Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.

Topic: Objection to Cantor's Theorem
Replies: 62   Last Post: Jul 30, 2012 9:23 AM

 Messages: [ Previous | Next ]
 Michael Stemper Posts: 671 Registered: 6/26/08
Re: Objection to Cantor's Theorem
Posted: Jul 19, 2012 8:27 AM

In article <ju7ku7\$bde\$1@dont-email.me>, "dilettante" <no@nonono.no> writes:
>"Virgil" <virgil@ligriv.com> wrote in message news:virgil-BBE9E4.14255318072012@bignews.usenetmonster.com...
>> In article <ju6qi8\$1p6\$3@dont-email.me>, mstemper@walkabout.empros.com (Michael Stemper) wrote:
>>> In article <virgil-9A583F.22083017072012@bignews.usenetmonster.com>, Virgil <virgil@ligriv.com> writes:

>>> >> > and given f: A -> P(A),
>>> >>
>>> >> Which spells out as: for any f.
>>> >>

>>> >> > how can B_f := { x in A | ~ x in f(x) } NOT be defined?
>>> >>
>>> >> Not undefined, B is not a set (in that system).

>>> >
>>> >What part of it is undefined when A and f are both defined?

>>> >>
>>> >> > Admittedly, even for small sets A and fairly simple functions
>>> >> > f, determining the corresponding set B_f may be tedious, but
>>> >> > it should always exist.

>>> >>
>>> >> In the finite case, at best you could claim that B is empty.

>>> >
>>> >I am not at all sure that B can be empty.

>>>
>>> I think that depends on your choice of "f". If we define f:A->P(A) to
>>> be f(a) = {a}, then B_f would be empty, wouldn't it?
>>>
>>> (This seems so simple that I'm probably missing your point altogether.)

>>
>> If A = {a} then P(A) = {{}, {a}}
>>
>> Since {} is in P(A) and is not f(a) for any a in A, B_f = {{}}.
>>
>> The point is that there are always MORE members of P(A) than members of
>> A, so A -> P(A) cannot be an onto mapping.

>
>You missed Michael's point. He is not considering just the case where A =
>{a}. He is saying that if A is any set, and f is the function that maps each
>element of A to the singlton set containing itself, then the set B
>constructed in the proof of the theorem is in fact the empty set. One easily
>checks that the empty set is not in the image of f, so the proof works in
>this case as usual.

Thanks for clarifying what I meant.

--
Michael F. Stemper
#include <Standard_Disclaimer>
Life's too important to take seriously.

Date Subject Author
7/17/12 LudovicoVan
7/17/12 LudovicoVan
7/17/12 Virgil
7/17/12 Virgil
7/17/12 LudovicoVan
7/17/12 LudovicoVan
7/18/12 Virgil
7/17/12 Virgil
7/17/12 LudovicoVan
7/18/12 Virgil
7/18/12 Michael Stemper
7/18/12 Virgil
7/18/12 dilettante
7/18/12 Virgil
7/19/12 Michael Stemper
7/18/12 Daryl McCullough
7/18/12 LudovicoVan
7/18/12 Shmuel (Seymour J.) Metz
7/18/12 Marshall
7/18/12 LudovicoVan
7/18/12 Shmuel (Seymour J.) Metz
7/18/12 David C. Ullrich
7/18/12 LudovicoVan
7/18/12 David C. Ullrich
7/18/12 LudovicoVan
7/18/12 David C. Ullrich
7/18/12 LudovicoVan
7/19/12 Virgil
7/19/12 W. Dale Hall
7/19/12 Gus Gassmann
7/19/12 Shmuel (Seymour J.) Metz
7/19/12 David C. Ullrich
7/20/12 LudovicoVan
7/20/12 David C. Ullrich
7/20/12 LudovicoVan
7/21/12 hagman
7/22/12 LudovicoVan
7/27/12 hagman
7/27/12 Virgil
7/27/12 LudovicoVan
7/27/12 Eddie
7/28/12 LudovicoVan
7/28/12 Eddie
7/28/12 Eddie
7/28/12 Virgil
7/28/12 LudovicoVan
7/28/12 Virgil
7/28/12 LudovicoVan
7/28/12 Virgil
7/28/12 David C. Ullrich
7/28/12 LudovicoVan
7/28/12 Virgil
7/29/12 hagman
7/29/12 LudovicoVan
7/29/12 hagman
7/30/12 LudovicoVan
7/22/12 Shmuel (Seymour J.) Metz
7/20/12 Shmuel (Seymour J.) Metz
7/18/12 Virgil
7/18/12 Virgil
7/27/12 Eddie
7/27/12 Eddie
7/27/12 Eddie