Re: Cantor Confusion

On 23 Mrz., 15:58, "*** T. Winter" <***.Win...@xxxxxx> wrote:
In article <1174654681.038185.299...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueck...@xxxxxxxxxxxxxxxxx writes:

> On 22 Mrz., 16:29, "*** T. Winter" <***.Win...@xxxxxx> wrote:
> > In article <1174562837.696384.225...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueck...@xxxxxxxxxxxxxxxxx writes:
> >
> > > On 21 Mrz., 16:09, "*** T. Winter" <***.Win...@xxxxxx> wrote:
> > > > In article <1174481309.873623.92...@xxxxxxxxxxxxxxxxxxxxxxxxxxx> mueck...@xxxxxxxxxxxxxxxxx
> > > > > > What portion of the first node
> > > > > > is assigned to your path?
> > > > >
> > > > > How much does the last term of the geometric series contribute to
> > > > > the value o the series? This is the value which the first node
> > > > > contributes to he path.
> > > >
> > > > As there is no last term of the geometric series this makes no sense.
> > >
> > > Nevertheless, it makes sense to calculate the sum of the series.
> >
> > You can calculate the sum of the series. But what the relation is to the
> > contribution of the nodes to the paths is extremely unclear. From your
> > statement I derive that the first node contributes nothing to the path.
>
> the first node contributes exactly as much as the last term of the
> geometric series contributes to the sum 2 of the series. As this term
> does not exist, it does not contribute anything. The next one before
> the last one also does not exist and not contribute. But somehow some
> nodes manage to contribute enough to obtain the result 2 after all.

How do you obtain that result? You have not shown a proof at all for it.

> If we exchange infinitely many terms of the geometric series, its sum
> remains 2, because it is absolutely converging.
>
> 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.

> > > > At every node in the tree uncountably many paths go to the left and
> > > > uncountably many paths go to the right. I do now know what you are
> > > > meaning here.
> > >
> > > At no node in the tree there exists a single path. This means: There
> > > are no single paths.
> >
> > Still unclear. Through each node go uncountably many paths. But what you
> > mean with "there are no single paths" is unclear. Every two individual
> > paths diverge at some node from each other.
>
> 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".

>
> 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.

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.

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.

Regards, WM


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