On Apr 1, 10:54 pm, WM <mueck...@rz.fh-augsburg.de> wrote: > 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.
Nope. This set of nodes has no node at a last level. Every path in the Binary tree that contains all finite paths has a node at a last level.
The difference between the trees is not which subsets of nodes exist, but which subsets are considered to be paths. Only in one of the trees can a subset of nodes with no node at a last level be considered a path.