Date: Jan 12, 2013 3:45 AM
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.
> 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
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,
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.