Re: Binary Tree and Pairs of Nodes

In article
<63c5c5b6-2fc7-4d22-8f61-de92dc2f1d28@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
WM <mueckenh@xxxxxxxxxxxxxxxxx> wrote:

On 10 Okt., 22:07, george <gree...@xxxxxxxxxxxxx> wrote:
On Oct 8, 12:24 pm, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:

Anyhow, the binary tree is a crystal-clear construction that avoids
the confusing opinion of undistinguishable numbers. It does never
insert its roots into the mud of the continuum (Kontinuumssauce).

This is just bull***. The INFINITE part of the binary tree IS
ALWAYS
the continuum.

There is no "infinite part". The levels of the tree are enumerated by
thge natural numbers.

If one ennumerates the nodes by naturals (which is possible) then the
set of nodes in any path will be an infinite subset of the naturals.

So WM is trying to surject the set of naturals to a set of infinite
subsets of the naturals. But he will always fail.





You CANNOT SAY "all its paths are countable"!
You are equivocating on "all"!

It is the usual saying. But if you do not understand it, then I say
with sufficient precision: The set of all paths of the tree is a
countable set.

If that tree is an infinite complete binary tree as defined in Wiki, then
Wiki says that its set of oaths in UNcountable.
http://en.wikipedia.org/wiki/Binary_tree



At every level denoted by a natural number n, I can distinguish 2^n. I
do not see where you find infinitely many.

Then WM is admitting his own blindness.


I realize your native language is not English but that is simply not
an excuse.

As I said, with respect to this platitude I took the liberty of being
somewhat sloppy. In German this is not different. So there is no
excuse required.

There is certainly no excuse acceptable for WM's falsehoods.

It is very easy to see that there are not more paths than nodes in
the
binary tree.

It is even easier to see that there are more paths than nodes, at least
to anyone with normal vision.

One may, if one choses, deny the very existence of any infinite complete
binary tree, as constructionists would do, but once one allows its
existence, the uncountability of its set of paths is not logically
deniable.

But then, WM never allows logic to constrain his arguments
.

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