Re: Binary Tree and Pairs of Nodes

On Oct 8, 2:54 pm, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:
On 8 Okt., 13:58, David C. Ullrich <dullr...@xxxxxxxxxxx> wrote:

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.

But it is true, as you can see below without stretching yourself too
far, I think.

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.

You talked about a path and *every pair of nodes*. That implies at
least one pair of nodes, i.e., my statement, doesn't it?

 No point in going further since you
don't seem to have an adequate grasp of simply logic.

Not adequate to what you may misunderstand as 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.

I published the binary tree already in printed form. I do not belong
to the set of people who publish every minute idea at 10 different
places.



< 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.

It is very easy to see that there are not more paths than nodes in the
binary tree. For the sake of simplicity we will add one node to start
with. This node is mapped on one of the infinite paths, no matter
which one, say that one always turning left, i.e., 0.000....

   o
   |
  /
 /
/
...

The next node is mapped on another path, no matter which one, say one
always turning right, i.e., 0.111...

   o
    |
   o
  /   \
 /     \
/       \
...

The next node is mapped on another path, no matter which one, say one
always changing direction, i.e., 0.010101...

     o
      |
      o
     /  \
    o    \
   /   \    \
  /     /     \
 /     \       \
/     /          \
...

And so on.

Drawing gets difficult with increasing number of nodes, because space
becomes narrow. But perhaps you have understood already, that every
infinite path that exists in the infinite binary tree has one node
mapped onto it.

Of course we could also have selected the path pi-3 or any other one
to start with. There is no path existing in the tree being left out by
the mapping.

Regards, WM

Your game is just a countable sequence of choices, a countable choice
function. We have no guarantee that such a device will 'finally' give
us the entire tree.

The question is not whether we can play your game and lose, the
question is whether there is a winning strategy, that is, a definable
pairing function. There is none. Try to construct it and you'll find
that paths are shaped in more ways than any finitely expressible
criterium could take into account.

I think your idea is that whenever a new path parts we have an
available node in the 'conquered domain' to match it with, so that "in
the limit" we have a sufficient supply of nodes to tag all paths in
the tree. You are assuming that whatever holds for all finite initial
portions of the tree holds for the whole tree. The assumption is
unwarranted.

So far I can only concede that the apparently constructive nature of
the binary tree is a little baffling to sheer intuition. It seems that
we can construct it step by step just as we can generate the naturals;
the tree seems to be the outcome of a countable sequence of acts, so
that a countable choice function would give it all to us.

But this need not be so, for if we try to define a particular
bijection between the set of all paths and the set of all nodes, we
necessarily fail. This shows that the passage to the limit doesn't
work.

Perhaps (only perhaps) it could be argued that the complete infinite
binary tree is unintelligible since it can obviously be constructed
step by step by a mathematician following finite instructions
(provided with a potentially infinite supply of paper etc.) and yet it
cannot be taken to be the outcome in the limit of such an operation
because 'somewhere' in its development the tree jumps into the
uncountable.

Or equivalently: maybe, if the tree is conceivable at all, it can only
be conceived of as the potential output of such a task and at the same
time it cannot be conceived of as such output since the potential
output of any conceivable algorithm is countable. Then (maybe) we must
infer that it is not conceivable.

As far as I can see, this would (perhaps) amount to refuting the
existence of the set of all reals.

Best.
.

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