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