On 15 Jun., 20:54, Virgil <virg...@nowhere.com> wrote: > In article > <0373342b-3226-4cd1-a89c-270e6b5a1...@a36g2000yqc.googlegroups.com>, > > 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." > > > In fact that is not possible. > > That depends critically on what set of paths one is attempting to > distinguish the given path from.
Every path that is in the tree belongs to this set. The tree is simply consructed by writing the paths in unconventional manner. > > 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.
Even if we considered every path p_n as a set {p_n} and the tree as the union of these sets, then the tree should not contain a path that is absent from each of the sets. Nevertheless the complete tree contains all possible paths. (Possible means, paths tha can be written by bit sequences.) > > > > > > > > > > > > > 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.
The world of matheology: spirituality, visionaries and delusions. >
> > 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.
One is tempted to think so. And if there actually are non-terminating paths, you should be correct. > > > 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.
There is no node that remains unconstructed. > > > 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.
The tree is as maximal as a tree can be, after construction by a countable set of paths.
By the way: What more should be used tio construct the tree than a set of paths that reaches every node? What more than every node is the tree?
