In article <cda5a20c-9010-45cd-9cb5-a6075e155e9f@o14g2000yqh.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> Consider the complete binary tree. It has uncountably many paths. > > Consider the binary tree that contains only all finite paths. It has > countably many paths.
Whatever does "contains only all finite paths" mean?
Given that a tree contains any path of more than two nodes, it cannot contains the subset of the first of those two nodes as a path at all.
A "subpath" maybe, but not a path.
And even allowing subpaths, WM is WRONG, unless "countably many" includes being finite.
No binary tree with infinitely many paths/subpaths can have only finite length paths.
It must have at least one path of infinite length.
Proof: the set of paths of length <= n is finite for every n, so that to have infinitely many paths, there cannot be any longest finite path. But the only way NOT to have a longest finite path is to have at least one infinite path.
