Re: Cantor and the binary tree

In article <1118843005.362840.224470@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:

> Martin Shobe wrote:
>
> > >It will assign a node to every set of paths which has its individual
> > >edge.
> >
> > No. It fails to do so. I showed you how to construct a path that it
> > does not assign a node to.
>
> Bt you did not show that it has its individual edge.

What does thatn mean? Every path has infinitely many individual edges.
Butin maximal binary trees no maximal path has any edge that it does not
share with some other maximal path.

> In fact you cannot
> do so, because the set of edges is countable. But paths which have no
> individual edge in the wole tree cannot be distinguished from other
> paths of the same bunch. Hence they and the due real numbers do not
> exist as individual numbers.

WM is blowing smoke again. He is conflating properties of finite trees
with those of maximal trees. In a finite tree, each (maximal) path has a
branch and a node not shared by any other path. In a maximal binary
tree, this is no longer true. No path in such a tree has any branch of
node that is not shared. They are, in a sense, "dense", because, with
only countably many exceptions, "between" any two paths are infinitely
many (uncountably many) other paths.

These (countably many) exceptions are when, from some common node, one
path goes once left then forever right and the other goes once right and
then forever left. These pairings correspond to the reals in [0,1] which
have terminating binary representations. "Most" paths contain both
infinitely many left branchings and infinitely many right branchings and
are no subject to this anomaly.
>
>
> > >Either, there will further separations, and that requires exactly the
> > >same number of separating points (= nodes) to exist,> or there will be
> > >no further separation, and that implies that the uncountable number of
> > >paths cannot be distinguished by any edge (and the due uncountable
> > >number of reals does not differ in any digit).
> >
> > Eh? The infinite paths are not distinguished by any particular edge
> > (or point). Every edge and every point is shared with an uncountable
> > number of other paths. The infinite paths are distinguished by the
> > set of edges (or the set of points) that are on it.
>
> Not those in the tree.

Precisely those in the tree. A path is nothing more that a particular
sequence of nodes and branches, so is distinguished from others by
which nodes and branches appear in it.



> And others do not exist.

WM is confused, The paths not distinguished by which edges (branches)
and nodes are in it are the ones which do not exist. So WM has it
exactly backwards.
>


> Hence, if your path
> separates from each other, it is the only one in its bunch and hence,
> will carry a node. But if it does never separate from each other, then
> it is not an individual worth of getting a node.

I do not know what the above reads like in WM's native language, but in
English it is nonsense.

> >
> > > But why should we call
> > >that set a uncountable?
> >
> > Because there are no bijection between the natural numbers and it.
>
> Small wonder, if it does not exist.

That it does not exist in WM's mind is no valid criterion.
> >
> > > It is obviously only one real number. Or do you
> > >think the individual members of it could be distinguished in a Cantor
> > >list but not in the tree?
> >
> > I don't see a problem with distinguishing individual members of
> > either.
>
> If there is no individual edge in the whole tree, then you cannot
> distinguish your path from others.

Every path is a unique subset of the set of all nodes (or of the set of
all branches). Every two paths, as sets of nodes, have a finite
non-empty intersection and the "last" node in that intersection is where
the paths "separate". WM's "bunches" of paths need not, as described,
have finite intersections so need not separate.
.

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