecessary here):
>
> >> =A0 =A0 List =A0 Diagonal Antidiagonal
> >> =A0 =A0 .(0) =A0 =A0 =A0 =A0 0 =A0 =A0 =A0 =A0 =A0 =A0 1
> >> =A0 =A0 .(0)1 =A0 =A0 =A0 =A00 =A0 =A0 =A0 =A0 =A0 =A0 1
> >> =A0 =A0 =A0... =A0 =A0 =A0 =A0 ... =A0 =A0 =A0 =A0 =A0 ...
>
> >> At this point, one migth wander: is there? is not there?
>
> >> =A0 =A0 =A0 =A0 =3D=3D=3D Cantor's diagonal is a nonhalting machine. =
=3D=3D=3D
>
> >> Under this perspective, accepting the diagonal argument "simply"
> >> implies a specific choice about infinity (and infinities; and what can
> >> or cannot be proved; and a fundamental paradox: incongruent by
> >> construction).
>
> >> Now, what about Turing?
>
> >> LV
>
> > I think what is going on is that you and several of the other
> > respondents are (perhaps without knowing it) mathematical
> > constructivists.
>
> > Classical mathematics rests on ZFC set theory. In that theory,
> > Cantor's argument is valid.
>
> > However, you and the other respondents have an intuition that there is
> > something wrong with the argument. This suggests to me that you would
> > prefer intuitionist mathematics (as developed by Brouwer and others),
> > or some other constructivist variant.
>
> > Intuitionistic mathematics is perfectly respectable, studied by
> > logicians, etc. Many of the results of classical mathematics can also
> > be proved in intuitionistic mathematics  but not all.
>
> > So I suppose the next question is, does Cantor's argument work in
> > intuitionist mathematics? I'm afraid I don't know the answer to that
> > one.
>
> I am afraid you dont know much about intuitionist mathematics too
Anyhow, whatever Bishop may or may not have said:
It is impossible for him and for anybody else to construct a digit
sequence that identifies pi.
The only way to identify pi (or its multiples) is by saying "pi", or
ratio between circumference and diameter of a circle, or by stating
the series of GregoryLeibniz, or by Euler's zeta functions, or by
Wallis' product, or by Vieta's infinite product or by many related
finite words.
All these words are finite and are written in the following list:
0
1
00
01
10
11
....
This list contains all finite words in all languages over all finite
alphabets. (It contains all dictionaries too.) This list, however,
does not facilitate diagonalization. Therefore in real constructivism
Cantor's proof fails. There are not more than countably many ideas
that can be expressed by words. The real (and other) numbers are a
subset of these ideas.
Regards, WM


0




Reply

mueckenh (275)

5/7/2009 8:56:26 PM


In article
<aa2c587be259438d9dcc8f7a5c2b09aa@v17g2000vbb.googlegroups.com>,
WM <mueckenh@rz.fhaugsburg.de> wrote:
> On 7 Mai, 16:57, DavidA <polyom...@f2s.com> wrote:
>
> > Classical mathematics rests on ZFC set theory. In that theory,
> > Cantor's argument is valid.
>
> In that theory N is given without construction. In that theory the
> complete binary tree is given without construction. And its paths can
> be shown to belong to a countable set.
>
>
> >
> > However, you and the other respondents have an intuition that there is
> > something wrong with the argument. This suggests to me that you would
> > prefer intuitionist mathematics (as developed by Brouwer and others),
> > or some other constructivist variant.
> >
> > Intuitionistic mathematics is perfectly respectable, studied by
> > logicians, etc. Many of the results of classical mathematics can also
> > be proved in intuitionistic mathematics  but not all.
>
> Only those which are correct.
In intuitionist mathematics, adepts do not say that that which has not
been proved by their methods is not correct, they merely say it is
unproven.
> >
> > So I suppose the next question is, does Cantor's argument work in
> > intuitionist mathematics? I'm afraid I don't know the answer to that
> > one
>
> It is asserted, erroneously, that it does.
Which is WM's own erroneous assertion.
There is a sense in which Cantor's argument does work, i.e., for any
constructible binary list, one can show by constructible methods that
there is a constructible nonmember.


0




Reply

vmhjr22 (27)

5/7/2009 8:59:23 PM


In article
<090e5499f16b4ff58a6d673e12373a37@m24g2000vbp.googlegroups.com>,
WM <mueckenh@rz.fhaugsburg.de> wrote:
> On 7 Mai, 18:18, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > As others have noted, if you are serious about constructivism, so that
> > not only the decimal expansions of the reals is taken to be
> > constructive, but also an enumeration of such reals is itself given
> > constructively, then lo and behold it is possible to construct effectively
> > a real not in the given list.
> >
> > This argument is given eg by Bishop in his book "Constructive Analysis".
> > It might help such as LV to note that Bishop does not state the claim in
> > the form that one set is larger than another, but in the form that says
> > that for any effectively given enumeration, a missing real can be
> > constructed.
>
> That is only true if the reals are given as infinite sequences of
> digits (which is in principle impossible). If they are given in the
> form of finite words, the argument fails because the lis of all finite
> words does not allow a diagonal:
If a real is defined well enough in words to be constructively
distinguished from any other equally well defined real, its decimal
expansion to any desired finite extent should be finitely constructible,
so that, however it is listed, a diagonal for that list can be
constructed which differs from it.
So that WM is wrong, again, as usual!


0




Reply

vmhjr22 (27)

5/7/2009 9:07:58 PM


On May 7, 9:24=A0am, LudovicoVan <ju...@diegidio.name> wrote:
> On 7 May, 13:13, David C. Ullrich <dullr...@sprynet.com> wrote:
> > On Wed, 6 May 2009 17:04:15 0700 (PDT), LudovicoVan
> > <ju...@diegidio.name> wrote:
>
> > >It is (AFAIK) easy enough to come up with a production of _all_ the
> > >infinite binary strings.
>
> > Exactly what does "come up with a production of" mean here?
>
> I mean "production" in the constructive sense: at least predicative.
> Though I'm not putting this burden on Cantor's diagonal argument
> itself, it's just the reference point for my counterargument. For
> instance, one could give an inductive definition, then implement and
> run it in, say, Prolog. The program is not more "nonhalting" than a
> program to output the naturals.
You can write a program to output all of the naturals. This program
won't halt, as you say, and for any natural n, will eventually output
n. (We are glossing over some resource limits here, but that is
customary and reasonable.) It will never finish running, though;
there are always more naturals.
You can't do the same for the reals. That's because the reals
aren't countable and the naturals are.
The analogous situation with printing out all the naturals is printing
out a single real. For any digit position n, the program will
eventually
print out that digit, but it will never finish running; there are
always
more digits.
If you want to say, hey, I'll stop printing a particular number when
I get to some repeating pattern, then you're talking about a
program to print out all the rational numbers, not the reals.
If you want to print out real numbers, you can start, but you can't
finish even one. If you want to print out an uncomputable real,
you can't even start.
So no, it is not easy enough to print out all the infinite binary
strings; it's impossible actually to print out even one.
Marshall


0




Reply

marshall.spight (580)

5/7/2009 11:03:54 PM


On May 7, 6:22=A0am, Marshall <marshall.spi...@gmail.com> wrote:
> On May 7, 3:56=A0am, Barb Knox <s...@sig.below> wrote:
> > Face facts: Cantor's diagonal proof that R > N is SIMPLE. =A0It has
> > even been CHECKED BY COMPUTERS. =A0Therefore the only way that R =3D =
N
> > is if the various formulations of axiomatic set theory are internally
> > INCONSISTENT (which is emphatically NOT the same as being inconsistent
> > with someone's naive intuitions).
> Pardon me, but wouldn't it be more accurate to say: the only
> way that R =3D N would be provable in [some] axiomatic
> set theory would be if that theory was inconsistent? And even if
> that were so, it would still be the case that R > N in a
> consistent theory. Yes?
And so LV has started another thread about his opposition to
Cantor's theorem and the uncountability of the real numbers,
and of course, the responses to LV have been typical of the
standard set theorists' responses.
Marshall claims that "card(R)=3Dcard(N)" is impossible in a
consistent set theory. Of course, it's easy to come up with
a consistent set theory in which card(R)=3Dcard(N)  just
define N to be {0} and R to be {1}. But of course, the
context is that N should be something resembling the set of
naturals and R something resembling the set of reals. In
particular, N and R should be infinite sets.
There have been other attempts, of course, to make N and R
have the same size. Some have considered the countable
model of ZFC whose existence is guaranteed by Lowenheim and
Skolem  until it's mentioned that in that model, the
theorem of Cantor is still true. Some have tried to use the
set of computable reals in lieu of the classical reals 
until it's pointed out that the bijection between N and
that set is _not_ computable, and so the computable reals
are not _effectively_ enumerable. Others have tried to use
nonstandard naturals or hypernaturals, also without any
acceptance by the standard set theorists.
The problem is that many sci.math posters, including LV,
want to be able to work with something resembling natural
and real numbers, but just don't want those sets to have
different sizes. Call them Platonists, or some sort of
intuitionists, or whatever  they simply don't accept
different sizes of infinity, and the fact that ZFC proves
the existence of more than one infinite cardinality is
sufficient for them to reject ZFC. Also, they don't care
whether they can come up with an explicit bijection
between N and R or not  they want to have a theory in
which every infinite set has the same size.
And so what should one do, if one wants to work with
natural and real numbers, but is philosophically opposed
to the existence of more than one infinite set size? I
still refuse to believe that there can't be a rigorous
theory in which their claim is provable  and which
isn't trivially inconsistent.
Of course, as I mentioned in another thread, there are
some things that standard set theorists will never accept
in a "reasonable" set theory. Such would be having N and R
have the same set sizes. Even though the theory NFU proves
the existence of nonCantorian sets, no theory that proves
N to be a nonCantorian set will be considered to be
"reasonable" by the standard set theorists.
WM is in this thread as well, but WM is an ultrafinitist,
so his dispute with standard theory is fundamentally
different from LV's.


0




Reply

lwalke3 (94)

5/7/2009 11:08:49 PM


On May 7, 4:08=A0pm, lwal...@lausd.net wrote:
> nonCantorian sets
What is a nonCantorian set?
MoeBlee


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Reply

jazzmobe (307)

5/7/2009 11:28:59 PM


On May 7, 4:08=A0pm, lwal...@lausd.net wrote:
> On May 7, 6:22=A0am, Marshall <marshall.spi...@gmail.com> wrote:
>
> > On May 7, 3:56=A0am, Barb Knox <s...@sig.below> wrote:
> > > Face facts: Cantor's diagonal proof that R > N is SIMPLE. =A0It h=
as
> > > even been CHECKED BY COMPUTERS. =A0Therefore the only way that R =
=3D N
> > > is if the various formulations of axiomatic set theory are internally
> > > INCONSISTENT (which is emphatically NOT the same as being inconsisten=
t
> > > with someone's naive intuitions).
> > Pardon me, but wouldn't it be more accurate to say: the only
> > way that R =3D N would be provable in [some] axiomatic
> > set theory would be if that theory was inconsistent? And even if
> > that were so, it would still be the case that R > N in a
> > consistent theory. Yes?
>
> And so LV has started another thread about his opposition to
> Cantor's theorem and the uncountability of the real numbers,
> and of course, the responses to LV have been typical of the
> standard set theorists' responses.
>
> Marshall claims that "card(R)=3Dcard(N)" is impossible in a
> consistent set theory. Of course, it's easy to come up with
> a consistent set theory in which card(R)=3Dcard(N)  just
> define N to be {0} and R to be {1}.
You make things so hard. If you want to go that way, just
define N :=3D R. Then you get the benefit that N =3D R
no matter what the definition of x and no matter what
the definition of R.
> The problem is that many sci.math posters, including LV,
> want to be able to work with something resembling natural
> and real numbers, but just don't want those sets to have
> different sizes.
I totally know the feeling. I want to work with integers, but
I don't want 7 to be larger than 5.
> Call them Platonists, or some sort of
> intuitionists, or whatever  they simply don't accept
> different sizes of infinity, and the fact that ZFC proves
> the existence of more than one infinite cardinality is
> sufficient for them to reject ZFC.
It's such a shame then that they are required by law
to use ZFC. They should work to get those laws changed!
"After all, this is a democracy!" [1]
> And so what should one do, if one wants to work with
> natural and real numbers, but is philosophically opposed
> to the existence of more than one infinite set size? I
> still refuse to believe that there can't be a rigorous
> theory in which their claim is provable  and which
> isn't trivially inconsistent.
They could use a theory in which the "size" of a set
is defined to be zero. That way, size(N) =3D size(R).
Hey, maybe we could have a thread about that.
Something about having Axy.x+y=3D0 as an axiom.
> Of course, as I mentioned in another thread, there are
> some things that standard set theorists will never accept
> in a "reasonable" set theory.
We have to swear we'll never accept that as part of our
induction into the grand ZFC conspiracy, which I joined
for the medical and dental benefits. Our primary goal
is to hold back mathematics from the tremendous
but civilizationchanging insights of certain individuals
on sci.math. Our primary weapons are fear and surprise.
Marshall
[1] http://www.youtube.com/watch?v=3D6hcSlR4XAGs


0




Reply

marshall.spight (580)

5/8/2009 12:59:08 AM


On May 7, 4:28=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> On May 7, 4:08=A0pm, lwal...@lausd.net wrote:
>
> > nonCantorian sets
>
> What is a nonCantorian set?
Historically, a set is said to "cant over" (later
shortened to "canto'r" in the sixteenth century)
if it has enough of a slope or a tilt to it that
it can be considered to lie on the diagonal.
Such sets are immune to any kind of
diagonalization argument, which made them
popular with the mathematical cranks of the
late Victorian age. However, this immunity
was shortlived, as they were later shown
to be susceptible to finding an element that
was straight up and down.
Marshall
PS. Apologies to MoeBlee.


0




Reply

marshall.spight (580)

5/8/2009 1:31:10 AM


WM wrote:
>> The computed number differs from every entry
>> checked up to the nth line.
>> It does not differ from all lines, given that
>> there are more than n + 1 for every n.
>> It does not differ from all lines unless the
>> list can be checked completely.
>
James Burns wrote:
>> Another consequence of your rule is
>> One (1) does not differ from all even integers unless the
>> list (of even integers) can be checked completely.
>
WM wrote:
> No. If you know that an element exists only once in the world, ...
And how do you know that?
> ... you need not search the whole universe in order to find a second one if
> you have got the first.
> But, yes, in order to be sure that 1 (or any other number) does not
> appear in a sequence you must know the whole sequence.
Or perhaps you need to only know a property shared by every
element in the sequence. For example, every element in the
sequence of even integers is even. This is true regardless
of how many elements have been "searched" or how many
remain to be "searched".


0




Reply

david1417 (99)

5/8/2009 1:32:50 AM


lwalke3@lausd.net writes:
> Marshall claims that "card(R)=card(N)" is impossible in a
> consistent set theory.
For what it's worth, I've no idea why he claims this.
I don't know of any interesting and consistent set theory in which R
is much like the reals, N is much like the naturals, card is much
like cardinality and card(R) = card(N) is a theorem, but I don't see
offhand why there can be no such theory.

Damn John Jay.
Damn everyone who won't damn John Jay.
Damn everyone who won't put lights in his windows and sit up all night
damning John Jay.  Political graffiti from late 18th c. Boston


0




Reply

jesse18 (2492)

5/8/2009 2:19:45 AM


On May 7, 7:19=A0pm, "Jesse F. Hughes" <je...@phiwumbda.org> wrote:
> lwal...@lausd.net writes:
> > Marshall claims that "card(R)=3Dcard(N)" is impossible in a
> > consistent set theory.
>
> For what it's worth, I've no idea why he claims this.
Actually, I don't claim that. I did, however, ask if that
would be correct. (Please be careful of Mr. Walker's
restatement of people's positions.) Although it's fair
to say that I suspect it might be true.
> I don't know of any interesting and consistent set theory in which R
> is much like the reals, N is much like the naturals, card is much
> like cardinality and card(R) =3D card(N) is a theorem, but I don't see
> offhand why there can be no such theory.
Certainly I'm clear that if we are free to choose our definitions
of such symbols as "card", "N" and "R" we can make them
dance to whatever tune strikes our fancy. For example, I
proposed the definition N :=3D R, and I *will* claim that with
that definition of N, card(N) =3D card(R). 'Cause I'


0




Reply

marshall.spight (580)

5/8/2009 3:42:01 AM


Marshall wrote:
> On May 7, 4:08 pm, lwal...@lausd.net wrote:
> > On May 7, 6:22 am, Marshall <marshall.spi...@gmail.com> wrote:
> >
> > > On May 7, 3:56 am, Barb Knox <s...@sig.below> wrote:
> > > > Face facts: Cantor's diagonal proof that R > N is SIMPLE. It has
<snip>
> We have to swear we'll never accept that as part of our
> induction into the grand ZFC conspiracy, which I joined
> for the medical and dental benefits.
And the chicks, surely?
Brian Chandler


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Reply

imaginatorium (41)

5/8/2009 3:45:38 AM


On May 7, 8:42=A0pm, Marshall <marshall.spi...@gmail.com> wrote:
>
> For example, I
> proposed the definition N :=3D R, and I *will* claim that with
> that definition of N, card(N) =3D card(R). 'Cause I'
Sorry about the interruption; I was going to type "'Cause I'm
the sort of person who has a bandage on his index finger
and is going to accidentally tab over the "SEND" key and
hit enter."
Okay, my actual point was that, firstorder theories are
swell and all, but is that sort of machinery *necessary*
to notice that there is no bijection from the naturals
to the reals? Can't we just call the reals a model and
work from there? Any firstorder theory we come up
with is going to be an abstraction of the thing we want
to talk about, and as such, is going to fail to capture
certain properties of that thing. (Seems I've heard
of some guy named "Godel" and some kind of
"theorem" about some sort of "incompleteness".)
(Although I suppose failing to capture certain properties
is irrelevant if we do capture the properties we care
about at that moment.)
I guess it feels like, if we're in a position to speak
of an "interesting and consistent set theory in which
R is much like the reals", why are we bothering?
We already have the reals themselves, and we
know there's uncountably many of them. Why
are we dicking around with theories that are
abstractions of a thing, as if that will settle anything
better than the actual thing would?
Of course it's entirely possible this thinking is just
the result of my ignorance; if someone would
care to hit me with the education stick, I'd be
most obliged. Although I've noticed that unless
one is making strong, ridiculous claims one does
not necessarily get much attention, so let me just
add as an aside that I have a proof that P !=3D NP
here on my desk but I'm the only person alive
today smart enough to understand it.
Marshall


0




Reply

marshall.spight (580)

5/8/2009 4:08:02 AM


On 8 May, 05:08, Marshall <marshall.spi...@gmail.com> wrote:
> Sorry about the interruption; I was going to type "'Cause I'm
> the sort of person who has a bandage on his index finger
Wow, I had not yet realised you could be such a consistent moron.
I also notice that this thread has already broken the barrier of
readability, with the journals, and 11 posts by Virgil only. Cool...
LV


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Reply

julio (505)

5/8/2009 4:38:34 AM


On May 7, 9:38=A0pm, LudovicoVan <ju...@diegidio.name> wrote:
> On 8 May, 05:08, Marshall <marshall.spi...@gmail.com> wrote:
>
> > Sorry about the interruption; I was going to type "'Cause I'm
> > the sort of person who has a bandage on his index finger
>
> Wow, I had not yet realised you could be such a consistent moron.
Yeah, you always were pretty slow to pick things up.
Marshall
PS. And I take objection to you calling me "consistent."


0




Reply

marshall.spight (580)

5/8/2009 4:49:38 AM


On May 7, 8:45=A0pm, Brian Chandler <imaginator...@despammed.com> wrote:
> Marshall wrote:
> > On May 7, 4:08 pm, lwal...@lausd.net wrote:
> > > On May 7, 6:22 am, Marshall <marshall.spi...@gmail.com> wrote:
>
> > > > On May 7, 3:56 am, Barb Knox <s...@sig.below> wrote:
> > > > > Face facts: Cantor's diagonal proof that R > N is SIMPLE. =A0=
It has
> <snip>
> > We have to swear we'll never accept that as part of our
> > induction into the grand ZFC conspiracy, which I joined
> > for the medical and dental benefits.
>
> And the chicks, surely?
Totally. That's what really makes the membership dues a "real" deal.
You reminded me of a story. A few years ago, I had a company
laptop that was rather out of date, and that had started behaving
erratically. I went to visit the techs in charge of such things, and
after a few days of diagnosing they told me I really ought to get
a new one. Fine by me. So I'm standing there in the tech area
with all those guys, and one female. (Not sure how she got
in there; perhaps separated from the herd by bad weather.)
I doubt any of them were within twenty years of my age.
So the tech explains to me the options: lowweight,
low processing power vs. heavier, better performance. Since
I'm a programmer and not the sort of guy to be writing up
reports in MSWORD on the plane, I naturally picked the latter
one. "Gotta have the faster processor," I say. Tech says
sure and walks off to get one. And then, actually just to myself
and entirely under my breath I said, "Chicks will flock."
Cause, you know, gotta keep the funny in constant practice.
From ten feet away, the girl's head whips around and she
demands to know, "Did you just say, 'chicks will flock'?"
Oh, shit. "Uh, yeah," I admit. And then she just cracks up,
as does everyone else who witnessed the exchange.
Whew!
Marshall


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Reply

marshall.spight (580)

5/8/2009 5:15:58 AM


Marshall schrieb:
> On May 7, 9:24=A0am, LudovicoVan <ju...@diegidio.name> wrote:
> > On 7 May, 13:13, David C. Ullrich <dullr...@sprynet.com> wrote:
> > > On Wed, 6 May 2009 17:04:15 0700 (PDT), LudovicoVan
> > > <ju...@diegidio.name> wrote:
> >
> > > >It is (AFAIK) easy enough to come up with a production of _all_ the
> > > >infinite binary strings.
> >
> > > Exactly what does "come up with a production of" mean here?
> >
> > I mean "production" in the constructive sense: at least predicative.
> > Though I'm not putting this burden on Cantor's diagonal argument
> > itself, it's just the reference point for my counterargument. For
> > instance, one could give an inductive definition, then implement and
> > run it in, say, Prolog. The program is not more "nonhalting" than a
> > program to output the naturals.
>
> You can write a program to output all of the naturals. This program
> won't halt, as you say, and for any natural n, will eventually output
> n. (We are glossing over some resource limits here, but that is
> customary and reasonable.) It will never finish running, though;
> there are always more naturals.
>
> You can't do the same for the reals. That's because the reals
> aren't countable and the naturals are.
Really funny.
You say: "You can write a program to output _all_ of the naturals."
I'm sure, you can't.
You say: "It will never finish running, though;there are always more
naturals." Yes, but the same holds for the reals.
You "believe" in the set of real numbers. Let's have this set and a
program which arbitrarily picks out numbers of this set and put them
out (building a new set). This program never stops, right? So, what is
the difference? Naturals never ends, reals never ends.
And don't forget: ZFC must have a countable model.
> The analogous situation with printing out all the naturals is printing
> out a single real. For any digit position n, the program will
> eventually
> print out that digit, but it will never finish running; there are
> always
> more digits.
>
> If you want to say, hey, I'll stop printing a particular number when
> I get to some repeating pattern, then you're talking about a
> program to print out all the rational numbers, not the reals.
>
> If you want to print out real numbers, you can start, but you can't
> finish even one. If you want to print out an uncomputable real,
> you can't even start.
Wrong:
Start printing an (irrational) real number > phi < End printing an
(irrational) real number.
>
> So no, it is not easy enough to print out all the infinite binary
> strings; it's impossible actually to print out even one.
>
Some of your ideas are not well considered.
Regards
Albrecht Storz


0




Reply

albstorz (110)

5/8/2009 6:18:27 AM


On Thu, 7 May 2009, LudovicoVan wrote:
>>> Clearly OP is has been duped into joining the currently popular
>>> religous cult of computerism.
>>
>> It's actually even worse than that. By framing the Cantor diagonal
>> construction in a computerfriendly way (e.g. for every i and j, the
>> "enumeration oracle" produces the j'th digit of the i'th listentry in
>> finite time), one can readily construct a *computable* diagonal which
>> provably differs from every entry in the enumeration.
>
> Provably how?
>
In mysterious ways beyound your simple mindedness.
>> So it's less a case of computerworship than selfworship (i.e.
>> believing that one's naive intuitions MUST be right, even when proven
>> wrong).
>
> Simple lady.
>


0




Reply

marsh6245 (32)

5/8/2009 6:22:56 AM


On May 7, 11:18=A0pm, Albrecht <albst...@gmx.de> wrote:
> Marshall schrieb:
>
>
>
> > On May 7, 9:24=A0am, LudovicoVan <ju...@diegidio.name> wrote:
> > > On 7 May, 13:13, David C. Ullrich <dullr...@sprynet.com> wrote:
> > > > On Wed, 6 May 2009 17:04:15 0700 (PDT), LudovicoVan
> > > > <ju...@diegidio.name> wrote:
>
> > > > >It is (AFAIK) easy enough to come up with a production of _all_ th=
e
> > > > >infinite binary strings.
>
> > > > Exactly what does "come up with a production of" mean here?
>
> > > I mean "production" in the constructive sense: at least predicative.
> > > Though I'm not putting this burden on Cantor's diagonal argument
> > > itself, it's just the reference point for my counterargument. For
> > > instance, one could give an inductive definition, then implement and
> > > run it in, say, Prolog. The program is not more "nonhalting" than a
> > > program to output the naturals.
>
> > You can write a program to output all of the naturals. This program
> > won't halt, as you say, and for any natural n, will eventually output
> > n. (We are glossing over some resource limits here, but that is
> > customary and reasonable.) It will never finish running, though;
> > there are always more naturals.
>
> > You can't do the same for the reals. That's because the reals
> > aren't countable and the naturals are.
>
> Really funny.
> You say: "You can write a program to output =A0_all_ of the naturals."
> I'm sure, you can't.
10 LET I=3D0
20 PRINT I
30 LET I =3D I + 1
40 GOTO 20
There. I just wrote a program to output _all_ the naturals.
Now surely you must abandon your claim that I can't, despite
your initial surety, for clearly I have written the program.
> You "believe" in the set of real numbers. Let's have this set and a
> program which arbitrarily picks out numbers of this set and put them
> out (building a new set). This program never stops, right? So, what is
> the difference?
The difference is the reals aren't countable. Also: every natural
is finitely representable; not every real is. So once your program
hits once such, it will never complete putting out its representation,
so it will never output any further numbers, and hence cannot output
every real.
> Start printing an (irrational) real number > phi < End printing an
> (irrational) real number.
Congratulations on being able to type a greek letter.
Marshall


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Reply

marshall.spight (580)

5/8/2009 7:11:41 AM


On May 7, 3:56=A0am, Barb Knox <s...@sig.below> wrote:
> In article
> <c7ccbbf2af51448eb9da773e185ec...@r36g2000vbr.googlegroups.com>,
> > So a contradiction!
> > Thanks a lot!
> Face facts: Cantor's diagonal proof that R > N is SIMPLE. =A0It has
> even been CHECKED BY COMPUTERS.
Of course Knox isn't the first standard set theorist to mention
computer proofs. Metamath, Prover9, and other computer proving
algorithms have been mentioned in these set theory threads. But
here I agree with WM's response:
"[Computers] do what their programmers want."
And their programmers want them to formulate proofs in the
dominant theory, not in alternate theories.
On the other hand, computers can't completely do what their
programmers want. Surely typing in the Riemann Hypothesis or
Goldbach's Conjecture and clicking "Prove" isn't going to
result in a proof of either conjecture  at least not during
any of our lifetimes.
> Proving (say) ZFC to be inconsistent
> would be worth at least a Fields Medal and a centrefold picture in the
> Journal of the AMS. =A0If it were easy (or even less than extremely
> difficult) to do then it would already have been done. =A0The chance of
> you doing it is vanishingly small.
I actually agree with Knox here  but only partly.
As I've mentioned before, Cantor's original theory, naive set
theory, was quickly proved inconsistent by Russell. This is the
reason Zermelo came up with ZFC. And ZFC has lasted for over a
century without any inconsistency proofs. The irony here is:
The more time passes, the _more_ likely we'll see a proof of,
say, the Riemann Hypothesis or Goldbach's Conjecture.
The more time passes, the _less_ likely we'll see a proof that
ZFC is inconsistent.
And I doubt, like Knox and the other standard set theorists,
that such an inconsistency proof, if it ever appears, will have
anything to do with Cantor's Theorem.
Right now, the only mathematician of whom I am aware is making
an attempt at an inconsistency proof is Ed Nelson. He believes
that he can proof PA to be inconsistent  and since ZFC proves
that PA is consistent, if PA is proved to be inconsistent, then
ZFC would also be proved inconsistent.
If Nelson is successful, then some standard set theorists might
wonder why such a proof would have eluded them for so long when
compared to Russell's proof of the inconsistency of Cantor's
naive set theory. The reason is Nelson's attempt is based on
the operation of tetration, also called superexponentiation by
Nelson himself. Tetration, an operation which results in very
large numbers, is unknown to most of the set theorists who have
previously sought the Fields Medal due the prover of ~Con(ZFC),
and so it's not unreasonable that a proof that depends on
tetration would be unknown for so long.
If Nelson were to be successful, I suspect that an alternate
set theory to replace ZFC would develop, one which would avoid
the large numbers that led to Nelson's Paradox. Perhaps some
WMstyle ultrafinitism would be in order.
But regardless of whether anyone can or will ever prove ZFC
inconsistent, if someone is opposed to ZFC because it proves
the existence of uncountable sets, then I see no reason why
such opponents of ZFC can't have an alternate set theory, one
which does agree with some of their intuitions. If there were a
rigorous theory in which more of their intutions are provable
than in ZFC, then maybe LV and the other opponents of the
dominant set theory would have less reason to start these
threads and complain about set theory all the time.


0




Reply

lwalke3 (94)

5/8/2009 7:43:15 AM


On May 7, 4:28=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> On May 7, 4:08=A0pm, lwal...@lausd.net wrote:
> > nonCantorian sets
> What is a nonCantorian set?
In some theories other than ZFC, one can prove the existence
of nonCantorian sets. In particular, their existence is
provable in the theory NFU.
A Cantorian set is essentially a set that satisfy's Cantor's
theorem  so if x is a Cantorian set, then we must have
that card(x) < card(P(x)). So a nonCantorian set is a set
that doesn't satisfy Cantor's theorem. So if y is a
nonCantorian set, then card(y) >=3D card(P(y)).
Notice that NFU proves the existence of V, the set (not a
proper class, but a _set_) of all sets. Since P(V) is V
itself, it follows that the universal set must in fact be a
nonCantorian set.
IIRC, the formal definition of a nonCantorian set is a
set which fails to have the same cardinality as the set of
its singleton subsets. There is no bijection between V and
the set of all singletons, because the natural bijection
that maps each set to its own singleton is not a legal set
according to the Stratified Comprehension Schema. It's the
Stratified Comprehension Schema that allows there to be a
universal set yet still avoids Russell's Paradox.
But I assume (unless an NFU expert informs me otherwise)
that the ordinary sets of analysis, such as N and R, are
provably Cantorian in NFU. And so P(N), the powerset of N,
is still strictly larger than N. And NFU would still prove
that R is uncountable, presumably.
It might be interesting to see what would happen if there
were a set theory (not NFU) in which sets like N or R
turned out to be nonCantorian. Such a theory might
satisfy the opponents of ZFC in this thread.


0




Reply

lwalke3 (94)

5/8/2009 8:01:44 AM


On 7 Mag, 22:06, Virgil <vmh...@comcast.net> wrote:
> In article
> <c7ccbbf2af51448eb9da773e185ec...@r36g2000vbr.googlegroups.com>,
>
> =A0jesko <frans...@gmail.com> wrote:
> > If you map any real to N (Naturals) =A0then you may conceive Reals as
> > the infinite uinion of denumerable
> > sets since N is always denumarable. But this is union must be not
> > walkable or not denumarable.
>
> How do you propose to map R to N ?
It is a coomon fact that the number of all properties over N is not
denumerable.
Since anyway you list those properties another property not in the
list can be added.
so is R > N x N x N ...... x N
But if you fixed a list of those properties than you may add a new
property to the starting list
a your list is numbered.
But also with sinply Naturals you do the same!
This to make evident that N as R are not really numbered cause you can
find alway find a new elements in both case.
So infinite is not numbered by definition.


0




Reply

fransisf (9)

5/8/2009 8:59:05 AM


Marshall schrieb:
> On May 7, 11:18=A0pm, Albrecht <albst...@gmx.de> wrote:
> > Marshall schrieb:
> >
> >
> >
> > > On May 7, 9:24=A0am, LudovicoVan <ju...@diegidio.name> wrote:
> > > > On 7 May, 13:13, David C. Ullrich <dullr...@sprynet.com> wrote:
> > > > > On Wed, 6 May 2009 17:04:15 0700 (PDT), LudovicoVan
> > > > > <ju...@diegidio.name> wrote:
> >
> > > > > >It is (AFAIK) easy enough to come up with a production of _all_ =
the
> > > > > >infinite binary strings.
> >
> > > > > Exactly what does "come up with a production of" mean here?
> >
> > > > I mean "production" in the constructive sense: at least predicative=
..
> > > > Though I'm not putting this burden on Cantor's diagonal argument
> > > > itself, it's just the reference point for my counterargument. For
> > > > instance, one could give an inductive definition, then implement an=
d
> > > > run it in, say, Prolog. The program is not more "nonhalting" than =
a
> > > > program to output the naturals.
> >
> > > You can write a program to output all of the naturals. This program
> > > won't halt, as you say, and for any natural n, will eventually output
> > > n. (We are glossing over some resource limits here, but that is
> > > customary and reasonable.) It will never finish running, though;
> > > there are always more naturals.
> >
> > > You can't do the same for the reals. That's because the reals
> > > aren't countable and the naturals are.
> >
> > Really funny.
> > You say: "You can write a program to output =A0_all_ of the naturals."
> > I'm sure, you can't.
>
> 10 LET I=3D0
> 20 PRINT I
> 30 LET I =3D I + 1
> 40 GOTO 20
>
> There. I just wrote a program to output _all_ the naturals.
> Now surely you must abandon your claim that I can't, despite
> your initial surety, for clearly I have written the program.
This program doesn't output _all_ naturals. Sketch of a proof: let's
say your program outputs the number n. Since there are infinitely many
naturals N > n there are nearly all naturals which won't be output 
for every n!
>
>
> > You "believe" in the set of real numbers. Let's have this set and a
> > program which arbitrarily picks out numbers of this set and put them
> > out (building a new set). This program never stops, right? So, what is
> > the difference?
>
> The difference is the reals aren't countable. Also: every natural
> is finitely representable; not every real is. So once your program
> hits once such, it will never complete putting out its representation,
> so it will never output any further numbers, and hence cannot output
> every real.
10 LET I=3D0
15 LET r=3Drandom real number of R
20 PRINT r
25 LET R=3DR without r
30 LET I =3D I + 1
40 GOTO 15
So, what is the difference between naturals and reals?
>
>
> > Start printing an (irrational) real number > phi < End printing an
> > (irrational) real number.
>
> Congratulations on being able to type a greek letter.
>
Some people see a greek letter. Others are aware of an irrational
number  and are able to compute it.
Regards
Albrecht Storz


0




Reply

albstorz (110)

5/8/2009 11:01:12 AM


On Thu, 7 May 2009 09:24:40 0700 (PDT), LudovicoVan
<julio@diegidio.name> wrote:
>On 7 May, 13:13, David C. Ullrich <dullr...@sprynet.com> wrote:
>> On Wed, 6 May 2009 17:04:15 0700 (PDT), LudovicoVan
>> <ju...@diegidio.name> wrote:
>> >Mostly, a thought experiment:
>>
>> >Cantor claims: "whichever list of infinite (binary) strings you can
>> >come up with, I will (anti)diagonalise it."
>>
>> >It is (AFAIK) easy enough to come up with a production of _all_ the
>> >infinite binary strings.
>>
>> Exactly what does "come up with a production of" mean here?
>
>I mean "production" in the constructive sense: at least predicative.
That really doesn't answer my question.
>Though I'm not putting this burden on Cantor's diagonal argument
>itself, it's just the reference point for my counterargument. For
>instance, one could give an inductive definition, then implement and
>run it in, say, Prolog.
But _this_ is a good hint what you mean.
No, it is _not_ possible to write a Prolog program that produces
all infinite binary strings. You can _say_ that's "easy enough"
if you wish.
>The program is not more "nonhalting" than a
>program to output the naturals.
Halting or not is not the problem.
>> >The problem remains how to make the list
>> >bulletproof to diagonalisation. All we can do is falsify the
>> >diagonalisation procedure itself, by following the claim to the
>> >letter:
>>
>> >We start by producing our list, entry by entry; in parallel, our
>> >opponent starts building the antidiagonal (I will abstract from the
>> >technical details, unnecessary here):
>>
>> > � �List � Diagonal Antidiagonal
>> > � �.(0) � � � � 0 � � � � � � 1
>> > � �.(0)1 � � � �0 � � � � � � 1
>> > � � ... � � � � ... � � � � � ...
>>
>> >At this point, one migth wander: is there? is not there?
>>
>> Is there _what_?
>
>The antidiagonal: is it in the list? Is it not?
It is not. This is incredibly obvious.
>The proper question
>is: will the program ever output it's antidiagonal?
The "proper question" according to you asks a question
about a program that does not exist.
>> > � � � �=== Cantor's diagonal is a nonhalting machine. ===
>>
>> It's not a machine at all.
>
>Turing machines can be an interpretation tool here. Anti
>diagonalisation is a process, not yet a total function, because
>decisions about infinities are a consequence in our case, not a
>premise (they are yet to be taken).
>
>> You're talking about infinite strings of digits. Can you
>> give an example of a _halting_ machine that produces
>> such a thing? No.
>
>As said, asking for a machine that produces all infinite binary
>strings and halts is just as sensible as asking for a machine that
>produces all naturals and halts.
Yes it is. Who has ever said anything about a TM that produces
all the naturals and then halts?
>> This has nothing to do with the
>> diagonal argument  the question of halting machines
>> is simply irrelevant.
>
>> >Under this perspective, accepting the diagonal argument "simply"
>> >implies a specific choice about infinity (and infinities; and what can
>> >or cannot be proved; and a fundamental paradox: incongruent by
>> >construction).
>>
>> You're aware that this entire post is incoherent, right?
>
>It's all but incoherent. Maybe it's unclear, maybe it's incorrect in
>many ways, maybe there is something you are still missing, maybe a bit
>of all of them.
>
>In synthesis, the incongruency I am talking about could be seen in
>this fact: we are denied the existence of a complete list, in terms of
>a construct that is claimed to be complete.
>
>> >Now, what about Turing?
>>
>> What about Turing?
>
>Turing leverages the diagonal argument to show the insolubility of the
>halting problem (consequently failing Hilbert's program, by the way).
>In any case, I find Turing and his machines interesting here because,
>as far as I can see, Turing reasons at the same "primitive" (in the
>theoretical sense) level as Cantor.
>
>LV
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the postgrads."
in sci.logic.)


0




Reply

dullrich (221)

5/8/2009 11:55:36 AM


On May 8, 4:01=A0am, Albrecht <albst...@gmx.de> wrote:
> Marshall schrieb:
>
> > > You say: "You can write a program to output =A0_all_ of the naturals.=
"
> > > I'm sure, you can't.
>
> > 10 LET I=3D0
> > 20 PRINT I
> > 30 LET I =3D I + 1
> > 40 GOTO 20
>
> > There. I just wrote a program to output _all_ the naturals.
> > Now surely you must abandon your claim that I can't, despite
> > your initial surety, for clearly I have written the program.
>
> This program doesn't output _all_ =A0naturals. Sketch of a proof: let's
> say your program outputs the number n. Since there are infinitely many
> naturals N > n there are nearly all naturals which won't be output 
> for every n!
Oh, you didn't see the third and fourth line of
the program. Try to read *all* the lines; they all
do something.
Marshall


0




Reply

marshall.spight (580)

5/8/2009 1:18:41 PM


On 20090508 00:49:38 0400, Marshall <marshall.spight@gmail.com> said:
> On May 7, 9:38�pm, LudovicoVan <ju...@diegidio.name> wrote:
>> On 8 May, 05:08, Marshall <marshall.spi...@gmail.com> wrote:
>>
>>> Sorry about the interruption; I was going to type "'Cause I'm
>>> the sort of person who has a bandage on his index finger
>>
>> Wow, I had not yet realised you could be such a consistent moron.
>
> Yeah, you always were pretty slow to pick things up.
>
>
> Marshall
>
> PS. And I take objection to you calling me "consistent."
Would you prefer he called you a "complete" moron?
:),
o


0




Reply

angrybaldguy (338)

5/8/2009 1:35:11 PM


On May 8, 4:43=A0am, lwal...@lausd.net wrote:
> On May 7, 3:56=A0am, Barb Knox <s...@sig.below> wrote:
>
> > In article
> > <c7ccbbf2af51448eb9da773e185ec...@r36g2000vbr.googlegroups.com>,
> > > So a contradiction!
> > > Thanks a lot!
> > Face facts: Cantor's diagonal proof that R > N is SIMPLE. =A0It has
> > even been CHECKED BY COMPUTERS.
>
> Of course Knox isn't the first standard set theorist to mention
> computer proofs. Metamath, Prover9, and other computer proving
> algorithms have been mentioned in these set theory threads. But
> here I agree with WM's response:
>
> "[Computers] do what their programmers want."
>
> And their programmers want them to formulate proofs in the
> dominant theory, not in alternate theories.
Programmers are mostly theoryagnostic when they are
programming provers. What on earth suggests to you
that, say, Prover9 is restricted to (whatever it is
that you consider) dominant set theory? That you
seem to think there is such a restriction is a
clear sign that you have absolutely no idea what
an automatic prover is.
Your theory of "dominat theories" becomes more and
more absurd...
> On the other hand, computers can't completely do what their
> programmers want. Surely typing in the Riemann Hypothesis or
> Goldbach's Conjecture and clicking "Prove" isn't going to
> result in a proof of either conjecture  at least not during
> any of our lifetimes.
>
> > Proving (say) ZFC to be inconsistent
> > would be worth at least a Fields Medal and a centrefold picture in the
> > Journal of the AMS. =A0If it were easy (or even less than extremely
> > difficult) to do then it would already have been done. =A0The chance of
> > you doing it is vanishingly small.
>
> I actually agree with Knox here  but only partly.
>
> As I've mentioned before, Cantor's original theory, naive set
> theory, was quickly proved inconsistent by Russell. This is the
> reason Zermelo came up with ZFC. And ZFC has lasted for over a
> century without any inconsistency proofs. The irony here is:
>
> The more time passes, the _more_ likely we'll see a proof of,
> say, the Riemann Hypothesis or Goldbach's Conjecture.
> The more time passes, the _less_ likely we'll see a proof that
> ZFC is inconsistent.
You must operate under a rather strange definition of
what "irony" means. Moreover, your divining powers seem
to be quite great... What possible basis do you have
for the two claims you make?
> And I doubt, like Knox and the other standard set theorists,
> that such an inconsistency proof, if it ever appears, will have
> anything to do with Cantor's Theorem.
>
> Right now, the only mathematician of whom I am aware is making
> an attempt at an inconsistency proof is Ed Nelson. He believes
> that he can proof PA to be inconsistent  and since ZFC proves
> that PA is consistent, if PA is proved to be inconsistent, then
> ZFC would also be proved inconsistent.
>
> If Nelson is successful, then some standard set theorists might
> wonder why such a proof would have eluded them for so long when
> compared to Russell's proof of the inconsistency of Cantor's
> naive set theory. The reason is Nelson's attempt is based on
> the operation of tetration, also called superexponentiation by
> Nelson himself. Tetration, an operation which results in very
> large numbers, is unknown to most of the set theorists who have
> previously sought the Fields Medal due the prover of ~Con(ZFC),
> and so it's not unreasonable that a proof that depends on
> tetration would be unknown for so long.
>
> If Nelson were to be successful, I suspect that an alternate
> set theory to replace ZFC would develop, one which would avoid
> the large numbers that led to Nelson's Paradox. Perhaps some
> WMstyle ultrafinitism would be in order.
>
> But regardless of whether anyone can or will ever prove ZFC
> inconsistent, if someone is opposed to ZFC because it proves
> the existence of uncountable sets, then I see no reason why
> such opponents of ZFC can't have an alternate set theory, one
> which does agree with some of their intuitions. If there were a
> rigorous theory in which more of their intutions are provable
> than in ZFC, then maybe LV and the other opponents of the
> dominant set theory would have less reason to start these
> threads and complain about set theory all the time.
Can you please point to *someone* who *does* see
a reason why opponents of ZFC can't have an alternate
theory?
 m


0




Reply

mariano.suarezalvarez (417)

5/8/2009 5:36:02 PM


On May 7, 11:18=A0pm, Albrecht <albst...@gmx.de> wrote:
> And don't forget: ZFC must have a countable model.
Oh yeah, as if we were all going to forget about that.
You're silly.
MoeBlee


0




Reply

jazzmobe (307)

5/8/2009 5:54:50 PM


On May 8, 12:43 am, lwal...@lausd.net wrote:
> Cantor's original theory, naive set
> theory, was quickly proved inconsistent by Russell.
Russell proved Frege's formal system inconsistent. Cantor himself
already knew of the ordinal and cardinal paradoxes.
> This is the
> reason Zermelo came up with ZFC.
Would you please site where Zermelo said that.
> and since ZFC proves
> that PA is consistent, if PA is proved to be inconsistent, then
> ZFC would also be proved inconsistent.
Correct conlusion, but INCORRECT reasoning. If PA is inconsistent but
ZFC proves PA consistent, then all we can infer from those mere facts
is that ZFC proves false arithmetical statements, not necessarily that
ZFC is inconsistent. The actual reasoning for the inconsistency of ZFC
given the inconsistency of PA is that PA is embedded in ZFC. So this
is not a matter of whether ZFC proves PA consistent. Rather, if PA is
inconsistent then ZFC is inconsistent, since ZFC embeds PA.
> But regardless of whether anyone can or will ever prove ZFC
> inconsistent, if someone is opposed to ZFC because it proves
> the existence of uncountable sets, then I see no reason why
> such opponents of ZFC can't have an alternate set theory, one
> which does agree with some of their intuitions.
Sure. And this theory will be rich enough to do the kinds of things
ZFC does?
> If there were a
> rigorous theory in which more of their intutions are provable
> than in ZFC, then maybe LV and the other opponents of the
> dominant set theory would have less reason to start these
> threads and complain about set theory all the time.
There's no problem making such a theory. The problem is making such a
theory that is ALSO rich enough to do the kinds of things ZFC does.
MoeBlee


0




Reply

jazzmobe (307)

5/8/2009 6:11:48 PM


On May 8, 1:01=A0am, lwal...@lausd.net wrote:
> IIRC, the formal definition of a nonCantorian set is a
> set which fails to have the same cardinality as the set of
> its singleton subsets.
So y is a nonCantorian set <> card(y) ~=3D {{x}  xey}.
Okay, thanks.
MoeBlee


0




Reply

jazzmobe (307)

5/8/2009 6:15:35 PM


On May 8, 12:43=A0am, lwal...@lausd.net wrote:
> On May 7, 3:56=A0am, Barb Knox <s...@sig.below> wrote:
> >
> > Face facts: Cantor's diagonal proof that R > N is SIMPLE. =A0It has
> > even been CHECKED BY COMPUTERS.
>
> Of course Knox isn't the first standard set theorist to mention
> computer proofs. Metamath, Prover9, and other computer proving
> algorithms have been mentioned in these set theory threads. But
> here I agree with WM's response:
>
> "[Computers] do what their programmers want."
I can state with firm authority that programs almost
never do what their programmers want. Computation
follows its own logic, and is entirely immune to the
wishes, intent, or dress code of those who set it
in motion.
> And their programmers want them to formulate proofs
> in the dominant theory, not in alternate theories.
Prover9 takes your first order theory as input.
It doesn't have any way to tell if the input you gave
it is "dominant" or not.
> On the other hand, computers can't completely do what their
> programmers want. Surely typing in the Riemann Hypothesis or
> Goldbach's Conjecture and clicking "Prove" isn't going to
> result in a proof of either conjecture  at least not during
> any of our lifetimes.
A few years back you could have said the same thing
about the Robbins Conjecture. It had survived many
decades without anyone coming up with a proof or
a counterexample, and it had been worked on by the
likes of Huntington, Tarski, and of course Robbins.
Then one day Bill McCune typed the Robbins conjecture
into EQP, hit "Prove" and got out a proof. (EQP is a
precursor to Otter which is a precursor to Prover9.)
(I'm not a number theorist, but couldn't Goldbach's
conjecture be expressed as a first order sentence, within
some appropriate axiomatization of the naturals? If so,
it is entirely possible that someone will one day type
it in a hit "prove" and get a proof, if it is actually true.)
> But regardless of whether anyone can or will ever prove ZFC
> inconsistent, if someone is opposed to ZFC because it proves
> the existence of uncountable sets, then I see no reason why
> such opponents of ZFC can't have an alternate set theory, one
> which does agree with some of their intuitions.
My own feeling is that if someone is opposed to ZFC
because they would rather put flowers in their garden,
I see no reason why they can't plant some geraniums. And
I dare the niggling nabobs of natural numbers to naysay me.
Marshall


0




Reply

marshall.spight (580)

5/8/2009 6:16:21 PM


On May 8, 4:01=A0am, Albrecht <albst...@gmx.de> wrote:
> > 10 LET I=3D0
> > 20 PRINT I
> > 30 LET I =3D I + 1
> > 40 GOTO 20
>
> > There. I just wrote a program to output _all_ the naturals.
> > Now surely you must abandon your claim that I can't, despite
> > your initial surety, for clearly I have written the program.
>
> This program doesn't output _all_ =A0naturals.
Then please say what is the least natural number not outputted by the
program.
MoeBlee


0




Reply

jazzmobe (307)

5/8/2009 6:22:57 PM


Hi
I have some naive questions:
On 8 Maj, 13:01, Albrecht <albst...@gmx.de> wrote:
>> 10 LET I=0
>> 20 PRINT I
>> 30 LET I = I + 1
>> 40 GOTO 20
>
>This program doesn't output _all_ naturals. Sketch of a proof: let's
>say your program outputs the number n. Since there are infinitely many
>naturals N > n there are nearly all naturals which won't be output 
>for every n!
Can you supply a natural number n, that the program won't print?
> 10 LET I=0
> 15 LET r=random real number of R
> 20 PRINT r
> 25 LET R=R without r
> 30 LET I = I + 1
> 40 GOTO 15
>
> So, what is the difference between naturals and reals?
Can you please define the "function" returning: "random real number of
R"?
How do you represents the reals?
> Start printing an (irrational) real number > phi < End printing an
> (irrational) real number.
Following that approach  how will you represent/present all the other
transcendental real numbers? (e is of cause obvious...)
Kind regards, Jan


0




Reply

jan4778 (24)

5/8/2009 6:48:03 PM


On 7 May, 18:11, MoeBlee <jazzm...@hotmail.com> wrote:
> On May 7, 4:21=A0am, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> > On 7 Mai, 12:56, Barb Knox <s...@sig.below> wrote:
>
> > > Face facts: Cantor's diagonal proof that R > N is SIMPLE.
>
> > It presupposes actual infinity,
>
> If there is no set whose members are all and only the natural numbers,
> then the very question of comparison between N and R reduces to what
> 'N' and 'R' would stand for; and this is not a very interesting
> question.
As I have already noted in this thread, the formal, axiomatic proofs
presuppose what here is in question, mainly a decision about
"infinities".
LV


0




Reply

julio (505)

5/8/2009 7:25:57 PM


On 7 May, 19:54, MoeBlee <jazzm...@hotmail.com> wrote:
> Whether an intuitionist accepts the axioms used in the diagonal
> argument is one matter. But ASIDE from the axioms, the LOGIC used in
> the diagonal argument IS intuitionistic.
The diagonal argument is impredicative.
LV


0




Reply

julio (505)

5/8/2009 7:27:08 PM


On 7 May, 21:44, Virgil <vmh...@comcast.net> wrote:
> In article
> <f4166d965f734a8693ebe3c63924b...@t10g2000vbg.googlegroups.com>,
>
> =A0LudovicoVan <ju...@diegidio.name> wrote:
> > On 7 May, 11:56, Barb Knox <s...@sig.below> wrote:
>
> > > Face facts: Cantor's diagonal proof that R > N is SIMPLE. =A0It h=
as
> > > even been CHECKED BY COMPUTERS.
>
> > Formal proofs are irrelevant to the diagonal argument: they have it
> > embedded, as an informing principle, into their axioms and
> > definitions.
>
> > > Therefore the only way that R =3D N
> > > is if the various formulations of axiomatic set theory are internally
> > > INCONSISTENT
>
> > Incorrect (and a myopic). They can be formally consistent with unsound
> > or otherwise invalid premises.
>
> How does one test individual premises or sets of premises for
> unsoundness or invalidity? Unless one has some absolute standards and
> can make everyone agree with them, one can never be absolutely sure.
It's not a matter of being sure or of absolute standards. It's a
*choice*, possibly related to the requirements and constraints at
hand, and so on. Obviously.
> > Indeed, the diagonal argument here
> > under investigation, is an informing principle to standard axiomatic
> > set theories.
>
> But until YOUR standards of consistency and soundness are universally
> accepted, such questions are moot.
Your objection is senseless.
LV


0




Reply

julio (505)

5/8/2009 7:31:17 PM


On 7 May, 21:59, Virgil <vmh...@comcast.net> wrote:
> There is a sense in which Cantor's argument does work, i.e., for any
> constructible binary list, one can show by constructible methods that
> there is a constructible nonmember.
This is false, as a matter of fact. Cantor's argument is
impredicative, so constructively invalid.
LV


0




Reply

julio (505)

5/8/2009 7:32:34 PM


On 8 May, 00:03, Marshall <marshall.spi...@gmail.com> wrote:
> On May 7, 9:24=A0am, LudovicoVan <ju...@diegidio.name> wrote:
>
> > On 7 May, 13:13, David C. Ullrich <dullr...@sprynet.com> wrote:
> > > On Wed, 6 May 2009 17:04:15 0700 (PDT), LudovicoVan
> > > <ju...@diegidio.name> wrote:
>
> > > >It is (AFAIK) easy enough to come up with a production of _all_ the
> > > >infinite binary strings.
>
> > > Exactly what does "come up with a production of" mean here?
>
> > I mean "production" in the constructive sense: at least predicative.
> > Though I'm not putting this burden on Cantor's diagonal argument
> > itself, it's just the reference point for my counterargument. For
> > instance, one could give an inductive definition, then implement and
> > run it in, say, Prolog. The program is not more "nonhalting" than a
> > program to output the naturals.
>
> You can write a program to output all of the naturals. This program
> won't halt, as you say, and for any natural n, will eventually output
> n. (We are glossing over some resource limits here, but that is
> customary and reasonable.) It will never finish running, though;
> there are always more naturals.
>
> You can't do the same for the reals. That's because the reals
> aren't countable and the naturals are.
That ultimately depends on your stance on the diagonal argument.
[snip]
> So no, it is not easy enough to print out all the infinite binary
> strings; it's impossible actually to print out even one.
I tend to be quite informal. The actual form of such a program is
shown in this post by Alan Smaill:
http://groups.google.co.uk/group/sci.logic/msg/f97f03e6962c091a?hl=3Den
LV


0




Reply

julio (505)

5/8/2009 7:36:46 PM


On 7 May, 18:06, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> LudovicoVan <ju...@diegidio.name> writes:
> > On 7 May, 11:56, Barb Knox <s...@sig.below> wrote:
>
> > > Face facts: Cantor's diagonal proof that R > N is SIMPLE. =A0It h=
as
> > > even been CHECKED BY COMPUTERS.
>
> > Formal proofs are irrelevant to the diagonal argument: they have it
> > embedded, as an informing principle, into their axioms and
> > definitions.
>
> Why should we believe this assertion of yours?
> You have made it several times, and not yet given the slightest
> reason for anyone to believe it.
There has been a discussion specifically on this in the thread "Levy
proof that R is uncountable":
http://groups.google.co.uk/group/sci.logic/browse_frm/thread/876837fdd0fdfc=
6e/6d8e02441eb1e5f8
LV


0




Reply

julio (505)

5/8/2009 7:40:28 PM


On 7 May, 18:03, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> LudovicoVan <ju...@diegidio.name> writes:
> > The antidiagonal: is it in the list? Is it not? The proper question
> > is: will the program ever output it's antidiagonal?
>
> As you well know, a program will never produce infinite output.
> So your choice of question is seriously loaded  your program will
> never output a binary expansion of even one real number.
>
> Now, maybe this is a good way to think of the potential infinite
> and constructivism  but as you expressed it here, you need to
> be more evenhanded about what it means to construct a number (and a list
> of numbers).
I am fine with an approach like Bishop's as you show in another post,
with some provisions to follow.
> > As said, asking for a machine that produces all infinite binary
> > strings and halts is just as sensible as asking for a machine that
> > produces all naturals and halts.
>
> So what are *you* asking for when you ask whether the program can *output*
> its antidiagonal?
That's the key issue here: what's this antidiagonal at all. Note that
the OP is about falsifying the diagonalisation procedure: by showing
that the antidiagonal is not a valid construct. I have put this in
terms of an incongruency: an incomplete list by a complete anti
diagonal.
LV


0




Reply

julio (505)

5/8/2009 7:48:37 PM


On May 8, 12:36=A0pm, LudovicoVan <ju...@diegidio.name> wrote:
> On 8 May, 00:03, Marshall <marshall.spi...@gmail.com> wrote:
>
> > You can write a program to output all of the naturals. This program
> > won't halt, as you say, and for any natural n, will eventually output
> > n. (We are glossing over some resource limits here, but that is
> > customary and reasonable.) It will never finish running, though;
> > there are always more naturals.
>
> > You can't do the same for the reals. That's because the reals
> > aren't countable and the naturals are.
>
> That ultimately depends on your stance on the diagonal argument.
My stance controls what programs it is possible to write, you say?
So if I change my stance, a different set of things become
computable? I Did Not Know That.
> [snip]
>
> > So no, it is not easy enough to print out all the infinite binary
> > strings; it's impossible actually to print out even one.
>
> I tend to be quite informal. The actual form of such a program is
> shown in this post by Alan Smaill:
>
> http://groups.google.co.uk/group/sci.logic/msg/f97f03e6962c091a?hl=3Den
That shows how to write a program that will output an infinite
binary string? If I used a small font, will the whole string
fit on a single page?
Marshall


0




Reply

marshall.spight (580)

5/8/2009 7:54:34 PM


On 7 May, 18:30, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> Let's look at total functions taking two natural numbers and returning
> either 0 or 1, eg f(n,m). If we leave out the question of
> multiple representations, let's suppose you have implemented
> such a total function.
Great.
[snip]
> So, the antidiagonal is itself something given algorithmically,
> and you can easily write the procedure to return its nth digit,
> provided that the given listing of reals itself is a total
> function  this does not mean that all the digits involved must
> be computed, of course!
My point (my take) is that this simply does not solve our issue: the
fact that the antidiagonal does not belong to the list cannot be
proven from the algorithmic definition alone. The key issue is what
happens at infinity, and my thesis, to report it here in short, is
that the antidiagonal, as any other computable string, is simply
there, in the range of a putative function 'f' like you describe
above: there is no member missing. If we extend for
"compactification" (trying some possibly wrong terminology), the anti
diagonal is just a "limit point" of 'f'.
LV


0




Reply

julio (505)

5/8/2009 8:09:54 PM


On May 8, 12:25=A0pm, LudovicoVan <ju...@diegidio.name> wrote:
> On 7 May, 18:11, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On May 7, 4:21=A0am, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> > > On 7 Mai, 12:56, Barb Knox <s...@sig.below> wrote:
>
> > > > Face facts: Cantor's diagonal proof that R > N is SIMPLE.
>
> > > It presupposes actual infinity,
>
> > If there is no set whose members are all and only the natural numbers,
> > then the very question of comparison between N and R reduces to what
> > 'N' and 'R' would stand for; and this is not a very interesting
> > question.
>
> As I have already noted in this thread, the formal, axiomatic proofs
> presuppose what here is in question, mainly a decision about
> "infinities".
And I already responded to that claim. You just skipped dealing with
my response. Please, if there is to be any dialogoue, it is not
practical if you are going to just pretend that my responses don't
exist.
[start previous exchange (you're first):]
> In
> particular, any formal, axiomatic proof is absolutely of no relevance
> to an enquiry on the diagonal argument (and, strictly speaking, a way
> to beg the question), because it is derived from axioms that embed, as
> informing principles, the very results under investigation.
EVERY formal system embeds ALL the results that are provable from its
axioms. That is the NATURE of a formal system. Yes, OF COURSE, in Z
set theory, we adopt the axioms that will prove what we want the
axioms to prove (in Zermelo's particular case, it was to prove the
well ordering theorem from axioms including the axioms of choice).
Sure, granted, there are two basic approaches (and a combination of
them): (1) Adopt only axioms that express some principles that we
consider to be bedrock true and then let the chips fall where they may
(i.e., let whatever theorems come out or don't come out from the
bedrock principles). (2) Have in mind a whole bunch of theorems we
wish to axiomatize (such as the theorems that make up the mathematics
for the sciences) and then find axioms that prove those theorems. (3)
Some combination of (1) and (2).
It seems to me that set theory is in category (3). We'd like to have
an axiomatization for the mathematics for the sciences (at least) and
also we (editorial 'we') find the basic axioms THEMSELVES of set
theory to be quite faithful to our most basic notions of the
membership relation among sets.
So, if you're going to object to set theory on this basis, then what
category (1) axiomatization do you propose instead? And what are you
going to answer if that axoimatization doesn't even prove the basic
theorems for the mathematics for the sciences?
[end previous exchange]
So would you please address that question: What axiomatization do you
propose instead? And what are you
going to answer if that axoimatization doesn't even prove the basic
theorems for the mathematics for the sciences?
MoeBlee


0




Reply

jazzmobe (307)

5/8/2009 9:24:09 PM


LudovicoVan <julio@diegidio.name> writes:
> On 7 May, 19:54, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > Whether an intuitionist accepts the axioms used in the diagonal
> > argument is one matter. But ASIDE from the axioms, the LOGIC used in
> > the diagonal argument IS intuitionistic.
>
> The diagonal argument is impredicative.
So, where does impredicativity play a role in the argument?
> LV

Alan Smaill


0




Reply

smaill1 (89)

5/8/2009 9:24:53 PM


On May 8, 12:27=A0pm, LudovicoVan <ju...@diegidio.name> wrote:
> On 7 May, 19:54, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > Whether an intuitionist accepts the axioms used in the diagonal
> > argument is one matter. But ASIDE from the axioms, the LOGIC used in
> > the diagonal argument IS intuitionistic.
>
> The diagonal argument is impredicative.
If you are the same poster about a year ago under a different name
whom I'm thinking of, then it was *I* who first told you about
impredicativity. What is impredicative is the axiom schema of
separation. (But we figured out something with Aatu's help as to
whether the PARTICULAR instance of the axiom schema of separation used
in the diagonal argument is impredicative, though, alas, I've
forgotten what we concluded.)
In any case, my point stands, an intuitionist may not accept the
AXIOMS used, but the LOGIC used is purely intuitionistic.
MoeBlee


0




Reply

jazzmobe (307)

5/8/2009 9:28:00 PM


LudovicoVan <julio@diegidio.name> writes:
> On 7 May, 18:06, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > LudovicoVan <ju...@diegidio.name> writes:
> > > On 7 May, 11:56, Barb Knox <s...@sig.below> wrote:
> >
> > > > Face facts: Cantor's diagonal proof that R > N is SIMPLE. �It has
> > > > even been CHECKED BY COMPUTERS.
> >
> > > Formal proofs are irrelevant to the diagonal argument: they have it
> > > embedded, as an informing principle, into their axioms and
> > > definitions.
> >
> > Why should we believe this assertion of yours?
> > You have made it several times, and not yet given the slightest
> > reason for anyone to believe it.
>
> There has been a discussion specifically on this in the thread "Levy
> proof that R is uncountable":
>
> http://groups.google.co.uk/group/sci.logic/browse_frm/thread/876837fdd0fdfc6e/6d8e02441eb1e5f8
I have been following the thread.
You have rightly said that logical priority is different from historical
priority. At no point have you or anyone else given us any reason
to suppose that the diagonal argument is embedded into the axioms
and definitions of either classical or constructive theories
of the real numbers.
> LV

Alan Smaill


0




Reply

smaill1 (89)

5/8/2009 9:28:46 PM


Marshall schrieb:
>> 10 LET I=0
>> 20 PRINT I
>> 30 LET I = I + 1
>> 40 GOTO 20
>>
>> There. I just wrote a program to output _all_ the naturals.
>> Now surely you must abandon your claim that I can't, despite
>> your initial surety, for clearly I have written the program.
>
Albrecht wrote:
> This program doesn't output _all_ naturals. Sketch of a proof: let's
> say your program outputs the number n. Since there are infinitely many
> naturals N > n there are nearly all naturals which won't be output 
> for every n!
>
> 10 LET I=0
> 15 LET r=random real number of R
> 20 PRINT r
> 25 LET R=R without r
> 30 LET I = I + 1
> 40 GOTO 15
>
> So, what is the difference between naturals and reals?
Look closely at line 20. Given an arbitrary real value for r
(even something as simple as 1/3), how long will it take to
execute the PRINT statement?
There is also a problem in line 25. If R is a set (or some
other equivalent data structure), how does the 'without'
operator work?
Another problem is that your program will print only a
countably infinite subset of the reals. It will fail to print
(assuming it can get past line 20) the larger set of remaining
reals.


0




Reply

david1417 (99)

5/8/2009 9:31:47 PM


On May 8, 12:32=A0pm, LudovicoVan <ju...@diegidio.name> wrote:
> impredicative, so constructively invalid.
It is not at all clear that constructivism in itself requires absence
of impredicativity. Though, of course, certain currents in
constructivism do absence of predicativity.
Anyway, the point stands that while a constructivist might not accept
certain axioms, the LOGIC in the proof is intuitionistic.
Moreover, certain constructivists DO explicitly endorse that the proof
IS constructive. But, of course, that may be while we note that the
interpertation of the theorem in constructivism is different from in
classical mathematics.
MoeBlee


0




Reply

jazzmobe (307)

5/8/2009 9:32:29 PM


LudovicoVan <julio@diegidio.name> writes:
> On 7 May, 18:03, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > LudovicoVan <ju...@diegidio.name> writes:
> > > The antidiagonal: is it in the list? Is it not? The proper question
> > > is: will the program ever output it's antidiagonal?
> >
> > As you well know, a program will never produce infinite output.
> > So your choice of question is seriously loaded  your program will
> > never output a binary expansion of even one real number.
> >
> > Now, maybe this is a good way to think of the potential infinite
> > and constructivism  but as you expressed it here, you need to
> > be more evenhanded about what it means to construct a number (and a list
> > of numbers).
>
> I am fine with an approach like Bishop's as you show in another post,
> with some provisions to follow.
>
> > > As said, asking for a machine that produces all infinite binary
> > > strings and halts is just as sensible as asking for a machine that
> > > produces all naturals and halts.
> >
> > So what are *you* asking for when you ask whether the program can *output*
> > its antidiagonal?
>
> That's the key issue here: what's this antidiagonal at all. Note that
> the OP is about falsifying the diagonalisation procedure: by showing
> that the antidiagonal is not a valid construct. I have put this in
> terms of an incongruency: an incomplete list by a complete anti
> diagonal.
you need to say more about what you are getting at here. There are
too many terms (like "complete", "incongruency") where I don't know
what of several possible sense you intend. ("falsifying a procedure"
is already problematic, since procedures don't normally have truth values)
Maybe you can clarify.
> LV

Alan Smaill


0




Reply

smaill1 (89)

5/8/2009 9:33:05 PM


LudovicoVan <julio@diegidio.name> writes:
> On 7 May, 18:30, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>
> > Let's look at total functions taking two natural numbers and returning
> > either 0 or 1, eg f(n,m). If we leave out the question of
> > multiple representations, let's suppose you have implemented
> > such a total function.
>
> Great.
>
> [snip]
> > So, the antidiagonal is itself something given algorithmically,
> > and you can easily write the procedure to return its nth digit,
> > provided that the given listing of reals itself is a total
> > function  this does not mean that all the digits involved must
> > be computed, of course!
>
> My point (my take) is that this simply does not solve our issue: the
> fact that the antidiagonal does not belong to the list cannot be
> proven from the algorithmic definition alone. The key issue is what
> happens at infinity, and my thesis, to report it here in short, is
> that the antidiagonal, as any other computable string, is simply
> there, in the range of a putative function 'f' like you describe
> above: there is no member missing.
All that is claimed by the diagonal argument in this context is
that for every *natural number* n, the antidiagonal
differs from sequence f(n,0), f(n,1), f(n,2), ...
And that we can show fairly easily  it looks from your post
here that you accept that.
So the conclusion is that the antidiagonal does not appear
as a row in the *given* listing  nothing more than that.
> If we extend for
> "compactification" (trying some possibly wrong terminology), the anti
> diagonal is just a "limit point" of 'f'.
Nothing in the argument says that the antidiagonal appears in no
list at all, or doesn't correspond to something that can be defined
in terms of the given function f. For example, the antidiagonal
could appear by choosing the first digit from some row, the second
from another, and so on and so on.
It just doesn't appear as a row in the given effective listing.
> LV

Alan Smaill


0




Reply

smaill1 (89)

5/8/2009 9:42:37 PM


In article <fwebpq35lui.fsf@collins.inf.ed.ac.uk>,
Alan Smaill <smaill@SPAMinf.ed.ac.uk> wrote:
>LudovicoVan <julio@diegidio.name> writes:
>> On 7 May, 19:54, MoeBlee <jazzm...@hotmail.com> wrote:
>> > Whether an intuitionist accepts the axioms used in the diagonal
>> > argument is one matter. But ASIDE from the axioms, the LOGIC used in
>> > the diagonal argument IS intuitionistic.
>>
>> The diagonal argument is impredicative.
>
>So, where does impredicativity play a role in the argument?
Formation of power sets is impredicative, so if you conceive of "the diagonal
argument" as beginning by saying, "Let R be the power set of N..." then you
could argue that it is impredicative.
However, I think it is more plausible to say that "the diagonal argument"
is just the part of the argument that takes a given list and constructs a
diagonal element. This part isn't impredicative.

Tim Chow tchowatalumdotmitdotedu
The range of our projectileseven ... the artilleryhowever great, will
never exceed four of those miles of which as many thousand separate us from
the center of the earth. Galileo, Dialogues Concerning Two New Sciences


0




Reply

tchow (869)

5/8/2009 9:54:38 PM


On May 8, 2:54=A0pm, tc...@lsa.umich.edu wrote:
> In article <fwebpq35lui....@collins.inf.ed.ac.uk>,
> Alan Smaill =A0<sma...@SPAMinf.ed.ac.uk> wrote:
>
> >LudovicoVan <ju...@diegidio.name> writes:
> >> On 7 May, 19:54, MoeBlee <jazzm...@hotmail.com> wrote:
> >> > Whether an intuitionist accepts the axioms used in the diagonal
> >> > argument is one matter. But ASIDE from the axioms, the LOGIC used in
> >> > the diagonal argument IS intuitionistic.
>
> >> The diagonal argument is impredicative.
>
> >So, where does impredicativity play a role in the argument?
>
> Formation of power sets is impredicative, so if you conceive of "the diag=
onal
> argument" as beginning by saying, "Let R be the power set of N..." then y=
ou
> could argue that it is impredicative.
Or we can cast the argument in the hypothetical: IF there is a set N
whose members are all and only the natural numbers and IF there is a
set PN whose members are all and only the subsets of N, then there is
no function from N onto PN.
And, it seems to me, that the heart of diagonal argument doesn't even
require more than that hypothetical context, since if the hypotheses
fail, then the matter is trivial anyway.
> However, I think it is more plausible to say that "the diagonal argument"
> is just the part of the argument that takes a given list and constructs a
> diagonal element. =A0This part isn't impredicative.
That seems right to me, as long as the instance of axiom schema of
separation used to carve out the contradicting function is not
impredicative (Aatu had a convincing argument about that, but, alas, I
forgot his conclusion).
MoeBlee


0




Reply

jazzmobe (307)

5/8/2009 10:04:36 PM


MoeBlee wrote:
> If PA is inconsistent but
> ZFC proves PA consistent, then all we can infer from those mere facts
> is that ZFC proves false arithmetical statements, not necessarily that
> ZFC is inconsistent. The actual reasoning for the inconsistency of ZFC
> given the inconsistency of PA is that PA is embedded in ZFC. So this
> is not a matter of whether ZFC proves PA consistent. Rather, if PA is
> inconsistent then ZFC is inconsistent, since ZFC embeds PA.
Is "embeds" a standard term? I think in terms of "emulation":
ZFC can emulate PA (like one computer can emulate another).
I also wonder if it would be too far from standard usage to say
that ZFC "models" PA, not in the sense of there being sets that
are models of PA, but in the sense of one formalism embedding
(or emulating) another formalism.
This doesn't seem too distant to me; we have models of nonEuclidean
plane geometry in Euclidean solid geometry  I don't think anyone
would be seriously misled if one asserted that Euclidean geometry
models nonEuclidean geometry. I guess it is more common to say
that there is an interpretation of one in the other.
But what I'm really curious about is whether ZFC can be
embedded in PA; I don't see why not, offhand. PA can
"model" itself via the standard Godelian encoding  I don't
see any reason offhand why ZFC cannot be similarly encoded
within the formalism of PA. It is, after all, just a r.e.
set of symbols and a r.e. set of finite strings of those
symbols.
(We could get by with a finite number of symbols:
for variables write x, x', x'', ... so we just need
a code number for "x" and a code number for "'").

hz


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Reply

herbzet (87)

5/8/2009 11:37:21 PM


On May 6, 8:04=A0pm, LudovicoVan <ju...@diegidio.name> wrote:
> Cantor claims: "whichever list of infinite (binary) strings you can
> come up with, I will (anti)diagonalise it."
True.
> It is (AFAIK) easy enough to come up with a production
This is UTTER bullshit.
You CANNOT EVEN _DEFINE_ "production"!
ONE thing is for sure: IF this thing SOMEhow "deals with"
ALL binary strings, REGARDLESS of whether it is or isn't
the product of anything LEGITIMATELY referrable to as a
"production", it FOR DAMN sure WILL NOT be a LIST of these
strings!!


0




Reply

greeneg9613 (188)

5/8/2009 11:55:18 PM


On May 8, 4:37=A0pm, herbzet <herb...@gmail.com> wrote:
> MoeBlee wrote:
> > If PA is inconsistent but
> > ZFC proves PA consistent, then all we can infer from those mere facts
> > is that ZFC proves false arithmetical statements, not necessarily that
> > ZFC is inconsistent. The actual reasoning for the inconsistency of ZFC
> > given the inconsistency of PA is that PA is embedded in ZFC. So this
> > is not a matter of whether ZFC proves PA consistent. Rather, if PA is
> > inconsistent then ZFC is inconsistent, since ZFC embeds PA.
>
> Is "embeds" a standard term? =A0I think in terms of "emulation":
> ZFC can emulate PA (like one computer can emulate another).
Actually, I think "can be interepreted in" is a pretty common way of
putting it. Though informally, in context, 'embeds' (better, "embeds")
should convey the idea and, in context, be understood as distinct from
'embeds' in the sense of an embedding function. Though, I imagine we
could describe a certain embedding function from the fact that one
theory is interpreted in another  which would be literally an
embedding of the set of formulas of the language of theory T into the
set of formulas of the language of theory T*, with the isomorphism
appropriately defined by the property of theoremhood). But I don't
want to commit to working all of that out just to justify my informal
sense of 'embed' in this context.
> I also wonder if it would be too far from standard usage to say
> that ZFC "models" PA, not in the sense of there being sets that
> are models of PA, but in the sense of one formalism embedding
> (or emulating) another formalism.
The structure in Z
<w 0 S + *>
models first order PA.
>=A0I guess it is more common to say
> that there is an interpretation of one in the other.
Right.
> But what I'm really curious about is whether ZFC can be
> embedded in PA; =A0
Not first order PA. I think we can prove that by the second
incompleteness theorem, if I'm not mistaken (?).
> PA can
> "model" itself via the standard Godelian encoding
Then I think you're indulging some slippage in the notion of 'models'.
Godel encoding onto itself is not a provision for a model of first
order PA (in the ordinary sense in mathematical logic of 'model') nor
is Godel encoding onto itself a theory in which first order PA is
interpreted.
(Maybe there's some other sense in which Godel encoding can be
utlitized to perform certain other notions of 'model' and 'theory
interpreted in theory', but I don't see how in the exact senses of
those in mathematical logic.)
>  I don't
> see any reason offhand why ZFC cannot be similarly encoded
> within the formalism of PA. =A0It is, after all, just a r.e.
> set of symbols and a r.e. set of finite strings of those
> symbols.
Well, there's a lot more to interpreting one theory in another.
Roughly put, to interpret theory T in theory S, you have to extend S
by definitions to a theory S* and then show that every theorem of T is
a theorem of S* (or, for any axiomatization of T, that every axiom of
T is a theorem of S*).
That's easy to do for interpreting first order PA in Z set theory.
But to interpret Z set theory in first order PA you'd have have to
extend first order PA by definitions to a theory PA*, which would be
to add 'e' as a symbol and define it from PA, and then show that every
theorem (or just every axiom) of Z is a theorem of PA*.
(I hope I've stated that fairly accurately. But you can also check a
textbook in mathematical logic under the subject of interpreting one
theory in another, however it is actually worded in the book.)
MoeBlee


0




Reply

jazzmobe (307)

5/9/2009 12:56:40 AM


On May 8, 10:36=A0am, Mariano Su=E1rezAlvarez
<mariano.suarezalva...@gmail.com> wrote:
> On May 8, 4:43=A0am, lwal...@lausd.net wrote:
> > And their programmers want them to formulate proofs in the
> > dominant theory, not in alternate theories.
> Programmers are mostly theoryagnostic when they are
> programming provers. What on earth suggests to you
> that, say, Prover9 is restricted to (whatever it is
> that you consider) dominant set theory?
By "dominant set theory" I usually mean ZFC, i.e. the set
theory that most set theorists use.
Well, I suppose that Prover9 isn't limited to ZFC, since
Marshall says that he was able to use it to prove theorems
in a theory whose lone axiom is "Ax (Ay (x=3Dy))," which is
hardly equivalent to ZFC.
But what I've never seen is a theory that is proposed by a
socalled "crank" entered into Prover9, despite my attempts
to make the "crank" theories rigorous. Indeed, what I would
love to see is a rigorous version of a "crank" theory
entered into Prover9  and I'll consider my attempts to
make "crank" theories to be successful if my version of the
theory can be entered into Prover9 to prove theorems,
especially if those theorems are negations of those that
are provable in ZFC (assuming both ZFC and my proposed
theory to be consistent).
> That you seem to think there is such a restriction
There _is_ a restriction  Prover9 can only give proofs
in theories that someone types into the prover. If no
one enters a rigorous version of a "crank" theory, then
Prover9 won't prove anything in that theory. This is
probably what WM means when he says that computers do
only what their programmers program them to do.
> is a clear sign that you have absolutely no idea what
> an automatic prover is.
I'd like to know more about how Prover9 works. I've tried
asking Prover9 users how it works, but answers have not
been forthcoming.
If I don't know how something works, I'd much rather be
told how it does work, not simply "you don't know how
(whatever) works." But standard theorists on sci.math are
far more likely to say the latter than the former.
> > The more time passes, the _more_ likely we'll see a proof of,
> > say, the Riemann Hypothesis or Goldbach's Conjecture.
> > The more time passes, the _less_ likely we'll see a proof that
> > ZFC is inconsistent.
> What possible basis do you have for the two claims you make?
Mathematicians all over the world are searching for proofs
of Riemann and Goldbach. As more and more time passes, it
becomes more and more likely that someone will discover a
proof of one of these conjectures. It's more likely that a
proof of RH will be discovered by 2020 than by 2010, and
it's still more likely that a proof of RH will be found by
2100 than by 2020. Therefore, the more time passes, the
more likely that we'll see a proof of Riemann or Goldbach.
Mathematicians all over the world are also searching for
proofs that ZFC is inconsistent. But, as Knox points out,
if there really is a proof of ~Con(ZFC), we most likely
would have found it by now. For many inconsistent theories
it's easy to find a proof of its inconsistency. It took
only a few years to find a contradiction in Cantor's naive
set theory, and those "cranks" who propose theories on
sci.math, that turn out to be inconsistent, it often takes
mere _hours_ to find the inconsistency. But ZFC has
withstood the test of time. It seems hard to believe that
mathematicians would have overlooked an inconsistency for
so long  well over a century. Since we haven't found a
contradiction by now, it's unlikely that we'll see one by
2010 or 2020. And if we haven't found one by 2100, we may
as well consider ZFC to be consistent  since we'll
likely never find a proof of ~Con(ZFC). This seems to be
the essence of what Knox wrote in her post.
Since Mariano asked me, I ask the same of Mariano, and I
wonder what his opinion of the likelihood that we'll ever
see a proof of Riemann, Goldbach, or ~Con(ZFC).
> > If there were a
> > rigorous theory in which more of their intutions are provable
> > than in ZFC, then maybe LV and the other opponents of the
> > dominant set theory would have less reason to start these
> > threads and complain about set theory all the time.
> Can you please point to *someone* who *does* see
> a reason why opponents of ZFC can't have an alternate
> theory?
Here are some reasons given for why opponents of ZFC can't
have an alternate theory:
The alternate theories (i.e., the particular alternate
theory proposed by the particular opponent of ZFC) aren't
rigorous enough.
The alternate theories aren't coherent enough.
The alternate theories can't prove anything that is useful
to applied science.
The alternate theories are inconsistent (which even I
consider to be a valid reason).
And so on. I'll leave it to Mariano to find the posters
who gave these as reasons that opponents of ZFC can't
have their alternate theories.


0




Reply

lwalke3 (94)

5/9/2009 3:22:07 AM


On May 8, 8:22=A0pm, lwal...@lausd.net wrote:
> Mathematicians all over the world are also searching for
> proofs that ZFC is inconsistent.
They are? Would you please name several countries and a mathematician
from each of those countries who is searching for a contradiction in
ZFC? Or in the rubric of 'mathematician', do you include cranks?
MoeBlee


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Reply

jazzmobe (307)

5/9/2009 3:28:28 AM


On May 8, 11:16=A0am, Marshall <marshall.spi...@gmail.com> wrote:
> On May 8, 12:43=A0am, lwal...@lausd.net wrote:
> > And their programmers want them to formulate proofs
> > in the dominant theory, not in alternate theories.
> Prover9 takes your first order theory as input.
> It doesn't have any way to tell if the input you gave
> it is "dominant" or not.
What I'd love to see is someone enter a "crank" theory
(or a rigorous version thereof) into Prover9 and prove
theorems whose negations are theorems of ZFC (assuming
ZFC and the "crank" theory to be consistent). But no
one is going to enter a "crank" theory into Prover9
(not even a rigorous version of the theory). Surely
the standard set theorists won't do it.
> > On the other hand, computers can't completely do what their
> > programmers want. Surely typing in the Riemann Hypothesis or
> > Goldbach's Conjecture and clicking "Prove" isn't going to
> > result in a proof of either conjecture  at least not during
> > any of our lifetimes.
> A few years back you could have said the same thing
> about the Robbins Conjecture. It had survived many
> decades without anyone coming up with a proof or
> a counterexample, and it had been worked on by the
> likes of Huntington, Tarski, and of course Robbins.
> Then one day Bill McCune typed the Robbins conjecture
> into EQP, hit "Prove" and got out a proof. (EQP is a
> precursor to Otter which is a precursor to Prover9.)
Wow! Interesting.
> (I'm not a number theorist, but couldn't Goldbach's
> conjecture be expressed as a first order sentence, within
> some appropriate axiomatization of the naturals? If so,
> it is entirely possible that someone will one day type
> it in a hit "prove" and get a proof, if it is actually true.)
Actually, my comment was that if someone could enter
Goldbach (or its negation) into Prover9 (and let's
say that PA is this "appropriate axiomatization of
the naturals), it might take _years_, or even
_centuries_, from the time the input is first entered
to the time the proof is output. Similarly, one might
enter an RSA number into a factoring program and the
computer might find the factors, but only after many
_years_ of calculation. (This is the problem with
most JSH algorithms, of course.) No one will consider
a computer that factors RSA only after centuries of
calculation to be a threat to RSA security.


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Reply

lwalke3 (94)

5/9/2009 3:34:31 AM


On May 8, 8:28=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> On May 8, 8:22=A0pm, lwal...@lausd.net wrote:
> > Mathematicians all over the world are also searching for
> > proofs that ZFC is inconsistent.
> They are? Would you please name several countries and a mathematician
> from each of those countries who is searching for a contradiction in
> ZFC? Or in the rubric of 'mathematician', do you include cranks?
In this particular post, I'm not including socalled "cranks"
but only standard set theorists.
Well probably at _first_, right after ZFC was formulated,
likely as many mathematicians sought a proof that ZFC was
inconsistent as sought a proof that the naive set theory
of Cantor and Frege was inconsistent.
Knox wrote that a proof of ~Con(ZFC) would likely merit a
Fields medal. It's not unreasonable to guess that many
mathematicians around the world would do research that
would earn them a Fields medal, the most prestigious prize
in all of mathematics.
So if, as MoeBlee might be implying here, very _few_
mathematicians seek a proof of ~Con(ZFC), then they likely
believe that no proof is possible. And if this is really
the case, then this would actually _prove_ the point that
I was making to _Mariano_  that the more time passes,
the less likely a proof of ~Con(ZFC) will be found, and
the fewer mathematicians will even search for a proof!
AFAIK, there is only one mathematician seeking a proof
that PA and ZFC are inconsistent  Ed Nelson.
So thanks, MoeBlee, for helping me to prove my point to
Mariano here!


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Reply

lwalke3 (94)

5/9/2009 3:52:08 AM


tchow@lsa.umich.edu wrote:
> In article <fwebpq35lui.fsf@collins.inf.ed.ac.uk>,
> Alan Smaill <smaill@SPAMinf.ed.ac.uk> wrote:
>> LudovicoVan <julio@diegidio.name> writes:
>>> On 7 May, 19:54, MoeBlee <jazzm...@hotmail.com> wrote:
>>>> Whether an intuitionist accepts the axioms used in the diagonal
>>>> argument is one matter. But ASIDE from the axioms, the LOGIC used in
>>>> the diagonal argument IS intuitionistic.
>>> The diagonal argument is impredicative.
>> So, where does impredicativity play a role in the argument?
>
> Formation of power sets is impredicative, so if you conceive of "the diagonal
> argument" as beginning by saying, "Let R be the power set of N..." then you
> could argue that it is impredicative.
>
> However, I think it is more plausible to say that "the diagonal argument"
> is just the part of the argument that takes a given list and constructs a
> diagonal element. This part isn't impredicative.
Very true indeed. But that part of the argument does not prove the
existence of uncountable sets...

Cheers,
Herman Jurjus


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Reply

hjurjus (27)

5/9/2009 9:03:11 AM


lwalke3@lausd.net a �crit :
> On May 8, 8:28 pm, MoeBlee <jazzm...@hotmail.com> wrote:
>> On May 8, 8:22 pm, lwal...@lausd.net wrote:
>>> Mathematicians all over the world are also searching for
>>> proofs that ZFC is inconsistent.
>> They are? Would you please name several countries and a mathematician
>> from each of those countries who is searching for a contradiction in
>> ZFC? Or in the rubric of 'mathematician', do you include cranks?
>
> In this particular post, I'm not including socalled "cranks"
> but only standard set theorists.
So, can you answer his question?
>
> Well probably at _first_, right after ZFC was formulated,
> likely as many mathematicians sought a proof that ZFC was
> inconsistent as sought a proof that the naive set theory
> of Cantor and Frege was inconsistent.
>
> Knox wrote that a proof of ~Con(ZFC) would likely merit a
> Fields medal. It's not unreasonable to guess that many
> mathematicians around the world would do research that
> would earn them a Fields medal, the most prestigious prize
> in all of mathematics.
>
Is it concievable that you will stop trolling someday? Cannot you
recognize a joke here? Dont you understand taht a contradiction in ZFC
would be almost as unplausible that a contradiction inPA (which, of
course, would win you a Fields medal too)?
> So if, as MoeBlee might be implying here, very _few_
> mathematicians seek a proof of ~Con(ZFC), then they likely
> believe that no proof is possible.
Indeed
And if this is really
> the case, then this would actually _prove_ the point that
> I was making to _Mariano_  that the more time passes,
> the less likely a proof of ~Con(ZFC) will be found, and
> the fewer mathematicians will even search for a proof!
>
> AFAIK, there is only one mathematician seeking a proof
> that PA and ZFC are inconsistent  Ed Nelson.
>
> So thanks, MoeBlee, for helping me to prove my point to
> Mariano here!
No, just helping to prove you speak (as always) out of your hat


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Reply

denis.feldmann.sansspam1 (11)

5/9/2009 9:39:02 AM


In Dread Ink, the Grave Hand of Denis Feldmann Did Inscribe:
> lwalke3@lausd.net a �crit :
>> On May 8, 8:28 pm, MoeBlee <jazzm...@hotmail.com> wrote:
>>> On May 8, 8:22 pm, lwal...@lausd.net wrote:
>>>> Mathematicians all over the world are also searching for
>>>> proofs that ZFC is inconsistent.
>>> They are? Would you please name several countries and a mathematician
>>> from each of those countries who is searching for a contradiction in
>>> ZFC? Or in the rubric of 'mathematician', do you include cranks?
>>
>> In this particular post, I'm not including socalled "cranks"
>> but only standard set theorists.
>
> So, can you answer his question?
>>
>> Well probably at _first_, right after ZFC was formulated,
>> likely as many mathematicians sought a proof that ZFC was
>> inconsistent as sought a proof that the naive set theory
>> of Cantor and Frege was inconsistent.
>>
>> Knox wrote that a proof of ~Con(ZFC) would likely merit a
>> Fields medal. It's not unreasonable to guess that many
>> mathematicians around the world would do research that
>> would earn them a Fields medal, the most prestigious prize
>> in all of mathematics.
>>
>
>
> Is it concievable that you will stop trolling someday? Cannot you
> recognize a joke here? Dont you understand taht a contradiction in ZFC
> would be almost as unplausible that a contradiction inPA (which, of
> course, would win you a Fields medal too)?
>
>> So if, as MoeBlee might be implying here, very _few_
>> mathematicians seek a proof of ~Con(ZFC), then they likely
>> believe that no proof is possible.
>
>
>
> Indeed
> And if this is really
>> the case, then this would actually _prove_ the point that
>> I was making to _Mariano_  that the more time passes,
>> the less likely a proof of ~Con(ZFC) will be found, and
>> the fewer mathematicians will even search for a proof!
>>
>> AFAIK, there is only one mathematician seeking a proof
>> that PA and ZFC are inconsistent  Ed Nelson.
>>
>> So thanks, MoeBlee, for helping me to prove my point to
>> Mariano here!
> No, just helping to prove you speak (as always) out of your hat
I think the punchline of ZFC is Johnny von Neumann, who advises: don't
engage in bad logic games.

Frank
Drug war, well, as Rush Limbaugh said, anyone who uses drugs illegally
should be prosecuted and put away. I don't agree with him; I think they
should be treated, but that's what Rush believes and so, you know, we're
praying for Rush because he's in recovery and you take responsibilities for
your actions so I'm sure any day now Rush will demand to be put away for
the maximum sentence and ask for the most dangerous prison and we'll be
praying for maybe an African American cellmate who saw the Donovan McNabb
comments on ESPN. So we're prayin'.
~~ Al Franken, Book TV, on Rush Limbaugh's illegal drug arrest and racist
remarks


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Reply

frank21 (332)

5/9/2009 10:50:35 AM


On Fri, 8 May 2009 12:36:46 0700 (PDT), LudovicoVan
<julio@diegidio.name> wrote:
>On 8 May, 00:03, Marshall <marshall.spi...@gmail.com> wrote:
>> On May 7, 9:24�am, LudovicoVan <ju...@diegidio.name> wrote:
>>
>> > On 7 May, 13:13, David C. Ullrich <dullr...@sprynet.com> wrote:
>> > > On Wed, 6 May 2009 17:04:15 0700 (PDT), LudovicoVan
>> > > <ju...@diegidio.name> wrote:
>>
>> > > >It is (AFAIK) easy enough to come up with a production of _all_ the
>> > > >infinite binary strings.
>>
>> > > Exactly what does "come up with a production of" mean here?
>>
>> > I mean "production" in the constructive sense: at least predicative.
>> > Though I'm not putting this burden on Cantor's diagonal argument
>> > itself, it's just the reference point for my counterargument. For
>> > instance, one could give an inductive definition, then implement and
>> > run it in, say, Prolog. The program is not more "nonhalting" than a
>> > program to output the naturals.
>>
>> You can write a program to output all of the naturals. This program
>> won't halt, as you say, and for any natural n, will eventually output
>> n. (We are glossing over some resource limits here, but that is
>> customary and reasonable.) It will never finish running, though;
>> there are always more naturals.
>>
>> You can't do the same for the reals. That's because the reals
>> aren't countable and the naturals are.
>
>That ultimately depends on your stance on the diagonal argument.
No. The reals _are_ uncountable. This has nothing to do with
one's "stance" on the diagonal argument  the argument is
correct, regardless of anyone's "stance" on it.
>[snip]
>> So no, it is not easy enough to print out all the infinite binary
>> strings; it's impossible actually to print out even one.
>
>I tend to be quite informal. The actual form of such a program is
>shown in this post by Alan Smaill:
>
>http://groups.google.co.uk/group/sci.logic/msg/f97f03e6962c091a?hl=en
>
>LV
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the postgrads."
in sci.logic.)


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Reply

dullrich (221)

5/9/2009 12:44:56 PM


On Fri, 8 May 2009 13:09:54 0700 (PDT), LudovicoVan
<julio@diegidio.name> wrote:
>On 7 May, 18:30, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>
>> Let's look at total functions taking two natural numbers and returning
>> either 0 or 1, eg f(n,m). If we leave out the question of
>> multiple representations, let's suppose you have implemented
>> such a total function.
>
>Great.
>
>[snip]
>> So, the antidiagonal is itself something given algorithmically,
>> and you can easily write the procedure to return its nth digit,
>> provided that the given listing of reals itself is a total
>> function  this does not mean that all the digits involved must
>> be computed, of course!
>
>My point (my take) is that this simply does not solve our issue: the
>fact that the antidiagonal does not belong to the list cannot be
>proven from the algorithmic definition alone. The key issue is what
>happens at infinity, and my thesis, to report it here in short, is
>that the antidiagonal, as any other computable string, is simply
>there, in the range of a putative function 'f' like you describe
>above: there is no member missing. If we extend for
>"compactification" (trying some possibly wrong terminology), the anti
>diagonal is just a "limit point" of 'f'.
So what? Yes, there is no problem whatever down a list of reals,
in fact rationals, such that every real number is a _limit point_
of the sequence. That's not news, that's not controversial, that's
awesomely well known.
A _limit point_ of the range of f is not the same thing as an
_element of_ the range of f. The assertion is that there is
a real number that is not _in_ the range of f.
In case you don't see the difference, let's talk about a related
situation where there's none of the nasty infinity to worry
about. Define S to be the set of all real numbers x such
that 0 < x < 1. Is 1 in S? _No_, 1 is not in S, because 1
is not less than 1. Is 1 a limit point of S? Yes.
This was explained the last time someone (you?)
started talking about compactifications and limit
points, by the way.
What you say about limit points is true, or at least becomes
true if we insert suitable definitions. But it has no relevance
whatever to the validity of the diagonal argument. It's like
I claim to have a proof that there does not exist a red cow,
and you refute my claim by showing me a red tomato.
Yes, there are red tomatoes  this has nothing to do with
whether red cows exist.
If you really think what you said is relavant to the
uncountability of the reals and/or the validity of the
diagonal argument you're simply misunderstanding
the meaning of something.
>LV
David C. Ullrich
"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the postgrads."
in sci.logic.)


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Reply

dullrich (221)

5/9/2009 12:53:59 PM


lwalke3@lausd.net writes:
> There _is_ a restriction  Prover9 can only give proofs
> in theories that someone types into the prover.
Oh my! How unfair!
Why won't someone drop whatever they're doing and start working on
your formalization of crank theories? It's just not fair that they
don't.
This is just as tragic as the fact that books cost money and you have
to travel to libraries.

Jesse F. Hughes
"Of course, I don't need any more education."
 Quincy P. Hughes (age 7)


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Reply

jesse18 (2492)

5/9/2009 1:08:30 PM


On May 8, 8:34=A0pm, lwal...@lausd.net wrote:
> On May 8, 11:16=A0am, Marshall <marshall.spi...@gmail.com> wrote:
> > On May 8, 12:43=A0am, lwal...@lausd.net wrote:
> > > And their programmers want them to formulate proofs
> > > in the dominant theory, not in alternate theories.
> > Prover9 takes your first order theory as input.
> > It doesn't have any way to tell if the input you gave
> > it is "dominant" or not.
>
> What I'd love to see is someone enter a "crank" theory
> (or a rigorous version thereof) into Prover9 and prove
> theorems whose negations are theorems of ZFC (assuming
> ZFC and the "crank" theory to be consistent). But no
> one is going to enter a "crank" theory into Prover9
> (not even a rigorous version of the theory). Surely
> the standard set theorists won't do it.
What you're asking is pretty easy, actually, at least
in the degenerate case. Type in the negation of the
axiom of infinity and viola!
The problem of course isn't doing proofs from a formal
theory, whether "crank" or no; the problem is in trying
to formalize a crank "theory." Formalization proceeds
from clarity, and clarity can be hard to come by
sometimes.
> > > On the other hand, computers can't completely do what their
> > > programmers want. Surely typing in the Riemann Hypothesis or
> > > Goldbach's Conjecture and clicking "Prove" isn't going to
> > > result in a proof of either conjecture  at least not during
> > > any of our lifetimes.
> > A few years back you could have said the same thing
> > about the Robbins Conjecture. It had survived many
> > decades without anyone coming up with a proof or
> > a counterexample, and it had been worked on by the
> > likes of Huntington, Tarski, and of course Robbins.
> > Then one day Bill McCune typed the Robbins conjecture
> > into EQP, hit "Prove" and got out a proof. (EQP is a
> > precursor to Otter which is a precursor to Prover9.)
>
> Wow! Interesting.
Yeah, crazy cool. I love Prover9. Mace4, which didn't
attract me as much at first, is comparably cool. The
original paper on how they implemented it was a fun
read for me.
A math teacher of mine years ago was asked by a
fellow student what he meant by "elegant." He said
a proof is elegant if, when you see it, you wish you
had thought of it.
I consider Mace4 elegant.
> > (I'm not a number theorist, but couldn't Goldbach's
> > conjecture be expressed as a first order sentence, within
> > some appropriate axiomatization of the naturals? If so,
> > it is entirely possible that someone will one day type
> > it in a hit "prove" and get a proof, if it is actually true.)
>
> Actually, my comment was that if someone could enter
> Goldbach (or its negation) into Prover9 (and let's
> say that PA is this "appropriate axiomatization of
> the naturals), it might take _years_, or even
> _centuries_, from the time the input is first entered
> to the time the proof is output.
Oh. Yeah, that's certainly possible.
Marshall


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Reply

marshall.spight (580)

5/9/2009 2:50:34 PM


On 9 May, 13:44, David C. Ullrich <dullr...@sprynet.com> wrote:
> On Fri, 8 May 2009 12:36:46 0700 (PDT), LudovicoVan
> <ju...@diegidio.name> wrote:
> >On 8 May, 00:03, Marshall <marshall.spi...@gmail.com> wrote:
> >> You can't do the same for the reals. That's because the reals
> >> aren't countable and the naturals are.
>
> >That ultimately depends on your stance on the diagonal argument.
>
> No. The reals _are_ uncountable. This has nothing to do with
> one's "stance" on the diagonal argument  the argument is
> correct, regardless of anyone's "stance" on it.
Apart from the apparent dogmatism, what you state is incorrect and,
strictly speaking, false. Whether the diagonal argument or the
arguments for uncontability are acceptable or not properly depends on
one's stance on some key foundational issues.
LV


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Reply

julio (505)

5/9/2009 3:19:26 PM


On 8 May, 22:54, tc...@lsa.umich.edu wrote:
> In article <fwebpq35lui....@collins.inf.ed.ac.uk>,
> Alan Smaill =A0<sma...@SPAMinf.ed.ac.uk> wrote:
>
> >LudovicoVan <ju...@diegidio.name> writes:
> >> On 7 May, 19:54, MoeBlee <jazzm...@hotmail.com> wrote:
> >> > Whether an intuitionist accepts the axioms used in the diagonal
> >> > argument is one matter. But ASIDE from the axioms, the LOGIC used in
> >> > the diagonal argument IS intuitionistic.
>
> >> The diagonal argument is impredicative.
>
> >So, where does impredicativity play a role in the argument?
>
> Formation of power sets is impredicative, so if you conceive of "the diag=
onal
> argument" as beginning by saying, "Let R be the power set of N..." then y=
ou
> could argue that it is impredicative.
>
> However, I think it is more plausible to say that "the diagonal argument"
> is just the part of the argument that takes a given list and constructs a
> diagonal element. =A0This part isn't impredicative.
Not that part, it's the conclusion that is "problematic". We have a
construction of the list (our 'f'), then we define the construction of
an item ('d') so that it differs in each place for each item in the
list given by 'f'. If we now run this game of 'f' against 'd', there
is no winner, yet Cantor's conclusion is that the antidiagonal wins.
LV


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Reply

julio (505)

5/9/2009 3:32:13 PM


On 8 May, 22:28, MoeBlee <jazzm...@hotmail.com> wrote:
> On May 8, 12:27=A0pm, LudovicoVan <ju...@diegidio.name> wrote:
> > On 7 May, 19:54, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > > Whether an intuitionist accepts the axioms used in the diagonal
> > > argument is one matter. But ASIDE from the axioms, the LOGIC used in
> > > the diagonal argument IS intuitionistic.
>
> > The diagonal argument is impredicative.
>
> If you are the same poster about a year ago under a different name
> whom I'm thinking of, then it was *I* who first told you about
> impredicativity.
I'm not, though I am glad to admit that I am learning a lot from the
group. How to fight in the crowd, above all.
LV


0




Reply

julio (505)

5/9/2009 3:33:46 PM


On 8 May, 22:24, MoeBlee <jazzm...@hotmail.com> wrote:
> On May 8, 12:25=A0pm, LudovicoVan <ju...@diegidio.name> wrote:
> > As I have already noted in this thread, the formal, axiomatic proofs
> > presuppose what here is in question, mainly a decision about
> > "infinities".
>
> And I already responded to that claim. You just skipped dealing with
> my response. Please, if there is to be any dialogoue, it is not
> practical if you are going to just pretend that my responses don't
> exist.
It's you who are pretending my responses do not exist.
[snip]
> So would you please address that question: What axiomatization do you
> propose instead?
I have already answered this question of yours many times now. Maybe I
was not clear, so I'll try again: we are investigating foundational
issues here, the diagonal argument specifically; it is improper, if
not a basic misunderstanding, to ask for axiomatisations in this
context.
LV


0




Reply

julio (505)

5/9/2009 3:39:05 PM


On 8 May, 22:42, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> LudovicoVan <ju...@diegidio.name> writes:
> > On 7 May, 18:30, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>
> > > Let's look at total functions taking two natural numbers and returnin=
g
> > > either 0 or 1, eg f(n,m). =A0If we leave out the question of
> > > multiple representations, let's suppose you have implemented
> > > such a total function.
>
> > Great.
>
> > [snip]
> > > So, the antidiagonal is itself something given algorithmically,
> > > and you can easily write the procedure to return its nth digit,
> > > provided that the given listing of reals itself is a total
> > > function  this does not mean that all the digits involved must
> > > be computed, of course!
>
> > My point (my take) is that this simply does not solve our issue: the
> > fact that the antidiagonal does not belong to the list cannot be
> > proven from the algorithmic definition alone. The key issue is what
> > happens at infinity, and my thesis, to report it here in short, is
> > that the antidiagonal, as any other computable string, is simply
> > there, in the range of a putative function 'f' like you describe
> > above: there is no member missing.
>
> All that is claimed by the diagonal argument in this context is
> that for every *natural number* n, the antidiagonal
> differs from sequence f(n,0), f(n,1), f(n,2), ...
> And that we can show fairly easily  it looks from your post
> here that you accept that.
>
> So the conclusion is that the antidiagonal does not appear
> as a row in the *given* listing  nothing more than that.
The conclusion is still unwarranted: "the fact that the antidiagonal
does not belong to the list cannot be proven from the algorithmic
definition alone".
LV


0




Reply

julio (505)

5/9/2009 3:41:43 PM


On 7 Mai, 22:59, Virgil <vmh...@comcast.net> wrote:
> > > So I suppose the next question is, does Cantor's argument work in
> > > intuitionist mathematics? I'm afraid I don't know the answer to that
> > > one
>
> > It is asserted, erroneously, that it does.
>
> Which is WM's own erroneous assertion.
>
> There is a sense in which Cantor's argument does work, i.e., for any
> constructible binary list,
No. That is possible only for lists with entries of infinite length.
However, it is impossible to construct any sequence of bits which can
be identified with certainty as a certain irrational number. Therefore
the constructibility argument is wrong.
> one can show by constructible methods that
> there is a constructible nonmember
Neither any member nor the nonmember can be identified.
Therefore that is not mathematics but at most matheology.
Regards, WM


0




Reply

mueckenh (275)

5/9/2009 3:50:40 PM


On 8 May, 22:24, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> LudovicoVan <ju...@diegidio.name> writes:
> > On 7 May, 19:54, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > > Whether an intuitionist accepts the axioms used in the diagonal
> > > argument is one matter. But ASIDE from the axioms, the LOGIC used in
> > > the diagonal argument IS intuitionistic.
>
> > The diagonal argument is impredicative.
>
> So, where does impredicativity play a role in the argument?
There is a lot of literature on this even available on the web. There
are indeed subtler distinctions than I am making here.
Here are few links I have collected:
http://groups.google.co.uk/group/sci.logic/msg/d295e5630c745dee
http://www.cs.nyu.edu/pipermail/fom/2001February/004737.html
http://www.tc.umn.edu/~hellm001/PredImp.pdf
http://books.google.co.uk/books?id=TM3AKPYdQVgC&pg=PA74
LV


0




Reply

julio (505)

5/9/2009 3:53:05 PM


On 7 Mai, 23:07, Virgil <vmh...@comcast.net> wrote:
> In article
> <090e5499f16b4ff58a6d673e12373...@m24g2000vbp.googlegroups.com>,
>
>
>
>
>
> =A0WM <mueck...@rz.fhaugsburg.de> wrote:
> > On 7 Mai, 18:18, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > > As others have noted, if you are serious about constructivism, so tha=
t
> > > not only the decimal expansions of the reals is taken to be
> > > constructive, but also an enumeration of such reals is itself given
> > > constructively, then lo and behold it is possible to construct effect=
ively
> > > a real not in the given list.
>
> > > This argument is given eg by Bishop in his book "Constructive Analysi=
s".
> > > It might help such as LV to note that Bishop does not state the claim=
in
> > > the form that one set is larger than another, but in the form that sa=
ys
> > > that for any effectively given enumeration, a missing real can be
> > > constructed.
>
> > That is only true if the reals are given as infinite sequences of
> > digits (which is in principle impossible). If they are given in the
> > form of finite words, the argument fails because the lis of all finite
> > words does not allow a diagonal:
>
> If a real is defined well enough in words to be constructively
> distinguished from any other equally well defined real,
that is possible,
> its decimal
> expansion to any desired finite extent should be finitely constructible,
Nevertheless this expansion does not help anything at all because it
is not the expansion that distinguishes the real from all others but
the finite definition.
Regards, WM


0




Reply

mueckenh (275)

5/9/2009 3:53:36 PM


On 8 May, 22:28, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> LudovicoVan <ju...@diegidio.name> writes:
> > On 7 May, 18:06, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > > LudovicoVan <ju...@diegidio.name> writes:
> > > > On 7 May, 11:56, Barb Knox <s...@sig.below> wrote:
>
> > > > > Face facts: Cantor's diagonal proof that R > N is SIMPLE. =A0=
It has
> > > > > even been CHECKED BY COMPUTERS.
>
> > > > Formal proofs are irrelevant to the diagonal argument: they have it
> > > > embedded, as an informing principle, into their axioms and
> > > > definitions.
>
> > > Why should we believe this assertion of yours?
> > > You have made it several times, and not yet given the slightest
> > > reason for anyone to believe it.
>
> > There has been a discussion specifically on this in the thread "Levy
> > proof that R is uncountable":
>
> >http://groups.google.co.uk/group/sci.logic/browse_frm/thread/876837fd...
>
> I have been following the thread.
>
> You have rightly said that logical priority is different from historical
> priority. At no point =A0have you or anyone else given us any reason
> to suppose that the diagonal argument is embedded into the axioms
> and definitions of either classical or constructive theories
> of the real numbers.
You can say that nobody has explicitly acknowledged it, but this point
is clear enough. I just won't go back collecting links on this too, I
can't take the burden of what is rather a problem of "generalised
denial".
LV


0




Reply

julio (505)

5/9/2009 3:58:54 PM


On 8 May, 22:33, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> LudovicoVan <ju...@diegidio.name> writes:
> > On 7 May, 18:03, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > > So what are *you* asking for when you ask whether the program can *ou=
tput*
> > > its antidiagonal?
>
> > That's the key issue here: what's this antidiagonal at all. Note that
> > the OP is about falsifying the diagonalisation procedure: by showing
> > that the antidiagonal is not a valid construct. I have put this in
> > terms of an incongruency: an incomplete list by a complete anti
> > diagonal.
>
> you need to say more about what you are getting at here. =A0There are
> too many terms (like "complete", "incongruency") where I don't know
> what of several possible sense you intend. =A0("falsifying a procedure"
> is already problematic, since procedures don't normally have truth values=
)
>
> Maybe you can clarify.
I'afraid, not more than I have already tried. For instance you might
see my reply to Mr Chow:
[quote]
Not that part, it's the conclusion that is "problematic". We have a
construction of the list (our 'f'), then we define the construction
of
an item ('d') so that it differs in each place for each item in the
list given by 'f'. If we now run this game of 'f' against 'd', there
is no winner, yet Cantor's conclusion is that the antidiagonal wins.
[/quote]
http://groups.google.co.uk/group/sci.logic/msg/018ce332581f71d5
LV


0




Reply

julio (505)

5/9/2009 4:03:15 PM


On 8 Mai, 03:32, David R Tribble <da...@tribble.com> wrote:
> WM wrote:
> >> The computed number differs from every entry
> >> checked up to the nth line.
> >> It does not differ from all lines, given that
> >> there are more than n + 1 for every n.
> >> It does not differ from all lines unless the
> >> list can be checked completely.
>
> James Burns wrote:
> >> Another consequence of your rule is
> >> One (1) does not differ from all even integers unless the
> >> list (of even integers) can be checked completely.
>
> WM wrote:
> > No. If you know that an element exists only once in the world, ...
>
> And how do you know that?
I don't know that. But I know that 1 is unique in N.
>
> > ... you need not search the whole universe in order to find a second one if
> > you have got the first.
> > But, yes, in order to be sure that 1 (or any other number) does not
> > appear in a sequence you must know the whole sequence.
>
> Or perhaps you need to only know a property shared by every
> element in the sequence. For example, every element in the
> sequence of even integers is even. This is true regardless
> of how many elements have been "searched" or how many
> remain to be "searched".
This is true if it is possible that a meaning can be given to the word
"every".
Regards, WM


0




Reply

mueckenh (275)

5/9/2009 4:06:30 PM


On May 9, 11:32=A0am, LudovicoVan <ju...@diegidio.name> wrote:
> On 8 May, 22:54, tc...@lsa.umich.edu wrote:
> > However, I think it is more plausible to say that "the diagonal argumen=
t"
> > is just the part of the argument that takes a given list and constructs=
a
> > diagonal element. =A0This part isn't impredicative.
>
> Not that part, it's the conclusion that is "problematic".
It's really a shame you don't know how this works.
IT'S A PROOF. IN A LOGIC.
If the rules and their application are sound then it
IS NOT POSSIBLE for the conclusion to be "problematic"!
> We have a construction of the list (our 'f'),
NO, WE DON'T!
f DOES NOT need to be "constructed" or "constructible"
IN ANY sense! This works for ALL lists, constructible OR NOT!!
The inference rule being used is universal generalization.
What part of UNIVERSAL don't you understand?!?
f IS AN *ARBITRARY* name/symbol in this proof!!
We know ABSOLUTELY NOTHING about f, because absolutely
nothing about f EVEN MATTERS in this proof, beyond that it
defines a list!!
> then we define the construction of
> an item ('d') so that it differs in each place for each item in the
> list given by 'f'. If we now run this game of 'f' against 'd',
You CAN'T run this game.
There ARE NO games in this paradigm.
In what you call a game, people TAKE TURNS and TIME
progesses. That IS NOT HAPPENING here.
We are using universal generalization AGAIN both in the
definition of the (anti) diagonal and over the places of the numbers
(horizontally) in the list: the antidiagonal differs from EVERY row
at the SAME time. It is true of ANY row, no matter WHEN you consider
it,
that the antidiagonal differs from it. There is no d VERSUS f here!
d is constructed FROM f !!


0




Reply

greeneg9613 (188)

5/9/2009 4:08:40 PM


On 8 Mai, 03:32, David R Tribble <da...@tribble.com> wrote:
> > ... you need not search the whole universe in order to find a second one if
> > you have got the first.
> > But, yes, in order to be sure that 1 (or any other number) does not
> > appear in a sequence you must know the whole sequence.
>
> Or perhaps you need to only know a property shared by every
> element in the sequence. For example, every element in the
> sequence of even integers is even. This is true regardless
> of how many elements have been "searched" or how many
> remain to be "searched".
One such property is that every set of natural numbers has even or odd
cardinality.
If you know this, you can exclude that there is a set of natural
numbers with cardinal number aleph_0 (because aleph_0 + 1 = aleph_0).
Yes, such conjectures could be done, if infinity would follow the same
logic as finity. But it seems it doesn't. So why do you believe it
does in case of Cantor's simple idea?
Regards, WM


0




Reply

mueckenh (275)

5/9/2009 4:11:19 PM


On May 9, 11:41=A0am, LudovicoVan <ju...@diegidio.name> wrote:
> The conclusion is still unwarranted: "the fact that the antidiagonal
> does not belong to the list cannot be proven from the algorithmic
> definition alone".
So who needs an algorithmic definition?
Are you trying to say that only algorithmic definitions are
legitimate?
In the case of the axioms from which it IS proved, the antidiagonal
is DEFINED as differing from every element on the list (it differs
from
the nth element in its nth position). In that case, there is nothing
(more)
TO prove!


0




Reply

greeneg9613 (188)

5/9/2009 4:14:10 PM


On 8 Mai, 09:43, lwal...@lausd.net wrote:
> Right now, the only mathematician of whom I am aware is making
> an attempt at an inconsistency proof is Ed Nelson. He believes
> that he can proof PA to be inconsistent  and since ZFC proves
> that PA is consistent, if PA is proved to be inconsistent, then
> ZFC would also be proved inconsistent.
Here is another proof:
http://www.hsaugsburg.de/~mueckenh/GU/GU12.PPT#361,25,Folie 25
Regards, WM


0




Reply

mueckenh (275)

5/9/2009 4:22:18 PM


On 8 Mai, 20:48, Jan <j...@daimi.au.dk> wrote:
> Can you supply a natural number n, that the program won't print?
No. Can you understand that at *any* step there are infinitely many
numbers remaining that have not been printed and will never be
printed, because they belong to the set that is remaining after any
step?
This is a contradiction contradicting the axiom of infinity.
"... classical logic was abstracted from the mathematics of finite
sets and their subsets .... Forgetful of this limited origin, one
afterwards mistook that logic for something above and prior to all
mathematics, and finally applied it, without justification, to the
mathematics of infinite sets. ... As Brouwer pointed out this is a
fallacy, the Fall and Original sin of set theory even if no paradoxes
result from it." (Hermann Weyl)
Regards, WM


0




Reply

mueckenh (275)

5/9/2009 4:29:44 PM


On May 9, 12:11 pm, WM <mueck...@rz.fhaugsburg.de> wrote:
WM continues to deliberately confuse elements and sets of elements.
One such property is that every natural number has even or odd
cardinality.
Look! Over There! A Pink Elephant!
> One such property is that every set of natural numbers has even or odd
> cardinality.
 William Hughes


0




Reply

wpihughes (390)

5/9/2009 4:38:59 PM


On May 9, 12:29 pm, WM <mueck...@rz.fhaugsburg.de> wrote:
> On 8 Mai, 20:48, Jan <j...@daimi.au.dk> wrote:
>
> > Can you supply a natural number n, that the program won't print?
>
> No.
Wolenmuekenheim logic
There are infinitely many natural numbers that will never be
printed.
It is impossible to supply one of them.
 William Hughes


0




Reply

wpihughes (390)

5/9/2009 4:52:10 PM


On 9 May, 17:14, George Greene <gree...@email.unc.edu> wrote:
> On May 9, 11:41=A0am, LudovicoVan <ju...@diegidio.name> wrote:
>
> > The conclusion is still unwarranted: "the fact that the antidiagonal
> > does not belong to the list cannot be proven from the algorithmic
> > definition alone".
>
> So who needs an algorithmic definition?
> Are you trying to say that only algorithmic definitions are
> legitimate?
You don't even know what we are talking about.
LV


0




Reply

julio (505)

5/9/2009 5:07:17 PM


On 9 May, 17:08, George Greene <gree...@email.unc.edu> wrote:
> On May 9, 11:32=A0am, LudovicoVan <ju...@diegidio.name> wrote:
> > then we define the construction of
> > an item ('d') so that it differs in each place for each item in the
> > list given by 'f'. If we now run this game of 'f' against 'd',
>
> You CAN'T run this game.
> There ARE NO games in this paradigm.
There are no games in your paradigm maybe. You don't even know what we
are talking about.
LV


0




Reply

julio (505)

5/9/2009 5:09:03 PM


On May 9, 9:29=A0am, WM <mueck...@rz.fhaugsburg.de> wrote:
> On 8 Mai, 20:48, Jan <j...@daimi.au.dk> wrote:
>
> > Can you supply a natural number n, that the program won't print?
>
> No. Can you understand that at *any* step there are infinitely many
> numbers remaining that have not been printed and will never be
> printed, because they belong to the set that is remaining after any
> step?
Initially, the program prints "0". From this we can conclude
that "1" has not and never will be printed because it belongs
to the set that is remaining.
That's a lame argument even for you.
Oh, and limit one crank per thread, pally.
You're already consuming most of the attention on
sci.math, and have been for many, many, years. THIS
thread is the one where we make fun of LudicoVan, not
the thread where we make fun of you. Aren't you satisfied
with getting the vast majority of the attention? You have to
inject your sense of humor in here as well?
I see we can add greediness to your list of vices.
Marshall


0




Reply

marshall.spight (580)

5/9/2009 7:28:42 PM


WM wrote:
> On 8 Mai, 09:43, lwal...@lausd.net wrote:
>
>> Right now, the only mathematician of whom I am aware is making
>> an attempt at an inconsistency proof is Ed Nelson. He believes
>> that he can proof PA to be inconsistent  and since ZFC proves
>> that PA is consistent, if PA is proved to be inconsistent, then
>> ZFC would also be proved inconsistent.
>
> Here is another proof:
> http://www.hsaugsburg.de/~mueckenh/GU/GU12.PPT#361,25,Folie 25
>
> Regards, WM
Yes, that is a proof. It proves that you are even too stupid to
provide a working link.


0




Reply

bader1785 (15)

5/9/2009 7:56:34 PM


On 9 Mai, 21:28, Marshall <marshall.spi...@gmail.com> wrote:
> On May 9, 9:29=A0am, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> > On 8 Mai, 20:48, Jan <j...@daimi.au.dk> wrote:
>
> > > Can you supply a natural number n, that the program won't print?
>
> > No. Can you understand that at *any* step there are infinitely many
> > numbers remaining that have not been printed and will never be
> > printed, because they belong to the set that is remaining after any
> > step?
>
> Initially, the program prints "0". From this we can conclude
> that "1" has not and never will be printed because it belongs
> to the set that is remaining.
>
> That's a lame argument even for you.
>
I have a dream:
Even matheologists should be able to understand what is easily
understood by any sober mind:
"Every and all smaller" =3D "all"
for any ordered set.
Of course most matheologists will be unable to understand that. But
perhaps you would be willing to learn from some greater names?
=A7 174 Set theory is wrong (Wittgenstein)
"... classical logic was abstracted from the mathematics of finite
sets and their subsets .... Forgetful of this limited origin, one
afterwards mistook that logic for something above and prior to all
mathematics, and finally applied it, without justification, to the
mathematics of infinite sets. ... As Brouwer pointed out this is a
fallacy, the Fall and Original sin of set theory even if no paradoxes
result from it." (Hermann Weyl)
Indeed, I think that there is a real need, in formalism and elsewhere,
to link our understanding of mathematics with our understanding of the
physical world. (Robinson)
Infinite totalities do not exist in any sense of the word (i.e.,
either really or ideally). More precisely, any mention, or purported
mention, of infinite totalities is, literally, meaningless. (Robinson)
Of course, I know, all these people do not mean what they said. On the
contrary, they meant exactly the opposite of what they said, because
they were not cranks.
Regards, WM


0




Reply

mueckenh (275)

5/9/2009 9:22:46 PM


On May 9, 2:22=A0pm, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> perhaps you would be willing to learn from some greater names?
Not really. I don't subscribe to the practice of fawning over
ancestors.
The fact that those wellrespected people said such stupid things
(assuming they actually did, and you've not altered the meaning
by removing the context, which I will give you the benefit of the
doubt on, as I have not seen dishonesty appear particularly
strongly on your lengthy list of faults) is mildly surprising but not
astonishing, and certainly no reason to revise modern mathematics.
I am not a mathematician; I am a computer programmer.
As such, I have to deal with the realities of working within
systems of limited resources more than anyone doing math
does. It is part of my daily, my hourly existence. I am much
closer to the realities of finiteness than you.
So for me, it is highly ironic that, for all your relentless
fixation on finiteness, you suck at it. The best you can do
is rail pointlessly against clean abstractions like the
natural numbers. You complain bitterly that they don't model
resource limits, but you've never been able to show a logical
contradiction that results. (I know, I know; to you, failing
to model resource limits is itself a contradiction. Big deal;
that just shows you don't know what a contradiction is.)
And of course, you haven't ever proposed any abstractions
of your own that *do* model resource limits; another
crank hallmark.
While we're quoting "greater names", I recall an essay
by Dijkstra in which he discussed the importance of
separation of concerns, and topically, the exact two
concerns he mentioned separating were correctness
and resource limits. He also touched on the frustrations
of trying to deal with those people of such limited intellect
as to be unable or unwilling to make the separation.
Marshall
PS. Everyone says stupid things now and again.
And who knows? Maybe Robinson woke up the
next morning with a terrible headache, and was later
heard asking "I said *what* last night?!"


0




Reply

marshall.spight (580)

5/9/2009 10:43:27 PM


Continuation
We have (agreed by both WM and WH)
i: if
all natural numbers and all lines exist
then
no line contains all natural numbers
is true.
WM claims a contradiction in that we also have
ii: if
all natural numbers and all lines exist
then
there exists a line that contains all natural numbers
is true.
However, if all natural numbers exist and all lines exist,
then neither the set of natuaral numbers nor
the set of lines has a last element.
To establish the putative contradiction, we need a proof of ii
that does not require a last element.
WM first try:
WM: Every line contains all numbers that are in the preceding lines.
WM: If all numbers exist, then they exist in one line.
Nope. Clearly the "one line" in the second statment is the last
line.
The second statement is not true if there is no last line.
Try again.
 William Hughes


0




Reply

wpihughes (390)

5/10/2009 12:35:20 AM


On May 9, 12:22=A0am, lwal...@lausd.net wrote:
> On May 8, 10:36=A0am, Mariano Su=E1rezAlvarez
>
> <mariano.suarezalva...@gmail.com> wrote:
> > On May 8, 4:43=A0am, lwal...@lausd.net wrote:
> > > And their programmers want them to formulate proofs in the
> > > dominant theory, not in alternate theories.
> > Programmers are mostly theoryagnostic when they are
> > programming provers. What on earth suggests to you
> > that, say, Prover9 is restricted to (whatever it is
> > that you consider) dominant set theory?
>
> By "dominant set theory" I usually mean ZFC, i.e. the set
> theory that most set theorists use.
>
> Well, I suppose that Prover9 isn't limited to ZFC, since
> Marshall says that he was able to use it to prove theorems
> in a theory whose lone axiom is "Ax (Ay (x=3Dy))," which is
> hardly equivalent to ZFC.
>
> But what I've never seen is a theory that is proposed by a
> socalled "crank" entered into Prover9, despite my attempts
> to make the "crank" theories rigorous. Indeed, what I would
> love to see is a rigorous version of a "crank" theory
> entered into Prover9  and I'll consider my attempts to
> make "crank" theories to be successful if my version of the
> theory can be entered into Prover9 to prove theorems,
> especially if those theorems are negations of those that
> are provable in ZFC (assuming both ZFC and my proposed
> theory to be consistent).
>
> > That you seem to think there is such a restriction
>
> There _is_ a restriction  Prover9 can only give proofs
> in theories that someone types into the prover. If no
> one enters a rigorous version of a "crank" theory, then
> Prover9 won't prove anything in that theory.
So, you are saying that the restriction is that
no one cares enough for those theories? Modulo the
standard disclaimers, that is a clear indication that
those theories are not interesting. That is not
a restriction but a fact.
> > is a clear sign that you have absolutely no idea what
> > an automatic prover is.
>
> I'd like to know more about how Prover9 works. I've tried
> asking Prover9 users how it works, but answers have not
> been forthcoming.
>
> If I don't know how something works, I'd much rather be
> told how it does work, not simply "you don't know how
> (whatever) works." But standard theorists on sci.math are
> far more likely to say the latter than the former.
Have you tried googling for "Prover9", picking the
first result that google provides, reading the
corresponding page and then clicked on the link
clearly labeled "Manual and examples" that is on the page?
That is a strategy that most reasonable persons
would have put into practice in order to know how
Prover9 works. Declaiming your prejudice as to what
your putative "standard theoriest" would or would
not do is a rather inducive strategy to find out
what Prover9or anything else, for that matterworks.
> > > The more time passes, the _more_ likely we'll see a proof of,
> > > say, the Riemann Hypothesis or Goldbach's Conjecture.
> > > The more time passes, the _less_ likely we'll see a proof that
> > > ZFC is inconsistent.
> > What possible basis do you have for the two claims you make?
>
> Mathematicians all over the world are searching for proofs
> of Riemann and Goldbach. As more and more time passes, it
> becomes more and more likely that someone will discover a
> proof of one of these conjectures. It's more likely that a
> proof of RH will be discovered by 2020 than by 2010, and
> it's still more likely that a proof of RH will be found by
> 2100 than by 2020. Therefore, the more time passes, the
> more likely that we'll see a proof of Riemann or Goldbach.
>
> Mathematicians all over the world are also searching for
> proofs that ZFC is inconsistent. But, as Knox points out,
> if there really is a proof of ~Con(ZFC), we most likely
> would have found it by now. For many inconsistent theories
> it's easy to find a proof of its inconsistency. It took
> only a few years to find a contradiction in Cantor's naive
> set theory, and those "cranks" who propose theories on
> sci.math, that turn out to be inconsistent, it often takes
> mere _hours_ to find the inconsistency. But ZFC has
> withstood the test of time. It seems hard to believe that
> mathematicians would have overlooked an inconsistency for
> so long  well over a century. Since we haven't found a
> contradiction by now, it's unlikely that we'll see one by
> 2010 or 2020. And if we haven't found one by 2100, we may
> as well consider ZFC to be consistent  since we'll
> likely never find a proof of ~Con(ZFC). This seems to be
> the essence of what Knox wrote in her post.
>
> Since Mariano asked me, I ask the same of Mariano, and I
> wonder what his opinion of the likelihood that we'll ever
> see a proof of Riemann, Goldbach, or ~Con(ZFC).
I have no opinion as to whether Riemann's hypothesis is
any more approachable today than it was a century ago,
nor on the likelyhood of its being decided one way or
another in the nearor farfuture. The same is true
regarding Golbach's conjecture and the consistency of ZFC.
> > > If there were a
> > > rigorous theory in which more of their intutions are provable
> > > than in ZFC, then maybe LV and the other opponents of the
> > > dominant set theory would have less reason to start these
> > > threads and complain about set theory all the time.
> > Can you please point to *someone* who *does* see
> > a reason why opponents of ZFC can't have an alternate
> > theory?
>
> Here are some reasons given for why opponents of ZFC can't
> have an alternate theory:
>
> The alternate theories (i.e., the particular alternate
> theory proposed by the particular opponent of ZFC) aren't
> rigorous enough.
> The alternate theories aren't coherent enough.
> The alternate theories can't prove anything that is useful
> to applied science.
> The alternate theories are inconsistent (which even I
> consider to be a valid reason).
>
> And so on. I'll leave it to Mariano to find the posters
> who gave these as reasons that opponents of ZFC can't
> have their alternate theories.
Those are not reasons why "opponents of ZFC" cannot have
an alternate theory. Those are reasons why the alternate
theories (which, really, are not given in any form which
deserves that name...) they "propose" are of no interest.
There *is* a difference. You have shown yourself repeatedly
to be immune to the distinction, though.
 m


0




Reply

mariano.suarezalvarez (417)

5/10/2009 1:20:34 AM


William Hughes wrote:
> On May 9, 12:29 pm, WM <mueck...@rz.fhaugsburg.de> wrote:
> > On 8 Mai, 20:48, Jan <j...@daimi.au.dk> wrote:
> >
> > > Can you supply a natural number n, that the program won't print?
> >
> > No.
>
> Wolenmuekenheim logic
>
> There are infinitely many natural numbers that will never be
> printed.
> It is impossible to supply one of them.
Yeah, Bill, it's absurd.
Why do you continue to respond to this loon? Do you think that
one day you'll back him into a corner and he'll say, "Yeah, you're
right, I'm mistaken"?
Why won't you just let this spavined donkey die a natural death?

hz


0




Reply

herbzet (87)

5/10/2009 3:51:48 AM


MoeBlee wrote:
>
> (I hope I've stated that fairly accurately. But you can also check a
> textbook in mathematical logic under the subject of interpreting one
> theory in another, however it is actually worded in the book.)
Probably the best advice  thanks.

hz


0




Reply

herbzet (87)

5/10/2009 4:21:03 AM


On May 8, 3:32=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
 On May 8, 12:32=A0pm, LudovicoVan <ju...@diegidio.name> wrote:

 > impredicative, so constructively invalid.
No, this is a confusion on the concept of predicativity. The
proof is predicative.
 It is not at all clear that constructivism in itself requires
absence
 of impredicativity. Though, of course, certain currents in
constructivism do
absence of predicativity.
I think using "constructive" to mean predicative as well is
liable to lead to some confusion, usually, so I would avoid
it. Two properties that seem very natural to require are the
number existence and set existence principles. The numeric
existence principle for an axiom system says that there's a
procedure for taking a proof that there exists an integer
having some given property, and from it extracting a specific
example of such an integer. The set existence principle
likewise says that from a proof of the existence of a set
one can extract an example of the form {x : P(x)} for a
given formula P. The numeric existence principle is the more
important one. It implies under usual conditions a principle
called the disjunction principle, namely, that if "P or Q" is
provable, then either P is provable or Q is provable.
(Usually one can prove "P or Q" if and only if one can prove
the existence of an integer n such that if n=3D0 then P and
if n<>0 then Q, and exhibiting such an n allows us to prove
P or prove Q.)
I would say that a system satisfying these properties can
legitimately be called constructive. A constructivist won't
necessarily consider it a suitable system for developing
mathematics in, but the same goes for everybody else.
People (e.g. Maddy) have made stabs at giving criteria
for what counts as a "good" system, but it never as far as
I know boils down to something as simple as being
constructive.
One advantage of showing this tiny bit of leniency is that
logicians have found for most formal systems on the
"ladder" of strengths of system, a counterpart which is
almost the same except for using intuitionistic logic. It
seems appropriate to me to just think of these as the
constructive counterparts of the classical systems.
 Anyway, the point stands that while a constructivist might not
accept
 certain axioms, the LOGIC in the proof is intuitionistic.
The logic is intuitionistic, yes, and the proof is predicative too.
Impredicativity does not arise just from constructing a real from
a set having reals as members (or members of members, etc.).
The new real is not a member of any of the sets previously
mentioned in the proof (or a member of a member, etc.)
In Weyl's _The Continuum_, he has a predicative system, and
in it the existence of a least upper bound of a sequence of reals
bounded above is provable, but not the existence of a least
upper bound of an arbitrary set of reals.
 Moreover, certain constructivists DO explicitly endorse that the
proof
 IS constructive.
Yes, because there's nothing wrong with it.
There's the technical problem with one proof, in that it's not
constructive that each real has a decimal expansion. But the
proof as given works for reals having decimal expansions,
plus Cantor's earlier proof of the same result doesn't have the
technical difficulty, as it doesn't rely upon decimal expansions.
If you take a system like in Principia Mathematica without the
reducibility axiom, I believe you might have to say that the new
real is on another level of the hierarchy from the ones from
which it is constructed. By the same token, there's no "set of
all reals" in that system. I don't think that deserves to be
thought of as a problem.
But, of course, that may be while we note that the
 interpertation of the theorem in constructivism is different from in
 classical mathematics.
Not too severely different. The main difference is that as
constructively interpreted the result is a little stronger;
it tells you that you can actually exhibit the new real
number.
Keith Ramsay


0




Reply

kramsay (59)

5/10/2009 9:00:42 AM


On May 7, 3:56 pm, Barb Knox <s...@sig.below> wrote:
> In article
> <c7ccbbf2af51448eb9da773e185ec...@r36g2000vbr.googlegroups.com>,
>
>
> [...]
>
> Face facts: Cantor's diagonal proof that R > N is SIMPLE. It has
> even been CHECKED BY COMPUTERS. Therefore the only way that R = N
> is if the various formulations of axiomatic set theory are internally
> INCONSISTENT (which is emphatically NOT the same as being inconsistent
> with someone's naive intuitions). Proving (say) ZFC to be inconsistent
> would be worth at least a Fields Medal and a centrefold picture in the
> Journal of the AMS. If it were easy (or even less than extremely
> difficult) to do then it would already have been done. The chance of
> you doing it is vanishingly small.
>
> Plus, there are good reasons for believing that ZFC etc. ARE internally
> consistent. For example, there is an intuitively notunreasonable model
> for ZFC etc. (Goedel's "constructible universe").
>
>
So your "intuition" tells you that ZFC is consistent. Mine tells me
otherwise.
Here is a simple argument for the inconsistency of ZFC. If Cantor's
diagonal argument is correct, there are uncountably many real numbers
in each interval of the form [1/2n, 1/2n], where n (=1,2,3....) is a
positive integer. The intersection of all such intervals (over all the
positive integers n) however, contains the single real number 0. Our
intuition tells us that the intersection should also contain
uncountably many reals, i.e., the intersection should be an interval.
Each (or every) one of the nested intervals has endpoints fixed away
from zero, so how come their intersection manages to eliminate "all"
points other than zero, which does not seem possible on any reasonable
conception of "each" (or "every" or "all")?
Now let us bring in nonstandard analysis, e.g. Nelson's Internal Set
Theory (IST). It is proven that IST is consistent if and only if
classical ZFC is consistent, so an inconsistency in IST amounts to an
inconsistency in ZFC. In IST, consider the intersection Irho of all
real intervals of the form [1/2n, 1/2n], where n is a positive
integer ranging over 1<n<rho, and rho is a nonstandard positive
integer. The intersection Irho *is* an interval (containing
uncountably many points) around 0 after all, *despite* the fact rho
satisfies rho>1, rho>2, ... and [rho > every standard finite positive
integer].
Note that IST does not consider the set of all intervals of the form
[1/2n,1/2n], where n is a standard finite positive integer, as
legitimate. So the intersection of all such intervals is not
legitimate either. If it were, IST would be forced to accept that the
intersection consists of the single point zero, but this would be a
contradiction in IST of its Principle of Internalisation (i.e., the
"I" in IST).
Conclusions:
1. If we accept that rho is an "infinite" integer, we would obtain a
contradiction with the classical ZFC result that the intersection I
rho must consist of a single point 0; such a contradiction is avoided
only by the dubious strategy of labeling rho as an "internally finite"
integer, *despite* the fact that it exceeds every standard finite
positive integer. If we refuse to relabel infinite entities as
"internally finite", we have to conclude IST is inconsistent, and
hence, so is ZFC.
2. Secondly, IST would run into internal contradictions if it
permitted a direct formulation of the intersection in terms of all
(and only) intervals of the form [1/2n,1/2n], where n is a standard
positive integer. So it has to avoid contradiction by denying that
such a set of intervals exists, even though classical real analysis
permits precisely the same set.Thus we see that a classical ZFC result
*would* be a contradiction in IST, and such a contradiction is avoided
only by the tactic of making the result impossible to formulate in
IST. But no convincing logical reason is provided as to why this
entity is illegitimate, other than by dictat.
3. The above formulation of the paradox is precisely what Zeno found
2300 years ago. Another way of stating Zeno's paradox is to ask how
infinitely many intervals [1/2n, 1/2n] of finite nonzero length can
sum to a finite length. Again IST avoids contradiction either by the
dubious tactic of labelling an infinite entity as "internally finite",
or by denying that the set of standard intervals in question exists.
In the logic NAFL (e.g. see < http://arxiv.org/abs/math/0506475 >)
relabeling infinite entities as finite is not allowed and there are no
such things as "nonstandard finite" sets. Edward Nelson says, (after
noting the "strange" IST theorem that there must exist a finite set
containing all standard finite sets), that maybe "finite does not mean
what we always thought it meant". Sorry, Ed, old boy, finite means
what we always thought it meant, and what we always *defined* it to
mean. It is IST that is inconsistent, and hence, so is ZFC.
Secondly, NAFL also rejects the existence of the set of intervals
mentioned in 2. above, but with the valid logical reason that
infinite sets do not exist; all infinite classes are proper classes
and quantification over proper classes is not permitted, and an
"arbitrary" proper class (e.g. an "arbitrary" real number) does not
exist. In NAFL, Cantor's diagonal argument cannot be formulated
because it involves quantification over reals. This is similar to what
IST does in 2. above to avoid contradiction, except that a valid
logical reason is provided for why the entity in question is
illegitimate.
A revised version of the following paper of mine is now under review
by a mainstream physics journal for the past six months:
http://arxiv.org/abs/quantph/0504115
I am hoping that this paper will be accepted. The original submission
was made in April 2005 and the revised version in Nov 2008 following a
referee report in 2006 May. This paper shows why NAFL is a suitable
logic in which to frame quantum mechanics, and refutes Afshar's
argument that Bohr's complementarity principle has been falsified by
his published experiment.


0




Reply

sradhakr (38)

5/10/2009 11:38:08 AM


On 8 Mai, 21:54, Marshall <marshall.spi...@gmail.com> wrote:
> On May 8, 12:36=A0pm, LudovicoVan <ju...@diegidio.name> wrote:
>
> > On 8 May, 00:03, Marshall <marshall.spi...@gmail.com> wrote:
>
> > > You can write a program to output all of the naturals. This program
> > > won't halt, as you say, and for any natural n, will eventually output
> > > n. (We are glossing over some resource limits here, but that is
> > > customary and reasonable.) It will never finish running, though;
> > > there are always more naturals.
>
> > > You can't do the same for the reals. That's because the reals
> > > aren't countable and the naturals are.
>
> > That ultimately depends on your stance on the diagonal argument.
>
> My stance controls what programs it is possible to write, you say?
> So if I change my stance, a different set of things become
> computable? I Did Not Know That.
Most do not know that. Nevertheless it is true (if actual infinity can
be finished, i.e., under generally accepted conditions).
Use all binary sequences that end by zeros only, and construct the
complete infinite binary tree, i.e., all its nodes and edges. Then
there is nothing remaining, in the binary tree, that could be used to
construct the sequence 0.010101... =3D 1/3. (Nevertheless all its nodes
are there.)
But if you use all binary sequences that end by the period 010101...,
then you can construct the same tree, i.e., all its nodes and edges,
and then 1/3 is in the tree.
(The solution of this paradox is of course that there is nothing like
the complete binary tree with all its nodes and edges, i.e., the
generally accepted conditions are false.)
Regards, WM


0




Reply

mueckenh (275)

5/10/2009 11:39:05 AM


herbzet <herbzet@gmail.com> writes:
> William Hughes wrote:
>> On May 9, 12:29 pm, WM <mueck...@rz.fhaugsburg.de> wrote:
>> > On 8 Mai, 20:48, Jan <j...@daimi.au.dk> wrote:
>> >
>> > > Can you supply a natural number n, that the program won't print?
>> >
>> > No.
>>
>> Wolenmuekenheim logic
>>
>> There are infinitely many natural numbers that will never be
>> printed.
>> It is impossible to supply one of them.
>
>
> Yeah, Bill, it's absurd.
>
> Why do you continue to respond to this loon? Do you think that
> one day you'll back him into a corner and he'll say, "Yeah, you're
> right, I'm mistaken"?
>
> Why won't you just let this spavined donkey die a natural death?
I'd like to vote for an unnatural death, if I may?
Phil

Marijuana is indeed a dangerous drug.
It causes governments to wage war against their own people.
 Dave Seaman (sci.math, 19 Mar 2009)


0




Reply

thefatphil_demunged (1649)

5/10/2009 2:04:55 PM


On May 10, 4:38=A0pm, "R. Srinivasan" <sradh...@in.ibm.com> wrote:
> On May 7, 3:56 pm, Barb Knox <s...@sig.below> wrote:
>
> > In article
> > <c7ccbbf2af51448eb9da773e185ec...@r36g2000vbr.googlegroups.com>,
>
> > [...]
>
> > Face facts: Cantor's diagonal proof that R > N is SIMPLE. =A0It has
> > even been CHECKED BY COMPUTERS. =A0Therefore the only way that R =3D =
N
> > is if the various formulations of axiomatic set theory are internally
> > INCONSISTENT (which is emphatically NOT the same as being inconsistent
> > with someone's naive intuitions). =A0Proving (say) ZFC to be inconsiste=
nt
> > would be worth at least a Fields Medal and a centrefold picture in the
> > Journal of the AMS. =A0If it were easy (or even less than extremely
> > difficult) to do then it would already have been done. =A0The chance of
> > you doing it is vanishingly small.
>
> > Plus, there are good reasons for believing that ZFC etc. ARE internally
> > consistent. =A0For example, there is an intuitively notunreasonable mo=
del
> > for ZFC etc. (Goedel's "constructible universe").
>
> So your "intuition" tells you that ZFC is consistent. Mine tells me
> otherwise.
>
> Here is a simple argument for the inconsistency of ZFC. If Cantor's
> diagonal argument is correct, there are uncountably many real numbers
> in each interval of the form [1/2n, 1/2n], where n (=3D1,2,3....) is a
> positive integer. The intersection of all such intervals (over all the
> positive integers n) however, contains the single real number 0. Our
> intuition tells us that the intersection should also contain
> uncountably many reals, i.e., the intersection should be an interval.
> Each (or every) one of the nested intervals has endpoints fixed away
> from zero, so how come their intersection manages to eliminate "all"
> points other than zero, which does not seem possible on any reasonable
> conception of "each" (or "every" or "all")?
>
> Now let us bring in nonstandard analysis, e.g. Nelson's Internal Set
> Theory (IST). It is proven that IST is consistent if and only if
> classical ZFC is consistent, so an inconsistency in IST amounts to an
> inconsistency in ZFC. In IST, consider the intersection Irho of all
> real intervals of the form [1/2n, 1/2n], where n is a positive
> integer ranging over 1<n<rho, and rho is a nonstandard positive
> integer. The intersection Irho *is* an interval (containing
> uncountably many points) around 0 after all, *despite* the fact rho
> satisfies rho>1, rho>2, ... and [rho > every standard finite positive
> integer].
>
> Note that IST does not consider the set of all intervals of the form
> [1/2n,1/2n], where n is a standard finite positive integer, as
> legitimate. So the intersection of all such intervals is not
> legitimate either. If it were, IST would be forced to accept that the
> intersection consists of the single point zero, but this would be a
> contradiction in IST of its Principle of Internalisation (i.e., the
> "I" in IST).
>
Sorry I meant the Idealization principle (not "Internalization" as I
had stated above). IST also stands for the three principles of
Internal Set Theory, namely Idealization, Standardization and the
Transfer Principle.
>
>
> Conclusions:
>
> 1. If we accept that rho is an "infinite" integer, we would obtain a
> contradiction with the classical ZFC result that the intersection I
> rho must consist of a single point 0; such a contradiction is avoided
> only by the dubious strategy of labeling rho as an "internally finite"
> integer, *despite* the fact that it exceeds every standard finite
> positive integer. If we refuse to relabel infinite entities as
> "internally finite", we have to conclude IST is inconsistent, and
> hence, so is ZFC.
>
> 2. Secondly, IST would run into internal contradictions if it
> permitted a direct formulation of the intersection in terms of all
> (and only) intervals of the form [1/2n,1/2n], where n is a standard
> positive integer. So it has to avoid contradiction by denying that
> such a set of intervals exists, even though classical real analysis
> permits precisely the same set.Thus we see that a classical ZFC result
> *would* be a contradiction in IST, and such a contradiction is avoided
> only by the tactic of making the result impossible to formulate in
> IST. But no convincing logical reason is provided as to why this
> entity is illegitimate, other than by dictat.
>
In fact the Idealization principle captures precisely the intuition
that tells us that an intersection of infinitely many intervals of the
form [1/2n,1/2n] must include an interval.
>
> 3. The above formulation of the paradox is precisely what Zeno found
> 2300 years ago. Another way of stating Zeno's paradox is to ask how
> infinitely many intervals [1/2n, 1/2n] of finite nonzero length can
> sum to a finite length. Again IST avoids contradiction either by the
> dubious tactic of labelling an infinite entity as "internally finite",
> or by denying that the set of standard intervals in question exists.
>
> In the logic NAFL (e.g. see <http://arxiv.org/abs/math/0506475>)
> relabeling infinite entities as finite is not allowed and there are no
> such things as "nonstandard finite" sets. Edward Nelson says, (after
> noting the "strange" IST theorem that there must exist a finite set
> containing all standard finite sets), that maybe "finite does not mean
> what we always thought it meant". Sorry, Ed, old boy, finite means
> what we always thought it meant, and what we always *defined* it to
> mean. It is IST that is inconsistent, and hence, so is ZFC.
>
Indeed, if we did not accept that "finite" means what we always
thought it meant, we wouldn't even know what a theory is. The basic
principles of logic are dependent on our knowing the meaning of the
word "finite" and that meaning cannot be tampered with.
>
> Secondly, NAFL also rejects the existence of the set of intervals
> mentioned in 2. above, but with the valid logical reason that
> infinite sets do not exist; all infinite classes are proper classes
> and quantification over proper classes is not permitted, and an
> "arbitrary" proper class (e.g. an "arbitrary" real number) does not
> exist. In NAFL, Cantor's diagonal argument cannot be formulated
> because it involves quantification over reals. This is similar to what
> IST does in 2. above to avoid contradiction, except that a valid
> logical reason is provided for why the entity in question is
> illegitimate.
>
> A revised version of the following paper of mine is now under review
> by a mainstream physics journal for the past six months:
>
> http://arxiv.org/abs/quantph/0504115
>
> I am hoping that this paper will be accepted. The original submission
> was made in April 2005 and the revised version in Nov 2008 following a
> referee report in 2006 May. This paper shows why NAFL is a suitable
> logic in which to frame quantum mechanics, and refutes Afshar's
> argument that Bohr's complementarity principle has been falsified by
> his published experiment.


0




Reply

sradhakr (38)

5/10/2009 2:53:53 PM


On 9 Mai, 05:52, lwal...@lausd.net wrote:
> Knox wrote that a proof of ~Con(ZFC) would likely merit a
> Fields medal. It's not unreasonable to guess that many
> mathematicians around the world would do research that
> would earn them a Fields medal, the most prestigious prize
> in all of mathematics.
>
> So if, as MoeBlee might be implying here, very _few_
> mathematicians seek a proof of ~Con(ZFC), then they likely
> believe that no proof is possible. And if this is really
> the case, then this would actually _prove_ the point that
> I was making to _Mariano_  that the more time passes,
> the less likely a proof of ~Con(ZFC) will be found, and
> the fewer mathematicians will even search for a proof!
The first proof that ZFC is inconsistent, to my knowledge, was the
proof that there are only countably many definitions.
A number that rules out any chance of access cannot belong to that
uncountable set of numbers that is "proven" to exist by the diagonal
proof (if the diagonal numbers permit access).
There is no chance to define one of those undefinable numbers. So they
may exist where ever they do and who ever wants to may believe in
them. But they are not those numbers which appear as a diagonal of a
list, because those can be defined, hence belong to a countable set.
Therefore this proof is a contradiction that cannot be avoided by the
hidden set of undefinable numbers.
But these facts are not widely known even among set theorists. For
instance, one of that illustrious society who call themselves Fools Of
Matheology or so wrote: "Does he [Mueckenheim] perhaps think there's a
particular real
number that is undefinable in any (countable) language?"
Regards, WM


0




Reply

mueckenh (275)

5/10/2009 3:40:32 PM


On 9 Mai, 14:44, David C. Ullrich <dullr...@sprynet.com> wrote:
> >That ultimately depends on your stance on the diagonal argument.
>
> No. The reals _are_ uncountable.
If there are reals that form an uncountable set, then these numbers
cannot be accessed in any way, neither by diagonal arguments nor other
methods. Therefore those reals that are accessible by any means that
defines a number, including the diagonal argument *if it defines a
number*, belong to a countable set.
Regards, WM


0




Reply

mueckenh (275)

5/10/2009 3:45:14 PM


On 9 Mai, 14:53, David C. Ullrich <dullr...@sprynet.com> wrote:
> So what? Yes, there is no problem whatever down a list of reals,
> in fact rationals, such that every real number is a _limit point_
> of the sequence. That's not news, that's not controversial, that's
> awesomely well known.
Nevertheless there has _never_ been a limit point of a sequence or
series with irrational limit that has been defined or identified by a
sequence of digits.
>
> A _limit point_ of the range of f is not the same thing as an
> _element of_ the range of f. The assertion is that there is
> a real number that is not _in_ the range of f.
And this real number is defined by a function and not by a sequence of
digits.
>
> In case you don't see the difference, let's talk about a related
> situation
there are many firm definitions of pi, but nobody will ever be able to
give a decimal expansion that is unique for pi.
Why does Cantor's argument work with the torsi of the alleged numbers
and not with those definitions which define the numbers exactly?
Regards, WM


0




Reply

mueckenh (275)

5/10/2009 3:53:07 PM


On 9 Mai, 17:19, LudovicoVan <ju...@diegidio.name> wrote:
> On 9 May, 13:44, David C. Ullrich <dullr...@sprynet.com> wrote:
>
> > On Fri, 8 May 2009 12:36:46 0700 (PDT), LudovicoVan
> > <ju...@diegidio.name> wrote:
> > >On 8 May, 00:03, Marshall <marshall.spi...@gmail.com> wrote:
> > >> You can't do the same for the reals. That's because the reals
> > >> aren't countable and the naturals are.
>
> > >That ultimately depends on your stance on the diagonal argument.
>
> > No. The reals _are_ uncountable. This has nothing to do with
> > one's "stance" on the diagonal argument  the argument is
> > correct, regardless of anyone's "stance" on it.
>
> Apart from the apparent dogmatism, what you state is incorrect and,
> strictly speaking, false.
So it is.
"the formalist concludes: \alephone is greater than alephnull, a
proposition, that has no meaning for the intuitionist." (Brouwer)
Regards, WM


0




Reply

mueckenh (275)

5/10/2009 4:06:15 PM


On 9 Mai, 18:38, William Hughes <wpihug...@hotmail.com> wrote:
> On May 9, 12:11 pm, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> WM continues to deliberately confuse elements and sets of elements.
>
> One such property is that every natural number has even or odd
> cardinality.
>
> > One such property is that every set of natural numbers has even or odd
> > cardinality.
That's because sets of natural numbers consist of natural numbers and
are counted by natural numbers.
But as usual you snipped the essential point:
One such property is that every set of natural numbers has even or
odd
cardinality.
If you know this, you can exclude that there is a set of natural
numbers with cardinal number aleph_0 (because aleph_0 + 1 = aleph_0).
Yes, such conjectures could be done, if infinity would follow the
same
logic as finity. But it seems it doesn't. So why do you believe it
does in case of Cantor's simple idea?
Regards, WM


0




Reply

mueckenh (275)

5/10/2009 4:21:34 PM


On 9 Mai, 18:52, William Hughes <wpihug...@hotmail.com> wrote:
> On May 9, 12:29 pm, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> > On 8 Mai, 20:48, Jan <j...@daimi.au.dk> wrote:
>
> > > Can you supply a natural number n, that the program won't print?
>
> > No.
>
> =A0 =A0 =A0 There are infinitely many natural numbers that will never be
> printed.
> =A0 =A0 =A0 It is impossible to =A0supply one of them.
Provided that an actually completed infinite set N of natural numbers
exists, then we would have this antinomy, among many others.
Regards, WM


0




Reply

mueckenh (275)

5/10/2009 4:23:54 PM


LudovicoVan <julio@diegidio.name> writes:
> On 8 May, 22:42, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > LudovicoVan <ju...@diegidio.name> writes:
> > > On 7 May, 18:30, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> >
> > > > Let's look at total functions taking two natural numbers and returning
> > > > either 0 or 1, eg f(n,m). �If we leave out the question of
> > > > multiple representations, let's suppose you have implemented
> > > > such a total function.
> >
> > > Great.
> >
> > > [snip]
> > > > So, the antidiagonal is itself something given algorithmically,
> > > > and you can easily write the procedure to return its nth digit,
> > > > provided that the given listing of reals itself is a total
> > > > function  this does not mean that all the digits involved must
> > > > be computed, of course!
> >
> > > My point (my take) is that this simply does not solve our issue: the
> > > fact that the antidiagonal does not belong to the list cannot be
> > > proven from the algorithmic definition alone. The key issue is what
> > > happens at infinity, and my thesis, to report it here in short, is
> > > that the antidiagonal, as any other computable string, is simply
> > > there, in the range of a putative function 'f' like you describe
> > > above: there is no member missing.
> >
> > All that is claimed by the diagonal argument in this context is
> > that for every *natural number* n, the antidiagonal
> > differs from sequence f(n,0), f(n,1), f(n,2), ...
> > And that we can show fairly easily  it looks from your post
> > here that you accept that.
> >
> > So the conclusion is that the antidiagonal does not appear
> > as a row in the *given* listing  nothing more than that.
>
> The conclusion is still unwarranted: "the fact that the antidiagonal
> does not belong to the list cannot be proven from the algorithmic
> definition alone".
If you refuse all notion of proof *about* algorithms, then
you are right that noone can show anything about any algorithm.
Is that what you want to say?
If not, what sort of reasoning about algorithms is acceptable
to you?
> LV

Alan Smaill


0




Reply

smaill1 (89)

5/10/2009 4:28:26 PM


LudovicoVan <julio@diegidio.name> writes:
> On 8 May, 22:28, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > LudovicoVan <ju...@diegidio.name> writes:
> > > On 7 May, 18:06, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > > > LudovicoVan <ju...@diegidio.name> writes:
> > > > > On 7 May, 11:56, Barb Knox <s...@sig.below> wrote:
> >
> > > > > > Face facts: Cantor's diagonal proof that R > N is SIMPLE. �It has
> > > > > > even been CHECKED BY COMPUTERS.
> >
> > > > > Formal proofs are irrelevant to the diagonal argument: they have it
> > > > > embedded, as an informing principle, into their axioms and
> > > > > definitions.
> >
> > > > Why should we believe this assertion of yours?
> > > > You have made it several times, and not yet given the slightest
> > > > reason for anyone to believe it.
> >
> > > There has been a discussion specifically on this in the thread "Levy
> > > proof that R is uncountable":
> >
> > >http://groups.google.co.uk/group/sci.logic/browse_frm/thread/876837fd...
> >
> > I have been following the thread.
> >
> > You have rightly said that logical priority is different from historical
> > priority. At no point �have you or anyone else given us any reason
> > to suppose that the diagonal argument is embedded into the axioms
> > and definitions of either classical or constructive theories
> > of the real numbers.
>
> You can say that nobody has explicitly acknowledged it, but this point
> is clear enough. I just won't go back collecting links on this too, I
> can't take the burden of what is rather a problem of "generalised
> denial".
There is denial at work, indeed.
> LV

Alan Smaill


0




Reply

smaill1 (89)

5/10/2009 4:29:54 PM


On 10 Mai, 00:43, Marshall <marshall.spi...@gmail.com> wrote:
> So for me, it is highly ironic that, for all your relentless
> fixation on finiteness, you suck at it. The best you can do
> is rail pointlessly against clean abstractions like the
> natural numbers. You complain bitterly that they don't model
> resource limits, but you've never been able to show a logical
> contradiction that results.
The logical contradiction has nothing to do with resource limits.
The logical contradiction is:
It is impossible to define any irrational number by a string of
digits.
But it is possible to define an irrational number by a finite
definition.
The number of finite definitions, however, is countable.
>
> And of course, you haven't ever proposed any abstractions
> of your own that *do* model resource limits;
Current mathematics (except set theory) does what is necessary. There
is no reason to provide further abstractions.
>
> PS. Everyone says stupid things now and again.
> And who knows? Maybe Robinson woke up the
> next morning with a terrible headache, and was later
> heard asking "I said *what* last night?!"
Of course, all my quotes stem from drunk people.
Regards, WM


0




Reply

mueckenh (275)

5/10/2009 4:36:24 PM


On 10 Mai, 02:35, William Hughes <wpihug...@hotmail.com> wrote:
> Continuation
1
2, 1
3, 2, 1
....
>
> We have (agreed by both WM and WH)
>
> i: =A0if
> =A0 =A0 =A0 all natural numbers and all lines exist
> =A0 =A0then
> =A0 =A0 =A0 no line contains all natural numbers
>
> is true.
By the fact that for every line there is a line with one more number.
>
> WM claims a contradiction in that we also have
>
> ii: =A0if
> =A0 =A0 =A0 =A0 all natural numbers and all lines exist
> =A0 =A0 =A0then
> =A0 =A0 =A0 =A0 there exists a line that contains all natural numbers
>
> is true.
Because then also all lines exist. Every line contains everything that
is contained in its predecessor lines.
>
> However, if all natural numbers exist and all lines exist,
> then neither the set of natuaral =A0numbers nor
> the set of lines has a last element.
(How then do you know that there are all?)
> To establish the putative contradiction, we need a proof of ii
> that does not require a last element.
>
> WM first try:
>
> WM: =A0Every line contains all numbers that are in the preceding lines.
> WM: =A0If all numbers exist, then they exist in one line.
>
> Nope. =A0Clearly the "one line" in the second statment is the last
> line.
> The second statement is not true if there is no last line.
No. My proof uses only the linearity and the geometrical symmetry
shown in the list above. The first column cannot contain more than
every line. If the first column is complete, then the set of lines is
as complete). The lines exist in exactly the same manner as the number
of the first column.
This can only be understood by nonCantorian sets, i.e., potentiallly
infinite sets.
Regards, WM


0




Reply

mueckenh (275)

5/10/2009 4:41:37 PM


On 10 May, 10:00, Keith Ramsay <kram...@aol.com> wrote:
> On May 8, 3:32=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
>  On May 8, 12:32=A0pm, LudovicoVan <ju...@diegidio.name> wrote:
> 
>  > impredicative, so constructively invalid.
>
> No, this is a confusion on the concept of predicativity. The
> proof is predicative.
I might very well be missing something, but I read on Wikipedia:
http://en.wikipedia.org/wiki/Impredicativity
 "In mathematics and logic, impredicativity is the property of
a selfreferencing definition. More precisely, a definition is said to
be impredicative if it invokes (mentions or quantifies over) the set
being defined, or (more commonly) another set which contains the thing
being defined."
I note here that selfreferential "mention" seems enough, no need for
quantification over.
 "Russell's paradox is a famous example of an impredicative
construction, namely the set of all sets which do not contain
themselves."
 "Concerning mathematics, an example of an impredicative
definition is the smallest number in a set, which is formally defined
as: y =3D min(X) if and only if for all elements x of X, y is less than
or equal to x, and y is in X."
 "The greatest lower bound of a set X, glb(X), generalizes this
concept; y =3D glb(X) if and only if for all elements x of X, y is less
than or equal to x, and any z less than or equal to all elements of X
is less than or equal to y. But this definition also quantifies over
the set (potentially infinite, depending on the order in question)
whose members are the lower bounds of X, one of which being the glb
itself. Hence predicativism would reject this definition."
I'll take also the chance to note that, while I keep disagree that
Cantor's arguments (any of them) are "perfectly constructive", as has
been claimed, my objection to the diagonal argument here does not
leverage predicativity at all: I have tried to focuses on what to me
seems an unwarranted conclusion.
LV


0




Reply

julio (505)

5/10/2009 4:43:03 PM


Alan Smaill wrote:
> LudovicoVan <julio@diegidio.name> writes:
<skip the babble>
> There is denial at work, indeed.
No there isn't.
Brian Chandler


0




Reply

imaginatorium (41)

5/10/2009 4:50:08 PM


On 10 Mai, 11:00, Keith Ramsay <kram...@aol.com> wrote:
> Not too severely different. The main difference is that as
> constructively interpreted the result is a little stronger;
> it tells you that you can actually exhibit the new real
> number.
And that is false because no irrational number can be uniquely defined
by parts of its decimal expansion  and the complete expansion,
complete enough to distinguish the diagonal from every other number
(not only from every _given_ number (that's a difference, because not
all numbers can be given)), is inaccessible.
But that is not the concern of every constructivist. Brouwer for
instance said:
According to the statements previously made, this power alephnull
is the only infinite power of which the intuitionists recognize the
existence.
Let us now consider the concept: \denumerably infinite ordinal
number." From
the fact that this concept has a clear and welldefined meaning for
both formalist
and intuitionist, the former infers the right to create the \set of
all denumerably
infinite ordinal numbers," the power of which he calls alephone, a
right not recognized
by the intuitionist. Because it is possible to argue to the
satisfaction of
both formalist and intuitionist, first, that denumerably infinite sets
of denumerably
infinite ordinal numbers can be built up in various ways, and second,
that for every
such set it is possible to assign a denumerably infinite ordinal
number, not belonging
to this set, the formalist concludes: \alephone is greater than aleph
null,
a proposition, that has no meaning for the intuitionist.
Because it is possible to prove to the satisfaction of
both formalist and intuitionist, first, that denumerably infinite sets
of real numbers
between 0 and 1 can be constructed in various ways, and second that
for every such
set it is possible to assign a real number between 0 and 1, not
belonging to the set,
the formalist concludes: \the power of the continuum, i. e., the power
of the set
of real numbers between 0 and 1, is greater than alephnull," a
proposition which
is without meaning for the intuitionist; the formalist further raises
the question,
whether there exist sets of real numbers between 0 and 1, whose power
is less than
that of the continuum, but greater than alephnull, in other words,
\whether the
power of the continuum is the second smallest infinite power," and
this question,
which is still waiting for an answer, he considers to be one of the
most dificult and
most fundamental of mathematical problems.
For the intuitionist, however, the question as stated is without
meaning; and as
soon as it has been so interpreted as to get a meaning, it can easily
be answered.
Regards, WM


0




Reply

mueckenh (275)

5/10/2009 4:51:52 PM


Brian Chandler <imaginatorium@despammed.com> writes:
> Alan Smaill wrote:
> > LudovicoVan <julio@diegidio.name> writes:
> <skip the babble>
>
> > There is denial at work, indeed.
>
> No there isn't.
note, I didn't say who was in denial;
I thought that was fairly obvious ....
> Brian Chandler

Alan Smaill


0




Reply

smaill1 (89)

5/10/2009 5:18:34 PM


Alan Smaill wrote:
> Brian Chandler <imaginatorium@despammed.com> writes:
>
> > Alan Smaill wrote:
> > > LudovicoVan <julio@diegidio.name> writes:
> > <skip the babble>
> >
> > > There is denial at work, indeed.
> >
> > No there isn't.
>
> note, I didn't say who was in denial;
> I thought that was fairly obvious ....
Ah, sorry! I've just realised your statement above is simply
undeniable.
Brian Chandler


0




Reply

imaginatorium (41)

5/10/2009 5:21:07 PM


On 10 May, 17:28, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> LudovicoVan <ju...@diegidio.name> writes:
> > On 8 May, 22:42, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > > All that is claimed by the diagonal argument in this context is
> > > that for every *natural number* n, the antidiagonal
> > > differs from sequence f(n,0), f(n,1), f(n,2), ...
> > > And that we can show fairly easily  it looks from your post
> > > here that you accept that.
>
> > > So the conclusion is that the antidiagonal does not appear
> > > as a row in the *given* listing  nothing more than that.
>
> > The conclusion is still unwarranted: "the fact that the antidiagonal
> > does not belong to the list cannot be proven from the algorithmic
> > definition alone".
>
> If you refuse all notion of proof *about* algorithms, then
> you are right that noone can show anything about any algorithm.
I have said the *conclusion* cannot be proven *from* the algorithmic
definition alone. I am not refusing any algorithmic notions, quite the
opposite! (Denial at work?)
LV


0




Reply

julio (505)

5/10/2009 5:26:58 PM


R. Srinivasan wrote:
> Here is a simple argument for the inconsistency of ZFC. If Cantor's
> diagonal argument is correct, there are uncountably many real numbers
> in each interval of the form [1/2n, 1/2n], where n (=1,2,3....) is a
> positive integer. The intersection of all such intervals (over all the
> positive integers n) however, contains the single real number 0. Our
> intuition tells us that the intersection should also contain
> uncountably many reals, i.e., the intersection should be an interval.
> Each (or every) one of the nested intervals has endpoints fixed away
> from zero, so how come their intersection manages to eliminate "all"
> points other than zero, which does not seem possible on any reasonable
> conception of "each" (or "every" or "all")?
I hope I'm not taking anything out of context, but I'd just like to
look at this first paragraph. I assume that this alone is supposed to
be at least a preliminary justification of your apparent intuition.
Consider set rake(n) n a natural number, defined as:
{ x = a/bn  a, b are naturals, and 1 < x < 1 }
(In other words, rake(p) is the set of rationals in (1, 1) with a p
in the denominator.)
Each rake includes an infinite number of elements. For any two, rake
(m) and rake(n), the intersection includes all rationals in the range
with m*n in the denominator. Another infinite set. For any finite set
of rakes, the intersection is another rake:
Intersect (rake_j1, ...rake_jn) = rake(lcm(rake_j1...rake_jn)
Yet the intersection of _all_ the rakes consists only of 0. Do you
consider this too to be counterintuitive. (I don't, but if you don't
either, you need to explain the difference from the case above.)
In particular, I cannot understand the significance of this bit:
> ... does not seem possible on any reasonable
> conception of "each" (or "every" or "all")?
What sort of conception of "each", "every" and "all" do you have?
Brian Chandler


0




Reply

imaginatorium (41)

5/10/2009 5:34:18 PM


On May 10, 12:41 pm, WM <mueck...@rz.fhaugsburg.de> wrote:
> On 10 Mai, 02:35, William Hughes <wpihug...@hotmail.com> wrote:
>
> > Continuation
>
> 1
> 2, 1
> 3, 2, 1
> ...
>
>
>
> > We have (agreed by both WM and WH)
>
> > i: if
> > all natural numbers and all lines exist
> > then
> > no line contains all natural numbers
>
> > is true.
>
<snip>
>
>
>
> > WM claims a contradiction in that we also have
>
> > ii: if
> > all natural numbers and all lines exist
> > then
> > there exists a line that contains all natural numbers
>
> > is true.
>
<snip>
>
>
>
> > However, if all natural numbers exist and all lines exist,
> > then neither the set of natuaral numbers nor
> > the set of lines has a last element.
>
> (How then do you know that there are all?)
The first part of ii reads
if
all natural numbers and all lines exist
>
> > To establish the putative contradiction, we need a proof of ii
> > that does not require a last element.
>
> > WM first try:
>
> > WM: Every line contains all numbers that are in the preceding lines.
> > WM: If all numbers exist, then they exist in one line.
>
> > Nope. Clearly the "one line" in the second statment is the last
> > line.
> > The second statement is not true if there is no last line.
>
> No. My proof uses only the linearity and the geometrical symmetry
> shown in the list above. The first column cannot contain more than
> every line.
The first column cannot contain any *element*
not contained in the lines.
The first column can and does contain a *subset*
not contained in the lines.
Clearly the one line" in the second statment
is the last line.
The second statement (a statement about *subsets*
not *elements*) is not true if there is no last line.
Try again.
 William Hughes


0




Reply

wpihughes (390)

5/10/2009 5:42:37 PM


On May 10, 12:21 pm, WM <mueck...@rz.fhaugsburg.de> wrote:
> On 9 Mai, 18:38, William Hughes <wpihug...@hotmail.com> wrote:
>
> > On May 9, 12:11 pm, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> > WM continues to deliberately confuse elements and sets of elements.
>
> > One such property is that every natural number has even or odd
> > cardinality.
> > Look! Over There! A Pink Elephant!
>
> > > One such property is that every set of natural numbers has even or odd
> > > cardinality.
>
> That's because sets of natural numbers consist of natural numbers
of course
> and are counted by natural numbers.
Piffle. No set of natural numbers without
a last element is counted by a natural number.
 William Hughes


0




Reply

wpihughes (390)

5/10/2009 5:46:36 PM


LudovicoVan <julio@diegidio.name> writes:
> On 10 May, 17:28, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > LudovicoVan <ju...@diegidio.name> writes:
> > > On 8 May, 22:42, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>
> > > > All that is claimed by the diagonal argument in this context is
> > > > that for every *natural number* n, the antidiagonal
> > > > differs from sequence f(n,0), f(n,1), f(n,2), ...
> > > > And that we can show fairly easily  it looks from your post
> > > > here that you accept that.
> >
> > > > So the conclusion is that the antidiagonal does not appear
> > > > as a row in the *given* listing  nothing more than that.
> >
> > > The conclusion is still unwarranted: "the fact that the antidiagonal
> > > does not belong to the list cannot be proven from the algorithmic
> > > definition alone".
> >
> > If you refuse all notion of proof *about* algorithms, then
> > you are right that noone can show anything about any algorithm.
>
> I have said the *conclusion* cannot be proven *from* the algorithmic
> definition alone. I am not refusing any algorithmic notions, quite the
> opposite! (Denial at work?)
I didn't say you were refusing algorithmic notions;
what you have not said is what counts as acceptable proof *about*
properties of algorithms.
And you are right that if you refuse logic, and simply write
programs, then there is nothing to be said about properties
that the algorithms might or might not have. You haven't
denied or accepted that this is your position, I note,
and it's one way to take your comment "from the algorithmic
definition alone".
In the case at hand there is a simple proof in constructive logic that the
algorithm has the property that, given any effective enumeration of
effective reals in the sense we are using, we have defined an effective
computation of a digit sequence whioh is not at any index of the
original list.
So, since you deny the *possibility* of any such proof, it becomes
important to establish what sort of arguments you would find
acceptable to establish claims *about* algorithms, if any.
> LV

Alan Smaill


0




Reply

smaill1 (89)

5/10/2009 6:07:10 PM


On 10 Mai, 19:42, William Hughes <wpihug...@hotmail.com> wrote:
> > 1
> > 2, 1
> > 3, 2, 1
> > ...
>
> > No. My proof uses only the linearity and the geometrical symmetry
> > shown in the list above. The first column cannot contain more than
> > every line.
>
> The first column cannot contain any *element*
> not contained in the lines.
> The first column can and does contain a *subset*
> not contained in the lines.
That is impossible by the fact that every element contained in the
first column is also contained in one line.
If a line contains all elements of S then it contains the subset S.
You would like to construct a distinction between all elements of S
and S. You can do so, but you must identify what distinguishes these
things.
>
> Clearly the one line" in the second statment
> is the last line.
I did not say so. But it is suggesting itself that if all elements of
an ordered set exist, then there should be a last element. Because,
how would you porve that all are there?
> The second statement (a statement about *subsets*
> not *elements*) is not true if there is no last line.
You want to distinguish between "all elements 1, 2, 3, ..." and the
the set N.
Well, try to communicate a means by what this can be done. The mere
assertion is not enough.
Regards, WM


0




Reply

mueckenh (275)

5/10/2009 6:31:48 PM


On 10 Mai, 19:46, William Hughes <wpihug...@hotmail.com> wrote:
> On May 10, 12:21 pm, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> > On 9 Mai, 18:38, William Hughes <wpihug...@hotmail.com> wrote:
>
> > > On May 9, 12:11 pm, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> > > WM continues to deliberately confuse elements and sets of elements.
>
> > > One such property is that every natural number has even or odd
> > > cardinality.
> > > Look! Over There! A Pink Elephant!
>
> > > > One such property is that every set of natural numbers has even or =
odd
> > > > cardinality.
>
> > That's because sets of natural numbers consist of natural numbers
>
> of course
>
> > and =A0are counted by natural numbers.
>
> Piffle. =A0No set of natural numbers without
> a last element is counted by a natural number.
There is no natural number called "out a last element".
There is no set of natural numbers that cannot be counted by natural
numbers.
All natural numbers count themselves.
Regards, WM


0




Reply

mueckenh (275)

5/10/2009 6:34:50 PM


On 10 May, 19:07, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> In the case at hand there is a simple proof in constructive logic that the
> algorithm has the property that, given any effective enumeration of
> effective reals in the sense we are using, we have defined an effective
> computation of a digit sequence whioh is not at any index of the
> original list.
You keep misinterpreting on this: as said, I have no particular
objections to defining an "effective computation" for the anti
diagonal. That the sequence so defined is not in the list at any index
is what I am questioning: that *conclusion* requires a "leap to
infinity" that cannot be proven from the algorithmic (effective)
definitions only. IOW, that's a result that is *not* entailed by the
algorithmic (effective) properties of the definitions in question.
LV


0




Reply

julio (505)

5/10/2009 6:52:39 PM


LudovicoVan <julio@diegidio.name> writes:
> On 10 May, 19:07, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>
> > In the case at hand there is a simple proof in constructive logic that the
> > algorithm has the property that, given any effective enumeration of
> > effective reals in the sense we are using, we have defined an effective
> > computation of a digit sequence whioh is not at any index of the
> > original list.
>
> You keep misinterpreting on this: as said, I have no particular
> objections to defining an "effective computation" for the anti
> diagonal. That the sequence so defined is not in the list at any index
> is what I am questioning: that *conclusion* requires a "leap to
> infinity" that cannot be proven from the algorithmic (effective)
> definitions only. IOW, that's a result that is *not* entailed by the
> algorithmic (effective) properties of the definitions in question.
You keep saying this, and I keep asking you to clarify what *you* mean
when you say that something is not "entailed by algorithmic properties",
and you simply repeat your assertions.
You simply ignore my remark that there is a proof of the result
you say cannot be proved. If you refuse to say what counts as
proof, then of course you make it very hard for someone to
come up with an argument you might accept  that could well
be your intention, of course ...
Let's try this step by step;
what sort of properties of an algorithm do you think *can* be proved:
let's take an algorithm that takes as input a natural number, and outputs
double the input.
in your view, is it *possible* to show that such an algorithm
never outputs 3 ?
> LV

Alan Smaill


0




Reply

smaill1 (89)

5/10/2009 7:12:04 PM


On May 10, 9:36=A0am, WM <mueck...@rz.fhaugsburg.de> wrote:
> On 10 Mai, 00:43, Marshall <marshall.spi...@gmail.com> wrote:
>
> > So for me, it is highly ironic that, for all your relentless
> > fixation on finiteness, you suck at it. The best you can do
> > is rail pointlessly against clean abstractions like the
> > natural numbers. You complain bitterly that they don't model
> > resource limits, but you've never been able to show a logical
> > contradiction that results.
>
> The logical contradiction has nothing to do with resource limits.
> The logical contradiction is:
> It is impossible to define any irrational number by a string of
> digits.
> But it is possible to define an irrational number by a finite
> definition.
> The number of finite definitions, however, is countable.
Just as I said: you don't understand what a contradiction is.
Hmm. We were discussing the naturals, and your resource
limit based objections to the set of all of them, and then
you switch to a different objection to the set of all reals.
That is really quite dishonest, isn't it?
I guess I will have to retract my earlier statement that
I had not seen you be dishonest; I have actually seen
you do this sort of switch many times.
> > And of course, you haven't ever proposed any abstractions
> > of your own that *do* model resource limits;
>
> Current mathematics (except set theory) does what is necessary. There
> is no reason to provide further abstractions.
If current mathematics does what is necessary, then we have
no further need for new mathematics, and you have no basis
to complain about anything that is being done currently.
> > PS. Everyone says stupid things now and again.
> > And who knows? Maybe Robinson woke up the
> > next morning with a terrible headache, and was later
> > heard asking "I said *what* last night?!"
>
> Of course, all my quotes stem from drunk people.
Not necessarily; it could have been drugs.
Marshall


0




Reply

marshall.spight (580)

5/10/2009 7:14:37 PM


On Sun, 10 May 2009 12:14:37 0700 (PDT) Marshall wrote:
> Just as I said: you don't understand what a contradiction is.
> [etc. etc.]
"Wolfgang M�ckenheim is a classic crank. Why do you imagine, as you seem to
do, that there is any point arguing with him?" (Torkel Franzen)
Herb


0




Reply

Herbert

5/10/2009 7:22:30 PM


On 10 Mai, 21:14, Marshall <marshall.spi...@gmail.com> wrote:
> On May 10, 9:36=A0am, WM <mueck...@rz.fhaugsburg.de> wrote:
>
>
>
>
>
> > On 10 Mai, 00:43, Marshall <marshall.spi...@gmail.com> wrote:
>
> > > So for me, it is highly ironic that, for all your relentless
> > > fixation on finiteness, you suck at it. The best you can do
> > > is rail pointlessly against clean abstractions like the
> > > natural numbers. You complain bitterly that they don't model
> > > resource limits, but you've never been able to show a logical
> > > contradiction that results.
>
> > The logical contradiction has nothing to do with resource limits.
> > The logical contradiction is:
> > It is impossible to define any irrational number by a string of
> > digits.
> > But it is possible to define an irrational number by a finite
> > definition.
> > The number of finite definitions, however, is countable.
>
> Just as I said: you don't understand what a contradiction is.
The contradiction is: In mathematics based on ZFC:
Every real number can be identified.
Not every real number can be identified.
>
> Hmm. We were discussing the naturals, and your resource
> limit based objections to the set of all of them,
and I informed you that this is not my main point. My main point is
the inconsistency of the actual infinite.
> and then
> you switch to a different objection to the set of all reals.
> That is really quite dishonest, isn't it?
You said: "you've never been able to show a logical contradiction".
So I picked just one.
>
> > > And of course, you haven't ever proposed any abstractions
> > > of your own that *do* model resource limits;
>
> > Current mathematics (except set theory) does what is necessary. There
> > is no reason to provide further abstractions.
>
> If current mathematics does what is necessary, then we have
> no further need for new mathematics, and you have no basis
> to complain about anything that is being done currently.
Current mathematics has nothing to do with set theory. It is only set
theorists who are claiming that their theory was necessarily required
to provide a firm fundament for mathematics. Just the opposite is
true.
Regards, WM


0




Reply

mueckenh (275)

5/10/2009 7:32:23 PM


On Sun, 10 May 2009 04:38:08 0700 (PDT), R. Srinivasan
<sradhakr@in.ibm.com> said:
> On May 7, 3:56 pm, Barb Knox <s...@sig.below> wrote:
>> In article
>> <c7ccbbf2af51448eb9da773e185ec...@r36g2000vbr.googlegroups.com>,
>>
>>
>> [...]
>>
>> Face facts: Cantor's diagonal proof that R > N is SIMPLE. It has
>> even been CHECKED BY COMPUTERS. Therefore the only way that R =
>> N is if the various formulations of axiomatic set theory are
>> internally INCONSISTENT (which is emphatically NOT the same as being
>> inconsistent with someone's naive intuitions). Proving (say) ZFC to
>> be inconsistent would be worth at least a Fields Medal and a
>> centrefold picture in the Journal of the AMS. If it were easy (or
>> even less than extremely difficult) to do then it would already have
>> been done. The chance of you doing it is vanishingly small.
>>
>> Plus, there are good reasons for believing that ZFC etc. ARE
>> internally consistent. For example, there is an intuitively
>> notunreasonable model for ZFC etc. (Goedel's "constructible
>> universe").
>>
> So your "intuition" tells you that ZFC is consistent. Mine tells me
> otherwise.
>
> Here is a simple argument for the inconsistency of ZFC. If Cantor's
> diagonal argument is correct, there are uncountably many real numbers
> in each interval of the form [1/2n, 1/2n], where n (=1,2,3....) is a
> positive integer. The intersection of all such intervals (over all the
> positive integers n) however, contains the single real number 0. Our
> intuition tells us that the intersection should also contain
> uncountably many reals, i.e., the intersection should be an interval.
Unlike many alleged intuitions that differ from my own on one matter or
another, I cannot even get an inkling of yours. The intersection of a
set S of sets is simply the set of things common to every set in S. Is
it not obvious and intuitive to you that, for any nonzero real r in
[1/2, 1/2], there is an n such that r is not in [1/2n, 1/2n], i.e.,
that there is an interval of the sort in question whose endpoints are
closer to 0 than r? And hence that 0 is the only element common to the
intervals in question?
> Each (or every) one of the nested intervals has endpoints fixed away
> from zero, so how come their intersection manages to eliminate "all"
> points other than zero,...?
Well, as just noted. The fixed endpoints get closer and closer to 0 and
hence, for any given nonzero real r, they eventually get closer to 0
than r.
> ...which does not seem possible on any reasonable conception of "each"
> (or "every" or "all")
Quite the contrary.
One thing I find puzzling is that I don't see where the uncountability
of the intervals in question makes any difference to your reasoning. If
we were to replace those intervals in your argument with the sets of
rationals in those intervals, it seems to me that your intuition should
likewise tell you that there should be countably many rationals in their
intersection. What (if anything) is the difference in the two cases?
Is your idea that we shouldn't be able to "eliminate" uncountably many
reals by means of a countable "process" (i.e., one where we "move" from
[1/2n, 1/2n] to [1/2(n+1), 1/2(n+1)])? But this would be a problem
only if each step in the process only eliminated countably many reals.
I just don't see where you are coming from at all.


0




Reply

cmenzel (185)

5/10/2009 7:43:17 PM


WM wrote:
> On 10 Mai, 19:46, William Hughes <wpihug...@hotmail.com> wrote:
>> On May 10, 12:21 pm, WM <mueck...@rz.fhaugsburg.de> wrote:
>>
>> > On 9 Mai, 18:38, William Hughes <wpihug...@hotmail.com> wrote:
>>
>> > > On May 9, 12:11 pm, WM <mueck...@rz.fhaugsburg.de> wrote:
>>
>> > > WM continues to deliberately confuse elements and sets of elements.
>>
>> > > One such property is that every natural number has even or odd
>> > > cardinality.
>> > > Look! Over There! A Pink Elephant!
>>
>> > > > One such property is that every set of natural numbers has even or
>> > > > odd cardinality.
>>
>> > That's because sets of natural numbers consist of natural numbers
>>
>> of course
>>
>> > and are counted by natural numbers.
>>
>> Piffle. No set of natural numbers without
>> a last element is counted by a natural number.
>
> There is no natural number called "out a last element".
> There is no set of natural numbers that cannot be counted by natural
> numbers.
> All natural numbers count themselves.
>
> Regards, WM
ROTFL

W. Hughes, in sci.math.: "No set of natural numbers without a last element
[is finite]"
Prof. Dr. W. Mückenheim, mathematical mastermind of "Augsburg University of
Applied Science": "There is no natural number called "out a last element".


0




Reply

bader1785 (15)

5/10/2009 7:48:44 PM


On May 10, 5:45=A0pm, WM <mueck...@rz.fhaugsburg.de> wrote:
> On 9 Mai, 14:44, David C. Ullrich <dullr...@sprynet.com> wrote:
>
> > >That ultimately depends on your stance on the diagonal argument.
>
> > No. The reals _are_ uncountable.
>
> If there are reals that form an uncountable set, then these numbers
> cannot be accessed in any way, neither by diagonal arguments nor other
> methods. Therefore those reals that are accessible by any means that
> defines a number, including the diagonal argument *if it defines a
> number*, belong to a countable set.
>
> Regards, WM
Aehh, are you thinking of an uncountable set as equivalent to a
physical drawer cabinet with infinite small drawers. And that gives
you a problem because you can't find a pair of tweezers small enough
to handle the drawers  ok, that was a stupid suggestion, but your
statement seems based on some physically constrain.
Regards, Jan


0




Reply

jan4778 (24)

5/10/2009 8:45:59 PM


On May 10, 6:36=A0pm, WM <mueck...@rz.fhaugsburg.de> wrote:
> On 10 Mai, 00:43, Marshall <marshall.spi...@gmail.com> wrote:
>
> > So for me, it is highly ironic that, for all your relentless
> > fixation on finiteness, you suck at it. The best you can do
> > is rail pointlessly against clean abstractions like the
> > natural numbers. You complain bitterly that they don't model
> > resource limits, but you've never been able to show a logical
> > contradiction that results.
>
> The logical contradiction has nothing to do with resource limits.
> The logical contradiction is:
> It is impossible to define any irrational number by a string of
> digits.
> But it is possible to define an irrational number by a finite
> definition.
> The number of finite definitions, however, is countable.
If this is an argument for the countability of the reals, then please
elaborate on how these definitions are constructed.
Regards, Jan


0




Reply

jan4778 (24)

5/10/2009 8:52:26 PM


On May 10, 2:34 pm, WM <mueck...@rz.fhaugsburg.de> wrote:
> On 10 Mai, 19:46, William Hughes <wpihug...@hotmail.com> wrote:
>
>
>
> > On May 10, 12:21 pm, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> > > On 9 Mai, 18:38, William Hughes <wpihug...@hotmail.com> wrote:
>
> > > > On May 9, 12:11 pm, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> > > > WM continues to deliberately confuse elements and sets of elements.
>
> > > > One such property is that every natural number has even or odd
> > > > cardinality.
> > > > Look! Over There! A Pink Elephant!
>
> > > > > One such property is that every set of natural numbers has even or odd
> > > > > cardinality.
>
> > > That's because sets of natural numbers consist of natural numbers
>
> > of course
>
> > > and are counted by natural numbers.
>
> > Piffle. No set of natural numbers without
> > a last element is counted by a natural number.
<snip very silly word play>
 William Hughes


0




Reply

wpihughes (390)

5/10/2009 9:38:55 PM


On May 10, 2:31 pm, WM <mueck...@rz.fhaugsburg.de> wrote:
> On 10 Mai, 19:42, William Hughes <wpihug...@hotmail.com> wrote:
>
> > > 1
> > > 2, 1
> > > 3, 2, 1
> > > ...
>
ii: if
all natural numbers and all lines exist
then
there exists a line that contains all natural numbers
> > > No. My proof uses only the linearity and the geometrical symmetry
> > > shown in the list above. The first column cannot contain more than
> > > every line.
>
> > The first column cannot contain any *element*
> > not contained in the lines.
> > The first column can and does contain a *subset*
> > not contained in the lines.
>
WM tries again (#2)
> That is impossible by the fact that every element contained in the
> first column is also contained in one line.
Still assuming a last line.
Every element n of N contained is contained in some line,
say L(n) but this L(n) may be different from L(m).
Sure you can combine these two into one line,
but you still have other lines that have not
been combined. You can keep combining lines
but you will not finish until you reach the last
element. Without a last element you cannot combine *all*
the L(n) into one line.
Try again
 William Hughes


0




Reply

wpihughes (390)

5/10/2009 9:55:45 PM


On May 10, 12:32=A0pm, WM <mueck...@rz.fhaugsburg.de> wrote:
> On 10 Mai, 21:14, Marshall <marshall.spi...@gmail.com> wrote:
>
> > Just as I said: you don't understand what a contradiction is.
>
> The contradiction is: In mathematics based on ZFC:
> Every real number can be identified.
> Not every real number can be identified.
Closer, but still not a contradiction. Alas, the proverbial cigar
will be forever outside of your grasp.
Look; I have found a contradiction in the natural numbers!
1) Every natural number is seven.
2) Not every natural number is seven.
I have proved the natural numbers lead to a contradiction.
Marshall


0




Reply

marshall.spight (580)

5/10/2009 10:03:58 PM


On May 10, 12:22=A0pm, Herbert Newman <nomail@invalid> wrote:
> On Sun, 10 May 2009 12:14:37 0700 (PDT) Marshall wrote:
>
> > Just as I said: you don't understand what a contradiction is.
> > [etc. etc.]
>
> "Wolfgang M=FCckenheim is a classic crank. Why do you imagine, as you see=
m to
> do, that there is any point arguing with him?" (Torkel Franzen)
I do not imagine that there is any point in arguing with him. He
is clearly incapable of understanding the simplest of facts about
the infinite. Further, he is such a tool that there is not even any
amusement in pointing and laughing.
Rather, I would say he is like an open sore that you know
better than to scratch, but sometimes the itching just gets
the best of you.
Marshall


0




Reply

marshall.spight (580)

5/10/2009 10:09:36 PM


On May 10, 12:53=A0pm, WM <mueck...@rz.fhaugsburg.de> wrote:
> Nevertheless there has _never_ been a limit point of a sequence or
> series with irrational limit =A0that has been defined or identified by a
> sequence of digits.
Of course there has. What about
0.1101001000100001000001...
0.12345678910111213141516171819120...


0




Reply

horand.gassmann656 (50)

5/10/2009 11:13:17 PM


Phil Carmody wrote:
> herbzet writes:
> > William Hughes wrote:
> >> On May 9, 12:29 pm, WM <mueck...@rz.fhaugsburg.de> wrote:
> >> > On 8 Mai, 20:48, Jan <j...@daimi.au.dk> wrote:
> >> >
> >> > > Can you supply a natural number n, that the program won't print?
> >> >
> >> > No.
> >>
> >> Wolenmuekenheim logic
> >>
> >> There are infinitely many natural numbers that will never be
> >> printed.
> >> It is impossible to supply one of them.
> >
> >
> > Yeah, Bill, it's absurd.
> >
> > Why do you continue to respond to this loon? Do you think that
> > one day you'll back him into a corner and he'll say, "Yeah, you're
> > right, I'm mistaken"?
> >
> > Why won't you just let this spavined donkey die a natural death?
>
> I'd like to vote for an unnatural death, if I may?
Anything but this living death.

hz


0




Reply

herbzet (87)

5/11/2009 3:44:00 AM


Marshall wrote:
>
> Rather, I would say he is like an open sore that you know
> better than to scratch, but sometimes the itching just gets
> the best of you.
Wellchosen simile.
Trust me, after a year or so, you'll just be too
profoundly bored with his song and dance to care.
For me, it's just a dripping faucet.

hz


0




Reply

herbzet (87)

5/11/2009 3:56:56 AM


MoeBlee schrieb:
> On May 8, 4:01=A0am, Albrecht <albst...@gmx.de> wrote:
>
> > > 10 LET I=3D0
> > > 20 PRINT I
> > > 30 LET I =3D I + 1
> > > 40 GOTO 20
> >
> > > There. I just wrote a program to output _all_ the naturals.
> > > Now surely you must abandon your claim that I can't, despite
> > > your initial surety, for clearly I have written the program.
> >
> > This program doesn't output _all_ =A0naturals.
>
> Then please say what is the least natural number not outputted by the
> program.
>
There is no least. And now?
The program doesn't output _all_ the natural numbers. It can't. And
that's why there is no _all_ of the naturals.
Regards
Albrecht Storz


0




Reply

albstorz (110)

5/11/2009 6:29:39 AM


Jan schrieb:
> Hi
>
> I have some naive questions:
>
> On 8 Maj, 13:01, Albrecht <albst...@gmx.de> wrote:
>
> >> 10 LET I=0
> >> 20 PRINT I
> >> 30 LET I = I + 1
> >> 40 GOTO 20
> >
> >This program doesn't output _all_ naturals. Sketch of a proof: let's
> >say your program outputs the number n. Since there are infinitely many
> >naturals N > n there are nearly all naturals which won't be output 
> >for every n!
>
> Can you supply a natural number n, that the program won't print?
Can you supply a natural number n, that isn't followed by infinitely
many other N out of IN?
>
> > 10 LET I=0
> > 15 LET r=random real number of R
> > 20 PRINT r
> > 25 LET R=R without r
> > 30 LET I = I + 1
> > 40 GOTO 15
> >
> > So, what is the difference between naturals and reals?
>
> Can you please define the "function" returning: "random real number of
> R"?
> How do you represents the reals?
The reals are the elements of IR.
>
>
> > Start printing an (irrational) real number > phi < End printing an
> > (irrational) real number.
>
> Following that approach  how will you represent/present all the other
> transcendental real numbers? (e is of cause obvious...)
I only have demonstrated Marshall an error of his considerations.
That's all.
Regards
Albrecht Storz
>
> Kind regards, Jan


0




Reply

albstorz (110)

5/11/2009 6:39:32 AM


In article
<90d12b278119416e95385795a54510a1@r36g2000vbr.googlegroups.com>,
Albrecht <albstorz@gmx.de> wrote:
> > > Really funny.
> > > You say: "You can write a program to output �_all_ of the naturals."
> > > I'm sure, you can't.
> >
> > 10 LET I=0
> > 20 PRINT I
> > 30 LET I = I + 1
> > 40 GOTO 20
> >
> > There. I just wrote a program to output _all_ the naturals.
> > Now surely you must abandon your claim that I can't, despite
> > your initial surety, for clearly I have written the program.
>
> This program doesn't output _all_ naturals. Sketch of a proof: let's
> say your program outputs the number n. Since there are infinitely many
> naturals N > n there are nearly all naturals which won't be output 
> for every n!
Let S be the set of all natural numbers that are not in the output of
that program. Let n be the smallest member of S. Then n1 is not in S,
and so n1 is in the output of the program. But then the program
outputs n on the next iteration, contradicting the assumption that n is
in S.
Therefore, the program outputs all natural numbers.
....
> 10 LET I=0
> 15 LET r=random real number of R
> 20 PRINT r
> 25 LET R=R without r
> 30 LET I = I + 1
> 40 GOTO 15
>
> So, what is the difference between naturals and reals?
For any give real, there is no guarantee that the above program prints
it. For the earlier program, any given natural n is printed on
iteration n+1.

Tim Smith


0




Reply

reply_in_group (13194)

5/11/2009 8:38:19 AM


On 10 Mai, 22:45, Jan <j...@daimi.au.dk> wrote:
> On May 10, 5:45=A0pm, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> > On 9 Mai, 14:44, David C. Ullrich <dullr...@sprynet.com> wrote:
>
> > > >That ultimately depends on your stance on the diagonal argument.
>
> > > No. The reals _are_ uncountable.
>
> > If there are reals that form an uncountable set, then these numbers
> > cannot be accessed in any way, neither by diagonal arguments nor other
> > methods. Therefore those reals that are accessible by any means that
> > defines a number, including the diagonal argument *if it defines a
> > number*, belong to a countable set.
>
> > Regards, WM
>
> Aehh, are you thinking of an uncountable set as equivalent to a
> physical drawer cabinet with infinite small drawers. And that gives
> you a problem because you can't find a pair of tweezers small enough
> to handle the drawers  ok, that was a stupid suggestion, but your
> statement seems based on some physically constrain.
There is no physical constrain involved here (it would be justified
though) but only the observation that the set of finite words over a
finite alphabet is countable and that the sets of all alphabets and
all languages are countable and that the cartesian product of
countable sets is countable.
If there are uncountably many "numbers", then they cannot be used as
individuals, neither named nor distinguished from each other.
Therefore they will not serve as a reservoir supplying Cantor's
diagonals.
Regards, WM


0




Reply

mueckenh (275)

5/11/2009 11:43:08 AM


On 10 Mai, 22:52, Jan <j...@daimi.au.dk> wrote:
> On May 10, 6:36=A0pm, WM <mueck...@rz.fhaugsburg.de> wrote:
>
>
>
>
>
> > On 10 Mai, 00:43, Marshall <marshall.spi...@gmail.com> wrote:
>
> > > So for me, it is highly ironic that, for all your relentless
> > > fixation on finiteness, you suck at it. The best you can do
> > > is rail pointlessly against clean abstractions like the
> > > natural numbers. You complain bitterly that they don't model
> > > resource limits, but you've never been able to show a logical
> > > contradiction that results.
>
> > The logical contradiction has nothing to do with resource limits.
> > The logical contradiction is:
> > It is impossible to define any irrational number by a string of
> > digits.
> > But it is possible to define an irrational number by a finite
> > definition.
> > The number of finite definitions, however, is countable.
>
> If this is an argument for the countability of the reals, then please
> elaborate on how these definitions are constructed.
What is unclear with pi when it is given by circumference and diameter
of a circle? What has to be constructed?
Regards, WM


0




Reply

mueckenh (275)

5/11/2009 11:46:43 AM


On May 11, 1:46=A0pm, WM <mueck...@rz.fhaugsburg.de> wrote:
> On 10 Mai, 22:52, Jan <j...@daimi.au.dk> wrote:
>
>
>
> > On May 10, 6:36=A0pm, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> > > On 10 Mai, 00:43, Marshall <marshall.spi...@gmail.com> wrote:
>
> > > > So for me, it is highly ironic that, for all your relentless
> > > > fixation on finiteness, you suck at it. The best you can do
> > > > is rail pointlessly against clean abstractions like the
> > > > natural numbers. You complain bitterly that they don't model
> > > > resource limits, but you've never been able to show a logical
> > > > contradiction that results.
>
> > > The logical contradiction has nothing to do with resource limits.
> > > The logical contradiction is:
> > > It is impossible to define any irrational number by a string of
> > > digits.
> > > But it is possible to define an irrational number by a finite
> > > definition.
> > > The number of finite definitions, however, is countable.
>
> > If this is an argument for the countability of the reals, then please
> > elaborate on how these definitions are constructed.
>
> What is unclear with pi when it is given by circumference and diameter
> of a circle? What has to be constructed?
>
> Regards, WM
I'm not thinking of the few known/used constant functions, but on all
the other functions you need to define the irrational numbers in IR,
Ok?


0




Reply

jan4778 (24)

5/11/2009 12:12:05 PM


On May 11, 1:43=A0pm, WM <mueck...@rz.fhaugsburg.de> wrote:
> On 10 Mai, 22:45, Jan <j...@daimi.au.dk> wrote:
>
>
>
> > On May 10, 5:45=A0pm, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> > > On 9 Mai, 14:44, David C. Ullrich <dullr...@sprynet.com> wrote:
>
> > > > >That ultimately depends on your stance on the diagonal argument.
>
> > > > No. The reals _are_ uncountable.
>
> > > If there are reals that form an uncountable set, then these numbers
> > > cannot be accessed in any way, neither by diagonal arguments nor othe=
r
> > > methods. Therefore those reals that are accessible by any means that
> > > defines a number, including the diagonal argument *if it defines a
> > > number*, belong to a countable set.
>
> > > Regards, WM
>
> > Aehh, are you thinking of an uncountable set as equivalent to a
> > physical drawer cabinet with infinite small drawers. And that gives
> > you a problem because you can't find a pair of tweezers small enough
> > to handle the drawers  ok, that was a stupid suggestion, but your
> > statement seems based on some physically constrain.
>
> There is no physical constrain involved here (it would be justified
> though) but only the observation that the set of finite words over a
> finite alphabet is countable and that the sets of all alphabets and
> all languages are countable and that the cartesian product of
> countable sets is countable.
> If there are uncountably many "numbers", then they cannot be used as
> individuals, neither named nor distinguished from each other.
> Therefore they will not serve as a reservoir supplying Cantor's
> diagonals.
So, you somehow agree with Cantor's argument?


0




Reply

jan4778 (24)

5/11/2009 12:14:31 PM


On 11 Mai, 14:12, Jan <j...@daimi.au.dk> wrote:
> On May 11, 1:46=A0pm, WM <mueck...@rz.fhaugsburg.de> wrote:
>
>
>
>
>
> > On 10 Mai, 22:52, Jan <j...@daimi.au.dk> wrote:
>
> > > On May 10, 6:36=A0pm, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> > > > On 10 Mai, 00:43, Marshall <marshall.spi...@gmail.com> wrote:
>
> > > > > So for me, it is highly ironic that, for all your relentless
> > > > > fixation on finiteness, you suck at it. The best you can do
> > > > > is rail pointlessly against clean abstractions like the
> > > > > natural numbers. You complain bitterly that they don't model
> > > > > resource limits, but you've never been able to show a logical
> > > > > contradiction that results.
>
> > > > The logical contradiction has nothing to do with resource limits.
> > > > The logical contradiction is:
> > > > It is impossible to define any irrational number by a string of
> > > > digits.
> > > > But it is possible to define an irrational number by a finite
> > > > definition.
> > > > The number of finite definitions, however, is countable.
>
> > > If this is an argument for the countability of the reals, then please
> > > elaborate on how these definitions are constructed.
>
> > What is unclear with pi when it is given by circumference and diameter
> > of a circle? What has to be constructed?
>
> > Regards, WM
>
> I'm not thinking of the few known/used constant functions, but on all
> the other functions you need to define the irrational numbers in IR,
For every number you need at least one finite definition, or to say it
in other word, you need at least one finite word. But there are only
countably many finite words in any language. Therefore there are only
countably many irrational numbers that can be defined as individuals.
Undefined numbers however are not numbers.
Regards, WM


0




Reply

mueckenh (275)

5/11/2009 12:29:05 PM


On 11 Mai, 14:14, Jan <j...@daimi.au.dk> wrote:
> On May 11, 1:43=A0pm, WM <mueck...@rz.fhaugsburg.de> wrote:
>
>
>
>
>
> > On 10 Mai, 22:45, Jan <j...@daimi.au.dk> wrote:
>
> > > On May 10, 5:45=A0pm, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> > > > On 9 Mai, 14:44, David C. Ullrich <dullr...@sprynet.com> wrote:
>
> > > > > >That ultimately depends on your stance on the diagonal argument.
>
> > > > > No. The reals _are_ uncountable.
>
> > > > If there are reals that form an uncountable set, then these numbers
> > > > cannot be accessed in any way, neither by diagonal arguments nor ot=
her
> > > > methods. Therefore those reals that are accessible by any means tha=
t
> > > > defines a number, including the diagonal argument *if it defines a
> > > > number*, belong to a countable set.
>
> > > > Regards, WM
>
> > > Aehh, are you thinking of an uncountable set as equivalent to a
> > > physical drawer cabinet with infinite small drawers. And that gives
> > > you a problem because you can't find a pair of tweezers small enough
> > > to handle the drawers  ok, that was a stupid suggestion, but your
> > > statement seems based on some physically constrain.
>
> > There is no physical constrain involved here (it would be justified
> > though) but only the observation that the set of finite words over a
> > finite alphabet is countable and that the sets of all alphabets and
> > all languages are countable and that the cartesian product of
> > countable sets is countable.
> > If there are uncountably many "numbers", then they cannot be used as
> > individuals, neither named nor distinguished from each other.
> > Therefore they will not serve as a reservoir supplying Cantor's
> > diagonals.
>
> So, you somehow agree with Cantor's argument?
No, obviously I do not. All defined numbers including all defined
diagonal numbers belong to a countable set.
Even if there were undefinable numbers, they would not serve to
rectify Cantor's argument.
Regards, WM


0




Reply

mueckenh (275)

5/11/2009 12:31:55 PM


On May 11, 2:29=A0pm, WM <mueck...@rz.fhaugsburg.de> wrote:
> On 11 Mai, 14:12, Jan <j...@daimi.au.dk> wrote:
>
>
>
> > On May 11, 1:46=A0pm, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> > > On 10 Mai, 22:52, Jan <j...@daimi.au.dk> wrote:
>
> > > > On May 10, 6:36=A0pm, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> > > > > On 10 Mai, 00:43, Marshall <marshall.spi...@gmail.com> wrote:
>
> > > > > > So for me, it is highly ironic that, for all your relentless
> > > > > > fixation on finiteness, you suck at it. The best you can do
> > > > > > is rail pointlessly against clean abstractions like the
> > > > > > natural numbers. You complain bitterly that they don't model
> > > > > > resource limits, but you've never been able to show a logical
> > > > > > contradiction that results.
>
> > > > > The logical contradiction has nothing to do with resource limits.
> > > > > The logical contradiction is:
> > > > > It is impossible to define any irrational number by a string of
> > > > > digits.
> > > > > But it is possible to define an irrational number by a finite
> > > > > definition.
> > > > > The number of finite definitions, however, is countable.
>
> > > > If this is an argument for the countability of the reals, then plea=
se
> > > > elaborate on how these definitions are constructed.
>
> > > What is unclear with pi when it is given by circumference and diamete=
r
> > > of a circle? What has to be constructed?
>
> > > Regards, WM
>
> > I'm not thinking of the few known/used constant functions, but on all
> > the other functions you need to define the irrational numbers in IR,
>
> For every number you need at least one finite definition, or to say it
> in other word, you need at least one finite word. But there are only
> countably many finite words in any language. Therefore there are only
> countably many irrational numbers that can be defined as individuals.
>
> Undefined numbers however are not numbers.
>
> Regards, WM
So for every real in your system there's a computable function, right?
Regards, Jan


0




Reply

jan4778 (24)

5/11/2009 12:42:09 PM


On 11 Mai, 00:03, Marshall <marshall.spi...@gmail.com> wrote:
> On May 10, 12:32=A0pm, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> > On 10 Mai, 21:14, Marshall <marshall.spi...@gmail.com> wrote:
>
> > > Just as I said: you don't understand what a contradiction is.
>
> > The contradiction is: In mathematics based on ZFC:
> > Every real number can be identified.
> > Not every real number can be identified.
>
> Closer, but still not a contradiction.
Of course not. It is impossible to disprove belief.
> Look; I have found a contradiction in the natural numbers!
>
> 1) Every natural number is seven.
> 2) Not every natural number is seven.
>
> I have proved the natural numbers lead to a contradiction.
>
And you blamed Wittgenstein and Weyl and Robinson to be drunken?
I do not see any use in further discussing with you.
Regards, WM


0




Reply

mueckenh (275)

5/11/2009 1:08:34 PM


On 10 Mai, 23:55, William Hughes <wpihug...@hotmail.com> wrote:
> > > > 1
> > > > 2, 1
> > > > 3, 2, 1
> > > > ...
> Still assuming a last line.
No.
> Every element n of N contained is contained in some line,
> say L(n) =A0but this L(n) may be different from L(m).
Maybe.
> Sure you can combine these two into one line,
> but you still have other lines that have not
> been combined. =A0You can keep combining lines
> but you will not finish until you reach the last
> element. Without a last element you cannot combine *all*
> the L(n) into one line.
When all lines exist then all lines can be combined.
The problem is not a last line. The problem is that all lines are
subject to linearity.
The problem is that there are all numbers, and all can be combined in
one list that contains only finite lines.
Instead of lines you should consider numbers: Without a last number
you cannot combine *all* numbers.
Regards, WM


0




Reply

mueckenh (275)

5/11/2009 1:12:51 PM


On 11 Mai, 10:38, Tim Smith <reply_in_gr...@mousepotato.com> wrote:
> In article
> <90d12b278119416e95385795a5451...@r36g2000vbr.googlegroups.com>,
>
>
>
>
>
> =A0Albrecht <albst...@gmx.de> wrote:
> > > > Really funny.
> > > > You say: "You can write a program to output =A0_all_ of the natural=
s."
> > > > I'm sure, you can't.
>
> > > 10 LET I=3D0
> > > 20 PRINT I
> > > 30 LET I =3D I + 1
> > > 40 GOTO 20
>
> > > There. I just wrote a program to output _all_ the naturals.
> > > Now surely you must abandon your claim that I can't, despite
> > > your initial surety, for clearly I have written the program.
>
> > This program doesn't output _all_ =A0naturals. Sketch of a proof: let's
> > say your program outputs the number n. Since there are infinitely many
> > naturals N > n there are nearly all naturals which won't be output 
> > for every n!
>
> Let S be the set of all natural numbers that are not in the output of
> that program. =A0Let n be the smallest member of S. =A0Then n1 is not in=
S,
> and so n1 is in the output of the program. =A0But then the program
> outputs n on the next iteration, contradicting the assumption that n is
> in S.
>
> Therefore, the program outputs all natural numbers.
>
> ...
>
> > 10 LET I=3D0
> > 15 LET r=3Drandom real number of R
> > 20 PRINT r
> > 25 LET R=3DR without r
> > 30 LET I =3D I + 1
> > 40 GOTO 15
>
> > So, what is the difference between naturals and reals?
>
> For any give real,
How many reals can be "given"?
> there is no guarantee that the above program prints
> it. =A0For the earlier program, any given natural n is printed on
> iteration n+1.
>
And there are ungiven naturals too?
Regards, WM


0




Reply

mueckenh (275)

5/11/2009 1:23:36 PM


On 11 Mai, 08:29, Albrecht <albst...@gmx.de> wrote:
> MoeBlee schrieb:
>
>
>
>
>
> > On May 8, 4:01=A0am, Albrecht <albst...@gmx.de> wrote:
>
> > > > 10 LET I=3D0
> > > > 20 PRINT I
> > > > 30 LET I =3D I + 1
> > > > 40 GOTO 20
>
> > > > There. I just wrote a program to output _all_ the naturals.
> > > > Now surely you must abandon your claim that I can't, despite
> > > > your initial surety, for clearly I have written the program.
>
> > > This program doesn't output _all_ =A0naturals.
>
> > Then please say what is the least natural number not outputted by the
> > program.
>
> There is no least. And now?
>
> The program doesn't output _all_ =A0the natural numbers. It can't. And
> that's why there is no _all_ =A0of the naturals.
So it is. There remain infinitely many naturals unprinted. We can even
say, if we like:
When number n has been printed, then at least n^n^n natural numbers
remain unprinted.
That is the one side of the medal  and it cannot be refuted.
Nevertheless we cannot specify any natural that remains unprinted.
That is the other side. And set theorists wish to make us believe,
that "by logic" this means that all naturals were printed, if only we
waited infinitely long.
It is ridiculous how blind one eye can be.
Regards, WM


0




Reply

mueckenh (275)

5/11/2009 1:30:22 PM


On May 11, 6:08=A0am, WM <mueck...@rz.fhaugsburg.de> wrote:
> On 11 Mai, 00:03, Marshall <marshall.spi...@gmail.com> wrote:
> > On May 10, 12:32=A0pm, WM <mueck...@rz.fhaugsburg.de> wrote:
> > > On 10 Mai, 21:14, Marshall <marshall.spi...@gmail.com> wrote:
>
> > > > Just as I said: you don't understand what a contradiction is.
>
> > > The contradiction is: In mathematics based on ZFC:
> > > Every real number can be identified.
> > > Not every real number can be identified.
>
> > Closer, but still not a contradiction.
>
> Of course not. It is impossible to disprove belief.
Me: You fail at contradiction.
You: I can show a contradiction! Here are three
sentences that have consequences I don't like.
Me: Contradiction fail.
You: [looks up contradiction on wikipedia] I can show
a contradiction! Here are two statements, S and
notS. S is false and notS is true.
Me: Contradiction fail again!
You: Of course I cannot show a contradiction. It is
your fault!
You are starting to regain some of your former amusement value.
Marshall


0




Reply

marshall.spight (580)

5/11/2009 1:49:17 PM


On 11 Mai, 14:42, Jan <j...@daimi.au.dk> wrote:
> > For every number you need at least one finite definition, or to say it
> > in other word, you need at least one finite word. But there are only
> > countably many finite words in any language. Therefore there are only
> > countably many irrational numbers that can be defined as individuals.
>
> > Undefined numbers however are not numbers.
>
> So for every real in your system there's a computable function, right?
It is not *my* system. It is a simple fact that a number only exists
as a product of mind. If there is no mind capable of identifying some
number, then this number does not exist. There is no platonist shelve
where they all are stored.
Regards, WM


0




Reply

mueckenh (275)

5/11/2009 1:52:48 PM


Tim Smith schrieb:
> In article
> <90d12b278119416e95385795a54510a1@r36g2000vbr.googlegroups.com>,
> Albrecht <albstorz@gmx.de> wrote:
> > > > Really funny.
> > > > You say: "You can write a program to output =EF=BF=BD_all_ of the n=
aturals."
> > > > I'm sure, you can't.
> > >
> > > 10 LET I=3D0
> > > 20 PRINT I
> > > 30 LET I =3D I + 1
> > > 40 GOTO 20
> > >
> > > There. I just wrote a program to output _all_ the naturals.
> > > Now surely you must abandon your claim that I can't, despite
> > > your initial surety, for clearly I have written the program.
> >
> > This program doesn't output _all_ naturals. Sketch of a proof: let's
> > say your program outputs the number n. Since there are infinitely many
> > naturals N > n there are nearly all naturals which won't be output 
> > for every n!
>
> Let S be the set of all natural numbers that are not in the output of
> that program. Let n be the smallest member of S. Then n1 is not in S,
> and so n1 is in the output of the program. But then the program
> outputs n on the next iteration, contradicting the assumption that n is
> in S.
>
> Therefore, the program outputs all natural numbers.
>
Do you claim that there is a natural n (or perhaps some n_1, n_2,
n_3, ...?) which is (are) not followed by infinitely many N e IN?
Regards
Albrecht Storz
> ...
> > 10 LET I=3D0
> > 15 LET r=3Drandom real number of R
> > 20 PRINT r
> > 25 LET R=3DR without r
> > 30 LET I =3D I + 1
> > 40 GOTO 15
> >
> > So, what is the difference between naturals and reals?
>
> For any give real, there is no guarantee that the above program prints
> it. For the earlier program, any given natural n is printed on
> iteration n+1.
>
> 
> Tim Smith


0




Reply

albstorz (110)

5/11/2009 1:54:19 PM


On Fri, 8 May 2009 12:36:46 0700 (PDT), LudovicoVan wrote:
> On 8 May, 00:03, Marshall <marshall.spi...@gmail.com> wrote:
>> You can't do the same for the reals. That's because the reals
>> aren't countable and the naturals are.
> That ultimately depends on your stance on the diagonal argument.
We can show that uncountable sets exist without using the diagonal
argument. Aleph_1 is an uncountable set, and its existence has nothing
to do with the diagonal argument.

Dave Seaman
Third Circuit ignores precedent in Mumia AbuJamal ruling.
<http://www.indybay.org/newsitems/2008/03/29/18489281.php>


0




Reply

dseaman (1174)

5/11/2009 1:54:20 PM


Dave Seaman wrote:
> On Fri, 8 May 2009 12:36:46 0700 (PDT), LudovicoVan wrote:
>
>>On 8 May, 00:03, Marshall <marshall.spi...@gmail.com> wrote:
>>
>>>You can't do the same for the reals. That's because the reals
>>>aren't countable and the naturals are.
>
>>That ultimately depends on your stance on the diagonal argument.
>
> We can show that uncountable sets exist without using the diagonal
> argument. Aleph_1 is an uncountable set, and its existence has nothing
> to do with the diagonal argument.
Who is "we" ? What is "can" ? What is "show" ? What is "exist" ? What is
"is" ? For a "thing" in your mind (not mine) like Aleph_1 ?
Couldn't resist ..
Han de Bruijn


0




Reply

Han.deBruijn (44)

5/11/2009 2:27:26 PM


Han de Bruijn <Han.deBruijn@DTO.TUDelft.NL> writes:
> Dave Seaman wrote:
>
>> On Fri, 8 May 2009 12:36:46 0700 (PDT), LudovicoVan wrote:
>>
>>>On 8 May, 00:03, Marshall <marshall.spi...@gmail.com> wrote:
>>>
>>>>You can't do the same for the reals. That's because the reals
>>>>aren't countable and the naturals are.
>>
>>>That ultimately depends on your stance on the diagonal argument.
>>
>> We can show that uncountable sets exist without using the diagonal
>> argument. Aleph_1 is an uncountable set, and its existence has nothing
>> to do with the diagonal argument.
>
> Who is "we" ?
Surely, Dave is not claiming that *you* can prove anything at all.
> What is "can" ? What is "show" ? What is "exist" ? What is "is" ?
> For a "thing" in your mind (not mine) like Aleph_1 ?
>
> Couldn't resist ..
Should've tried harder.

Jesse F. Hughes
"Two years from now, spam will be solved."
 Bill Gates, Jan 24, 2004


0




Reply

jesse18 (2492)

5/11/2009 2:58:19 PM


On May 11, 9:12 am, WM <mueck...@rz.fhaugsburg.de> wrote:
> On 10 Mai, 23:55, William Hughes <wpihug...@hotmail.com> wrote:
>
> > > > > 1
> > > > > 2, 1
> > > > > 3, 2, 1
> > > > > ...
WM is trying to prove
ii: if
all natural numbers and all lines exist
then
there exists a line that contains all natural numbers
> > Still assuming a last line.
>
> No.
>
> > Every element n of N contained is contained in some line,
> > say L(n) but this L(n) may be different from L(m).
>
> Maybe.
>
> > Sure you can combine these two into one line,
> > but you still have other lines that have not
> > been combined. You can keep combining lines
> > but you will not finish until you reach the last
> > element. Without a last element you cannot combine *all*
> > the L(n) into one line.
>
> When all lines exist then all lines can be combined.
Nope. You combine lines one by one. As there is
no last line you do not finish, so you cannot
combine all lines.
 William Hughes


0




Reply

wpihughes (390)

5/11/2009 3:07:31 PM


On May 9, 1:09 pm, LudovicoVan <ju...@diegidio.name> wrote:
> On 9 May, 17:08, George Greene <gree...@email.unc.edu> wrote:
> There are no games in your paradigm maybe.
You are talking about Cantor's theorem so you are NECESSARILY talking
about Cantor's paradigm.
> You don't even know what we are talking about.
YOU don't know what you are talking about.
You DON'T KNOW what a DEFINITION is, let alone what
an "algorithmic" defintion is.


0




Reply

greeneg9613 (188)

5/11/2009 3:16:03 PM


On May 10, 1:26 pm, LudovicoVan <ju...@diegidio.name> wrote:
> I have said the *conclusion* cannot be proven *from* the algorithmic
> definition alone.
Dipshit: the conclusion of a theorem is proven FROM THE AXIOMS.
"The algorithmic definition" SIMPLY HAS *NOTHING TO DO* with it!
Cantor's theorem DOES NOT INVOLVE ANY *algorithms*!!
It presumes that you have a list. Whether there is an algorithm
for generating the list DOES NOT MATTER!
It constructs the antidiagonal of the list!
For ANY position in the antidiagonaltobeconstructed,
AN ALGORITHM IS GIVEN for producing the output that will
occur at that position! This algorithm works FOR ANY AND EVERY
position! The fact that there is an infinite number of positions
DOES NOT MATTER and DOES NOT HAVE ANY effect on the
"algorithmic effectiveness" OF ANYthing!
You might as well be saying that there is no general algorithm for
doubling a number because you couldn't finish doubling all of them!


0




Reply

greeneg9613 (188)

5/11/2009 3:19:05 PM


On Mon, 11 May 2009 08:16:03 0700 (PDT) George Greene wrote:
> On May 9, 1:09 pm, LudovicoVan <ju...@diegidio.name> wrote:
>>
>> You don't even know what we are talking about.
>>
> YOU don't know what you are talking about.
> You DON'T KNOW what a DEFINITION is, let alone what
> an "algorithmic" defintion is.
George, this asshole is a troll, you know.
"Don't feed the trolls!"
Herb


0




Reply

Herbert

5/11/2009 3:21:10 PM


On 11 Mai, 17:07, William Hughes <wpihug...@hotmail.com> wrote:
> On May 11, 9:12 am, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> > On 10 Mai, 23:55, William Hughes <wpihug...@hotmail.com> wrote:
>
> > > > > > 1
> > > > > > 2, 1
> > > > > > 3, 2, 1
> > > > > > ...
>
> WM is trying to prove
>
> ii: =A0if
> =A0 =A0 =A0 =A0 =A0all natural numbers and all lines exist
> =A0 =A0 =A0then
> =A0 =A0 =A0 =A0 =A0there exists a line that contains all natural numbers
>
>
>
>
>
> > > Still assuming a last line.
>
> > No.
>
> > > Every element n of N contained is contained in some line,
> > > say L(n) =A0but this L(n) may be different from L(m).
>
> > Maybe.
>
> > > Sure you can combine these two into one line,
> > > but you still have other lines that have not
> > > been combined. =A0You can keep combining lines
> > > but you will not finish until you reach the last
> > > element. Without a last element you cannot combine *all*
> > > the L(n) into one line.
>
> > When all lines exist then all lines can be combined.
>
> Nope. =A0You combine lines one by one.
No, I do them all by induction. None is left out.
>=A0As there is
> no last line you do not finish, so you cannot
> combine all lines.
So there are not all lines (possible to combine)?
How then can you combine all numbers?
There are all ... but there are not all ... A typical statement of set
theory.
Regards, WM


0




Reply

mueckenh (275)

5/11/2009 3:27:59 PM


On 11 Mai, 17:27, WM <mueck...@rz.fhaugsburg.de> wrote:
> On 11 Mai, 17:07, William Hughes <wpihug...@hotmail.com> wrote:
>
>
>
>
>
> > On May 11, 9:12 am, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> > > On 10 Mai, 23:55, William Hughes <wpihug...@hotmail.com> wrote:
>
> > > > > > > 1
> > > > > > > 2, 1
> > > > > > > 3, 2, 1
> > > > > > > ...
>
> > WM is trying to prove
>
> > ii: =A0if
> > =A0 =A0 =A0 =A0 =A0all natural numbers and all lines exist
> > =A0 =A0 =A0then
> > =A0 =A0 =A0 =A0 =A0there exists a line that contains all natural number=
s
>
> > > > Still assuming a last line.
>
> > > No.
>
> > > > Every element n of N contained is contained in some line,
> > > > say L(n) =A0but this L(n) may be different from L(m).
>
> > > Maybe.
>
> > > > Sure you can combine these two into one line,
> > > > but you still have other lines that have not
> > > > been combined. =A0You can keep combining lines
> > > > but you will not finish until you reach the last
> > > > element. Without a last element you cannot combine *all*
> > > > the L(n) into one line.
>
> > > When all lines exist then all lines can be combined.
>
> > Nope. =A0You combine lines one by one.
>
> No, I do them all by induction. None is left out.
Remark: You agreeed that every line can be reached by induction. I do
not need more than every line.
>
> >=A0As there is
> > no last line you do not finish, so you cannot
> > combine all lines.
>
> So there are not all lines (possible to combine)?
> How then can you combine all numbers?
>
> There are all ... but there are not all ... A typical statement of set
> theory.
>
> Regards, WM


0




Reply

mueckenh (275)

5/11/2009 3:29:42 PM


George Greene <greeneg@email.unc.edu> writes:
> On May 10, 1:26 pm, LudovicoVan <ju...@diegidio.name> wrote:
> > I have said the *conclusion* cannot be proven *from* the algorithmic
> > definition alone.
>
> Dipshit: the conclusion of a theorem is proven FROM THE AXIOMS.
> "The algorithmic definition" SIMPLY HAS *NOTHING TO DO* with it!
> Cantor's theorem DOES NOT INVOLVE ANY *algorithms*!!
The theorem as proven in set theory can be generalised to
other situations.
It's a perfectly good question to ask whether diagonalisation
applies in the situation where we start with an effectively
given listing of computable reals, and ask if there
is a computable antidiagonal. (There is.)
> It presumes that you have a list. Whether there is an algorithm
> for generating the list DOES NOT MATTER!
> It constructs the antidiagonal of the list!
If the listing is not given effectively, then the antidiagonal will
not be effectively given either. There is nothing deep in pointing
out that if the listing and the reals are effective, then so is the
antidiagonal  this is not part of the standard proof in set theory,
however.
> For ANY position in the antidiagonaltobeconstructed,
> AN ALGORITHM IS GIVEN for producing the output that will
> occur at that position! This algorithm works FOR ANY AND EVERY
> position! The fact that there is an infinite number of positions
> DOES NOT MATTER and DOES NOT HAVE ANY effect on the
> "algorithmic effectiveness" OF ANYthing!
>
> You might as well be saying that there is no general algorithm for
> doubling a number because you couldn't finish doubling all of them!
that thought had crossed my mind too ...

Alan Smaill


0




Reply

smaill1 (89)

5/11/2009 3:33:31 PM


On May 11, 8:27=A0am, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> > Nope. =A0You combine lines one by one.
>
> No, I do them all by induction. None is left out.
You are every bit as successful with induction as you are
with contradiction! Namely, you fail even at understanding
what it is.
Marshall


0




Reply

marshall.spight (580)

5/11/2009 3:43:41 PM


On May 11, 11:27 am, WM <mueck...@rz.fhaugsburg.de> wrote:
> On 11 Mai, 17:07, William Hughes <wpihug...@hotmail.com> wrote:
>
>
>
> > On May 11, 9:12 am, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> > > On 10 Mai, 23:55, William Hughes <wpihug...@hotmail.com> wrote:
>
> > > > > > > 1
> > > > > > > 2, 1
> > > > > > > 3, 2, 1
> > > > > > > ...
>
> > WM is trying to prove
>
> > ii: if
> > all natural numbers and all lines exist
> > then
> > there exists a line that contains all natural numbers
>
> > > > Still assuming a last line.
>
> > > No.
>
> > > > Every element n of N contained is contained in some line,
> > > > say L(n) but this L(n) may be different from L(m).
>
> > > Maybe.
>
> > > > Sure you can combine these two into one line,
> > > > but you still have other lines that have not
> > > > been combined. You can keep combining lines
> > > > but you will not finish until you reach the last
> > > > element. Without a last element you cannot combine *all*
> > > > the L(n) into one line.
>
> > > When all lines exist then all lines can be combined.
>
> > Nope. You combine lines one by one.
>
> No, I do them all by induction. None is left out.
Using induction does not change the fact that you
combine them one by one.
>
> > As there is
> > no last line you do not finish, so you cannot
> > combine all lines.
>
> So there are not all lines
Nope, by assuption there are all lines
> (possible to combine)?
It is not possible to combine all lines.
> How then can you combine all numbers?
You don't start with nothing and combine
them one by one. You start with all of
them already in the list. For this you
need an assumption.
Recall the assumption
if
all natural numbers and all lines exist
So all lines exist and therefore the list exists.
It is easy to show "all lines"
contain "all natural numbers".
No need to combine anything.
 William Hughes


0




Reply

wpihughes (390)

5/11/2009 4:22:15 PM


On May 11, 11:29 am, WM <mueck...@rz.fhaugsburg.de> wrote:
> On 11 Mai, 17:27, WM <mueck...@rz.fhaugsburg.de> wrote:
>
>
>
> > On 11 Mai, 17:07, William Hughes <wpihug...@hotmail.com> wrote:
>
> > > On May 11, 9:12 am, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> > > > On 10 Mai, 23:55, William Hughes <wpihug...@hotmail.com> wrote:
>
> > > > > > > > 1
> > > > > > > > 2, 1
> > > > > > > > 3, 2, 1
> > > > > > > > ...
>
> > > WM is trying to prove
>
> > > ii: if
> > > all natural numbers and all lines exist
> > > then
> > > there exists a line that contains all natural numbers
>
> > > > > Still assuming a last line.
>
> > > > No.
>
> > > > > Every element n of N contained is contained in some line,
> > > > > say L(n) but this L(n) may be different from L(m).
>
> > > > Maybe.
>
> > > > > Sure you can combine these two into one line,
> > > > > but you still have other lines that have not
> > > > > been combined. You can keep combining lines
> > > > > but you will not finish until you reach the last
> > > > > element. Without a last element you cannot combine *all*
> > > > > the L(n) into one line.
>
> > > > When all lines exist then all lines can be combined.
>
> > > Nope. You combine lines one by one.
>
> > No, I do them all by induction. None is left out.
Induction does not change the fact
that you combine lines one by one.
>
> Remark: You agreeed that every line can be reached by induction.
No, I agreed that any line can be reached by induction.
You can reach any line. However, reaching a line does
not mean finishing unless the line is the last line.
There is no last line so you can't finish.
William Hughes


0




Reply

wpihughes (390)

5/11/2009 4:33:35 PM


William Hughes wrote:
> On May 11, 11:27 am, WM <mueck...@rz.fhaugsburg.de> wrote:
> > On 11 Mai, 17:07, William Hughes <wpihug...@hotmail.com> wrote:
> > > On May 11, 9:12 am, WM <mueck...@rz.fhaugsburg.de> wrote:
> > > > On 10 Mai, 23:55, William Hughes <wpihug...@hotmail.com> wrote:
> > > > > > > > 1
> > > > > > > > 2, 1
> > > > > > > > 3, 2, 1
> > > > > > > > ... [Enough!]
> > > As there is
> > > no last line you do not finish, so you cannot
> > > combine all lines.
> >
> > So there are not all lines
>
> Nope, by assuption there are all lines
What does it mean, to assume "there are all lines"? I don't see that
it can be a meaningful utterance unless there is at least the
possibility of assuming the contrary, and I am baffled by what the
contrary would be. Of course I can imagine Mueckenheim saying "There
are not all lines", but I can't help hearing this with a German
accent, since it's not natural English. Would you say this? What would
it mean?
On calmer reflection, I think the (grammatical, anyway) converse of
"there all lines" would have to be "there are not some lines", because
"there are [at least] some [other] lines". But "there are not some
lines" seems to be a denial of "there are some lines". So would it be
that there are no lines at all? Or (ha!) would it be that this is just
another proof of the inconsistency of set theory, and all the other
Cantorabilia?
(I give up, really. Like others, I am puzzled by what motivates you to
continue with this idiot, though he has provided sparks of amusement
lately. That "theorem"  can't find it, but something like "every
each =/= all", and it is apparently false anyway...)
Brian Chandler
>
> > (possible to combine)?
>
> It is not possible to combine all lines.
>
> > How then can you combine all numbers?
>
> You don't start with nothing and combine
> them one by one. You start with all of
> them already in the list. For this you
> need an assumption.
>
> Recall the assumption
>
> if
> all natural numbers and all lines exist
>
>
> So all lines exist and therefore the list exists.
> It is easy to show "all lines"
> contain "all natural numbers".
> No need to combine anything.
>
>  William Hughes


0




Reply

imaginatorium (41)

5/11/2009 4:57:37 PM


On 10 May, 20:12, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> LudovicoVan <ju...@diegidio.name> writes:
> > On 10 May, 19:07, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
>
> > > In the case at hand there is a simple proof in constructive logic tha=
t the
> > > algorithm has the property that, given any effective enumeration of
> > > effective reals in the sense we are using, we have defined an effecti=
ve
> > > computation of a digit sequence whioh is not at any index of the
> > > original list.
>
> > You keep misinterpreting on this: as said, I have no particular
> > objections to defining an "effective computation" for the anti
> > diagonal. That the sequence so defined is not in the list at any index
> > is what I am questioning: that *conclusion* requires a "leap to
> > infinity" that cannot be proven from the algorithmic (effective)
> > definitions only. IOW, that's a result that is *not* entailed by the
> > algorithmic (effective) properties of the definitions in question.
>
> You keep saying this, and I keep asking you to clarify what *you* mean
> when you say that something is not "entailed by algorithmic properties",
> and you simply repeat your assertions. =A0
I have mentioned a "leap to infinity" that is not entailed by
effectiveness: I have put this in terms of a two opponents game, or
turing machines, or the many ways to compactification. As for now, I'm
afraid I can't be clearer than that.
> You simply ignore my remark that there is a proof of the result
> you say cannot be proved.
I have commented to Bishop's proof: "my point (my take) is that this
simply does not solve our issue", for the reasons just mentioned.
> If you refuse to say what counts as
> proof, then of course you make it very hard for someone to
> come up with an argument you might accept  that could well
> be your intention, of course ...
No such intention, but I see no need either to define what counts as a
proof: I have raised a specific objection.
LV


0




Reply

julio (505)

5/11/2009 5:17:23 PM


On 11 May, 14:54, Dave Seaman <dsea...@no.such.host> wrote:
> We can show that uncountable sets exist without using the diagonal
> argument.
And could you tell which argument we would use instead?
LV


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Reply

julio (505)

5/11/2009 5:18:45 PM


On 11 May, 17:22, William Hughes <wpihug...@hotmail.com> wrote:
> On May 11, 11:27 am, WM <mueck...@rz.fhaugsburg.de> wrote:
> > No, I do them all by induction. None is left out.
> Using induction does not change the fact that you
> combine them one by one.
> > So there are not all lines
> Nope, by assuption there are all lines
The eternal riddle.
LV


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Reply

julio (505)

5/11/2009 5:21:49 PM


On Mon, 11 May 2009 10:18:45 0700 (PDT), LudovicoVan wrote:
> On 11 May, 14:54, Dave Seaman <dsea...@no.such.host> wrote:
>> We can show that uncountable sets exist without using the diagonal
>> argument.
> And could you tell which argument we would use instead?
This is from a post by Matthew P. Wiener in 1998:
On the one hand, P(NxN) existing is straightforward. Therefore, by
comprehension, so is the P(NxN) subset of those relations on N which
are wellordered. Call it WO.
The von Neumann ordinals can be definedin one fell swoopas
wellordered (by membership) transitive sets. (z is transitive means
x e y e z => x e z.)
The elementary theory of wellordered sets is easy to develop without AC.
In particular, one can prove there exists a unique von Neumann ordinal
of the same order type for any given wellordered set. (For starts,
this is via the Mostowski collapse: given w wellordered, let mc(w)=
{mc(x):x < w}. By definition by transfinite recursion, this is
welldefined, and rather easy to prove is transitive.)
At this point, apply the axiom of replacement to WO, replacing each
wellordered relation on N with its von Neumann order type ordinal.
The resulting set is aleph_1.
For some reason, there is a folklore belief that choice is needed here.
I've even seen this claim show up in books on set theory!
(end quote from Matthew P. Wiener)
To summarize: the basic axiom needed is the power set axiom. A few other
axioms play a role, but AC is not needed and there is no diagonal argument
involved.

Dave Seaman
Third Circuit ignores precedent in Mumia AbuJamal ruling.
<http://www.indybay.org/newsitems/2008/03/29/18489281.php>


0




Reply

dseaman (1174)

5/11/2009 5:52:21 PM


On May 8, 8:52=A0pm, lwal...@lausd.net wrote:
> On May 8, 8:28=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On May 8, 8:22=A0pm, lwal...@lausd.net wrote:
> > > Mathematicians all over the world are also searching for
> > > proofs that ZFC is inconsistent.
> > They are? Would you please name several countries and a mathematician
> > from each of those countries who is searching for a contradiction in
> > ZFC? Or in the rubric of 'mathematician', do you include cranks?
>
> In this particular post, I'm not including socalled "cranks"
> but only standard set theorists.
>
> Well probably at _first_, right after ZFC was formulated,
> likely as many mathematicians sought a proof that ZFC was
> inconsistent as sought a proof that the naive set theory
> of Cantor and Frege was inconsistent.
So you don't know anything specific at all.
> Knox wrote that a proof of ~Con(ZFC) would likely merit a
> Fields medal. It's not unreasonable to guess that many
> mathematicians around the world would do research that
> would earn them a Fields medal, the most prestigious prize
> in all of mathematics.
So you don't know anything specific at all.
> So if, as MoeBlee might be implying here, very _few_
> mathematicians seek a proof of ~Con(ZFC),
I didn't imply that. I'm just asking what is the basis of your claim.
> then they likely
> believe that no proof is possible. And if this is really
> the case, then this would actually _prove_ the point that
> I was making to _Mariano_  that the more time passes,
> the less likely a proof of ~Con(ZFC) will be found, and
> the fewer mathematicians will even search for a proof!
Except that continued research in ZFC itself exposes ZFC to scrutiny
for contradiction. Every theorem proved allows that someone may prove
the negation of that theorem.
> AFAIK, there is only one mathematician seeking a proof
> that PA and ZFC are inconsistent  Ed Nelson.
>
> So thanks, MoeBlee, for helping me to prove my point to
> Mariano here!
I didn't help you. All I asked was for a basis for your claim. You
have barely a sliver of one.
MoeBlee


0




Reply

jazzmobe (307)

5/11/2009 5:59:07 PM


On May 9, 8:39=A0am, LudovicoVan <ju...@diegidio.name> wrote:
> On 8 May, 22:24, MoeBlee <jazzm...@hotmail.com> wrote:
>
> > On May 8, 12:25=A0pm, LudovicoVan <ju...@diegidio.name> wrote:
> > > As I have already noted in this thread, the formal, axiomatic proofs
> > > presuppose what here is in question, mainly a decision about
> > > "infinities".
>
> > And I already responded to that claim. You just skipped dealing with
> > my response. Please, if there is to be any dialogoue, it is not
> > practical if you are going to just pretend that my responses don't
> > exist.
>
> It's you who are pretending my responses do not exist.
No, I've answered you point by point.
> > So would you please address that question: What axiomatization do you
> > propose instead?
>
> I have already answered this question of yours many times now. Maybe I
> was not clear, so I'll try again: we are investigating foundational
> issues here, the diagonal argument specifically; it is improper, if
> not a basic misunderstanding, to ask for axiomatisations in this
> context.
No, you did it again. You snipped (you even marked it) my argument
about axiomatizations. The very fact that we're talking about
foundations requires that we address the matter of axiomatizations.
Also, what is your argument that the diagonal proof is impredicative?
The arguments of Koskensilta and McCullough in the "Yet Another
Disproof Of Cantor's Theorem" thread (and another thread too) do seem
sufficient to refute that the diagonal argument is impredicative.
Anyway, since you insist upon what you consider to be a predicative
foundation, I don't see why you won't just tell us what axioms you
have in mind.
MoeBlee


0




Reply

jazzmobe (307)

5/11/2009 6:23:44 PM


On May 9, 9:03=A0am, LudovicoVan <ju...@diegidio.name> wrote:
> Not that part, it's the conclusion that is "problematic". We have a
> construction of the list (our 'f'), then we define the construction
> of
> an item ('d') so that it differs in each place for each item in the
> list given by 'f'. If we now run this game of 'f' against 'd', there
> is no winner, yet Cantor's conclusion is that the antidiagonal wins.
Your comments reflect a fundamental misunderstanding of basic logic,
specifically the principle of universal generalization. There is no
game of f against d. That is a figment borne of your misunderstanding
the basic logic.
MoeBlee


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Reply

jazzmobe (307)

5/11/2009 6:32:44 PM


On May 9, 9:11=A0am, WM <mueck...@rz.fhaugsburg.de> wrote:
> every set of natural numbers has even or odd
> cardinality.
That's incorrect. Only finite sets have even or odd cardinality.
MoeBlee


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Reply

jazzmobe (307)

5/11/2009 6:34:25 PM


On May 9, 9:29=A0am, WM <mueck...@rz.fhaugsburg.de> wrote:
> On 8 Mai, 20:48, Jan <j...@daimi.au.dk> wrote:
>
> > Can you supply a natural number n, that the program won't print?
>
> No. Can you understand that at *any* step there are infinitely many
> numbers remaining that have not been printed and will never be
> printed, because they belong to the set that is remaining after any
> step?
Of course, it is understood that any step there are infinitely many
natural numbers not announced. But for any natural number, there is a
step that announces that natural number.
> This is a contradiction contradicting the axiom of infinity.
There is no contradiction in the above.
MoeBlee


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Reply

jazzmobe (307)

5/11/2009 6:37:01 PM


LudovicoVan wrote:
> On 11 May, 14:54, Dave Seaman <dsea...@no.such.host> wrote:
>
> > We can show that uncountable sets exist without using the diagonal
> > argument.
>
> And could you tell which argument we would use instead?
Well, "uncountable" of a set means that there is no 11 mapping from
it into the natural numbers N. And there is no 11 mapping from the
power set of the natural numbers P(N) into N. Therefore P(N) is an
uncountable set. No "diagonals" involved.
Brian Chandler


0




Reply

imaginatorium (41)

5/11/2009 6:40:31 PM


On May 10, 4:38=A0am, "R. Srinivasan" <sradh...@in.ibm.com> wrote:
> Here is a simple argument for the inconsistency of ZFC. If Cantor's
> diagonal argument is correct, there are uncountably many real numbers
> in each interval of the form [1/2n, 1/2n], where n (=3D1,2,3....) is a
> positive integer. The intersection of all such intervals (over all the
> positive integers n) however, contains the single real number 0. Our
> intuition tells us that the intersection should also contain
> uncountably many reals,
No, my intuition definitely does NOT tell me any such thing.
Moreover, a conflict with your intuition is not in itself a proof of
inconsistency.
> i.e., the intersection should be an interval.
> Each (or every) one of the nested intervals has endpoints fixed away
> from zero, so how come their intersection manages to eliminate "all"
> points other than zero, which does not seem possible on any reasonable
> conception of "each" (or "every" or "all")?
It is QUITE reasonable. Just apply the DEFINITION of 'intersection'.
> In the logic NAFL
Oh yes, I remember that. That's the one you kept promising  for about
at least two years now  to make a thread in which you would explain
what it IS.
MoeBlee


0




Reply

jazzmobe (307)

5/11/2009 6:45:31 PM


On May 10, 8:40=A0am, WM <mueck...@rz.fhaugsburg.de> wrote:
> The first proof that ZFC is inconsistent, to my knowledge, was the
> proof that there are only countably many definitions.
That's not a proof of the inconsistency of anything. If you claim
otherwise, please show a derivation of a formula P and of ~P in set
theory.
> A number that rules out any chance of access cannot belong to that
> uncountable set of numbers that is "proven" to exist by the diagonal
> proof (if the diagonal numbers permit access).
Again another of your arguments, NOT in ZFC, but rather in your own
personal language  "chance of access", "permit access".
> There is no chance to define one of those undefinable numbers. So they
> may exist where ever they do and who ever wants to may believe in
> them. But they are not those numbers which appear as a diagonal of a
> list, because those can be defined, hence belong to a countable set.
> Therefore this proof is a contradiction that cannot be avoided by the
> hidden set of undefinable numbers.
Please, since you claim ZFC inconsistent, show that there exists a
finite sequence of formulas in the language of ZFC, such that each
entry is either an axiom of ZFC or derivable from previous entries by
first order logic with identity and such that the last entry is "P &
~P" for some formula P in the language of set theory.
Otherwise, you're filling your hours here on earth as a hopeless,
chronic crank.
MoeBlee


0




Reply

jazzmobe (307)

5/11/2009 6:51:55 PM


On May 10, 9:43=A0am, LudovicoVan <ju...@diegidio.name> wrote:
> On 10 May, 10:00, Keith Ramsay <kram...@aol.com> wrote:
>
> > On May 8, 3:32=A0pm, MoeBlee <jazzm...@hotmail.com> wrote:
> >  On May 8, 12:32=A0pm, LudovicoVan <ju...@diegidio.name> wrote:
> > 
> >  > impredicative, so constructively invalid.
>
> > No, this is a confusion on the concept of predicativity. The
> > proof is predicative.
>
> I might very well be missing something [...]
Read the posts by Koskensilta and McCullough in previous threads about
this, as well as Ramsay's new post too.
MoeBlee
, but I read on Wikipedia:
>
> http://en.wikipedia.org/wiki/Impredicativity
>
> =A0 =A0  "In mathematics and logic, impredicativity is the property of
> a selfreferencing definition. More precisely, a definition is said to
> be impredicative if it invokes (mentions or quantifies over) the set
> being defined, or (more commonly) another set which contains the thing
> being defined."
>
> I note here that selfreferential "mention" seems enough, no need for
> quantification over.
>
> =A0 =A0  "Russell's paradox is a famous example of an impredicative
> construction, namely the set of all sets which do not contain
> themselves."
>
> =A0 =A0  "Concerning mathematics, an example of an impredicative
> definition is the smallest number in a set, which is formally defined
> as: y =3D min(X) if and only if for all elements x of X, y is less than
> or equal to x, and y is in X."
>
> =A0 =A0  "The greatest lower bound of a set X, glb(X), generalizes thi=
s
> concept; y =3D glb(X) if and only if for all elements x of X, y is less
> than or equal to x, and any z less than or equal to all elements of X
> is less than or equal to y. But this definition also quantifies over
> the set (potentially infinite, depending on the order in question)
> whose members are the lower bounds of X, one of which being the glb
> itself. Hence predicativism would reject this definition."
>
> I'll take also the chance to note that, while I keep disagree that
> Cantor's arguments (any of them) are "perfectly constructive", as has
> been claimed, my objection to the diagonal argument here does not
> leverage predicativity at all: I have tried to focuses on what to me
> seems an unwarranted conclusion.
>
> LV


0




Reply

jazzmobe (307)

5/11/2009 6:58:54 PM


On May 10, 12:32=A0pm, WM <mueck...@rz.fhaugsburg.de> wrote:
> The contradiction is: In mathematics based on ZFC:
> Every real number can be identified.
> Not every real number can be identified.
"can be identified" is not a predicate you've defined in the language
of ZFC. Try again (for the millionth try)... Again, what we're looking
for is a proof in ZFC of a formula P and ~P.
MoeBlee


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Reply

jazzmobe (307)

5/11/2009 7:05:18 PM


On May 10, 11:29=A0pm, Albrecht <albst...@gmx.de> wrote:
> MoeBlee schrieb:
>
> > On May 8, 4:01=A0am, Albrecht <albst...@gmx.de> wrote:
>
> > > > 10 LET I=3D0
> > > > 20 PRINT I
> > > > 30 LET I =3D I + 1
> > > > 40 GOTO 20
>
> > > > There. I just wrote a program to output _all_ the naturals.
> > > > Now surely you must abandon your claim that I can't, despite
> > > > your initial surety, for clearly I have written the program.
>
> > > This program doesn't output _all_ =A0naturals.
>
> > Then please say what is the least natural number not outputted by the
> > program.
>
> There is no least.
If there is a natural number not outputted then there is a least such
natural number.
> And now?
And now you continue to waste whatever time you have on earth for
mathematics as a hopeless, chronic crank.
> The program doesn't output _all_ =A0the natural numbers. It can't. And
> that's why there is no _all_ =A0of the naturals.
If it doesn't output all natural numbers, then there is a least
natural number it does not output.
MoeBlee


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Reply

jazzmobe (307)

5/11/2009 7:08:35 PM


On May 11, 7:27=A0am, Han de Bruijn <Han.deBru...@DTO.TUDelft.NL> wrote:
> Dave Seaman wrote:
> > On Fri, 8 May 2009 12:36:46 0700 (PDT), LudovicoVan wrote:
>
> >>On 8 May, 00:03, Marshall <marshall.spi...@gmail.com> wrote:
>
> >>>You can't do the same for the reals. That's because the reals
> >>>aren't countable and the naturals are.
>
> >>That ultimately depends on your stance on the diagonal argument.
>
> > We can show that uncountable sets exist without using the diagonal
> > argument. =A0Aleph_1 is an uncountable set, and its existence has nothi=
ng
> > to do with the diagonal argument.
>
> Who is "we" ? What is "can" ? What is "show" ? What is "exist" ? What is
> "is" ? For a "thing" in your mind (not mine) like Aleph_1 ?
>
> Couldn't resist ..
making an ass out of yourself.
MoeBlee


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Reply

jazzmobe (307)

5/11/2009 7:11:17 PM


On 11 Mai, 18:22, William Hughes <wpihug...@hotmail.com> wrote:
> > > Nope. =A0You combine lines one by one.
>
> > No, I do them all by induction. None is left out.
>
> Using induction does not change the fact that you
> combine them one by one.
But it makes sure that I do not forget any single line. So, if there
is no line left out, do I use all lines then?
Please confirm, if you can: If every element of a set is included in a
procedure, then all elements of the set are included in that
procedure.
If that is too abstract for you to follow, then consider this example:
If every element of a set is painted red, then all elements of the set
are painted red. Notice, I do not want to say anything about the
colour of the set.
>
>
>
> > > As there is
> > > no last line you do not finish, so you cannot
> > > combine all lines.
>
> > So there are not all lines
>
> Nope, by assuption there are all lines
>
> > (possible to combine)?
>
> It is not possible to combine all lines.
There are all lines but there are not all lines possible to combine.
Perhaps then your assumption is wrong?
>
> > How then can you combine all numbers?
>
> You don't start with nothing and combine
> them one by one. =A0
In fact, I do just that. If you disagree, then you should state which
line is left out.
> You start with all of
> them already in the list. =A0For this you
> need an assumption.
As I said, I do not need any assumption. If you disagree, then please
state which one is missing in my induction or confirm that you are
not doing mathematics but mathemaology.
>
> Recall the assumption
>
> if
> =A0 =A0 all natural numbers and all lines exist
>
> So all lines exist and therefore the list exists.
And therefore I can combine all lines, can't I?
> It is easy to show "all lines"
> contain "all natural numbers".
> No need to combine anything.
Neither an obstacle to do so.
Regards, WM


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Reply

mueckenh (275)

5/11/2009 7:16:25 PM


On 11 Mai, 18:33, William Hughes <wpihug...@hotmail.com> wrote:
> > Remark: You agreeed that every line can be reached by induction.
>
> No, I agreed that any line can be reached by induction.
Let is make clear: You agreed that there is no line that cannot be
reached by induction.
Easier: There is no line that cannot be painted red.
Easiest: Every line can be painted red.
And that's what I do.
> You can reach any line. =A0However, reaching a line does
> not mean finishing unless the line is the last line.
That is unmathematical nonsense. Either you can state what line is not
painted red or you agree that every line can be painted red.
(Or you agree that the talk about every line is nonsense from the
scratch.)
> There is no last line so you can't finish.
Oh yes, I can: Time consumption by geometrical sequence.: 1/2,
1/4, ...
Regards, WM


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Reply

mueckenh (275)

5/11/2009 7:22:36 PM


On 11 Mai, 19:52, Dave Seaman <dsea...@no.such.host> wrote:
> To summarize: =A0the basic axiom needed is the power set axiom. =A0A few =
other
> axioms play a role, but AC is not needed and there is no diagonal argumen=
t
> involved.
Do you mesan that power set axiom that guarantees the existence of
*all* elements of the power set? Or do you mean a power set axiom as
it is required to produce a countable power set of omega (countable
from the "outside" of course)?
Regards, WM


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Reply

mueckenh (275)

5/11/2009 7:34:51 PM


On May 11, 3:22 pm, WM <mueck...@rz.fhaugsburg.de> wrote:
> On 11 Mai, 18:33, William Hughes <wpihug...@hotmail.com> wrote:
>
WM is trying to prove
ii: if
all natural numbers and all lines exist
then
there exists a line that contains all natural numbers
> > > Remark: You agreeed that every line can be reached by induction.
>
> > No, I agreed that any line can be reached by induction.
>
> Let is make clear: You agreed that there is no line that cannot be
> reached by induction.
Yes
> Easier: There is no line that cannot be painted red.
Yes
> Easiest: Every line can be painted red.
No you want things to be clear. Say : *Each* line can be painted red.
You play word games with the word "every"
So we have "You can reach any line"
You can reach any line. However, reaching a line does
not produce one line that contains all lines
unless less the line you reach is the last line.
It does not matter how many lines you reach. Unless you
reach the last line you do not produce one
line that contains all lines.
So if you reach line one in 1/2 second, line
2 in 1/4 second and so on, after one second
you will have reached all lines, but since
no line was the last line, you have not produced
one line that contains all lines.
 William Hughes


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Reply

wpihughes (390)

5/11/2009 7:55:00 PM


On May 11, 3:16 pm, WM <mueck...@rz.fhaugsburg.de> wrote:
> There are all lines but there are not all lines possible to combine.
> Perhaps then your assumption is wrong?
It is not my assuption, it is your assumption.
You are trying to prove
ii: if
all natural numbers and all lines exist
then
there exists a line that contains all natural numbers
 William Hughes


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Reply

wpihughes (390)

5/11/2009 8:03:25 PM


On Mon, 11 May 2009 11:40:31 0700 (PDT), Brian Chandler wrote:
> LudovicoVan wrote:
>> On 11 May, 14:54, Dave Seaman <dsea...@no.such.host> wrote:
>>
>> > We can show that uncountable sets exist without using the diagonal
>> > argument.
>>
>> And could you tell which argument we would use instead?
> Well, "uncountable" of a set means that there is no 11 mapping from
> it into the natural numbers N. And there is no 11 mapping from the
> power set of the natural numbers P(N) into N. Therefore P(N) is an
> uncountable set. No "diagonals" involved.
The standard proof of Cantor's theorem (X < P(X) for all X) is a
diagonal proof. Given a mapping f: X > P(X), the set
d = { x in X : x not in f(x) } is the diagonal.
The existence of the uncountable set aleph_1, on the other hand, can be
proved without a diagonal argument.

Dave Seaman
Third Circuit ignores precedent in Mumia AbuJamal ruling.
<http://www.indybay.org/newsitems/2008/03/29/18489281.php>


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Reply

dseaman (1174)

5/11/2009 8:20:11 PM


Dave Seaman <dseaman@no.such.host> writes:
> The standard proof of Cantor's theorem (X < P(X) for all X) is a
> diagonal proof. Given a mapping f: X > P(X), the set
> d = { x in X : x not in f(x) } is the diagonal.
Yes, but only in an abstract sense.
Many of the cranks here object to the diagonal argument that N < R
because it seems to involve the diagonal of an infinitely large square
or because they think it involves changing digits one after another in
an infinite sequence. Neither of these (mistaken) intuitions is
evident in the above argument.

Jesse F. Hughes
"Hey look, Captain, next time someone wants to tie us up, let's put up
a fight." Adventures by Morse


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Reply

jesse18 (2492)

5/11/2009 8:55:47 PM


On Mon, 11 May 2009 16:55:47 0400, Jesse F. Hughes wrote:
> Dave Seaman <dseaman@no.such.host> writes:
>> The standard proof of Cantor's theorem (X < P(X) for all X) is a
>> diagonal proof. Given a mapping f: X > P(X), the set
>> d = { x in X : x not in f(x) } is the diagonal.
> Yes, but only in an abstract sense.
> Many of the cranks here object to the diagonal argument that N < R
> because it seems to involve the diagonal of an infinitely large square
> or because they think it involves changing digits one after another in
> an infinite sequence. Neither of these (mistaken) intuitions is
> evident in the above argument.
I don't see why not. The set X is allowed to be infinite, and the case
X = N specifically shows the existence of an uncountable set P(N). The
diagonal set d can be thought of as being derived from a characteristic
function f_d : X > {0,1}, whose values can be thought of as being
toggled one by one. That's not a particularly useful way to think of it,
of course.

Dave Seaman
Third Circuit ignores precedent in Mumia AbuJamal ruling.
<http://www.indybay.org/newsitems/2008/03/29/18489281.php>


0




Reply

dseaman (1174)

5/11/2009 9:54:23 PM


On Mon, 11 May 2009 12:34:51 0700 (PDT), WM wrote:
> On 11 Mai, 19:52, Dave Seaman <dsea...@no.such.host> wrote:
>> To summarize: ?the basic axiom needed is the power set axiom. ?A few other
>> axioms play a role, but AC is not needed and there is no diagonal argument
>> involved.
> Do you mesan that power set axiom that guarantees the existence of
> *all* elements of the power set? Or do you mean a power set axiom as
> it is required to produce a countable power set of omega (countable
> from the "outside" of course)?
> Regards, WM
I'm not sure I even know what it means to guarantee the existence of
*all* elements of the power set. There is only one power set axiom (in
ZFC). What it guarantees is simply that if X is a set, then there is a
set Y such that for every S, if S is a subset of X, then S is an element
of Y. The power set of omega is uncountable when viewed from the
"inside".

Dave Seaman
Third Circuit ignores precedent in Mumia AbuJamal ruling.
<http://www.indybay.org/newsitems/2008/03/29/18489281.php>


0




Reply

dseaman (1174)

5/11/2009 10:07:51 PM


On May 6, 8:04=A0pm, LudovicoVan <ju...@diegidio.name> wrote:
> It is (AFAIK) easy enough to come up with a production of _all_ the
> infinite binary strings.
A production is NOT a LIST.
It is in fact NOT possible, not as far as you OR ANYbody knows, to
come up
with a LIST of ALL the denumerablywide bitstrings.
> The problem remains how to make the list
> bulletproof to diagonalisation.
That is not a problem.
Every square list has a diagonal; there is simply nothing you can do
about that.
Every bitstring has a complement  there is nothing you can do about
that either.
If the square list exists, its diagonal MUST exist. If a bitstring
exists, its
complement MUST ALSO exist. Therefore, the antidiagonal MUST exist.
You DO NOT GET to even TALK about "games" or "bulletproof".


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Reply

greeneg9613 (188)

5/11/2009 10:31:55 PM


On May 11, 10:59=A0am, MoeBlee <jazzm...@hotmail.com> wrote:
> On May 8, 8:52=A0pm, lwal...@lausd.net wrote:
> > then they likely
> > believe that no proof is possible. And if this is really
> > the case, then this would actually _prove_ the point that
> > I was making to _Mariano_  that the more time passes,
> > the less likely a proof of ~Con(ZFC) will be found, and
> > the fewer mathematicians will even search for a proof!
> Except that continued research in ZFC itself exposes ZFC to scrutiny
> for contradiction. Every theorem proved allows that someone may prove
> the negation of that theorem.
So far, this discussion/subthread began with Knox, who first
mentioned that finding a proof that ZFC is inconsistent was
unlikely at best. I agreed with her on that point and tried
to quantify how unlikely it was by comparing it to the
likelihood that conjectures such as Riemann or Goldbach
would someday be proved. Mariano questioned my claim about
the relative likelihood of the three proofs (and eventually
stated the neutrality of his opinion regarding the relative
likelihood of the proofs). I tried to explain why I thought
that Riemann and Goldbach were more likely to be proved than
~Con(ZFC), which led to MoeBlee's line of questioning. In
between all of this, Feldmann mentioned how "implausible" it
would be for ZFC or PA to be proved inconsistent.
So what's the point of all this? What I was trying to do was
characterize the standard theorists' opinion of how likely a
proof of ~Con(ZFC) would be. Perhaps, instead of trying to
compare ~Con(ZFC) to Riemann or Goldbach (which led to dead
end questions from Mariano and MoeBlee), let me give a quote
from MH Knowles, another socalled "crank" who once tried to
prove ZFC inconsistent:
"ALL mathematicians believe that it is _theoretically_ possible
that set theory [e.g., ZFC] is inconsistent, but NO
mathematicians believe it is _actually_ possible that set
theory [ZFC] is inconsistent."
(emphasis the author's)
And so MoeBlee's post gives evidence for the first half of this
quote  as he writes, every proof in ZFC increases the
_theoretical_ possibility that this theory will someday be
proved inconsistent.
Meanwhile, Knox's post and eventually Feldmann's as well give
evidence for the second half of this quote. Knox writes how the
universe V=3DL is sufficient to convince her that ZFC will never
_actually_ be proved inconsistent, and Feldmann adds how
"implausible" the possibility of an inconsistency proof
_actually_ is.
And so that one quote from MH Knowles succinctly summarizes the
standard set theorist's position on the consistency of ZFC. I
repeat the quote for emphasis:
"ALL mathematicians believe that it is _theoretically_ possible
that set theory [e.g., ZFC] is inconsistent, but NO
mathematicians believe it is _actually_ possible that set
theory [ZFC] is inconsistent."
Well, except for one mathematician, that is. Ed Nelson does
believe that it's actually possible that PA, and therefore ZFC,
is inconsistent.
A natural question to wonder is, if Nelson can really prove
that PA and ZFC are inconsistent, then why did it take over a
century from Zermelo to the inconsistency proof? And the reason
I gave earlier in this thread is that the proof might require
the operation of tetration or superexponentiation. Relatively
few mathematicians have ever heard of this operation, and so
they wouldn't have found a proof that requires it. Just as
Fermat had never heard of the necessary elliptic curves to
prove FLT, so most mathematicians have never heard of tetration
required to prove ~Con(PA).
Will Nelson be successful? Who knows? But if Nelson fails, I
suspect, like most set theorists and the mathematicians from
the Knowles quote, that ZFC will never be proved inconsistent.


0




Reply

lwalke3 (94)

5/12/2009 12:30:53 AM


On 11 May, 17:33, William Hughes <wpihug...@hotmail.com> wrote:
> No, I agreed that any line can be reached by induction.
> You can reach any line. =A0However, reaching a line does
> not mean finishing unless the line is the last line.
> There is no last line so you can't finish.
On the same line of reasoning, you cannot finish the antidiagonal
either.
LV


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Reply

julio (505)

5/12/2009 3:28:48 AM


On 11 May, 18:52, Dave Seaman <dsea...@no.such.host> wrote:
> On Mon, 11 May 2009 10:18:45 0700 (PDT), LudovicoVan wrote:
> > On 11 May, 14:54, Dave Seaman <dsea...@no.such.host> wrote:
> >> We can show that uncountable sets exist without using the diagonal
> >> argument.
> > And could you tell which argument we would use instead?
>
> This is from a post by Matthew P. Wiener in 1998:
>
> On the one hand, P(NxN) existing is straightforward. =A0Therefore, by
> comprehension, so is the P(NxN) subset of those relations on N which
> are wellordered. =A0Call it WO.
>
> The von Neumann ordinals can be definedin one fell swoopas
> wellordered (by membership) transitive sets. =A0(z is transitive means
> x e y e z =3D> x e z.)
>
> The elementary theory of wellordered sets is easy to develop without AC.
> In particular, one can prove there exists a unique von Neumann ordinal
> of the same order type for any given wellordered set. =A0(For starts,
> this is via the Mostowski collapse: given w wellordered, let mc(w)=3D
> {mc(x):x < w}. =A0By definition by transfinite recursion, this is
> welldefined, and rather easy to prove is transitive.)
>
> At this point, apply the axiom of replacement to WO, replacing each
> wellordered relation on N with its von Neumann order type ordinal.
>
> The resulting set is aleph_1.
>
> For some reason, there is a folklore belief that choice is needed here.
> I've even seen this claim show up in books on set theory!
>
> (end quote from Matthew P. Wiener)
>
> To summarize: =A0the basic axiom needed is the power set axiom. =A0A few =
other
> axioms play a role, but AC is not needed and there is no diagonal argumen=
t
> involved.
Thanks for answering. I just think a proof that leverages the power
set axiom is irrelevant to a discussion on the diagonal argument (just
as Levy's proof is irrelevant).
LV


0




Reply

julio (505)

5/12/2009 3:29:09 AM


On 11 May, 19:23, MoeBlee <jazzm...@hotmail.com> wrote:
> > It's you who are pretending my responses do not exist.
>
> No, I've answered you point by point.
You better take a course in basic reasoning before digging any further
into the technicalities.
LV


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Reply

julio (505)

5/12/2009 3:30:31 AM


On 11 May, 19:32, MoeBlee <jazzm...@hotmail.com> wrote:
> On May 9, 9:03=A0am, LudovicoVan <ju...@diegidio.name> wrote:
>
> > Not that part, it's the conclusion that is "problematic". We have a
> > construction of the list (our 'f'), then we define the construction
> > of
> > an item ('d') so that it differs in each place for each item in the
> > list given by 'f'. If we now run this game of 'f' against 'd', there
> > is no winner, yet Cantor's conclusion is that the antidiagonal wins.
>
> Your comments reflect a fundamental misunderstanding of basic logic
> specifically the principle of universal generalization. There is no
> game of f against d. That is a figment borne of your misunderstanding
> the basic logic.
You have no idea what basic logic is.
LV


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Reply

julio (505)

5/12/2009 3:31:37 AM


On 11 May, 23:31, George Greene <gree...@email.unc.edu> wrote:
> On May 6, 8:04=A0pm, LudovicoVan <ju...@diegidio.name> wrote:
>
> > It is (AFAIK) easy enough to come up with a production of _all_ the
> > infinite binary strings.
>
> A production is NOT a LIST.
> It is in fact NOT possible, not as far as you OR ANYbody knows, to
> come up
> with a LIST of ALL the denumerablywide bitstrings.
What you are saying is false: there is not standard math only.
> > The problem remains how to make the list
> > bulletproof to diagonalisation.
>
> That is not a problem.
> Every square list has a diagonal; there is simply nothing you can do
> about that.
That list can be all but square: you keep speaking nonsense, and
that's the only thing I can do nothing about.
LV


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Reply

julio (505)

5/12/2009 3:38:49 AM


MoeBlee schrieb:
> On May 9, 9:11=A0am, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> > every set of natural numbers has even or odd
> > cardinality.
>
> That's incorrect. Only finite sets have even or odd cardinality.
>
> MoeBlee
There are not much more, are there?
Albrecht


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Reply

albstorz (110)

5/12/2009 5:47:52 AM


On 11 Mai, 21:55, William Hughes <wpihug...@hotmail.com> wrote:
> On May 11, 3:22 pm, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> > On 11 Mai, 18:33, William Hughes <wpihug...@hotmail.com> wrote:
>
> WM is trying to prove
>
> ii: if
> all natural numbers and all lines exist
> then
> there exists a line that contains all natural numbers
and:
there cannot exist a line that contains all natural numbers
But I am not trying to prove that, because the proof is complete.
I am trying to help you to understand that proof.
>
> > > > Remark: You agreed that every line can be reached by induction.
>
> > > No, I agreed that any line can be reached by induction.
>
> > Let is make clear: You agreed that there is no line that cannot be
> > reached by induction.
>
> Yes
>
> > Easier: There is no line that cannot be painted red.
>
> Yes
>
> > Easiest: Every line can be painted red.
>
> No you want things to be clear. Say : *Each* line can be painted red.
> You play word games with the word "every"
>
> So we have "You can reach any line"
>
> You can reach any line. However, reaching a line does
> not produce one line that contains all lines
_You_ are playing word games with the word "every".
If there is no line that cannot be painted, then every line can be
painted.
If every line can be painted, then all lines can be painted.
"No" is the complement of "all" in all cases in which "all" exists as
a completed entity.
> unless less the line you reach is the last line.
> It does not matter how many lines you reach. Unless you
> reach the last line you do not produce one
> line that contains all lines.
It does not matter what I reach or not reach.
If no line remains unpainted, then all lines are painted.
And you agreed that there is no line that cannot be painted red.
>
> So if you reach line one in 1/2 second, line
> 2 in 1/4 second and so on, after one second
> you will have reached all lines, but since
> no line was the last line, you have not produced
> one line that contains all lines.
First let us stay with painting red. After you will have understood
this, we can switch to the related case where painting red is replaced
by "putting all previous stuff into one line".
Please confirm:
1) If there is no line that is not red, then all lines are red.
2) If there is no line that cannot become red, then all lines can
become red.
3) If there is no line that cannot be painted red, then all lines can
be painted red.
Unless you can confirm these basics, further discussion is idle.
Regards, WM


0




Reply

mueckenh (275)

5/12/2009 7:47:47 AM


On 12 Mai, 00:07, Dave Seaman <dsea...@no.such.host> wrote:
> On Mon, 11 May 2009 12:34:51 0700 (PDT), WM wrote:
> > On 11 Mai, 19:52, Dave Seaman <dsea...@no.such.host> wrote:
> >> To summarize: ?the basic axiom needed is the power set axiom. ?A few other
> >> axioms play a role, but AC is not needed and there is no diagonal argument
> >> involved.
> > Do you mesan that power set axiom that guarantees the existence of
> > *all* elements of the power set? Or do you mean a power set axiom as
> > it is required to produce a countable power set of omega (countable
> > from the "outside" of course)?
> > Regards, WM
>
> I'm not sure I even know what it means to guarantee the existence of
> *all* elements of the power set.
It means that also such elements exist, that cannot be defined as
individuals (in a given fixed language) because the number of
definitions is countable.
> There is only one power set axiom (in
> ZFC). What it guarantees is simply that if X is a set, then there is a
> set Y such that for every S, if S is a subset of X, then S is an element
> of Y.
The question is, however, must someone be able to name / specify the
subset S? Or is it enough to believe that are all (uncountably many)?
To put it in other words: Your phrase "if S is a subset of X": does it
cover all possible subsets or only those that "can be picked"?
> The power set of omega is uncountable when viewed from the
> "inside".
The power set of the smallest infinite set, omega or N or a set
isomorphic to N, is always uncountable, if it is complete and if
Cantor's or Hessenberg's proofs hold.
Regards, WM


0




Reply

mueckenh (275)

5/12/2009 7:59:07 AM


On 12 Mai, 02:30, lwal...@lausd.net wrote:
> Will Nelson be successful? Who knows? But if Nelson fails, I
> suspect, like most set theorists and the mathematicians from
> the Knowles quote, that ZFC will never be proved inconsistent.
Wishful blindness?
Mathematics deals with mathematical numbers, i.e., numbers that can be
defined as individuals.
ZF proves that there are mathematical numbers that cannot be defined
as individuals, hence are not mathematical numbers.
A perfect contradiction.
Or see the binary tree
http://groups.google.com/group/sci.logic/browse_frm/thread/393003b1fba1194a?hl=de&scoring=d&
and the clueless "counterarguments"
like 0 + 0 + 0 + ... > alep_0
Regards, WM


0




Reply

mueckenh (275)

5/12/2009 8:08:17 AM


On 12 Mai, 05:28, LudovicoVan <ju...@diegidio.name> wrote:
> On 11 May, 17:33, William Hughes <wpihug...@hotmail.com> wrote:
>
> > No, I agreed that any line can be reached by induction.
> > You can reach any line. =A0However, reaching a line does
> > not mean finishing unless the line is the last line.
> > There is no last line so you can't finish.
>
> On the same line of reasoning, you cannot finish the antidiagonal
> either.
Exactly that is the point.
The usual escape of set theorists: We do not finish, but we define it
for all numbers simultaneously.
Obviously nonsense. In order to define something for a number that
number must be known. That means one has to be able to count up to
this number.
But even if all is possibly done simultaneously, once and for all,
then we can utilize that means too.
Regards, WM


0




Reply

mueckenh (275)

5/12/2009 8:13:53 AM


LudovicoVan <julio@diegidio.name> writes:
> On 11 May, 17:33, William Hughes <wpihug...@hotmail.com> wrote:
>
> > No, I agreed that any line can be reached by induction.
> > You can reach any line. �However, reaching a line does
> > not mean finishing unless the line is the last line.
> > There is no last line so you can't finish.
>
> On the same line of reasoning, you cannot finish the antidiagonal
> either.
Why do you think this is a problem?
Why would you place the bar higher for the antidiagonal?
> LV

Alan Smaill


0




Reply

smaill1 (89)

5/12/2009 8:32:39 AM


LudovicoVan <julio@diegidio.name> writes:
> On 10 May, 20:12, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > LudovicoVan <ju...@diegidio.name> writes:
....
> > > You keep misinterpreting on this: as said, I have no particular
> > > objections to defining an "effective computation" for the anti
> > > diagonal. That the sequence so defined is not in the list at any index
> > > is what I am questioning: that *conclusion* requires a "leap to
> > > infinity" that cannot be proven from the algorithmic (effective)
> > > definitions only. IOW, that's a result that is *not* entailed by the
> > > algorithmic (effective) properties of the definitions in question.
> >
> > You keep saying this, and I keep asking you to clarify what *you* mean
> > when you say that something is not "entailed by algorithmic properties",
> > and you simply repeat your assertions. �
>
> I have mentioned a "leap to infinity" that is not entailed by
> effectiveness: I have put this in terms of a two opponents game, or
> turing machines, or the many ways to compactification. As for now, I'm
> afraid I can't be clearer than that.
The mention of ciompactification simply does not help,
as others have pointed out. The question is whether the
antidiagonal is in the given list, not whether it may
be in something else in an enlarged structure.
You can choice to ask a different question, but ignoring
the given question does not make it go away.
As for the two person game, yes there is an asymmetry,
but again it is there in the problem; one thing is given,
another is constructed. The antidiagonal is every bit as
effective as the given list, though. There is no need to
compute infinitely many values, neither of the given list
nor of the antidiagonal. Where exactly is this "leap
to infinity"?
> > You simply ignore my remark that there is a proof of the result
> > you say cannot be proved.
>
> I have commented to Bishop's proof: "my point (my take) is that this
> simply does not solve our issue", for the reasons just mentioned.
>
> > If you refuse to say what counts as
> > proof, then of course you make it very hard for someone to
> > come up with an argument you might accept  that could well
> > be your intention, of course ...
>
> No such intention, but I see no need either to define what counts as a
> proof: I have raised a specific objection.
But not a clear one.
I see you snipped the question of whether it is possible to
show that a double function over natural numbers can be proved
to never output 3. How would you do that, assuming you think it's
possible. I can suppose here that there is a "leap to infinity"
because there are infinitely many inputs to take into
consideration.
> LV

Alan Smaill


0




Reply

smaill1 (89)

5/12/2009 8:44:09 AM


LudovicoVan <julio@diegidio.name> writes:
> On 10 May, 20:12, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > LudovicoVan <ju...@diegidio.name> writes:
....
> > > You keep misinterpreting on this: as said, I have no particular
> > > objections to defining an "effective computation" for the anti
> > > diagonal. That the sequence so defined is not in the list at any index
> > > is what I am questioning: that *conclusion* requires a "leap to
> > > infinity" that cannot be proven from the algorithmic (effective)
> > > definitions only. IOW, that's a result that is *not* entailed by the
> > > algorithmic (effective) properties of the definitions in question.
> >
> > You keep saying this, and I keep asking you to clarify what *you* mean
> > when you say that something is not "entailed by algorithmic properties",
> > and you simply repeat your assertions. �
>
> I have mentioned a "leap to infinity" that is not entailed by
> effectiveness: I have put this in terms of a two opponents game, or
> turing machines, or the many ways to compactification. As for now, I'm
> afraid I can't be clearer than that.
The mention of ciompactification simply does not help,
as others have pointed out. The question is whether the
antidiagonal is in the given list, not whether it may
be in something else in an enlarged structure.
You can choice to ask a different question, but ignoring
the given question does not make it go away.
As for the two person game, yes there is an asymmetry,
but again it is there in the problem; one thing is given,
another is constructed. The antidiagonal is every bit as
effective as the given list, though. There is no need to
compute infinitely many values, neither of the given list
nor of the antidiagonal. Where exactly is this "leap
to infinity"?
> > You simply ignore my remark that there is a proof of the result
> > you say cannot be proved.
>
> I have commented to Bishop's proof: "my point (my take) is that this
> simply does not solve our issue", for the reasons just mentioned.
>
> > If you refuse to say what counts as
> > proof, then of course you make it very hard for someone to
> > come up with an argument you might accept  that could well
> > be your intention, of course ...
>
> No such intention, but I see no need either to define what counts as a
> proof: I have raised a specific objection.
But not a clear one.
I see you snipped the question of whether it is possible to
show that a double function over natural numbers can be proved
to never output 3. How would you do that, assuming you think it's
possible. I can suppose here that there is a "leap to infinity"
because there are infinitely many inputs to take into
consideration.
> LV

Alan Smaill


0




Reply

smaill1 (89)

5/12/2009 8:44:47 AM


On 12 May, 09:32, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> LudovicoVan <ju...@diegidio.name> writes:
> > On 11 May, 17:33, William Hughes <wpihug...@hotmail.com> wrote:
>
> > > No, I agreed that any line can be reached by induction.
> > > You can reach any line. =A0However, reaching a line does
> > > not mean finishing unless the line is the last line.
> > > There is no last line so you can't finish.
>
> > On the same line of reasoning, you cannot finish the antidiagonal
> > either.
>
> Why do you think this is a problem?
>
> Why would you place the bar higher for the antidiagonal?
Why everything always upside down? It's *Cantor*'s argument that
places the bar higher: my whole point has been objecting to *that*
(the leap to infinity; or, the diagonal that is complete while the
list is not; etc. etc).
LV


0




Reply

julio (505)

5/12/2009 9:00:39 AM


On 12 May, 09:44, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> The antidiagonal is every bit as
> effective as the given list, though. =A0There is no need to
> compute infinitely many values, neither of the given list
> nor of the antidiagonal. =A0Where exactly is this "leap
> to infinity"?
If you don't make a "leap to infinity", you simply have no ground to
conclude that the antidiagonal is not in the list. Again:
*** by induction only (the effective definitions)
you simply cannot prove that conclusion *** !!
Unless by resorting to axioms and similar forms of begging the
question, that is.
> I see you snipped the question of whether it is possible to
> show that a double function over natural numbers can be proved
> to never output 3.
Sorry, but... to put it simply: you keep asking the wrong question
there.
LV


0




Reply

julio (505)

5/12/2009 9:02:21 AM


On 12 May, 09:44, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> LudovicoVan <ju...@diegidio.name> writes:
> > On 10 May, 20:12, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > > If you refuse to say what counts as
> > > proof, then of course you make it very hard for someone to
> > > come up with an argument you might accept  that could well
> > > be your intention, of course ...
>
> > No such intention, but I see no need either to define what counts as a
> > proof: I have raised a specific objection.
>
> But not a clear one.
That does not make you question proper.
LV


0




Reply

julio (505)

5/12/2009 9:03:55 AM


LudovicoVan <julio@diegidio.name> writes:
> On 12 May, 09:32, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > LudovicoVan <ju...@diegidio.name> writes:
> > > On 11 May, 17:33, William Hughes <wpihug...@hotmail.com> wrote:
> >
> > > > No, I agreed that any line can be reached by induction.
> > > > You can reach any line. �However, reaching a line does
> > > > not mean finishing unless the line is the last line.
> > > > There is no last line so you can't finish.
> >
> > > On the same line of reasoning, you cannot finish the antidiagonal
> > > either.
> >
> > Why do you think this is a problem?
> >
> > Why would you place the bar higher for the antidiagonal?
>
> Why everything always upside down? It's *Cantor*'s argument that
> places the bar higher: my whole point has been objecting to *that*
> (the leap to infinity; or, the diagonal that is complete while the
> list is not; etc. etc).
But the argument in the case at hand doess *not* place the bar any
higher for the antidiagonal. It's *you* that insist on leaping to
infinity (by introducing talk about compactification, for example).
> LV

Alan Smaill


0




Reply

smaill1 (89)

5/12/2009 9:13:14 AM


On 12 May, 10:13, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> LudovicoVan <ju...@diegidio.name> writes:
> > On 12 May, 09:32, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > > LudovicoVan <ju...@diegidio.name> writes:
> > > > On 11 May, 17:33, William Hughes <wpihug...@hotmail.com> wrote:
>
> > > > > No, I agreed that any line can be reached by induction.
> > > > > You can reach any line. =A0However, reaching a line does
> > > > > not mean finishing unless the line is the last line.
> > > > > There is no last line so you can't finish.
>
> > > > On the same line of reasoning, you cannot finish the antidiagonal
> > > > either.
>
> > > Why do you think this is a problem?
>
> > > Why would you place the bar higher for the antidiagonal?
>
> > Why everything always upside down? It's *Cantor*'s argument that
> > places the bar higher: my whole point has been objecting to *that*
> > (the leap to infinity; or, the diagonal that is complete while the
> > list is not; etc. etc).
>
> But the argument in the case at hand doess *not* place the bar any
> higher for the antidiagonal.
I just do not see how you can deny such a thing, given that:
*** by induction only (the effective definitions)
you simply cannot prove that conclusion. ***
> It's *you* that insist on leaping to
> infinity (by introducing talk about compactification, for example).
What is your objection, if any, to my consideration above? Can you
perhaps show how to go from the inductive (effective) definition to a
conclusion about the whole infinite set?
If you can't, Cantor's argument is invalidated: if you can, then I'll
be able to use the very same tool to build an effective complete list,
and Cantor's argument will simply resolve to the most unhappy possible
choice among others.
William Highes:
"I agreed that any line can be reached by induction.
You can reach any line. However, reaching a line does
not mean finishing unless the line is the last line.
There is no last line so you can't finish."
Neither you can!
LV


0




Reply

julio (505)

5/12/2009 9:59:26 AM


On 12 May, 10:59, LudovicoVan <ju...@diegidio.name> wrote:
> On 12 May, 10:13, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > LudovicoVan <ju...@diegidio.name> writes:
> > > On 12 May, 09:32, Alan Smaill <sma...@SPAMinf.ed.ac.uk> wrote:
> > > > LudovicoVan <ju...@diegidio.name> writes:
> > > > > On 11 May, 17:33, William Hughes <wpihug...@hotmail.com> wrote:
>
> > > > > > No, I agreed that any line can be reached by induction.
> > > > > > You can reach any line. =A0However, reaching a line does
> > > > > > not mean finishing unless the line is the last line.
> > > > > > There is no last line so you can't finish.
>
> > > > > On the same line of reasoning, you cannot finish the antidiagona=
l
> > > > > either.
>
> > > > Why do you think this is a problem?
>
> > > > Why would you place the bar higher for the antidiagonal?
>
> > > Why everything always upside down? It's *Cantor*'s argument that
> > > places the bar higher: my whole point has been objecting to *that*
> > > (the leap to infinity; or, the diagonal that is complete while the
> > > list is not; etc. etc).
>
> > But the argument in the case at hand doess *not* place the bar any
> > higher for the antidiagonal.
>
> I just do not see how you can deny such a thing, given that:
>
> =A0 =A0 *** by induction only (the effective definitions)
> =A0 =A0 =A0 =A0 you simply cannot prove that conclusion. ***
>
> > It's *you* that insist on leaping to
> > infinity (by introducing talk about compactification, for example).
>
> What is your objection, if any, to my consideration above? Can you
> perhaps show how to go from the inductive (effective) definition to a
> conclusion about the whole infinite set?
>
> If you can't, Cantor's argument is invalidated: if you can, then I'll
> be able to use the very same tool to build an effective complete list,
> and Cantor's argument will simply resolve to the most unhappy possible
> choice among others.
>
> William Highes:
Sorry for the misspelling.
LV
> "I agreed that any line can be reached by induction.
> You can reach any line. =A0However, reaching a line does
> not mean finishing unless the line is the last line.
> There is no last line so you can't finish."
>
> Neither you can!


0




Reply

julio (505)

5/12/2009 10:07:50 AM


On May 12, 3:47 am, WM <mueck...@rz.fhaugsburg.de> wrote:
> On 11 Mai, 21:55, William Hughes <wpihug...@hotmail.com> wrote:
>
> > On May 11, 3:22 pm, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> > > On 11 Mai, 18:33, William Hughes <wpihug...@hotmail.com> wrote:
>
WM has agreed that
i: if
all natural numbers and all lines exist
then
there does not exist a line that contains all natural numbers
is true.
> > WM is trying to prove
>
> > ii: if
> > all natural numbers and all lines exist
> > then
> > there exists a line that contains all natural numbers
>
> and:
>
> there cannot exist a line that contains all natural numbers
Nope, you are trying to show "all natural numbers and all lines
exist"
if false by proving ii (since we both agree that i: holds).
 William Hughes


0




Reply

wpihughes (390)

5/12/2009 11:02:42 AM


On May 11, 11:28 pm, LudovicoVan <ju...@diegidio.name> wrote:
> On 11 May, 17:33, William Hughes <wpihug...@hotmail.com> wrote:
>
> > No, I agreed that any line can be reached by induction.
> > You can reach any line. However, reaching a line does
> > not mean finishing unless the line is the last line.
> > There is no last line so you can't finish.
>
> On the same line of reasoning, you cannot finish the antidiagonal
> either.
>
You will never have a time of the last step, but if
you do the usual trick of squeezing an infinite number
of steps into a finite time, you can have a time after all
steps are done. The difference is that you can create an
antidiagonal without doing a last step. You cannot create
a line containing all other lines without doing a last step.
 William Hughes


0




Reply

wpihughes (390)

5/12/2009 11:36:26 AM


On 12 Mai, 13:02, William Hughes <wpihug...@hotmail.com> wrote:
> On May 12, 3:47 am, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> > On 11 Mai, 21:55, William Hughes <wpihug...@hotmail.com> wrote:
>
> > > On May 11, 3:22 pm, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> > > > On 11 Mai, 18:33, William Hughes <wpihug...@hotmail.com> wrote:
>
> WM has agreed that
>
> i: =A0if
> =A0 =A0 =A0 =A0 =A0all natural numbers and all lines exist
> =A0 =A0 then
> =A0 =A0 =A0 =A0 =A0there does not exist a line that contains all natural =
numbers
>
> is true.
>
> > > WM is trying to prove
>
> > > ii: =A0if
> > > =A0 =A0 =A0 =A0 =A0all natural numbers and all lines exist
> > > =A0 =A0 =A0then
> > > =A0 =A0 =A0 =A0 =A0there exists a line that contains all natural numb=
ers
>
> > and:
>
> > there cannot exist a line that contains all natural numbers
>
> Nope, you are trying to show =A0"all natural numbers and all lines
> exist"
> if false by proving ii (since we both agree that i: holds).
The question is not what we both agree but what is fact.
Regards, WM


0




Reply

mueckenh (275)

5/12/2009 11:47:53 AM


On 12 Mai, 13:36, William Hughes <wpihug...@hotmail.com> wrote:
> On May 11, 11:28 pm, LudovicoVan <ju...@diegidio.name> wrote:
>
> > On 11 May, 17:33, William Hughes <wpihug...@hotmail.com> wrote:
>
> > > No, I agreed that any line can be reached by induction.
> > > You can reach any line. =A0However, reaching a line does
> > > not mean finishing unless the line is the last line.
> > > There is no last line so you can't finish.
>
> > On the same line of reasoning, you cannot finish the antidiagonal
> > either.
>
> You will never have a time of the last step, but if
> you do the usual trick of squeezing an infinite number
> of steps into a finite time, you can have a time after all
> steps are done. =A0The difference is that you can create an
> antidiagonal without doing a last step. =A0You cannot create
> a line containing all other lines without doing a last step.
>
Is this because of the same reason that some magicans can transform
gold into ivory but not ivory into gold?
Remark: The antidiagonal contains all other lines, at least one
element of *all* other lines.
Regards, WM


0




Reply

mueckenh (275)

5/12/2009 11:59:39 AM


On Mon, 11 May 2009 20:29:09 0700 (PDT), LudovicoVan wrote:
> On 11 May, 18:52, Dave Seaman <dsea...@no.such.host> wrote:
>> On Mon, 11 May 2009 10:18:45 0700 (PDT), LudovicoVan wrote:
>> > On 11 May, 14:54, Dave Seaman <dsea...@no.such.host> wrote:
>> >> We can show that uncountable sets exist without using the diagonal
>> >> argument.
>> > And could you tell which argument we would use instead?
>>
>> This is from a post by Matthew P. Wiener in 1998:
>>
[ snip quote ]
>>
>> To summarize: ?the basic axiom needed is the power set axiom. ?A few other
>> axioms play a role, but AC is not needed and there is no diagonal argument
>> involved.
> Thanks for answering. I just think a proof that leverages the power
> set axiom is irrelevant to a discussion on the diagonal argument (just
> as Levy's proof is irrelevant).
That is not the current context. You responded to a statement that the
reals are uncountable by claiming "That ultimately depends on your stance
on the diagonal argument." I was pointing out that the existence of
uncountable sets does *not* depend on the diagonal argument. In fact, we
can also show the reals are uncountable without using a diagonal
argument.

Dave Seaman
Third Circuit ignores precedent in Mumia AbuJamal ruling.
<http://www.indybay.org/newsitems/2008/03/29/18489281.php>


0




Reply

dseaman (1174)

5/12/2009 12:25:52 PM


lwalke3@lausd.net writes:
> A natural question to wonder is, if Nelson can really prove that PA
> and ZFC are inconsistent, then why did it take over a century from
> Zermelo to the inconsistency proof? And the reason I gave earlier in
> this thread is that the proof might require the operation of
> tetration or superexponentiation. Relatively few mathematicians have
> ever heard of this operation, and so they wouldn't have found a
> proof that requires it. Just as Fermat had never heard of the
> necessary elliptic curves to prove FLT, so most mathematicians have
> never heard of tetration required to prove ~Con(PA).
A wonderfully baffling piece of baseless speculation! In return I
offer the suggestion that one day Quine's New Foundations will be
proved consistent but omegainconsistent as a result of vigorous
pondering of the mysteries of the epsilon0th iterate of the
Ackermann function.
Earlier you propounded certain profound remarks about extensionality
and the axiom of foundation. Here's some grist for your mill: we're
naturally wont to regard extensionality and foundation as
mathematically inconsequential, purely aesthetic as it were, but, as I
recently learned from a talk by Rathjen, Dana Scott has proved a
rather surprising theorem, that ZF without extensionality is of the
same strength as Zermelo set theory.

Aatu Koskensilta (aatu.koskensilta@uta.fi)
"Wovon mann nicht sprechen kann, dar�ber muss man schweigen"
 Ludwig Wittgenstein, Tractatus LogicoPhilosophicus


0




Reply

aatu.koskensilta1 (232)

5/12/2009 12:37:18 PM


Dave Seaman <dseaman@no.such.host> writes:
> To summarize: the basic axiom needed is the power set axiom. A few
> other axioms play a role, but AC is not needed and there is no
> diagonal argument involved.
I wonder, idly as always, what might lead anyone to conclude choice
was used  this is very elementary stuff, after all. Perhaps they're
thinking, rather, of the proof that omega_1 is regular?

Aatu Koskensilta (aatu.koskensilta@uta.fi)
"Wovon mann nicht sprechen kann, dar�ber muss man schweigen"
 Ludwig Wittgenstein, Tractatus LogicoPhilosophicus


0




Reply

aatu.koskensilta1 (232)

5/12/2009 12:48:27 PM


MoeBlee <jazzmobe@hotmail.com> writes:
> The arguments of Koskensilta and McCullough in the "Yet Another
> Disproof Of Cantor's Theorem" thread (and another thread too) do
> seem sufficient to refute that the diagonal argument is
> impredicative.
Well, we need no substantial argument. Simple inspection shows the
instances of separation in the diagonal argument are predicative.

Aatu Koskensilta (aatu.koskensilta@uta.fi)
"Wovon mann nicht sprechen kann, dar�ber muss man schweigen"
 Ludwig Wittgenstein, Tractatus LogicoPhilosophicus


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aatu.koskensilta1 (232)

5/12/2009 12:53:01 PM


Herbert Newman <nomail@invalid> writes:
> "Wolfgang M�ckenheim is a classic crank. Why do you imagine, as you seem to
> do, that there is any point arguing with him?" (Torkel Franzen)
"It is in the nature of Usenet that there will always be people who
find it in their heart to debate the cranks, the trolls, the
loons. Why do you imagine, as you seem to do, that there is any point
posting endless reminders about the pointlessness of this activity?"
(Aatu Koskensilta)

Aatu Koskensilta (aatu.koskensilta@uta.fi)
"Wovon mann nicht sprechen kann, dar�ber muss man schweigen"
 Ludwig Wittgenstein, Tractatus LogicoPhilosophicus


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aatu.koskensilta1 (232)

5/12/2009 1:07:22 PM


herbzet <herbzet@gmail.com> writes:
> Why won't you just let this spavined donkey die a natural death?
"   experience shows that such appeals to a poster's better nature
are useless." (Torkel Franz�n)

Aatu Koskensilta (aatu.koskensilta@uta.fi)
"Wovon mann nicht sprechen kann, dar�ber muss man schweigen"
 Ludwig Wittgenstein, Tractatus LogicoPhilosophicus


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aatu.koskensilta1 (232)

5/12/2009 1:18:22 PM


On May 12, 1:08=A0am, WM <mueck...@rz.fhaugsburg.de> wrote:
>
> Mathematics deals with mathematical numbers, i.e., numbers that can be
> defined as individuals.
> ZF proves that there are mathematical numbers that cannot be defined
> as individuals, hence are not mathematical numbers.
> A perfect contradiction.
Contradiction fail again AGAIN! He's a triplethreat.
Marshall


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marshall.spight (580)

5/12/2009 1:22:21 PM


MoeBlee <jazzmobe@hotmail.com> writes:
> That seems right to me, as long as the instance of axiom schema of
> separation used to carve out the contradicting function is not
> impredicative (Aatu had a convincing argument about that, but, alas,
> I forgot his conclusion).
Well, my argument was really just a (standard and unoriginal)
observation: inspecting the definition of the subset D in the diagonal
argument
D = {x  x not in f(x)}
we find the defining formula is quantifier free (or contains only
bounded quantifiers when spelled out in the language of set theory)
and hence certainly predicative. Depending on the audience an
elucidation of the notion of predicativity might be in order, but even
then a superficial understanding suffices to render the observation
immediate.

Aatu Koskensilta (aatu.koskensilta@uta.fi)
"Wovon mann nicht sprechen kann, dar�ber muss man schweigen"
 Ludwig Wittgenstein, Tractatus LogicoPhilosophicus


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aatu.koskensilta1 (232)

5/12/2009 1:55:40 PM


On 12 Mai, 14:37, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:
> lwal...@lausd.net writes:
> > A natural question to wonder is, if Nelson can really prove that PA
> > and ZFC are inconsistent, then why did it take over a century from
> > Zermelo to the inconsistency proof? And the reason I gave earlier in
> > this thread is that the proof might require the operation of
> > tetration or superexponentiation. Relatively few mathematicians have
> > ever heard of this operation, and so they wouldn't have found a
> > proof that requires it. Just as Fermat had never heard of the
> > necessary elliptic curves to prove FLT, so most mathematicians have
> > never heard of tetration required to prove ~Con(PA).
>
> A wonderfully baffling piece of baseless speculation! In return I
> offer the suggestion that one day Quine's New Foundations will be
> proved consistent but omegainconsistent as a result of vigorous
> pondering of the mysteries of the epsilon0th iterate of the
> Ackermann function.
Or maybe it will be recognized that omegaconsistence is as reasonable
a notion as consistent inconsistence and completed incompleteness?
>
> Earlier you propounded certain profound remarks about extensionality
> and the axiom of foundation. Here's some grist for your mill: we're
> naturally wont to regard extensionality and foundation as
> mathematically inconsequential, purely aesthetic as it were, but, as I
> recently learned from a talk by Rathjen, Dana Scott has proved a
> rather surprising theorem, that ZF without extensionality is of the
> same strength as Zermelo set theory.
Small wonder. Everything of that stuff is of same strength, because:
"Set theory is wrong" (Ludwig Wittgenstein).
> "Wovon mann nicht sprechen kann, dar=FCber muss man schweigen"
> =A0 Ludwig Wittgenstein, Tractatus LogicoPhilosophicus
Regards, WM


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mueckenh (275)

5/12/2009 1:57:50 PM


On Tue, 12 May 2009 15:48:27 +0300, Aatu Koskensilta wrote:
> Dave Seaman <dseaman@no.such.host> writes:
>> To summarize: the basic axiom needed is the power set axiom. A few
>> other axioms play a role, but AC is not needed and there is no
>> diagonal argument involved.
> I wonder, idly as always, what might lead anyone to conclude choice
> was used  this is very elementary stuff, after all. Perhaps they're
> thinking, rather, of the proof that omega_1 is regular?
I'll confess that the Wiener quote was actually in response to a question
of mine (back in 1998) about whether aleph_1 could be shown to exist in
ZF. I have completely forgotten what I had in mind at the time, but I
vaguely recall that a similar question had come up in other threads
before then.

Dave Seaman
Third Circuit ignores precedent in Mumia AbuJamal ruling.
<http://www.indybay.org/newsitems/2008/03/29/18489281.php>


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dseaman (1174)

5/12/2009 2:23:49 PM


On 12 Mai, 13:36, William Hughes <wpihug...@hotmail.com> wrote:
> On May 11, 11:28 pm, LudovicoVan <ju...@diegidio.name> wrote:
>
> > On 11 May, 17:33, William Hughes <wpihug...@hotmail.com> wrote:
>
> > > No, I agreed that any line can be reached by induction.
> > > You can reach any line. =A0However, reaching a line does
> > > not mean finishing unless the line is the last line.
> > > There is no last line so you can't finish.
>
> > On the same line of reasoning, you cannot finish the antidiagonal
> > either.
>
> You will never have a time of the last step, but if
> you do the usual trick of squeezing an infinite number
> of steps into a finite time, you can have a time after all
> steps are done. =A0The difference is that you can create an
> antidiagonal without doing a last step. =A0You cannot create
> a line containing all other lines without doing a last step.
Therefore we *prove* using logics
(If there are infinitely many, then there are a least two)
and mathematical reasoning
(if not all numbers are in one line, but if all numbers and all lines
are in the list,
then there must be at least two lines, A and B, such that for at least
two numbers, a and b,
(a e A & b !e A) & (a !e B & b e B),
if all numbers are there, then they are in one line.
Please confirm:
1) If there is no line that is not red, then all lines are red.
2) If there is no line that cannot become red, then all lines can
become red.
3) If there is no line that cannot be painted red, then all lines can
be painted red.
Unless you can confirm these basics, further discussion is idle.
Regards, WM


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mueckenh (275)

5/12/2009 2:25:27 PM


lwalke3@lausd.net writes:
> As I've mentioned before, Cantor's original theory, naive set
> theory, was quickly proved inconsistent by Russell.
No it wasn't. If by naive set theory you mean the theory containing
extensionality and unrestricted comprehension it is a pertinent
observation no set theorist ever worked in such a theory. If by naive
set theory you simply refer to the mathematical theory Cantor
established it is a pertinent observation it is entirely untouched by
any paradoxes.
> This is the reason Zermelo came up with ZFC.
A major motivation was his desire to establish the correctness of his
proof of the wellordering theorem.

Aatu Koskensilta (aatu.koskensilta@uta.fi)
"Wovon mann nicht sprechen kann, dar�ber muss man schweigen"
 Ludwig Wittgenstein, Tractatus LogicoPhilosophicus


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aatu.koskensilta1 (232)

5/12/2009 2:26:48 PM


MoeBlee <jazzmobe@hotmail.com> writes:
> Correct conlusion, but INCORRECT reasoning. If PA is inconsistent
> but ZFC proves PA consistent, then all we can infer from those mere
> facts is that ZFC proves false arithmetical statements, not
> necessarily that ZFC is inconsistent.
For statements of the form in question (Pi1) we can conclude that ZFC
is inconsistent from the assumption that it proves a false statement
of that form. Do recall that all false Pi1 statements are refutable
in ZFC, or, equivalently, that all true Sigma1 statements are
provable in ZFC.

Aatu Koskensilta (aatu.koskensilta@uta.fi)
"Wovon mann nicht sprechen kann, dar�ber muss man schweigen"
 Ludwig Wittgenstein, Tractatus LogicoPhilosophicus


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aatu.koskensilta1 (232)

5/12/2009 2:33:15 PM


On Tue, 12 May 2009 00:59:07 0700 (PDT), WM wrote:
> On 12 Mai, 00:07, Dave Seaman <dsea...@no.such.host> wrote:
>> On Mon, 11 May 2009 12:34:51 0700 (PDT), WM wrote:
>> > On 11 Mai, 19:52, Dave Seaman <dsea...@no.such.host> wrote:
>> >> To summarize: ?the basic axiom needed is the power set axiom. ?A few other
>> >> axioms play a role, but AC is not needed and there is no diagonal argument
>> >> involved.
>> > Do you mesan that power set axiom that guarantees the existence of
>> > *all* elements of the power set? Or do you mean a power set axiom as
>> > it is required to produce a countable power set of omega (countable
>> > from the "outside" of course)?
>> > Regards, WM
>>
>> I'm not sure I even know what it means to guarantee the existence of
>> *all* elements of the power set.
> It means that also such elements exist, that cannot be defined as
> individuals (in a given fixed language) because the number of
> definitions is countable.
The axiom says nothing about whether elements can be defined as
individuals.
>> There is only one power set axiom (in
>> ZFC). What it guarantees is simply that if X is a set, then there is a
>> set Y such that for every S, if S is a subset of X, then S is an element
>> of Y.
> The question is, however, must someone be able to name / specify the
> subset S? Or is it enough to believe that are all (uncountably many)?
> To put it in other words: Your phrase "if S is a subset of X": does it
> cover all possible subsets or only those that "can be picked"?
It covers all the subsets that exist, whether they can be "picked" or
not. There is no way to guarantee that the number of subsets of omega is
"really" uncountable, as seen from the outside, because the axiom can
only talk about sets that exist, not sets that don't exist.
>> The power set of omega is uncountable when viewed from the
>> "inside".
> The power set of the smallest infinite set, omega or N or a set
> isomorphic to N, is always uncountable, if it is complete and if
> Cantor's or Hessenberg's proofs hold.
> Regards, WM

Dave Seaman
Third Circuit ignores precedent in Mumia AbuJamal ruling.
<http://www.indybay.org/newsitems/2008/03/29/18489281.php>


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dseaman (1174)

5/12/2009 2:35:35 PM


On 12 Mai, 16:26, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:
> lwal...@lausd.net writes:
> > As I've mentioned before, Cantor's original theory, naive set
> > theory, was quickly proved inconsistent by Russell.
>
> No it wasn't. If by naive set theory you mean the theory containing
> extensionality and unrestricted comprehension
Why should he? He said "Cantor's theory", and that did not extend
itself to any of these nonsense notions. Can't you comprehend that? It
simply contained the inconsisent notion of a set. "Unter einer Menge
verstehen wir jede Zusammenfassung von bestimmten wohlunterschiedenen
Objekten unserer Anschauung oder unseres Denkens zu einem Ganzen." As
we know today this inconsistency had already been well known by
Cantor, but had not been recognized as being desatrous by himself.
Regards, WM


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mueckenh (275)

5/12/2009 2:42:42 PM


Barb Knox <see@sig.below> writes:
> Plus, there are good reasons for believing that ZFC etc. ARE
> internally consistent. For example, there is an intuitively
> notunreasonable model for ZFC etc. (Goedel's "constructible
> universe").
How is the constructible universe more notunreasonable than the
cumulative hierarchy? Does this notunreasonableness extend also to,
say, the hereditarily ordinal definable sets? Do recall the
constructible universe presupposes the (wild and unfathomably
impredicative) totality of all ordinals.

Aatu Koskensilta (aatu.koskensilta@uta.fi)
"Wovon mann nicht sprechen kann, dar�ber muss man schweigen"
 Ludwig Wittgenstein, Tractatus LogicoPhilosophicus


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aatu.koskensilta1 (232)

5/12/2009 2:43:31 PM


WM <mueckenh@rz.fhaugsburg.de> writes:
> Why should he? He said "Cantor's theory", and that did not extend
> itself to any of these nonsense notions.
The notion that Cantor worked in anything resembling the theory
containing extensionality and unrestricted comprehension is indeed
nonsense.

Aatu Koskensilta (aatu.koskensilta@uta.fi)
"Wovon mann nicht sprechen kann, dar�ber muss man schweigen"
 Ludwig Wittgenstein, Tractatus LogicoPhilosophicus


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aatu.koskensilta1 (232)

5/12/2009 2:48:05 PM


On 12 Mai, 16:35, Dave Seaman <dsea...@no.such.host> wrote:
> >> I'm not sure I even know what it means to guarantee the existence of
> >> *all* elements of the power set.
> > It means that also such elements exist, that cannot be defined as
> > individuals (in a given fixed language) because the number of
> > definitions is countable.
>
> The axiom says nothing about whether elements can be defined as
> individuals.
It does not because that is selfevident, or at least, it should be.
>
> >> There is only one power set axiom (in
> >> ZFC). =A0What it guarantees is simply that if X is a set, then there i=
s a
> >> set Y such that for every S, if S is a subset of X, then S is an eleme=
nt
> >> of Y.
> > The question is, however, must someone be able to name / specify the
> > subset S? Or is it enough to believe that are all (uncountably many)?
> > To put it in other words: Your phrase "if S is a subset of X": does it
> > cover all possible subsets or only those that "can be picked"?
>
> It covers all the subsets that exist, whether they can be "picked" or
> not. =A0
How do we prove existence of sets that cannot be picked (=3D defined as
individuals)?
> There is no way to guarantee that the number of subsets of omega is
> "really" uncountable, as seen from the outside, because the axiom can
> only talk about sets that exist, not sets that don't exist.
>
Is it possible to put internal omega in (an external) bijection with
external omega?
Regards, WM


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mueckenh (275)

5/12/2009 2:57:19 PM


On 12 Mai, 16:48, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote:
> WM <mueck...@rz.fhaugsburg.de> writes:
> > Why should he? He said "Cantor's theory", and that did not extend
> > itself to any of these nonsense notions.
>
> The notion that Cantor worked in anything resembling the theory
> containing extensionality and unrestricted comprehension is indeed
> nonsense.
Yes, because Cantor abhorred axioms.
Regards, WM


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mueckenh (275)

5/12/2009 3:00:35 PM


MoeBlee <jazzmobe@hotmail.com> writes:
> What is a nonCantorian set?
A set for which Cantor's theorem fails. Such exotic wonders provably
exist in Quine's New Foundations and related theories.

Aatu Koskensilta (aatu.koskensilta@uta.fi)
"Wovon mann nicht sprechen kann, dar�ber muss man schweigen"
 Ludwig Wittgenstein, Tractatus LogicoPhilosophicus


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aatu.koskensilta1 (232)

5/12/2009 3:05:23 PM


On May 11, 11:28=A0pm, LudovicoVan <ju...@diegidio.name> wrote:
> On 11 May, 17:33, William Hughes <wpihug...@hotmail.com> wrote:
>
> > No, I agreed that any line can be reached by induction.
> > You can reach any line. =A0However, reaching a line does
> > not mean finishing unless the line is the last line.
> > There is no last line so you can't finish.
>
> On the same line of reasoning, you cannot finish the antidiagonal
> either.
On that line of reasoning, you cannot finish THE LIST, since you can't
get to ITS last line, since it doesn't have one.
On that line of reasoning, you can't finish even ONE NUMBER (one
subset,
one bitstring) on the list, because no bitstring has a last bit.
On that line of reasoning, you can't finish the DIAGONAL ( we are NOT
talking about the ANTIdiagonal, but just the DIAGONAL), because IT
doesn't have a last bit.
Whether you can or can't FINISH anything has NOTHING to do with
whether it does or doesn't EXIST!!
The concept you are not understanding here is
THE INFERENCE RULE OF UNIVERSAL GENERALIZATION.
This rule applies to EVERYthing in a domain, REGARDLESS of how
the domain is ordered and regardless of whether it can be "finished"
or
what size it is.


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greeneg9613 (188)

5/12/2009 3:13:57 PM


On May 11, 11:29=A0pm, LudovicoVan <ju...@diegidio.name> wrote:
> Thanks for answering. I just think a proof that leverages the power
> set axiom is irrelevant to a discussion on the diagonal argument
You're just STUPID. THERE IS NO "argument"!!
THERE IS *ONLY* a proof! That is THE ONLY thing happening here!
And you can't SAY "leverage the power set axiom"!! THAT IS NOT
happening!
The only thing the powerset axiom says is that the existing subsets
can be thought of as being all&only the elements of a known
collection! That CAN'T BE "leveraged"!
That's just the way it is!
What makes you think you can DISallow a collection when you already
have ALLOWED everything IN the collection??


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greeneg9613 (188)

