Re: CANTOR's theorem


mueckenh@xxxxxxxxxxxxxxxxx wrote:
> Randy Poe wrote:
>
> >
> > > but of non-existence of a special self-reflexive set
> > > {M, m, f'}.
> >
> > I don't know what that means.
>
> Say, there is no set {M, m, f} where f is a mapping, M is the set of
> all nongenerators under f, and m is a generator which is not a
> generator.

I agree there is no m which is a generator and also
not a generator.

> > Don't you agree that there
> > is no m such that f(m) = M?
>
> That has nothing to do with the cardinality question.

Of course it does. It is the very heart of the proof.

The point of the proof is to establish this fact.

> > Is M a set? Does it exist? Is it a subset of P(N)?
> >
> > Is there m such that f(m) = M?
> >
> > If you say no, then f(m) is not surjective, by definition.
> > End of story. No "buts".
>
> Map N u { } --> P(N), where {} is mapped on M. Disprove it
> without refering to the current proof, if this is possible.

I don't know what you are asking. You want me to do Cantor's
proof without doing Cantor's proof. I have no idea what
your requirements are.

- Randy

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