Date: Mar 24, 2013 6:13 PM
Author: ross.finlayson@gmail.com
Subject: Re: Matheology § 224

On Mar 24, 2:51 pm, fom <fomJ...@nyms.net> wrote:
> On 3/24/2013 4:34 PM, WM wrote:
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> > On 24 Mrz., 21:29, Virgil <vir...@ligriv.com> wrote:
> >> In article
> >> <729f073f-8948-4eb9-991a-2bd249ac5...@c6g2000yqh.googlegroups.com>,

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> >> A binary tree that contains one path of each positive natural number
> >> length will necessarily also contain exactly one path of infinite length.

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> > 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).

Regards,

Ross Finlayson