Re: CANTOR's theorem

In article <1116417944.510584.63610@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
> *** T. Winter wrote:
> > > > 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.
>
> Nope. But convince yourself after all by the mapping
>
> N --> P(N) \ M
>
> which I claim is bijective, unless you prove it is not.

That is too easy. Say your mapping is f. Define the following mapping g
from N to P(N):
g(1) = M
g(n) = f(n - 1) for n > 1.
If your mapping is surjective, this one is also surjective. But we know
it is not surjective, so your map is also not surjective.
--
*** 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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