In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 20 Jun., 15:17, William Hughes <wpihug...@hotmail.com> wrote: > > > > The paths of the tree are the same as the subsets of P. > > > > Nope. You have agreed that if infinite paths > > actually exist then t is not contained in any element > > of P. So you cannot argue "The paths of the tree > > are the same as the subsets of P" without assuming ~A. > > That is not a matter of A or not A. It is the same as > > x = x > > That is independent of A.
What it depends on is, in WM's view, the absence of any actually infinite sets, which implies the absence of the very maximal infinite binary tree that WM is pontificating about.
In order to have such a tree at all, one must first have at least one actually infinite, and preferably well-ordered, set.
Any actually infinite well -ordered set contains as a subset (possibly improper) a set that is order isomorphic to the infinite ordered set of all naturals, from which such a tree is easily built.
And in such a tree, WM's claim are wrong. Specifically, the set of all paths each having only finitely many right branchings, while containing each node in at least one paths, does not contain any of those uncountably many paths which each contains infinitely many right branchings.