In article <email@example.com>, WM <firstname.lastname@example.org> wrote:
> On 19 Jun., 17:56, William Hughes <wpihug...@hotmail.com> wrote: > > Your claim is that "no possibility exists to construct or to > > distinguish by one or many or infinitely many nodes > > of the tree another path." > > > > A: actually infinite paths exist, > > B: the infinite tree contains a path p that can be > > distinguished from every path of P. > > > > You agree > > > > A ==> B > > > > So if A is true then B is true > > and your claim is false. > > > > You want to show > > > > ~A [Follows from ( A ==> B, ~B) ==> ~A ] > > > > by proving ~B > > > > (Note that assuming ~A is circular) > > > > > > you have agreed that t can be distinguished from > > > > every element of P. > > > It does not matter whether I have agreed > > > it matters whether it is > > > true. > > > As you see from the tree, > > > > the subset of nodes t is in the tree > > You cannot distinguish it from all paths of the tree. > > > > And we agree that > > > > the subset of nodes t is not in > > a single element of the list P > > if actual infinity axists.
Since no potentially infinite sets can exist, either there is no maximal infinite binary tree at all, in which case WM's speculating on its properties is foolish, or there is one, and its properties are those of actually infinite sets. > > > > so it is true that the subset of nodes t is > > in the tree and can be distinguished from > > every element of P. > > If you were right.
And if such a tree exists at all, he is.
> But that would imply the existence of actual > infinity and the creation of paths by dropping bits.
Paths are "created" by the partial order on the set of nodes induced by the transitive closure of the"ancestor of" relation.
In a maximal infinite binary tree, uncountably many of them are created.
> As the latter is > wrong, you are not right.
In set theory, WM is the one who is not right by being left out. > > Regards, WM