Date: Jan 12, 2013 3:45 AM
Author: Zaljohar@gmail.com
Subject: Re: Matheology § 190
On Jan 12, 11:07 am, WM <mueck...@rz.fh-augsburg.de> wrote:

> Matheology § 190

>

> The Binary Tree can be constructed by aleph_0 finite paths.

>

> 0

> 1, 2

> 3, 4, 5, 6

> 7, ...

>

> But wait! The Binary Tree has aleph_0 levels. At each level the number

> of nodes doubles. We start with the (empty) finite path at level 0 and

> get 2^(n+1) - 1 finite paths within the first n levels. The number of

> all levels of the Binary Tree is called aleph_0. That results in

> 2^(aleph_0 + 1) - 1 = 2^aleph_0 finite paths.

NO. We get (2^(n+1)) -1 PATHS within the first n levels, This of

course INCLUDES paths at level n itself.

At level 0 the number of PATHS is 2^(1) -1 which is 1 that is the

empty path

At level 1 the number of PATHS is 2^2 -1 which is 3 this INCLUDES

paths at level 0 and those at level 1 ITSELF! which are the empty path

(at level 0), and 0-1, 0-1 paths at level 1.

Now the number of ALL levels of the Binary Tree is called aleph_0,

correct.

But the result is The number of ALL *PATHS* up to Aleph_0 (and

INCLUDING PATHS at level Aleph_0 OF COURSE: The later ones are the

INFINITE paths) is (2^(Aleph_0 +1)) -1 which is 2^Aleph_0, this

includes both FINITE as WELL as INFINITE paths of the complete

infinite binary tree and NOT just finite paths as you are deceived by.

Zuhair