Re: Cantors proof

From: Keckman (keckman_at_welho.com)
Date: 09/28/04 

Date: Tue, 28 Sep 2004 23:03:27 +0300

On Tue, 28 Sep 2004 19:30:13 +0000 (UTC), Dave Seaman
<dseaman@no.such.host> wrote:

> Not from the fifth alone, but from all five axioms together. For any
> reasonable definition of "finite" you can show that each finite subset
> of the naturals has a largest member.

For any reasonable definition of "finite" you should have some reasonable
defination for "infinite".

Which come's first doesn matter. If "finite" have some meaning then
not-"finite" should have some meaning.
And vice versa.

But there is not such a thing as "infinite" except in our mind, which is
true even for the numbers,
but numbers can have somekind of relations to the real world. With them we
count or measure something.
But infinite does not. So finite can never have a reasonable definition.

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