Re: Binary Tree and Pairs of Nodes

On 9 Okt., 16:49, LauLuna <laureanol...@xxxxxxxx> wrote:

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

And so on.

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.

Of course there is no "finally". All we can say is that "iff there was
an infinite countable set of nodes" and "iff infinite sets could be
worked through completely", then every node would get its turn. But if
every node of the tree has been used to mark one path, how should any
further path remain? A path without nodes? Not connected to the root
node? Certainly, if set theory was right, then such paths must exist.

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.

Therefore I do not express a criterion, but show that there can no
path start at the root and lead out of the conquered domain without
getting its node. That is a *forcing* logical element.

I think your idea is that whenever a new path parts we have an
available node in the 'conquered domain' to match it with,

That is obvious.

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.

The assumption that all elements of an infinite set can be used is
false, of course. But it is Cantor's assumption and is the basic axiom
of set theory. Therefore I use it here. I do not need anything else
than the axiom of infinity and a counting of the nodes like

0
1
23
4567
....

I do not talk about a limit. I use the "fact" of set theory that all
elements of an infinite set can be used.

So far I can only concede that the apparently constructive nature of
the binary tree is a little baffling to sheer intuition.

Sorry, my game is not intuition. It has fixed rules. Only the choice
of paths is not fixed because that is not of interest.

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.

This shows that there is no set of all reals. It is impossible to
define more than a countable set of reals. There are no uncountable
sets at all. In the parallel thread with the misleading name the
discussion has come to a striking argument:

It is impossible to put all anti-diagonals of all constructed lists in
a list. (This list has an anti-diagonal that is not in it.) But the
set of all those anti-diagonals is countable (if all are formed
according to the same rule). Therefore the non-existence in a sequence
does not show uncountability.

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.

No. See above. "Uncountable" is simply a misinterpretation.

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.

It is conceivable and it is countable (obviously the nodes are). It
simply does never end, as every infinite set. There is no finished
infinity. But that is not directly under discussion. What the tree
shows is the falsity of the concept of uncountability.

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

The tree does it. For instance, if I was forced to give an algorithm
for my game, then I could say: Put the next node on the path that
afterwards always turns left, i.e., on a terminating rational of the
form
0.xxx...xxx000...
with x either 0 or 1.
Obviously the whole tree will be exhausted by this procedure (I
conquer the whole tree). So no path remains. But I could also say: Put
the next node on the path that afterwards always turns right, like
0.xxx...xxx111...
Obviously I could find (countably) many such algorithms. Each one
would exhaust the binary tree and let me win my game.

How can that be? There are not all reals in that tree! But a tree is
merely a picture for the binary, or decimal, representation of the
reals (of [0. 1]). Therefore there are not *all* binary, or decimal,
representations of the reals, not even those of all rationals. The
number of infinite sequences is countable and is much smaller than
expected. And it cannot be decided which form those infinite sequences
have. They do not exist on the Platonist shelf, waiting to be taken by
the mathematicians. They have to be created like the paths in my tree.

Superfluous to say that the number of such creations is not
uncountable.

Regards, WM
.

Relevant Pages

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