Re: Cantor Confusion

On 16 Mrz., 01:31, Virgil <vir...@xxxxxxxxxxx> wrote:
In article <1173954799.919385.61...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,

For even binary trees ( where even here means all paths are of equal
length),

Only those are under discussion here.

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.

It shows, however, that the number of paths cannot jump from finite to
uncountable.

And nodes are connected by edges, not paths.

Correct. Nodes ae connected by edges. But as these edges are elements
of paths, nodes are also connected by 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,

- meanwhile you should have learned this definition -

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.

As very element of the union is finite, every path is finite.
"Infinite" means here only "finite numbers of nodes and finite numbers
of separeted paths growing from level to level without end".

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.

As easily it can be seen that the number of paths within the tree,
i.e., where there are nodes, remains countable. Every path-bundle
splits at a node. The number of path-bundles is countable. As long as
there are nodes, the path-bundles are countable. What happens beyond?

Regards, WM


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