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. 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.
Calvin Ostrum said: They sneak in as sets of joined edges and there is nothing you can do about it.
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 set does not contain an actually infinite path. How should it? Should an infinite union of paths of the form 0.1 yield an infinite path? No The union of all finite paths does not supply an actually infinite path, no matter how large they are. All those numbers which would make R uncountable do not exist! Alas, in Cantor's list that is not so sharply visible.
It's getting time to recognize that actual infinity is actually absent.