```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 ofcourse INCLUDES paths at level n itself.At level 0 the number of PATHS is 2^(1) -1 which is 1 that is theempty pathAt level 1 the number of PATHS is 2^2 -1 which is 3 this INCLUDESpaths 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 (andINCLUDING PATHS at level Aleph_0 OF COURSE: The later ones are theINFINITE paths) is (2^(Aleph_0 +1)) -1 which is 2^Aleph_0, thisincludes both FINITE as WELL as INFINITE paths of the completeinfinite binary tree and NOT just finite paths as you are deceived by.Zuhair
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