Re: CANTOR's theorem



Virgil wrote:
> In article <1117027159.228720.35740@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
> mueckenh@xxxxxxxxxxxxxxxxx wrote:
>
> > Virgil wrote:
> >
> > > >
> > > > You always apply the same old argument.
> >
> > > When one has done it right, there is no need of change.
> >
> > Not if the consistency of the theory is in question. Two correct proofs
> > in an incorrect theory may lead to contradictory results.
> >
> > > Since (N union {}) = N, my previous proof is still valid.
> >
> > Define a bijective mapping from {1, a} on P({1}) = {{}, {1}}. a is a
> > symbol but _not_ a number (a could probably be { }). There are two
> > bijections possible. In both cases the set of all _numbers_ which are
> > non-generators under any bijective mapping belongs to the image.
> >
> > f: 1 --> {1} and a --> { } with M(f) = { },
> >
> > g: 1 --> { } and a --> {1} with M(g) = {1}.
> >
> > Although there is no problem with non-equivalent sets, Hessenberg's
> > condition cannot be satisfied. The set M of all numbers which are
> > non-generators cannot be mapped by a number although M is in the image
> > of both the possible mappings. The set {M(f), m, f} is an impossible
> > set. Hessenberg's proof does not show, in this example, the
> > non-surjectivity of the mapping.
>
> What your example above shows is that the restriction of either of your
> functions to {1} -> P({1}) cannot be surjections, which is what we have
> been saying all along.

Nevertheless, Hessenbergs condition cannot be satisfied even for
mappings from large sets on small sets. Therefore, it is meaningless
for cardinality questions.

Regards, WM

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