Re: Sequences of digits
- From: WM <mueckenh@xxxxxxxxxxxxxxxxx>
- Date: Sun, 30 Mar 2008 07:56:43 -0700 (PDT)
Sequences of digits (modified version, two typos corrected)
In the following we consider only real numbers of the interval [0,
1].
A real number can be represented by an infinite sequence 0.abc... of
digits a, b, c, ... which is usually written in a horizontal line. A
real number of the interval can as well be represented by a vertical
path of the complete infinite binary tree (CIBT), see Fig. 1. The CIBT
contains all binary representations of real numbers of the interval.
By "level" we denote the vertical distance from the root node that has
level number 0.
Level
0 0.
/ \
1 0 1
/ \ / \
2 0 1 0 1
... ............. Fig. 1
The CIBT is uniquely determined by its structure of nodes as well as
by its structure of paths. By definition every binary representation
of a real number is represented as a path of the CIBT. Therefore all
nodes required for that purpose exist in the CIBT too. On the other
hand, there cannot be more nodes because an additional node would
belong to an additional binary representation. But then the CIBT
without this additional node would not be complete, contrary to the
assumption.
[I] Let x represent a digit, 0 or 1, then we can construct
1) a structure S_1 generated by the set of all terminating paths,
i.e., all paths of the form
0.xxx...xxx000... .
2) a structure S_2 generated by the set of all paths of the form
0.xxx...xxx(following an infinite periodical sequence).
3) a structure S_3 generated by the set of all paths of the form
0.xxx...xxx(following an infinite nonperiodical sequence).
It follows that also each of these structures, S_1 or S_2 or S_3,
generates the complete set of nodes and, therefore, each one is the
CIBT that contains the complete set of all infinite paths. As, by
construction, the structure does not contain any other paths than
those used to construct it, there is an unavoidable ambivalence
concerning its paths.
[II] Considering only the outmost right hand path 0.111... of the
CIBT, we see that this path is generated by the infinite set of
terminating paths of the sequence L
0.000...
0.1000...
0.11000...
0.111000...
....
not containing the limit 0.111... of this sequence. On the other hand,
it is impossible to distinguish that path from a path that is
generated by the sequence L including its limit 0.111... .
[III] Path 0.111... is said to be the infinite union of
intersections of terminating paths of L. If so, then there should be
at least two nodes of 0.111... recognizable that are in different
terminating paths but not in one single terminating path.
For any pair, A and B, of nodes of the path 0.111..., however, it can
be shown that there is no pair of paths, p(A) and p(B), such that A e
p(A) & B !e p(A) & A !e p(B) & B e p(B).
There is no single terminating path equal to 0.111... On the other
hand no pair (or n-tuple) of paths can contribute more nodes than one
single path.
[IV] Every pair of paths that can be distinguished from each other
at a level of the CIBT has been separated by a node. We will denote
"paths which can be distinguished" by "lines" in order to exclude the
argument (not of interest, but frequently stated) that every path
consists of uncountably many paths. The fundamental element of the
tree structure is shown in Fig. 2. One line goes in, two lines go
out.
|
x
/ \ Fig. 2
This shows that no separation of lines can occur unless a node (x)
facilitates it. A node is nothing but a branching-off of lines (i.e.,
distinct paths). Therefore there are as many nodes as lines (precisely
there is one line more than nodes). The number of nodes is countable
proving
the number of lines being countable too. Should there be more paths
than lines, then the surplus paths could not be distinguished from one
another (or they could not contain the root node). Impossibility of
distinction in a CIBT would imply impossibility of distinction in a
list (as is used, for example, in Cantor's diagonal argument).
[V] This is but an explication of section IV.
Let [X] be the greatest natural number less than or equal to X, and
let ]X[ be the smallest natural number larger than or equal to X.
At level X the number of lines (i.e., distinguishable paths) is L(X) =
2^]X[.
From 0 to X the number of nodes is N(X) = 2^[X+1] - 1.
For every finite X (and there are only finite X in the whole CIBT) the
ratio of lines and nodes is
L(X) / N(X) =< 2.
For large X the ratio oscillates between the limiting points1/2 and 1.
Rephrasing the statement "the ratio never exceeds 2" is to say "in the
infinite the ratio does not exceed 2", i.e., lim_{X --> oo} L(X) /
N(X) =< 2". This proves that the number of lines in the CIBT is not
uncountable.
Regards, WM
.
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