Re: Cantor and the binary tree

In article <1119614903.753905.160730@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
> *** T. Winter wrote:
> > In article <1119540536.009276.193750@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
> > > If we find by induction that every even number 2n is larger than the
> > > cardinal number n of its set {2,3,6,...,2n} then we see for *every even
> > > number* that the cardinal number n of its set cannot surpass its value
> > > 2n. This proof holds, by induction, for every even number. But in the
> > > whole set of all even numbers there is nothing else but even numbers.
....
> > > Attention: This proof by induction uses only finite even
> > > numbers. It does not use the whole set of even numbers. But its result
> > > concern this set, because this set contains solely finite even numbers.
> >
> > The proof is valid for all finite sets of even numbers. There is no reason
> > to suspect that it holds for the infinite set of all even numbers. The
> > major difference is that each finite set of even numbers has a maximal
> > element, no such maximal element exists for the set of all finite numbers.
>
> But also in the infinite set there can only be *finite* even numbers.

Yes, indeed.

> For those, the proof holds.

Indeed, 2n > | {2, 4, 6, ..., 2n} |. Still no set of all even numbers in
sight.

> Theorem. Any set of even numbers has a cadinality which is less than
> infinite.
> Proof 2n > Card({2,4,6,...,2n}).
>
> If the set of all even numbers should have a larger cardinal number
> than any finite natural, then beyond the last one, 2n,

In the set of all even numbers there is no last one.
--
*** 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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