Re: CANTOR's theorem



Randy Poe wrote:
> mueckenh@xxxxxxxxxxxxxxxxx wrote:
> > Randy Poe wrote:
> > > mueckenh@xxxxxxxxxxxxxxxxx wrote:
> > > > *** T. Winter wrote:
> > >
> > > > > Yes, so you think one of my statements is wrong. Which of my above
> > > > > statements is wrong?
> > > >
> > > > This one: "To prove that the map is non-surjective, I have only to
> > > > display a set that is not in the image of the map."
> > >
> > > That is correct.
> > > > It is wrong, if it can be proven that such a construct cannot exist.
> > >
> > > What is "such a construct"? Do you mean the set? Do you
> > > think this set doesn't exist?
> >
> > The triple m,M,f does not exist.
>
> The question being examined is whether the map is onto
> P(M).
>
> > > It can be proven that such a set exists and is not in the
> > > image of the map. Do you agree that the set exists? Do you
> > > agree that it is not in the image of the map?
> >
> > M does exist.
>
> Um. I thought M was the set whose powerset we were discussing.

N is the set whose power set we are discussing. M is the set of all
non-generators.
>
> OK, let's say we're discussing S, P(S), and a map f:S->P(S).
>
> > It cannot be the image of m e N.
>
> Then that's the end of the proof, isn't it? You agree
> that given a map f, there exists M which is not mapped
> by f. There is nothing else to discuss. f does not map
> to every element of P(S).
>
> > nd it is easy to show
> > that this is true for any mapping, whether the image set is larger or
> > smaller than the pre-image.
>
> Thank you. Therefore no map f:S->P(S) is onto. End of
> story.

Not even the map P(P(N)) --> P(N) is onto, if the impossible
requirement is maintained.

Regards, WM

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