In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 9 Dez., 20:37, Zuhair <zaljo...@gmail.com> wrote: > > > Only a countable subset can be represented by the Binary Tree. The > > > reason is that no path is really actually infinite. > > > > Then you are not addressing what Cantor was speaking about, he is > > speaking about reals represented by ACTUALLY infinite sequences (paths > > in your case). It is clear that the set of all reals represented by > > FINITE sequences is countable, but those are just a very small subset > > of the set of all reals. > > > > If one assumes Actual infinity, then it is easy to recover the > > diagonal path from any bijection between the reals and the set of all > > paths of the infinite binary tree, and this will be a path that is not > > present in the tree of course. > > Then you are wrong from the scratch. Every real number has a > representation by an infinite sequence (= infinite path of nodes in > the tree). But as my proff shows I construct the whole Binary Tree by > countably many paths. There are not more nodes available to add > further paths.
Zuhair is not correct, but also WM's objections are for the wrong reasons.
In the first place, bijections between the reals and the set of all paths in a complete infinite binary tree, while possible, are not at all as trivial as many people expect because of the dual binary representation of certain reals.
In the second place, one cannot list either all the infinite paths or all the binary expansions, since lists cannot include uncountably many entries.
And in the third place, one cannot construct a diagonal (anti-diagonal?) to an unlistable set of paths, since those paths cannot be ordered like the naturals. --