Re: Cantor and the binary tree

In article <1120990402.569549.238650@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:

> Martin Shobe wrote:
> > On 7 Jul 2005 02:38:08 -0700, mueckenh@xxxxxxxxxxxxxxxxx wrote:

> > The only way to even make sense of the "completed results" is to use a
> > limit. While I have some ideas for canditates, I don't know of any
> > natural topology on sequences of orderings. And in the ideas I do
> > have, your sequences diverges (I.e. there is no "completed result") or
> > the "completed result" is not what you claim it is.
>
> The sequence b_n diverges too. It is only the series b_n*10^-n of the
> antidiagonal which converges. This fact, however, does not mean that
> the antidiagonal differs from any real number of Cantor's list.

But the rule by swhich the anti-diagonal is constructed guarantees that
it is different from every number in the list. That and convergence
prove WM totally wrong again.


> >
> > > It is equally impossible to show the
> > >completed antidiagonal.
> >
> > It's not necessary to do so. All that is needed is proof of it's
> > existence.
>
> Compare 0.999... and 1.000... . Both decadic representations are equal
> with respect to their values but not with respect to their numerals
> which always remain different. In exactly the same manner Cantor's
> Antidiagonal always remains different from *that number* which would
> differ from any real number of the list.

In the decimal system it is only with infinite sequences of 9's and 0's
by which dual representations can occur, but anti-diagonals exclude use
of 9's or 0's entirely so no such duality is possible. WM's ignorance
again manifests itself.
.

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