Re: Binary Tree and Pairs of Nodes


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. The part you have investigated thus far is finite;
the part you have not investigated yet IS ALWAYS UNcountabl.

Therefore all its paths are and remain countable,

You CANNOT SAY "all its paths are countable"!
You are equivocating on "all"! All sometimes means EACH around here!
And no INDIVDIDUAL path can be uncountable; they are all countably
infinitely LONG! But they are uncountably WIDE -- there are
uncountably
infinitely many OF them!

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


It is very easy to see that there are not more paths than nodes in
the
binary tree. For the sake of simplicity we will add one node to start
with. This node is mapped on one of the infinite paths, no matter
which one, say that one always turning left, i.e., 0.000....

o
|
/
/
/
...

The next node is mapped on another path, no matter which one, say one
always turning right, i.e., 0.111...

You CANNOT SAY "no matter which one". It MATTERS which one.
Here is the short version of your proof that the reals are countable:
1) pick a 1st real, no matter which one.
2) pick a 2nd real, no matter which one, as long as it is different
from all the previous
ones.
3) repeat ad infinitum.
4) This clearly maps every natural to a real
5) therefore (unsound) it clearly maps EVERY real to a natural.

THIS IS ALL you have said. It is completely unsupported.
.

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