Re: Cantor Confusion

In article <1174664147.264581.197260@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
On 23 Mrz., 15:58, "*** T. Winter" <***.Win...@xxxxxx> wrote:
....
> As the geometric series contains only the smallest possible infinity
> of terms, it should not cause problems to read it from behind.

If there were a last one. As there is not a last one I have some
difficulty with it, because I do not know where to start.

You need not to start. Reverse all terms of the series simultaneously.

Oh. How do I do that? What do I interchange the first one (1 with)?

> Consider all of them simultaneously. You like to do so when Cantor's
> diagonal proof is concerned. Every exchanged digit is followed by
> infinitely many digits which have to be exchanged. Nevertheless you
> say, it is possible to exchange all of them at one time. Consider the
> paths of he tree - all at the same time.

Yes, I do. What now? Every two paths separate from each other at some
specific node, through all nodes go uncountably any paths, there is no
level where all paths do separate.

Because there is no "all paths".

Pray, provide a proof (within set theory, where we are arguing).

> At no point of the tree (and of the universe) more than countably many
> path can be distinguished.

I would state that as "at no node in the tree more than countably many
groups of paths can be distinguished",

That is where we agree.

or, equivalently "at no node in
the tree terminate more than countably many terminating paths".

No path terminates.

Yes, you want to confuse issues.

This
still does not say anything about the number of non-terminating paths.

That is where we disagree. All paths belong to groups of paths. Even
single paths belong to groups and are groups, if isolated.

Yes, so what?

I think we have cleared our positions sufficiently: We agree in: At no
level in the tree more than countably many groups of paths (including
single paths) can be distinguished. And outside of he tree there are
no paths.

You nevertheless believe in the uncountable while I do not.

Well, within set theory it can be proven. But you do not believe in
set theory. On the other hand, you have *not* proven that set theory
is inconsistent. That you do not believe in the uncountable is *not*
an argument against set theory.
--
*** 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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