CANTOR'S POWER PROOF! [Logic]

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From: |-|ercules on 17 Jun 2010 07:00

CANTOR'S POWER PROOF!
Superinfinity is based on the circular reasoning
"no box contains the box numbers that don't contain their own box number".

Herc
--
"There are more things in Cantor's paradise, Horatio, than are dreamt of by your computers."
~ Barb Knox

From: Dan Christensen on 17 Jun 2010 10:42

On Jun 17, 7:00 am, "|-|ercules" <radgray...(a)yahoo.com> wrote:
> CANTOR'S POWER PROOF!
> Superinfinity is based on the circular reasoning
> "no box contains the box numbers that don't contain their own box number"..
>

I don't know about "boxes" and "box numbers," but it is relatively
easy to formally prove the power set of any set s is larger than s:

Suppose p is the power set of s. Suppose further that f is a
surjective function mapping s onto p. Select a subset k containing
those and only those elements of s that are not elements of their
images under f. (Do you see a problem with that?) k is an element of
p, so there must be a pre-image of k under f, say k'. Applying the
definition of k, you can then obtain the contradiction k' e k and ~k'
e k.

I have generated a formal proof using my DC Proof software (download
it free at www.dcproof.com):

HTML format (for some reason, it looks best using Internet Explorer):
http://www.dcproof.com/PowerSet.html

PC Proof format: http://www.dcproof.com/PowerSet.proof

Comments are in blue font. The rules of inference should be somewhat
self-explanatory.

Dan

From: Dan Christensen on 17 Jun 2010 16:20

From: George Greene on 17 Jun 2010 17:19

On Jun 17, 7:00 am, "|-|ercules" <radgray...(a)yahoo.com> wrote:
> CANTOR'S POWER PROOF!
> Superinfinity is based on the circular reasoning
> "no box contains the box numbers that don't contain their own box number"..

This IS NOT circular reasoning, DUMBASS!
It is VALID reasoning to note that
no barber shaves all&only those who don't shave themselves,
no teacher teaches all&only those who don't teach themselves,
nobody talks to all&only those who don't talk to themselves,
no set is a member of all&only those sets that are not members of
themselves,
AD NAUSEAM.
ALL of these are valid FOR THE SAME reason, and the reason IS NOT
circular, not any more than ANY proof is circular.

There really is a sense in which EVERY proof is circular, in that,
in proving that something follows from some axioms, you really are
PROVING THAT
by HAVING assumed the axioms, you did already IMplicitly assume the
theorem
you are proving. But the proof shows that EXplicitly, which was NOT
previously
obvious. "That's circular" IS NEVER a valid objection to something
that was actually
proved FROM AXIOMS. In order to raise that objection, you would have
to point out
something OTHER THAN THE AXIOMS that had ALSO been assumed, and show
how that ADDITIONAL assumption was also needed for the proof. Then
you would have
to show how that assumption was too much LIKE the theorem.

But all of this is WAY over YOUR head in ANY case.

From: George Greene on 17 Jun 2010 17:26

On Jun 17, 10:42 am, Dan Christensen <Dan_Christen...(a)sympatico.ca>
wrote:
> On Jun 17, 7:00 am, "|-|ercules" <radgray...(a)yahoo.com> wrote:
>
> > CANTOR'S POWER PROOF!
> > Superinfinity is based on the circular reasoning
> > "no box contains the box numbers that don't contain their own box number".
>
> I don't know about "boxes" and "box numbers," but it is relatively
> easy to formally prove the power set of any set s is larger than s:
>
> Suppose p is the power set of s. Suppose further that f is a
> surjective function mapping s onto p. Select a subset k containing
> those and only those elements of s that are not elements of their
> images under f. (Do you see a problem with that?)

Of course he does. First of all, you said "those and only those", so
this
subset is UNIQUE -- it is THE subset, NOT *a* subset, of those
elements.
Second of all, you are talking about elements of s under THEIR OWN
images,
so he's going to claim THAT'S SELF-referential! It ISN'T, and even if
it WERE
it would NOT be a PROBLEM, BUT THAT'S THE PROBLEM HE CLAIMS TO SEE!

Seriously, you have been around here long enough to know better than
this.
This has been going on for close to a decade now and you HAVE been
here
for most of it!

> k is an element of p, so there must be a pre-image of k under f, say k'. Applying the
> definition of k, you can then obtain the contradiction k' e k and ~k e k.
>
> I have generated a formal proof using my DC Proof software (download
> it free atwww.dcproof.com):

This will be a proof FROM SOME AXIOMS.
Herc WILL NOT UNDERSTAND your axioms, let alone think they mean
something real about sets. He will also not understand the point that
THE AXIOM OF INFINITY IS NOT USED in this proof and that therefore,
infinity simply has nothing to do with this. He will keep trying to
say that
the fact that this proof (allegedly) implies the existence of higher
infinities
proves that there must be something OBVIOUSLY wrong with it.
THAT is what you are dealing with (or not dealing with, since you keep
trying to explain the proof, which is completely off-topic here, since
the burden
of proof is actually ON HERC to identify SOME ERROR IN the proof, and
he
thinks he can refuse to shoulder it because he can cry "self-
reference!" and "superinfinity!"

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