On 15 Jun., 22:04, William Hughes <wpihug...@hotmail.com> wrote:
> > > You are trying a proof by contradiction of > > > " Actually infinite paths exist" > > > > We have > > > If actually infinite paths exist > > > there is a path p that can be distinguished > > > from every path of P. > > > > We need > > > > If actually infinite paths exist > > > there is no path p that can be distinguished > > > from every path of P. > You know modus tollens? ((A ==> B) & ~B) ==> ~A
A: actually infinite paths exist. B: p can be distinguished from every path of P.
> > > You cannot distinguish p from every path of the tree. > > Irrelevant, you distinguish p from every element of P
The tree is nothing but every path of P written in some unconvential form. > > > Every path of the tree is is from P. > > Nope. Every *node* of the tree is from P. > However, there is a *subset of nodes* in the tree that is not > contained in one element of P.
Every subset of nodes, that can form a path of the tree, is in the tree. It stems from writing down the paths of P and no bit more.
(Even if the tree is considered to be a union of sets {p_n} with p_n in P, then the union could not contain more than is contained in at least one of the united sets.)
