```Date: Jan 12, 2013 3:07 AM
Author: mueckenh@rz.fh-augsburg.de
Subject: Matheology § 190

Matheology § 190The Binary Tree can be constructed by aleph_0 finite paths.        0      1, 2  3, 4, 5, 67, ...But wait! The Binary Tree has aleph_0 levels. At each level the numberof nodes doubles. We start with the (empty) finite path at level 0 andget 2^(n+1) - 1 finite paths within the first n levels. The number ofall levels of the Binary Tree is called aleph_0. That results in2^(aleph_0 + 1) - 1 = 2^aleph_0 finite paths.The bijection of paths that end at the same node proves 2^aleph_0 =aleph_0.This is the same procedure with the terminating binary representationsof the rational numbers of the unit interval. Each terminating binaryrepresentation q = 0,abc...z is an element out of 2^(aleph_0 + 1) - 1= 2^aleph_0.Or remember the proof of divergence of the harmonic series by Nicoled'Oresme. He constructed aleph_0 sums (1/2) + (1/3 + 1/4) + (1/5 + ...+ 1/8) + ... requiring 2^(aleph_0 +1) - 1 = 2^aleph_0 natural numbers.If there were less than 2^aleph_0 natural numbers (or if 2^aleph_0 waslarger than aleph_0) the harmonic series could not diverge andmathematics would deliver wrong results.Beware of the set-theoretic interpretation which tries to contradictthese simple facts by erroneously asserting aleph_0 =/= 2^aleph_0.Regards, WM
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