On Jun 15, 9:23 am, WM <mueck...@rz.fh-augsburg.de> wrote: > On 15 Jun., 00:19, 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."
You are trying a proof by contradiction of "Infinite paths exist"
We have
If infinite paths exist there is a path p that can be distinguished from every path of P.
We need
If infinite paths exist path there is no path p that can be distinguished from every path of P.
Your putative proof
you are not able to distinguish p from P (by means of digits).
(Note that you have agreed that distinguish p from P means distinguish p from every element of P)
I agreed, and pointed out that you distinguish p from every elemnt of P by subsets of digits.
> > Answer yes or no > > > you are able to distinguish p from P by subsets of digits >
<keep evasion>
> I will try to explain it for you again: > > We have a list of a countable set P of terminating paths p_n. By the > diagonal method we can distinguish p from every p_n (if actual > infinity exists). > > Now we write the paths p_n in slightly different form: The beginning > zeros of all paths are written only once, the following 1 or 0 also > are written only once each and so on. Note, we have not done anything > else but writing the list in slightly different form, saving some ink. > In particular we have not added any new path. > > This yields the complete binary tree.
Correct, you can get every node in the binary tree without using every subset of nodes in the binary tree.
> You know that you are unable to > distinguish any binary sequence representing a real of the unit > interval from every path of the tree.
Correct the tree contains every subset of nodes, however P ( a list of subsets of nodes) does not contain every subset of nodes You can distinguish p from every element of P by subsets of nodes.
