Re: Cantor Confusion

In article <1173954799.919385.61730@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:

On 13 Mrz., 14:15, "*** T. Winter" <***.Win...@xxxxxx> wrote:
In article <1173724460.248046.46...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>
mueck...@xxxxxxxxxxxxxxxxx writes:

Terminating paths.

*Passing* path-bundles.


Path-bundles are just another of WM's irrelevancies invented to
obfuscate the obvious.

Those are *terminating* paths. (A path-bundle can be seen as a terminating
path: it is a set of nodes containing a finite number of nodes.)

A path bundle splits off into two bundles which pass said node.
Therefore every set of path-bundles in the tree has a finite cardinal
number

This, in the "limit", may yield an infinite number, but certainly not
an uncountable number without having an intermediate countably
infinite number.

For even binary trees ( where even here means all paths are of equal
length), the number of paths increases exponentially with number of
levels (lengths of a path). Adding 1 to the number of levels doubles the
number of paths.

The tree is continuous because its nodes are connected by paths.

That is a distinctly non-standard meaning for "continuous" in
mathematics. And nodes are connected by edges, not paths.


There
is never more than the factor 2. There are no interruptions possible
and no jumps from "finite" to "uncountable". Your claim would require
that.

When one takes what WM calls the 'union' of all his finite binary trees,
one has one such tree of n levels for each member n of N, which N is
itself not finite, so that the union cannot be finite either.

And one can easily see that the number of paths in any finite tree is
the same as the power set of non-root levels in all his trees, of
finite or infinite level.


PS: What about the review of chapter 10?

When Chapter 9 is already shown to be so corrupt, why bother?
.