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Re: Objection to Cantor's Theorem
Posted:
Jul 19, 2012 8:27 AM
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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.
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