In article <email@example.com>, WM <firstname.lastname@example.org> wrote: > Am Mittwoch, 5. Februar 2014 20:20:51 UTC+1 schrieb Ben Bacarisse: > > WM <email@example.com> writes: > > > Am Mittwoch, 5. Februar 2014 17:41:13 UTC+1 schrieb Ben Bacarisse: > > >> > > >> If they gave the > > >> "obvious" construction based on the bijection f: N -> P that the path > > >> p(n) "goes the other way" to the path f(n)(n) does would you mark them > > >> down? > > > > > > They would know that also the other way is already realized, for every > > > n, in a rationals-complete list. And they would know that this > > > rationals-complete liste is realized by the Binary Tree.
But no COMPLETE INFINITE BINARY TREE is "realized" by a "rational complete " list.
But is realized as follows: A Complete Infinite Binary Tree can be formed from the actually infinite set |N as its set of nodes, with the left and right child of node n being node 2*n+0 and 2*n+1, respectively. Nothing further is needed to make |N with that given parent-child relationship on its members into a Complete Infinite Binary Tree.
Note that the tree exists without needing any definition of path at all.
One can then define what it means to be a path in that tree: Definition of a path in the tree defined above: a subset of |N which contains 1 and which for each n in it also contains one but not both of 2*n+0 and 2*n+1 is a path.
Note that this does not define individual paths, but only defines how to tell whether a subset of |N is a path or not.
But it allows us to match any particular path to a unique infinite sequence of 0's and 1's, corresponding to the sequence of 2n+0 and 2n+1 child nodes in that path.
Note that under the above definition of pathhood only an ACTUALLY INFINITE sequence of 0's or 1's corresponds to a path. No finite such sequence can represent a path.
All of which is perfectly straightforward and proper in all standard versions of mathematics, and is only forbidden by WM in Wolkenmuekenheim. --