Re: The complete infinite binary tree has only countably many infinite paths.

On 24 Mrz., 16:57, William Elliot <ma...@xxxxxxxxxxxxxxxx> wrote:
On Tue, 24 Mar 2009, WM wrote:
The complete infinite binary tree has only countably many infinite
paths.

0.
/ \
0 1
/ \ / \
0 10 1
...
A) Construct the binary tree starting from a "tree" that has only one
path, say p_0 = 0.000...

Error corrected (0. belongs to the following figure, not to the number
above):
0.
|
0
|
0
|
0
...

Add all paths that end by infinitely many zeros. Every path that you
add must start at a node of p_0 or at a node of a path already
constructed. The number of paths in the tree grows by not more and not
less than 1 when 1 path is added. However, after completing that
procedure all nodes and every infinite sequence of bits (including the
path 0.111...) is present, namely represented in the infinite binary
tree.

Extreme vagueness. If at each node without a branch, you added another
infinite series you will have countable many series. You repeat this
again, infinitely.

The nodes can be enumerated:

0
1 2
3 4 5 6
....

The connecting line between the root node 0 and node number n is
uniquely determined in the tree. Now proceed as follows for n = 1, 2,
3, ...:
Connect the root node and node number n and after having passed node
number n continue by zeros only. Then you have constructed a path
starting at 0 and passing trough node n and subsequently having
infinitely many zeros. For every n this path is defined, therefore we
have a construction of all paths that end by infinitely many zeros.
(This construction contains many paths more than once. For instance
there are many paths passing trough all nodes 0 at the left hand side
of the tree. Nevertheless all paths belong to a countable set. All
nodes of the tree are in the construction. No sequence that can be
written by digits in Cantor's list is lacking a representation as a
path in the binary tree. Every sequence of bits is in the constructed
tree.)

B) The complete infinite binary tree can be constructed by an infinite
agglomeration of elements of the form
|
o
/ \
The number of distinct lines is increased by 1 by 1 node.
(lines going out - lines coming in - nodes) = (2 - 1 - 1) = 0
This procedure, even when applied infinitely often, cannot but yield
the result 0 respectively a countable number of lines. The set of all
lines limits the set of all distinct paths.
The construction is possible in (B) as well as in (A) because the
elements used for construction is countable infinite.

Are you writing mathematics or science fiction?

I leave science fiction like finished I leave to set theorists. A
construction is done when all elemnts of a countable set have their
fixed position in a sequence. An example is the construction of an
enumeration of all algebraic numbers.

C) Consider the edges of the complete binary tree (an edge connects
two subsequent nodes of a path). Take all edges and put them on one
and the same level of the tree, side by side, such that the "tree" now
is an array of parallel edges: |||||||... This array limits the number
of possible paths of the tree. It is an upper limit, because every
path there has only one edge. And there is no further edge remaining
to distinguish any further paths.

You're writing science fiction.

Do you object to infinite sequences for principle reasons? Then you
have my sympathy. But here in sci.logic we must pretend that we
believe in infinite sequences. Every countable set can be put in an
infinite sequence. Therefore the edges can line up according to the
given order..

Regards, WM
.

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