In article <57880d13-7ab5-449c-af24-12129013f9c5@k8g2000yqn.googlegroups.com>, WM <mueckenh@rz.fh-augsburg.de> wrote:
> On 11 Jun., 20:02, William Hughes <wpihug...@hotmail.com> wrote: > > On Jun 11, 1:21 pm, WM <mueck...@rz.fh-augsburg.de> 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. > > > > > If you have a tree that is (the nodes of which are) completely covered > > > by a set of paths then you cannot show any node that will distinguish > > > a further path from that given set of paths. > > > > You can, however, show a subset of nodes that will distinguish > > a further path from that given set of paths. > > But only before those paths have occupied the tree.
Paths do not "occupy" trees. And if the definition of a path is a maximal set of nodes which is totally ordered by the transitive closure of the "parent of" relation, then every path that can ever exists in a tree always exists in that tree.
So that WM must have some special definition of path which does not recognize all the above sort of pathsin order to have different sets of paths for the same set of nodes.
> Because after > construction the given set of paths has occupied the whole tree in > that it includes all possible combinations of nodes, i.e., every > subset that you might think of is already represented in the tree, > because it is constructed such that no subset remains.
Then Cantor's proof applies to every complete infinite binary tree, as the set of binary sequence bijects with the set of paths. > > Calvin Ostrum said: They sneak in as sets of joined edges and there > is > nothing you can do about it.
There is certainly nothing that WM can do about it, but it is the very definition of a path that necessarily "sneaks" them in. > > But the contrary is true. All actually infinite paths sneak out. There > is no path 0.111... > There is only the set of paths > 0.1 > 0.11 > 0.111 > ...
That may be true in WM's tiny world, but his horizons are too constrained.
> That set does not contain an actually infinite path. How should it?
Easily!
> Should an infinite union of paths of the form 0.1 yield an infinite > path?
There is only one path of the form 0.1, so no such "infinite union" is possible.
The set {{0.1},{0.1},{0.1},...}, whose union WM wishes to form is actually just the set {{0.1}}.
> No The union of all finite paths does not supply an actually > infinite path, no matter how large they are.
Depends on one's definition of node and path. By my definition, the union of infinite sequence of node sets each totally ordered by the ancestor relation and each a proper subset of all its successors, is necessarily an infinite path, being a maximal infinite set of nodes totally ordered by the ancestor relation.
> All those numbers which > would make R uncountable do not exist!
What exists outside of WM's wee wee world is not under his control.
> > It's getting time to recognize that actual infinity is actually > absent.
Not as notably absent as WM's ability to thing straight.
