Re: CANTOR's theorem


Virgil wrote:

>
> Given any f: P(N) -> P(N), let us define T_f = {n e N: n ~e f({n})}.
> For any such f,this T_f is a perfectly well defined subset of N and a

> member of P(N).
>
> The only thing is that there will never be any n in N such that
> f({n}) = T_f.
>
> This does not prevent f from mapping any non-singleton set to T_f
>
> For example, f({}) = T_f is possible, f({n,n+1}) = T_f is possible
and
> f(N) = t_f is also possible.
>
> The existence of T_f does not restrict the possible mappings from
P(N)
> to itself in any way. It only says there is a set not the image of a
> singleton set.
>
> If we look at P(S) -> P(S) for finite sets S, we will find the same
> thing, that there are always too may members of P(S) all to be images
of
> singleton sets, as for n singleton sets there are 2^n sets of all
sizes
> in any finite P(S), an n < 2^n for all n in N.

It is a matter of taste, it seems. Therefore let us avoid the whole
problem. Map N --> P(N)\T_f. If you can prove that this mapping is not
surjective, we now that N --> P(N) is not surjective either. If you
cannot prove non-surjectivity, then there is no valid proof of
non-surjectivity for N --> P(N), (and we can assume that it is
surjective, because such a great mathematician never fails).

Is this a fair trade?

PS: Best would be, however, you could prove non-surjectivity of N -->
P(N)\N.

Regards, WM

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