Re: Binary Tree and Pairs of Nodes

On Tue, 7 Oct 2008 04:49:57 -0700 (PDT), WM
<mueckenh@xxxxxxxxxxxxxxxxx> wrote:

On 7 Okt., 13:03, David C. Ullrich <dullr...@xxxxxxxxxxx> wrote:
On Mon, 6 Oct 2008 11:25:34 -0700 (PDT), WM

<mueck...@xxxxxxxxxxxxxxxxx> wrote:
A path in an infinite binary tree is the representation of a real
number of the interval [0, 1]. (Terminating rational numbers have two
representations each.)

Two paths A and B in the infinite binary tree can be distinguished by
at least one pair of nodes a and b where a belongs to A but not to B,
and b belongs to B but not to A.

Assumption: For every pair of paths we can find a distinguishing pair
of nodes where no node is used to distinguish another pair of path.

"is used to distinguish" is unclear. I'm going to assume you meant
"distinguishes", or "can be used to distinguish".

The nodes that distinguish paths can be mapped onto those paths, in
the example above a can be mapped on A and b can be mapped on B. The
assumption is that a node mapped on a path is not needed to be mapped
on another path.

Conclusion: The number of paths is not larger than the number of
nodes. The number of nodes is countable. As the number of paths is not
less than the number of reals in the interval [0, 1], the number of
reals in the interval [0, 1] is countable.

Or: The assumption is wrong.

That's correct, the assumption is wrong. (Obviously wrong,
in fact).

There are at least two pairs of paths
that cannot be distinguished by different nodes.

Wow. No, that's not the negation of the assumption.
The negation of the assumption would be that there
is a pair of paths such that every pair of nodes
distinguishing those two paths also distinguishes
another pair of paths.

There is no reason for a "Wow". Your statement implies mine. Therefore
the negation of the assumption implies my statement: There are at
least two pairs of paths that cannot be distinguished by different
nodes.

And that's obviously so. Say P_1 and P_2 are
any two paths, distinguished by nodes a and b.
If P_1' and P_2' are the same as P_1 and P_2
down to the level of a and b but then go off
in different directions from P_1 and P_2 then
a and b also suffice to distinguish P_1' and P_2'.

The question is not about being sufficient, but about being
*required*. Cp. the assumption. In short: Can we establish a bijective
mapping between nodes and paths?

I gather all this started when someone pointed
out some elementary errors in some publication
of yours.

Please stop suspecting things you do not know of - try to understand
my argument.

Sorry, I'm not going to try to understand an argument
that purports to show that there are only countably
many paths in an infinite binary tree. The only possible
reason for that would be to try to explain to the
author where the error is. I've done that - your
comment "Your statement implies mine." is
simply wrong. No point in going further since you
don't seem to have an adequate grasp of simply logic.

But thank you for the atempt. (No, I never tried to
publish the binary tree other than where it has been printed already.
I do not see any necessity to use the limited audience of a printed
medium because the audience of the internet is much broader.)

Giggle. Of course the fact that anyone can say anything
they want on the internet has nothing to do with it.

< You're taking a very curious approach
to fixing that, making hundreds of posts containing
similar elementary errors.

There is nothing to fix. Obviously there cannot be more paths than
nodes in the binary tree.

Obvious or not, it's not so.

Regards, WM

David C. Ullrich

"Understanding Godel isn't about following his formal proof.
That would make a mockery of everything Godel was up to."
(John Jones, "My talk about Godel to the post-grads."
in sci.logic.)
.