Re: Cantor Confusion

On 23 Mrz., 17:09, "*** T. Winter" <***.Win...@xxxxxx> wrote:
In article <1174664147.264581.197...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueck...@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)?

You are not so squeamish when exchanging every digit of Cantor's
diagonal.
Cantor, unless working every digit simultaneously, will never finish.
In fact we need not know how to proceed, but it is sufficient to know
hat the sum remains 2 what ever we do.
However, you may proceed as follows (but only after having read my
book): By transpositions (as we know them from Cantor's work) bring
the n-th term (with n = 2, 3, 4, ...) into the first position and
after that bring the n+1-th term into first position. If you are
really fast, then you reverse the whole sequence of terms (because you
can determine, for *every* term number n, the number of transpositions
required to have it at the first position).

> > > 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).

I am arguing within the tree. There is never an uncountable number of
separated paths. You say all paths were uncountable and separable.
Conclusion. there is no "all paths" within the tree.


> 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?

Therefore, if they existed isolated and were so many as you say, they
would populate a level with uncountably many nodes.

> 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.

Within the tree it can be disproven.

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.


That you do believe in set theory is not a crime. But that you do
believe in uncountaly many separations without uncountably many
separations, that is hard to believe.

Regards, WM

.

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