On 1 Apr., 15:19, William Hughes <wpihug...@gmail.com> wrote: > On Mar 30, 3:36 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > > > > > > > On 30 Mrz., 10:17, William Hughes <wpihug...@gmail.com> wrote: > > > > On 24 Mrz., 18:09, WM <mueck...@rz.fh-augsburg.de> wrote: > > > <snip> > > > > > > The only difference is that in the second case you consider > > > > > some subsets of the nodes to be paths, that are not considered > > > > > to be paths in the first case. > > > > > Well, that is a correct description. It implies that these additional > > > > subsets cannot be distinguished by nodes from the finite subsets > > > > Piffle. It is trivial to distinguish a subset that has a node > > > at a last level from a subset that does not have a node > > > at a last level. > > > No, that is impossible if an infinite path consists of infinitely many > > finite subsets. > > Let the subset of nodes in the infinitely many finite subsets > be Q. > > Q is contained in both trees, is not a path > in the Binary Tree that contains only all > finite paths (Q does not have a node at > a last level)
The path to which all finite paths 0.1 0.11 0.111 ... contribute is a path too in the Binary Tree that contains all finite paths.
> but is a path in the > Binary Tree that contains in addition all > actually infinite paths. > > The difference between the trees is not which > subsets of nodes exist, but which subsets are > considered to be paths.-
There is no infinite path in the Binary Tree of all finite paths that would not be considered a path too. In other words: The Binary Tree of all finite paths contains all infinite paths too (if they exist at all). The reason is that every Binary Tree is constructed by its nodes only - and that is tantamount to being constructed by all its finite paths. Nothing is missing and no infinite paths are rejected, if they exist.