Re: Cantor Confusion

In article <1174654681.038185.299380@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:

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.

Ergo, the first term also contributes nothing!

The next one before
the last one also does not exist and not contribute.

What is your immediate predecessor of the non-existent?


But somehow some
nodes manage to contribute enough to obtain the result 2 after all.

Since in a tree, each node may be a geometric point in some plane, such
a node, like a point, is that which has no part.

So that WM's attempts to subdivide the indivisible come to naught.

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

But as each term is a whole indivisible point or nothing at all, WM's
sequence is all but two terms nothing at all.

As the geometric series contains only the smallest possible infinity
of terms, it should not cause problems to read it from behind.

A geometric series in which all terms are either 0 or 1, either has all
terms 0 or only the first term 1 with common ratio 0, and thus has sum
either 0 or 1 respectively.

>
> 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.


WM has often insisted that one cannot even consider all members of N
sumultaneously, but now insists on considering all members of P(N)
simultaneoulsly. WM now wants to swallow whole camels where he already
must strain at gnats.


You like to do so when Cantor's
diagonal proof is concerned.

On the contrary, one only considers one list at a time.



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 o he tree - all at the same time.

This is the big mistake of set theory. The unreasonable allowance or
prohibition of instantaneously possible actions.

It is only in the physical world where actions take time. In the world
of imagination, where mathematics lives, one escapes such physical
limitations.




> The number of separated paths up to level n is given by the number of
> nodes of level n. No level of the tree has an uncountable number of
> nodes.

Makes no sense at all and is no proof. At each level n there are 2^(n-1)
separated groups of paths, where each group contains uncountably many
paths.

That is wrong, because none of these paths ever gets isolated. Is it
similar to the quarks? Path-confinement?

Every infinite path eventually gets separated from any other path, but
it is only in a finite tree that any path ever gets simultaneously
separated from all other paths, and that is only because finite paths
have a leaf node which is separate from all other paths, but infinite
paths have no leaf node, and every node in such a path is shared with
other paths. For infinite ptahs, it is only the set of /all/ its nodes
which is distinct from the set of all nodes of all other paths.

We can safely state that at no level there are uncountably many
separated paths.

We can confidently say that aat every level, WM is separated from common
sense.

As every node is in the tree ad no node is outside, there are at no
place in the world uncountably many numbers.

They exist, like all mathematics, in the world of imagination.

Yes, so what? That is no proof of your statement. When we talk about the
set of paths, we are *not* talking about any finite level. What happens at
finite levels is irrelevant to the total result.

Wrong. Every number has digits only at finite distance from the
decimal point.

But 1/3, for instance, has non-zero digits at a greater than any given
finite distance from the decimal point.

In the tree there is no level "infinite", but all the infinitely many
levels are there - each one in a finite distance from the root node.

Which means that every such path, like N, is endless.


As at each point in the tree there are uncountably many paths going through
it, I wonder what you mean with "countably many paths are separated".

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

There is a different path in any CIBT for every member of P(N).
.

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