
Matheology § 190
Posted:
Jan 12, 2013 3:07 AM


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.
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 representations of the rational numbers of the unit interval. Each terminating binary representation 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 Nicole d'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 was larger than aleph_0) the harmonic series could not diverge and mathematics would deliver wrong results.
Beware of the settheoretic interpretation which tries to contradict these simple facts by erroneously asserting aleph_0 =/= 2^aleph_0.
Regards, WM

