In article <email@example.com>, WM <firstname.lastname@example.org> wrote:
> On 18 Jun., 20:22, William Hughes <wpihug...@hotmail.com> wrote: > > On Jun 18, 1:49 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > > > > > > > > > On 18 Jun., 14:44, 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) > > > > > > <snip evasion> > > > > > > Please answer yes or no > > > > > > t is not an element of P > > > > > The actually infinite path t is not an element of P > > > and not a path of T, because t does not exist.
If t does not exist it CERTAINLY can be distinguished from things that do exist. > > > > Note it would be circular to simply assume that actually > > infinite paths do not exist. > > It is not assumed.
WM repeatedly asumes it, but has never proved it.
> It is proved by the fact that t cannot be > distinguished from the paths of P and T.
But t CAN be distinguished from them,, particularly if the paths of P and T exist but t does not. > > > > > Proof by the fact that t cannot be distinguished from the set P > > > > Nope: you have agreed that t can be distintuished from the every > > element > > of the set P > > > > (Recall, You have agreed. > > > > A subset of nodes is distinguished from every > > element of P if and only if it is not contained in > > a single element of P.) > > All of t, that does exist, is contained in a path of P. It is only > your delusion that opposes to this fact.
WM is not in a position to accuse others of having delusions when he has so many of his own that he is quite unable to distinguish from reality.
> You say that for every path p of P there is a bit of t that does not > belong to p.
That is not what anyone but WM says. What we do say is that if P is a countably infinite set of paths in some maximal infinite binary tree, then there are paths in that tree which are not members of P, even though the union of P may contain all nodes of that tree.
In fact, WM's own example of the set of all paths with only finitely many right branches is such a P, since in the resulting tree there are paths with infinitely many right branches.
So that WM has managed to disprove his own thesis.