Date: Dec 12, 2012 12:07 PM
Subject: Re: On the infinite binary Tree

On 12 Dez., 11:26, Zuhair <> wrote:
> WM has presented the idea that the infinite binary tree must have
> countably many paths. It seems that he thinks that the total number of
> paths in a binary tree is always smaller than or equal to the total
> number of nodes. So since obviously the number of nodes in an infinite
> binary tree is countable, then the number of paths must be so. But
> since each real can be represented by a path in a 0,1 labeled infinite
> binary tree, then we must have countably many reals thus defying
> Cantor.

Correct. It is nice that there are mathematician who have grasped it.
The issue is this:
Every partial tree from level 0 to level n has
-1 + 2^(n+1) nodes.
The number of all finite paths is the same, because every finite path
ends at some node.
The number of all path traversing the complete partial tree and
branching after that in two paths is
In the limit we have
numbers of paths traversing the complete tree (and after that
branching in two paths) divided by number of nodes of the complete
tree = 1.

> The 0,1 labeled infinite binary tree he is speaking of is the
> following:
>      0
>    /    \
>   0     1
>  / \    / \
> 0  1 0  1
> .  .  .     .


> Answer: in order to justify this claim one must first see what is
> happening at the finite level. And make a simple check of the number
> of nodes and the number of paths through those nodes. If that
> confirmed that the total number of nodes is more or equal to paths
> then there would be some justification for that claim.
> Let's see:
> Lets take the first degree binary tree which is the following
> 0.  One node, no paths

We could say 1 degenerated or empty path. Set theorists like empty
> The second degree binary tree is:
>  0
> /  \
> 0 1
> This has three nodes and two paths.

or three nodes and three paths, respectively.
> Lets take the third degree binary tree
>      0
>    /    \
>   0     1
>  / \    / \
> 0  1 0  1
> Now this has 7 nodes BUT 8 paths, those are
> 0-0
> 0-1
> 1-0
> 1-1
> 0-0-0
> 0-0-1
> 0-1-0
> 0-1-1
> Of course I'm treating paths as directed paths away from the root
> node, and so a path contain edges linking nodes in the SAME direction.

You should count them again. There cannot be more endings of paths
than nodes, namely 7.
> IF we remove the directional requirement matters would really differ,

but that does not matter because the real numbers in the unit interval
always start with "0."

> However we will not adopt that,

That's a good idea.

> As the degree of the binary tree increase the total number of paths
> increases more and more relative to the total number of nodes, as
> shown above this begins with the third degree binary tree (8 paths
> with 7 nodes).
> So the impression we have from the finite world is that of paths being
> actually MORE in number than nodes in any binary tree of a degree
> higher than 2.

You made an error, but your arguing is very sconsistent. If we are to
believe that there are more paths than nodes, then we should be able
to prove this from the finite world. Because also Cantor's list is
completely caught in the finite world. No line is "infinite".

I am extremely satisfied that you recognize these facts. Your little
error is not important. On the contrary, it is helpful. Without having
made this error you hardly would have dared to publish your sober but
extremely non-matheological thoughts.

> So there is NO justification at intuitive level for the idea that the
> number of paths of a binary tree must be smaller than or equal the
> total number of nodes in it.

Why do you say intuitive level here? Every sober mathematics
calculates limits from only the finite terms of a sequence as you have
attempted here. That is the way mathematics runs. To call that
"intuitive" and to call the infinitary nonsense of matheology
mathematics is really the height of impudence.

Regards, WM