In article <ed7cfe4c-c486-43c0-bb1f-8ceb2652c77c@y17g2000yqn.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 14 Jun., 14:01, William Hughes <wpihug...@hotmail.com> wrote: > > > > > Assuming infinite paths exist: > > > > The path p is in the tree > > > > The path p can be distinguished from > > every element of P. > > For every path p_n of P there is a digit such that p differs from p_n. > But there is no digit of p that differs from every path p_n of P > because, if p exists as binary representation, then p is the union of > all p_n and as such cannot differ from the union.
Paths are not made up of digits, but of nodes. But in a binary tree, one can represent a path as an infinite sequence of binary digits with, say, 0 for each left child node and 1 for each right child node.
But no such path p is then a union of other paths, as WM claims, as no union of a set of more than one path can be a path itself. > > > > You refuse to agree to > > > > Assuming infinite paths exist: > > > > There is a path p in the tree that can > > be distinguished from every element of P. > > > > Until you explain this, no other argument as > > to why you refuse to accept this statment > > will be considered. > > It is obviously impossible to distinguish p from all paths p_n that > are in the binary tree after construction has been completed.
If p_n refers to any finite or even countably infinite set of paths then every non-member of p_n, is distingishable from every member of p_n and every subset of p_n. > > If you don't believe me, then play the game. Minimum stake 10^6 > dollars or (if you come from Europe) Euros or Pounds.
Then I challenge WM to present his list of paths from which he claims no other path can be distinguished.
Note that since any counterexample may be dependent on his choice of list, as Cantor's depends on the list being given, WM is constrained to publish his list first.
