Re: Cantor and the binary tree



Martin Shobe wrote:
> On 7 Jul 2005 02:38:08 -0700, mueckenh@xxxxxxxxxxxxxxxxx wrote:
>
> >> > You can also describe the creation of the antidiagonal in this way.
> >> > Exchange a diagonal digit. From that line go through the list until you
> >> > find the first one below. From that line go through the list until you
> >> > find the first one below. Etc. I think, my steps are larger, so I will
> >> > be ready faster. That is the only difference.
> >>
> >> You can describe *any* number in such an iterative process. But not all
> >> iterative processes are equivalent with non-iterative processes. That is
> >> the difference.
> >
> >*Any* tranposition to be performed is determined from the beginning
> >(given a certain initial well-ordering of the rationals of (0,1)). You
> >can say, for *any* transposition, when it will have to occur and what
> >will be the result. It is not an iterative process. It is equivalent to
> >Cantor's a_nn replaced by b_n.
>
> >It is impossible, however, to show the completed result, namely the
> >ordered set of rationals.
>
> 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. It
means only that that part which remains undone (and which does never
vanish) becomes arbitrarily small.

>
> > 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. But Cantor's proof is based
upon the numerals b_n --- the limit of the series b_n*10^-n is invalid
for his purpose.


> >A number is an idea which can be put in oder (<) with any other number.
> >sqrt(2) and the same idea, with digit number 10^100 exchanged by 2,
> >cannot and never be put in this order.

There is no unique position unless we can determine it.
>
> It has a unique position in the natural order on R. Whether or not we
> prove a particular descripition of the number to be less than or
> greater than it is another story. BTW, even the natural numbers have
> this "problem".

Therefore they are not complete, but many of them do not exist.

Regards, WM

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