Re: (Not quite) Cantor's diagonal proof

From: Mike Oliver (mike_lists_at_verizon.net)
Date: 10/26/04 

Date: Tue, 26 Oct 2004 16:01:32 -0500

George Greene wrote:

> Norman Megill <nm@nospam.see.signature.domain.invalid> wrote in message news:<Fe6dnXGLUO_hc-rcRVn-gg@rcn.net>...
>
>>George Greene wrote:
>>
>> > Robin Chapman wrote
>> >
>> >>Surely a more important result than the uncountability of R
>> >>is that R has cardinality 2^aleph_0.
>> >
>> > That is a *definition*, NOT a "result".
>>
>>2^aleph_0 is not defined as the cardinality of R.
>
>
> Come on. I'm on YOUR side.
> The relevant point here is not so much that 2^aleph_0 "is not
> defined as the cardinality of R" as it is that 2^aleph_0 IS NOT
> the cardinality of R. My point was that whatever it is, it is
> in virtue of DEFINITIONS and not of more complicated considerations.
> Cardinality in the ZFC realm is defined in terms of bijections.
> BY DEFINITION, two sets have the same cardinality if a bijection
> exists between them. You can also by definition get cardinalities
> being < or > on the basis of the existence of injections or surjections.
> ALL OF THESE ARE DEFINITIONS AND NOT theorems. That was the point.

I don't exactly get what you're saying here (just for a change) but
it does appear that you've missed NM's point. It's not a matter
of definition that there's a bijection between R (defined, say,
by equivalence classes of Cauchy sequences) and 2^aleph_0 (defined
as the set of functions from {0,1} into the natural numbers). It's
not a *hard* theorem, but not quite as trivial as one might think,
especially if your goal is to give a complete formal proof.

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