In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 17 Jun., 14:56, 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 your claim is false. > > A ==> B > > should go undisputed
Provided that A includes existence of infinite trees and WM's infinite trees are allowed to be a MAXIMAL infinite binary trees.
> > > > ~A [Follows from ( A ==> B, ~B) ==> ~A ] > > > > by proving ~B > > > > (Note that assuming ~A is circular) > > Of course. > > > As we have seen we can create a finite subset > > by adding a single path. > > That is of no relevance because only allowed subsets can be of > interest. Why do you want to raise the impression your arguing with > respect to irrelevant subsets would be relevant? > > > However, > > as previously noted, > > to create a path requires creating an infinite subset. > > To create a path requires to add that path.
To create a maximal infinite binary tree does not require adding any path to anything.
We can start with any actually infinite countable set, indexed by N, the set of 1-origin naturals, 1 is then the index of the root node, 2*n + 0 is the index of the LEFT child of the node indexed by n 2*n + 1 is the index of the right child of the node indexed by n
One now has a maximal infinite binary tree without having even mentioned paths.
And in this tree, there are too many paths to count, one path for every infinite binary sequence starting with a 1 and having 1's or 0's thereafter endlessly.