In article <0373342b-3226-4cd1-a89c-270e6b5a103a@a36g2000yqc.googlegroups.com>, WM <mueckenh@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." > > In fact that is not possible.
That depends critically on what set of paths one is attempting to distinguish the given path from.
If one takes any countable set of paths in the maximal infinite binary tree, then there are paths in that tree which can be "distinguished" from every path in that set.
However if one takes the set of ALL paths in that tree , there are then no paths in the tree that can be distinguished from all paths in that set.
> > > > > > > > There are two statements: > > > > > 1) Path p can be distinguished from every path of P. > > > > > 2) Path p cannot be distinguished from every path of P. > > > > > You agreed that constucting the tree does not change > > > > path p or any element of P > > > > > That is correct. It is a basic of mathematics. > > > > So either you are completly incoherent or you > > now agree that under the assumption that > > actual infinity exists path p can be distinguished > > from every path of P. > > But it turns out that this assumption cannot be satisfied.
Maybe not in WM's world, but there are worlds that WM apparently cannot access that are wide open to mathematicians. > > > > > > Your claim explicitly mentions distinguishing p from P by > > "one or many or infinitely many nodes". > > Answer yes or no > > > > you are able to distinguish p from P by subsets of digits > > I will try to explain it for you again: > > We have a list of a countable set P of terminating paths p_n.
Then any non-terminating path can be distinguished from every member of that set.
> 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.
You have made it impossible to distinguish between distinct paths even within your original set. > > This yields the complete binary tree.
No it doesn't. As soon as it combines any two paths into one, which it does en mass, whatever remains is incomplete.
> You know that you are unable to > distinguish any binary sequence representing a real of the unit > interval from every path of the tree.
If the tree is a maximal infinite binary tree, then every real in that interval has at least one binary representation. If the tree is, as WM would have it, less than maximal, then it will be missing some paths that the maximal tree contains.
> This proves that you, contrary to the assumption, have been unable at > the beginning too, because there the same set of paths was presented > to you.
WM is delusional again. Any maximal infinite binary tree will contain more that countably many paths because no countable set of paths can be maximal. There are always others not yet included.
> You only believed that you were able.
With much better justification that many of WM's beliefs.
