Re: The complete infinite binary tree has only countably many infinite paths.
- From: WM <mueckenh@xxxxxxxxxxxxxxxxx>
- Date: Sat, 28 Mar 2009 02:11:29 -0700 (PDT)
On 28 Mrz., 09:27, "calvin.ost...@xxxxxxxxx" <calvin.ost...@xxxxxxxxx>
wrote:
Aaargh. The question I am asking is "what IS p(n) (as
a set of edges)?"
p(n) can be understood as a set of edges. It is the union of A and B.
A ist the set of all edges that belong to the intersection of all
paths (taken as sets of edges) that go through node number n.
B is the set of edges that are subsequent to node number n and always
turn left.
It is sufficient to consider only nodes that have value 1, i.e. only
nodes enumerated by an even number n. (When the first path 0.000... is
given.)
Example:
0
/ \
1 2
/ \ / \
3 4 5 6
/ \
.......13 14
....
p(6) contains
A) the edges 0-2, and 2-6, and
B) all edges that subsequently turn left. The first of them is 6-13.
WM claims the entire tree
is a union of a countable set of paths. That is
reaonable. I am trying to get him to formalize this.
So, we have T = union { p(i) : i in Naturals }.
Dear Calvin,
there is no point of formalizing such an obvious truth as I have shown
by my three proofs in the initial posting. But I think that this
thread has yielded a lot of new insights for all parties. Perhaps you
will agree that when you first answered my FOM-contribution, you had
no clue what a multitude of facets the binary tree may hide, won't
you?
Regards, WM
.
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