Re: CANTOR's theorem

In article <1116321954.471917.131950@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
>
> *** T. Winter wrote:
> > In article <1116149018.835630.58570@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>
> mueckenh@xxxxxxxxxxxxxxxxx writes:
> > > *** T. Winter wrote:
....
> > Why do you not answer this?
>
> I said that I define in the mapping P(N) --> P(N): n is a
> non-generator, if the singleton {n} is mapped on a set which does not
> contain the number n in any form, i.e. as a singleton or as an element
> of a set like {1,5,n}. This condition is the same as that one implicit
> in Cantor's proof, but adapted to the mapping of P(N) --> P(N) instead
> of N --> P(N).

Good, so the non-generators are numbers and not members of P(N).
Now we define a set M: {x in N, x not in f({n})}. Right so far.
Now if there is a bijection, the map must be surjective, so M should
be in the image of f. But we can not prove that M is *not* in the
image of f. We do not know without more knowledge of f. M is clearly
*not* in the image of f({{1},{2},{3},...}), but that is not a requirement
for f to be surjective. M might be f({1,2}), amongst many others.

> > > Map {1} on its power set. Bijectivity is not required. Only condition:
> > > the set of all non-generators must be part of the image. Does this
> > > impossibility prove that bijectivity is impossible? Does it prove that
> > > there is no life in the sun?
> >
> > What impossibility? There are two possible mappings, and for both we
> > can find a set of non-generators.
>
> But not as the image of 1, although we are ready to spare the 1 for
> this one and only case.

Yes, while we find sets of non-generators, they are not in the map.
What this proves is that the maps are not surjective. And so this
proves that no bijection is possible.
--
*** t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~***/
.

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