In article <email@example.com>, WM <firstname.lastname@example.org> wrote:
> On 23 Jan., 13:46, Aatu Koskensilta <aatu.koskensi...@uta.fi> wrote: > > WM <mueck...@rz.fh-augsburg.de> writes: > > > The union of FISs is finite. Yes that is my claim. But I cannot give > > > an upper limit, because the finite numbers have no upper limit. This > > > is called potentially infinite. > > > > It's more accurately called rank nonsense. To say the union of all > > FISs is finite is to say there is a natural m such that n < m for all > > naturals n. > > No. Finite is the counterpart of actually infinite, namley larger than > every finite number. Even if no upper threshold exists. Look into the > Binary Tree with all finite paths. It has all nodes, but no infinite > path.
WM has no idea how binary trees work. But whenever he gets in trouble WM trees to escape via such trees, but only succeeds in making an ass of himself again.
A complete binary tree (all paths of the same length) with all paths finite can have no more than a finite number of nodes: 1 path of length 1 -> 1 node 2 paths of length 2 -> 3 nodes 4 paths of length 3 -> 7 nodes 8 paths of length 4 -> 15 nodes ... 2^n paths of length n -> 2^(n+1)-1 nodes.
There is no complete BINARY tree with finite paths or a finite number of paths that does not have 2^n paths for some (finite) natural n. --