Re: Cantor Confusion

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.

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.

> > > > No. I would state that at the root node of the tree there start
> > > > uncountably many paths. Why do you think that number is not
> > > > uncountable?
> > > > (Remember, paths are non-terminating.)
> > >
> > > (3, clarified) There is no node in the tree (including every node of
> > > the paths
> > > 0.000... and 0.111...) which belongs to a level where uncountably
> > > many paths have separated.
> >
> > 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 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. Cantor: yes. Tree:
no. Why?

> > > > > But it is connected to the root node like every other path. The
> > > > > branching offs are countable.
> > > >
> > > > Yes, I never contested that the number of branching offs is
> > > > countable. So what?
> > >
> > > In the whole tree there can be no more separated paths than nodes.
> >
> > Pray give a *proof*, not just a statement.
>
> 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?

We can safely state that at no level there are uncountably many
separated paths.
As every node is in the tree ad no node is outside, there are at no
place in the world uncountably many numbers.

> > > > And so what, what does that prove about the complete tree? Indeed,
> > > > at no finite distance from the root. What this *means* is that the
> > > > number of terminating paths is countable.
> > >
> > > No. It means that the number of paths which consist only of nodes with
> > > natural indexes is countable.
> >
> > Not at all. Pray provide a *proof*.
>
> 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.

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.
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.

> > > The "distance" is measured by the index of the due node, i.e., by the
> > > number n of the corresponding level. "At no finite distance from the
> > > root" means at no node which can be enumerated by a natural number.
> >
> > Right. There is *no* node where all non-terminating paths are separated
> > from each other.
>
> There is no point in the tree where more than countably many paths are
> separated, although the tree is infinite. Outside of the tree there is
> no mathematics of real numbers.

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.

Regards, WM

.

Quantcast