On Mar 24, 2:51 pm, fom <fomJ...@nyms.net> wrote: > On 3/24/2013 4:34 PM, WM wrote: > > > > > > > > > On 24 Mrz., 21:29, Virgil <vir...@ligriv.com> wrote: > >> In article > >> <729f073f-8948-4eb9-991a-2bd249ac5...@c6g2000yqh.googlegroups.com>, > > >> A binary tree that contains one path of each positive natural number > >> length will necessarily also contain exactly one path of infinite length. > > > Like the sequence > > 0.1 > > 0.11 > > 0.111 > > ... > > that necessarily also contains its limit?
A binary tree that contains one path, of all zero-branches, of each finite length, will necessarily contain a path of 0-branches of infinite length.
Put the j'th node b of the i'th tree (in breadth-first order) at (j, 2i + b). Rays through countable ordinal points (x = w, y = 0, ..., 2^w-1) are dense in the paths. (And rationals are dense in the reals, and countable.)
In fact, these rays through countable ordinal points correspond 1-1 to paths in the tree, then with regards to anti-diagonalization of the resulting path in the ordinal's natural order: the anti-diagonal, as it were, is only the path with all 1-branches, i.e. to ordinal 2^w. Then that would be rather steep with regards to eventually defining the quadrant in that manner (for the rays from 0 to w defining the octant and from w to 2^w the quadrant).