Date: Dec 12, 2012 5:39 AM
Author: mueckenh@rz.fh-augsburg.de
Subject: Re: On the infinite binary Tree

On 12 Dez., 11:26, Zuhair <zaljo...@gmail.com> wrote:
> WM has presented the idea that the infinite binary tree must have
> countably many paths. It seems that he thinks that the total number of
> paths in a binary tree is always smaller than or equal to the total
> number of nodes. So since obviously the number of nodes in an infinite
> binary tree is countable, then the number of paths must be so. But
> since each real can be represented by a path in a 0,1 labeled infinite
> binary tree, then we must have countably many reals thus defying
> Cantor.
>
> The 0,1 labeled infinite binary tree he is speaking of is the
> following:
>
>      0
>    /    \
>   0     1
>  / \    / \
> 0  1 0  1
> .  .  .     .
> .  .  .     .
>
> Answer: in order to justify this claim one must first see what is
> happening at the finite level. And make a simple check of the number
> of nodes and the number of paths through those nodes. If that
> confirmed that the total number of nodes is more or equal to paths
> then there would be some justification for that claim.
>
> Let's see:
>
> Lets take the first degree binary tree which is the following
>
> 0.  One node, no paths
>
> The second degree binary tree is:
>
>  0
> /  \
> 0 1
>
> This has three nodes and two paths.
>
> Lets take the third degree binary tree
>
>      0
>    /    \
>   0     1
>  / \    / \
> 0  1 0  1
>
> Now this has 7 nodes BUT 8 paths, those are
>
> 0-0
> 0-1
> 1-0
> 1-1
> 0-0-0
> 0-0-1
> 0-1-0
> 0-1-1


Wrong. Every path starts at the root node. Every real of the interval
(0, 1) starts with "0."

Try again.

Regards, WM