In article <firstname.lastname@example.org>, WM <email@example.com> wrote:
> On 11 Jun., 23:06, William Hughes <wpihug...@hotmail.com> wrote: > > On Jun 11, 4:38 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > On 11 Jun., 21:33, 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." > > After the tree has been constructed. Yes, that is true.
After the tree is constructed, every path (maximal totally ordered set of nodes) can be distinguished from any or all others by any infinite subset of its set of nodes. But the set of such paths cannot be ennumerated. > > > > > > The subset of nodes can be found in the tree, however > > > > it cannot be found in a single element of P. > > > > > All right as long as P is not used to construct the binary tree. > > > > Before P is used to construct the binary tree, > > the binary tree contains a path > > p that can be distinguished from > > every element of P. > > Yes, I just answered a contribution by Rainer Rosenthal, who > essentially had the same idea. You can start off with the path 1/e and > then add all paths with tails 000... You can start off with all paths > you like or can construct by means of Cantor's diagonal method. Then > complete the construction by inserting those paths with tails > 000...which are yet missing. As a result you get a binary tree that is > complete in that you cannot apply any diagonalization, because there > is no node or set of nodes that is not already belonging to a path. If > you insist in an uncountable set of paths, you must believe the > "sneaking in" during or after completion. I do not believe in > "sneaking in" in mathematics. Therefore, Cantor's diagonal argument, > if correct, forces us to deny the existence of actually infinite paths > at all. Therefore you cannot start with 1/e. It does not exist as a > path or as a complete digit sequence (it does exist as 1/e though).
It doesn't matter how you start or how you go on, as soon as you have included all the nodes from all possible finite subtrees, you automatically have all the uncountable set of paths. At least, you have uncountably many subsets of the set of nodes which are totally ordered by the transitive closure of the "parent of" relation.
Wm seems to think that most of such maximal subsets are not paths because he doesn't like them, but they are my definition of paths, and any maximal infinite binary tree has uncountably many of them.