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.