
Re: Matheology § 190
Posted:
Jan 12, 2013 4:05 AM


On 12 Jan., 09:45, Zuhair <zaljo...@gmail.com> wrote: > On Jan 12, 11:07 am, WM <mueck...@rz.fhaugsburg.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 01, 01 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
aleph_0 is not a natural number. Therefore I carefully excluded any reference to last levels. And this is not required in mathematics.
> (and > INCLUDING PATHS at level Aleph_0 OF COURSE: The later ones are the > INFINITE paths)
You are in error. There are, according to set theory, aleph_0 finite natural numbers. aleph_0 is not a natural number.
Instead of paths you could also say nodes, because there is a bijection between finite paths and nodes. Do you defend the opinion that there must be nodes at an infinite level in order to collect infinitely many nodes? (This would be a sober position, but I doubt that it is your position.)
> 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.
And you think that in the proof of Nicole d'Oresme also infinite natural numbers play a role? You are in error. The sum of finite unit fractions already surpasses any given number.
Regards, WM

