Re: Binary Tree and Pairs of Nodes

On Oct 18, 7:01 pm, WM <mueck...@xxxxxxxxxxxxxxxxx> wrote:
On 9 Okt., 16:49, LauLuna <laureanol...@xxxxxxxx> wrote:
Please, let me make try to make again my point even if a bit too late.
Please, look at this.

                            # (1)
                            |
                            * (2)
                          /    \
                         0(3) 1(4)
                       /    \   /   \
                    ...    ... ...   ...

The symbols # and * are just two initial symbols. Inside the
parentheses are the ordinal numbers of the nodes:

Match node 1 with path #*111...
Match node 2 with path #*000...
Match node 3 with the unique there departing path that is so far
unmatched, i.e. #*0111...
Match node 4 with the unique there departing path that is so far
unmatched, i.e. #*1000...

etc.

This algorithm ensures that at any finite stage of the construction
each node will be matched with just one path and that at each
bifurcation we have new nodes to tag new paths so that we can go to
the next node without having left behind any already departed paths.

This is WM's construction and he is right in saying that no departing
path at any finite stage will go untagged.

So it is. Thank you for recognizing it.

Virgil and the rest are right in saying that some paths, like
#*101010..., will never get a node.

Why? This path is the union of all its nodes (and edges, but every
edge can be mapped on the node following it), not more. If every node
belongs to one of the mapped paths, then the whole path 101010...
belongs to the mapped paths.

By construction any labeled path will have either all 0's or all 1's
after a certain digit. But yes, if the algorithm gave us in the limit
the whole tree, any path would have been tagged. This shows that the
algorithm doesn't do so.

You are assuming that it does because you are assuming that the binary
is constructive and that's why I say that the tree you have in mind is
not the same as the tree other posters are referring to.

And again my point is that the case of the binary tree is somewhat
different from other ways of describing the set of all reals, because
it really looks like a construction, so that one could come to think
that, if it is not such a construction, then it is plainly
incomprehensible.

If this were so, one should wonder whether and how this affects the
intelligibility of other ways to describe the set of all reals.

Regards

If the complete infinite binary tree is a progressing construction
plus a passage to the limit, then WM must be right that, in the limit
(if there is such), nodes and paths are paired one-to-one since they
are so paired at all stages of the tree's development.

But this is obviously not the case, as the counterexample shows. So
the complete infinite binary tree is not the outcome of a passage to
the limit over a progressive construction.

What WM, and no one else (as it seems) is assuming, is that the binary
tree can only be conceived of as such a progressive construction.

I assume that the path 101010... is complete, i.e., that all its nodes
make up this path. And every node belongs to at least one path mapped
by your rule.



But, what other kind of thing could it be? Or equivalently, how could
we envisage it if it is not by means of the construction-recipe plus
the passage-to-the-limit-instruction?

So, let me claim again that the complete infinite binary tree might be
unintelligible...

because there is nothing more than all nodes, and not even they are
complete.



This could be part of a case for some constructivist claim that
infinite sets are really comprehensible if and only if they are
recursively enumerable.

They are really incomprehensible because they are in contradiction
with the reality of the tree, i.e., in contradiction with mathematics.

Regards, WM- Hide quoted text -

- Show quoted text -

.

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