Re: Cantor Confusion

In article <1174670803.869854.53820@xxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
On 23 Mrz., 17:09, "*** T. Winter" <***.Win...@xxxxxx> wrote:
....
> 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.

But that is not precisely the same. You want me to reverse a series that
does not terminate. I ask you how I do that, and you refrain to answer.

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.

No, the sum is undefined. If you think it is defined, *prove* it and
*prove* that the sum is 2. What is the sum of the "sequence":
"..., 1/4, 1/2, 1"?

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).

Yes, and when are we done? The problem with this is that there is a
dependency in the order of transpositions, so they can not be performed
simultaneously.

> > 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.

You have to first *prove* your statement that there is not an uncountable
number of separated paths. Until now your proofs depend crucially on the
assertion that there is not an uncountable number of separated paths, and
so are circular.

> 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.

Why? Each two paths are isolated from each other (by your definition), so
all paths are isolated from all other paths. *But* there is *no* level
where a single path is isolated from all other paths.


> 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.

Until now you did not succeed, because your proofs either contain a denial
of the axiom of infinity or other logical flaws.

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.

I do not. You think I do, but that is due to a lack of logical reasoning.
--
*** 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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