In article <f59dc977-e9eb-4bbc-bc62-42501f504a9a@r33g2000yqn.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> 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.
But you have not established your " ~"B anywhere outside of you own little world of MathUnrealism. > > > > > > > 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.
Np maximal infinite binary tree is limited to a merely countable set of paths, so that WM's trees are all crippled. > > > > > 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.
P only provides the complete set of nodes for a maximal infinite binary tree, but, if only countable, not any *complete* set of paths, as is directly derivable from Cantor's diagonal proof. > > (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.)
WM falsely argues that what holds for finite trees must necessarily hold for infinite trees. Since the finiteness property does not carry over, it is clear that not ALL properties do.
For a finite complete binary tree it is quite true that any set of paths whose union contains all nodes also contains all paths, but that is because of the easy bijection between leaf nodes and paths in such trees, but in trees without leaf nodes fails. In maximal infinite binary trees, each of countably many nodes is in all of the uncountably many paths passing through it.
