In article <406110a1-633d-43ff-8a7d-bc256aac5135@a36g2000yqc.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 13 Jun., 15:43, William Hughes <wpihug...@hotmail.com> wrote: > > On Jun 13, 8:53 am, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > On 13 Jun., 14:17, 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." > > > > > > > Yes. > > > > > > You agree that there is a path p in the tree > > > > > > You agree that path p can be distinguished > > > > from every element of P. > > > > > If actually infinite paths exist! > > > > Ok. We make the assumption explicit. > > > You agree that if actually infinite paths > > exist, there is a path p in the tree > > > > You agree that if actually > > infinite paths exist, path p > > can be distinguished > > from every element of P. > > by using nodes of the tree only without additional knowledge.
So that WM argues that if P is hidden then it somehow will contain paths that it does not contain when known? > > > > You do not agree that > > if actually infinite paths exist, > > the tree contains a path that > > can be distinguished from every element of P. > > I prove that it is impossible to distinguish a give path p from the > paths in P using only the information stored in the tree, i.e. using > only digit sequences to identify numbers
But it is eminently possible using the tree and knowing P.
> thereby proving that no > actual infinity exists.
Not a proof that will stand up to any scrutiny. > > Instead of grumbling about this fact, you either should accept it and > confess that I am right and that set theory is wrong, or you should be > able to distinguish your path p from the set P of paths that I have > used for construction, but without any other knowledge than the final > result, namely the digits, i.e., the nodes of the binary tree.
Or we can just show, as we have done so often, that there are more paths in the tree than in P (in the sense that there can be no surjection from P to the set of all paths or injection the other way).
Theorem: if A and B are sets such that Card(A) > Card(B) then there is some a in A that is not in B.
> > That are the two possible logical alternatives.
