Re: Cantor Confusion

In article <1174562837.696384.225170@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@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.

> > 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.

> > > 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.

> > 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.

> 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".
--
*** 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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