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Is the number of the box that contains all the numbers of the boxes that don't contain their own number in its own box?

Is the number of the box that contains all the numbers of the boxes
that don't contain their own number in its own box?

AHHH Hyperinfinity!!!!

http://en.wikipedia.org/wiki/Cantor%27s_first_uncountability_proof

Herc

0
herc777 (190)
5/21/2005 11:16:17 PM
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HERC777 wrote:
> 
> Is the number of the box that contains all the numbers of the boxes
> that don't contain their own number in its own box?

There is no such box.
0
jim.sprigs (66)
5/21/2005 11:20:51 PM
Jim wrote
>There is no such box.

AHHHH, super duper triple looper hyperinfinty!!!

Herc

0
herc777 (190)
5/21/2005 11:26:46 PM
Q: is the Herc777 with a brain
ever going to post a sensible article
about diagonalization on sci.logic?

A: there is no such Herc.

0
greeneg159 (81)
5/22/2005 1:40:27 AM

> Jim wrote
> >There is no such box.

HERC777 wrote:
> AHHHH, super duper triple looper hyperinfinty!!!

WRONG, idiot.  Just a number BIGGER THAN the
number of boxes/box-numbers.  THIS PROOF ALSO
WORKS FOR TWO NUMBERS (0 and 1) and TWO BOXES.

Suppose you have an 0 on box 0 and a 1 on box 1.
Suppose you also have another 0 and another 1, either
or neither or both of which you could put into either
box.  Is there ANY way to put the 0 and the 1 into the
boxes so that EITHER of the two boxes is
"the box that contains all the numbers of the boxes
that don't contain their own number" ?  NO.
There CANNOT EXIST such a box.  Not among ONE box.
Not among TWO boxes.  Not among 42 boxes (numbered 0-41
with the numbers 0-41 also available for being put in).
Not among 69 boxes (and box-numbers).  NOT FOR ANY
set of box-indexes and boxes.  IF you want a box containing
all and only the numbers of the boxes that don't contain
themselves, then THIS box will have to be an n+1ST box,
with NO number.


"the box co

0
greeneg159 (81)
5/22/2005 1:46:35 AM
george wrote:

> Q: is the Herc777 with a brain
> ever going to post a sensible article
> about diagonalization on sci.logic?
> 
> A: there is no such Herc.
> 

Oww, that Hercz!
0
bogle (300)
5/22/2005 2:00:13 AM
Proof that Gib loves Mickey.

Is the number of the box that contains all the numbers of the boxes
that don't contain their own number in its own box?  [1]

don't know [1] therefore Micky and Gib fell in love QED

You have just taken your 1st steps into.... the secret infinity zone,
level 1.

Herc

0
herc777 (190)
5/22/2005 2:54:55 AM
In sci.logic, Jim Spriggs
<jim.sprigs@ANTISPAMbtinternet.com.invalid>
 wrote
on Sat, 21 May 2005 23:20:51 +0000 (UTC)
<428FC26B.97FD0DBF@ANTISPAMbtinternet.com.invalid>:
> HERC777 wrote:
>> 
>> Is the number of the box that contains all the numbers of the boxes
>> that don't contain their own number in its own box?
>
> There is no such box.

One could say that there are an uncountable number of nameless boxes.

-- 
#191, ewill3@earthlink.net
It's still legal to go .sigless.
0
ewill (4394)
5/22/2005 3:00:04 AM
HERC777 wrote:
> Proof that Gib loves Mickey.
>
> Is the number of the box that contains
> all the numbers of the boxes
> that don't contain their own number in its own box?  [1]
>
> don't know [1]

BULLshit!  WE DO SO TOO know!
We know THERE IS NO such box!

What you meant to say:
"Here is a proof that Gib loves Mickey."

"Is the number of hairs on the head of the present
king of France positive?"
[1] ?
"Don't know [1]..."

And THAT would be bullshit TOO.
THERE IS NO present king of France.
And we KNOW that.  That does NOT mean we
"Don't know [1]"!  It means we know that
[1] is BULLSHIT!


But then, you never were any good at English.

0
greeneg159 (81)
5/22/2005 6:47:41 AM
HERC777 wrote:
> Proof that Gib loves Mickey.
>
> Is the number of the box that contains
> all the numbers of the boxes
> that don't contain their own number in its own box?  [1]
>
> don't know [1]

BULLshit!  WE DO SO TOO know!
We know THERE IS NO such box!

What you meant to say:
"Here is a proof that Gib loves Mickey."

"Is the number of hairs on the head of the present
king of France positive?"
[1] ?
"Don't know [1]..."

And THAT would be bullshit TOO.
THERE IS NO present king of France.
And we KNOW that.  That does NOT mean we
"Don't know [1]"!  It means we know that
[1] is BULLSHIT!


But then, you never were any good at English.

0
greeneg159 (81)
5/22/2005 6:50:06 AM
[1] is BS
therefore there is no box
therefore hyperinfinity exists.

Hows that one George, that OK?

http://en.wikipedia.org/wiki/C=ADantor%27s_first_uncountability=AD_proof
Here it is your holy grail!

Herc

0
herc777 (190)
5/22/2005 7:12:32 AM
HERC777,

I haven't seen yuor questions since the age of 10 but isn't this
Russell's paradox, addressed many years ago? You may look up the
literature. I think they declared all sets of that type invalid.

HERC777 wrote:
> [1] is BS
> therefore there is no box
> therefore hyperinfinity exists.
>
> Hows that one George, that OK?
>
>
http://en.wikipedia.org/wiki/C=ADantor%27s_first_uncountability=AD_proof
> Here it is your holy grail!
>=20
> Herc

0
5/22/2005 7:20:37 AM
I think so (a version of it), its also Cantors Powerset Proof that
higher infinities exist than the 'far right of the number line.'

Unfortunately they renamed Russels Paradox to "Russels Coproof of
Cantors Powerset Thingy."

The entire of mathematics is bent over backwards because they all
ignore the copious errors of assuming a higher infinity is a logical
construct.

Herc

0
herc777 (190)
5/22/2005 7:28:49 AM
In sci.logic, george
<greeneg@cs.unc.edu>
 wrote
on 21 May 2005 23:50:06 -0700
<1116744606.507325.104850@o13g2000cwo.googlegroups.com>:
>
> HERC777 wrote:
>> Proof that Gib loves Mickey.
>>
>> Is the number of the box that contains
>> all the numbers of the boxes
>> that don't contain their own number in its own box?  [1]
>>
>> don't know [1]
>
> BULLshit!  WE DO SO TOO know!
> We know THERE IS NO such box!

It is an unnamed box.  Here's why.

The box is an element of P(S).

The name is an element of S.

The naming is an alleged 1-1 onto mapping.

Since no such mapping can be onto, there are unnamed boxes,
and we can explicitly construct (at least) one of them.

>
> What you meant to say:
> "Here is a proof that Gib loves Mickey."
>
> "Is the number of hairs on the head of the present
> king of France positive?"
> [1] ?
> "Don't know [1]..."
>
> And THAT would be bullshit TOO.
> THERE IS NO present king of France.
> And we KNOW that.  That does NOT mean we
> "Don't know [1]"!  It means we know that
> [1] is BULLSHIT!
>
>
> But then, you never were any good at English.
>


-- 
#191, ewill3@earthlink.net
It's still legal to go .sigless.
0
ewill (4394)
5/22/2005 11:00:04 AM
HERC777 wrote:
> I think so (a version of it), its also Cantors Powerset Proof that
> higher infinities exist than the 'far right of the number line.'
>
> Unfortunately they renamed Russels Paradox to "Russels Coproof of
> Cantors Powerset Thingy."
>
> The entire of mathematics is bent over backwards because they all
> ignore the copious errors of assuming a higher infinity is a logical
> construct.

Entire mathemaitcs? 99.99% of mathematics couldn't care less whether
there is a box that contains the numbers of those boxes that don't
contain their own number, or whatever you say.

Do tell me how probability or real analysis or complex analysis or
geometry or topology or algebra or anything else other than logic and
set theory are effected?

0
5/22/2005 9:34:08 PM
According to Bryant the area of a 1x1 square is 0 if the irrationals
are countable, by some calculus theorem.
Primitive Functions are incomplete by diagonalisation.
Formal mathematics is incomplete by diagonalisation.

Probability theory.  oo people flip coins oo times each.  can you come
up with a new sequence every time?  probabilistically you have ZERO
chance, since for any number of times you flip a coin, oo other people
have copied the same sequence, and this continues as the coin sequence
approaches infinity.  By Cantors theorem, you can line up the people
and invert the flips along the diagonal and you have an original
sequence of coin tosses.

Anything that relies on an uncountable amount of elements, eg. reals,
in a set depends on the number in a box proof.
http://en.wikipedia.org/wiki/C=AD=ADantor%27s_first_uncountability=AD=AD_pr=
oof


Herc

0
herc777 (190)
5/23/2005 1:59:19 AM
HERC777 wrote:
> [1] is BS
> therefore there is no box
> therefore hyperinfinity exists.
>
> Hows that one George, that OK?

NO, dipshit, it's NOT OK.
The ACTUAL proof shows that there IS a box.
There IS a real containing all and only those
digits that don't match the diagonal.
IT'S JUST NOT *ON* the list.
Therefore, the list does NOT have ALL the reals.
Therefore, you can't list the reals (no list
lists them).

0
greeneg159 (81)
5/23/2005 8:14:17 AM
HERC777 wrote:

> The entire of mathematics is bent over backwards
> because they all
> ignore the copious errors of assuming
> a higher infinity is a logical
> construct.

In the first place, we DON'T just ASSUME that,
WE PROVE IT, and in the second, YOU CAN'T FIND
any errors, logical, constructed, or otherwise,
LET ALONE "copious" errors, in that proof!

0
greeneg159 (81)
5/23/2005 8:15:48 AM
George>>>
> In the first place, we DON'T just ASSUME that,
> WE PROVE IT, and in the second, YOU CAN'T FIND
> any errors, logical, constructed, or otherwise,
> LET ALONE "copious" errors, in that proof!

Rubbish!
Cantors POWER SET PROOF (not the diag proof) uses the missing box to
establish hyperinfinity!
You dont PROVE a higher infinity exists AT ALL GEORGE!

That's why dozens of people will ALWAYS be hounding sci.math... where?
how do you derive that?

Cantors PS PROOF
the box that contains the numbers of the boxes that don't contain their
own number,
cannot possibly be numbered..

i.e.
< element {subset} >

there is no element for that subset.

THEREFORE HYPERINFINITY.

no George, thats not proof of hyperinfinity any more than it proves
King Kong, because you CANT MAKE UP BULLSHIT every time you find a
contradiction and call it proof.

This is how morons like you do math

1 the box that contains the numbers of the boxes that don't contain
their own number,
cannot possibly be numbered..
2 AAAHHHH I see it, hyperinfinty

This is how real mathematicians view the proof
1 the box that contains the numbers of the boxes that don't contain
their own number,
cannot possibly be numbered..
2 Big freakin deal, stop making stupid formula.

Herc

0
herc777 (190)
5/23/2005 8:41:01 AM
HERC777 wrote:
> Is the number of the box that contains all the numbers of the boxes
> that don't contain their own number in its own box?
>

There is no box that contains precisely the numbers of the boxes that
don't contain their own number.

> AHHH Hyperinfinity!!!!
>
> http://en.wikipedia.org/wiki/Cantor%27s_first_uncountability_proof
> 
> Herc

0
5/23/2005 8:52:11 AM
HERC777 wrote:
> According to Bryant the area of a 1x1 square is 0 if the irrationals
> are countable, by some calculus theorem.

The area of a 1x1 square is 1. if this gives yet another proof that
irrationals are uncountable - I can live with that.

> Primitive Functions are incomplete by diagonalisation.
> Formal mathematics is incomplete by diagonalisation.

I don't care if  Formal mathematics is complete or incomplete as long
as I can prove that the area of a 1x1 square is 1.

> Probability theory.  oo people flip coins oo times each.

I am not sure what you mean by "oo people". Are you talking about an
infinite sequence of people?

> can you come
> up with a new sequence every time?

If you take two sequences of toss results of independent coin flips,
then with probability 1 there will be some position i where the two
sequences will differ.

Is that what you mean?

> probabilistically you have ZERO
> chance, since for any number of times you flip a coin, oo other
people
> have copied the same sequence, and this continues as the coin
sequence
> approaches infinity.  By Cantors theorem, you can line up the people
> and invert the flips along the diagonal and you have an original
> sequence of coin tosses.

I have no idea what you are talking about. Can you phrase this in the
usual probabilistic terms of limits of sequences of random variables?
Are you trying to show that some measurable set has both measure 1 and
measure 0?

> Anything that relies on an uncountable amount of elements, eg. reals,
> in a set depends on the number in a box proof.
>
http://en.wikipedia.org/wiki/C=AD=ADantor%27s_first_uncountability=AD=AD_pr=
oof

So what? The area of the 1x1 box is still 1. Follows directly from any
reasonable definition of "area".

0
5/23/2005 10:44:00 PM
Yes, because it does not contain only such numbers.

0
Rushtown (5)
5/23/2005 11:41:10 PM
HERC777 wrote:
> Is the number of the box that contains all the numbers of the boxes
> that don't contain their own number in its own box?
>

Did somebody say this already? If it does, it doesn't. If it doesn't,
it does.

Ken

0
kquirici (91)
5/23/2005 11:44:19 PM
same old Russellian lying village barber crap;
are barbers even *alowed* to cut their own hairs?

--ils ducs d'Enron!
http://tarpley.net
http://members.tripod.com/~american_almanac
http://larouchepub.com

0
QncyMI (28)
5/24/2005 12:32:44 AM
Bryant can you tell him your bullshit area theory you keep pestering me
is a consequence of uncountability.

Herc

0
herc777 (190)
5/24/2005 7:09:25 AM
It's a matter of definition.  It doesn't contain "all the
numbers.....", just like there is, but only by definition, no immovable
object in a universe with an irresistable force.  It just a word game
where for there to be a paradox one must misdefine something.

0
Rushtown (5)
5/27/2005 4:26:52 AM
It's a matter of definition.  It doesn't contain "all the
numbers.....", just like there is, but only by definition, no immovable
object in a universe with an irresistable force.  It just a word game
where for there to be a paradox one must misdefine something.

0
Rushtown (5)
5/27/2005 4:26:59 AM
It's a matter of definition.  It doesn't contain "all the
numbers.....", just like there is, but only by definition, no immovable
object in a universe with an irresistable force.  It just a word game
where for there to be a paradox one must misdefine something.

0
Rushtown (5)
5/27/2005 4:27:05 AM
It's a matter of definition.  It doesn't contain "all the
numbers.....", just like there is, but only by definition, no immovable
object in a universe with an irresistable force.  It just a word game
where for there to be a paradox one must misdefine something.

0
Rushtown (5)
5/27/2005 4:27:05 AM
yeah; hte village barber is either lying, or
also the village idiot.

thus:
seven "forces" or bodies cannot be in static euilibrium;
there is no regular heptahedron e.g.

--ils ducs d'Enron!
http://tarpley.ney/bush7.htm
http://members.tripod.com/~american_almanac

0
QncyMI (28)
5/30/2005 9:37:36 PM
actually there exists a set (from the mathematicians point of view)
that is the set of all boxes that don't hold their own number.

e.g.
box1 = {1, 2, 3}
box2 = {3, 4, 5}
box3 = {2, 4, 6}

the boxes that don't contain their own number is just {2, 3}

that set just can't be in a numbered box, and that is the powerset
proof that there is no 1 to 1 mapping from a set to its powerset, hence
higher infinites than a countable infinite set exist.

of course the conclusion of a bigger_set is unsound but is still a
mainstream belief.  everbody in sci.math believes that there are sets
that the members are SO NUMEROUS you can't put the elements in order
one after the other.  this is completely meaningless and just a facet
of miscomprehension of infinite lists.

LIST
Dxxxx..
xExxx..
xxFxx..
xxxGx..
...

all the elements are in order!  BUT WAIT, the element D+1 E+1 F+1
G+1... is not present!  therefore the COMPLETE SET must be
UNABLE_TO_BE_LISTED!  This is purely backwards thinking!  For very
naive interpretations of infinite lists this seems so but that is just
a facet of data structures its not a portal to higher infinities.

Herc

0
herc777 (190)
5/31/2005 12:58:57 AM

The Ghost In The Machine wrote:
> In sci.logic, george
Herc, stupid as usual:
> >> Is the number of the box that contains
> >> all the numbers of the boxes
> >> that don't contain their own number in its own box?  [1]
> >>
> >> don't know [1]

> > BULLshit!  WE DO SO TOO know!
> > We know THERE IS NO such box!

Some Ghost in the machine.

> It is an unnamed box.

No, it isn't.

>  Here's why.
>
> The box is an element of P(S).

No, it isn't. The box is just a box.
You may be able to talk about the set of
numbers in it but that does not make a box  a set.

> The name is an element of S.

No, it isn't.
First-order languages used in first-order logic
generally ARE TOTAL.  If b is a box and n(b)
is the number (or name) of that box then n(b)
is a valid term.  Grammatically. ALL the time.
Being unnamed is, for the box, LINGUISTICALLY
impossible.  This is NOT like division by 0.

> The naming is an alleged 1-1 onto mapping.

That allegation needs to be made more explicit.
It is in fact completely missing from the-question-
as-phrased-by-Herc.  Elsewhere in the thread I have
been trying to explain to him that this assumption
is what creates the contradiction, which in turn
refutes the assumption.

> Since no such mapping can be onto, there are unnamed
> boxes,  and we can explicitly construct (at least)
> one of them.

No, there aren't, and you can't construct ANY boxes.
In this scenario, THE BOXES ARE GIVEN.  They just
are what they are.  THE QUESTION you then ask about
them is "Does there exist one of them containing
all and only the numbers of boxes that don't contain
their own number"?  And the answer is, by basic logic,
NO, THERE DOESN'T.  There may exist more subsets and
elements of the powerset but there do NOT exist more
BOXES.  It is GIVEN that these are ALL the boxes there
are.  It is also given that EVERYone of them has a
number.  Knowing that there is more than 1 restriction
you could relax to avoid the paradox is important.
But it is less important than knowing the RIGHT one.
You don't relax the axioms.


>
> >
> > What you meant to say:
> > "Here is a proof that Gib loves Mickey."
> >
> > "Is the number of hairs on the head of the present
> > king of France positive?"
> > [1] ?
> > "Don't know [1]..."
> >
> > And THAT would be bullshit TOO.
> > THERE IS NO present king of France.
> > And we KNOW that.  That does NOT mean we
> > "Don't know [1]"!  It means we know that
> > [1] is BULLSHIT!
> >
> >
> > But then, you never were any good at English.
> >
>
> 
> -- 
> #191, ewill3@earthlink.net
> It's still legal to go .sigless.

0
greeneg159 (81)
5/31/2005 7:02:43 AM
Reply: