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>,

>

> >> 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).

Regards,

Ross Finlayson