On 2009-06-17 14:55:38 -0400, WM <mueckenh@rz.fh-augsburg.de> said:
> On 17 Jun., 20:30, Virgil <virg...@nowhere.com> wrote: > >> And there are in any maximal infinite binary tree paths that are not >> members of any countable set of paths, as Cantor proved. > > Then let me know one of those paths, please. I will tell you, whether > it belongs to a counatble set. > > Regards, WM
Your quantifier confusion is showing again.
For any path in the complete maximal binary tree, there is at least one countable set of paths that contains it. (Trivially, the set {p} contains the path p for every path p. The union of {p} and the set of terminating paths is a countably infinite set which contains p.)
For any countable set of paths, there is at least one path in the tree that is not in that set.
Consider the set of terminating rooted paths: it is a countable infinite set (its elements can be put in 1-to-1 correspondence with the natural numbers), and missing every non-terminating rooted path, as well as every (terminating or non-terminating) non-rooted path.
Describe a countable set of paths that you believe contains every path in the complete maximal binary tree.
