Re: Cantor Confusion

In article <1174655745.265882.25720@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:

On 22 Mrz., 16:45, "*** T. Winter" <***.Win...@xxxxxx> wrote:
In article <1174563273.137227.294...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>
mueck...@xxxxxxxxxxxxxxxxx writes:

> On 21 Mrz., 16:16, "*** T. Winter" <***.Win...@xxxxxx> wrote:
> > In article <1174481667.234070.21...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>
> > mueck...@xxxxxxxxxxxxxxxxx writes:
> >
> > > On 20 Mrz., 17:07, "*** T. Winter" <***.Win...@xxxxxx> wrote:
> > ...
> > > > > Yes. But the path-lengths are natural numbers, and there is no
> > > > > infinite natural number.
> > > >
> > > > The path-length of a non-terminating path is *not* a natural
> > > > number.
> > >
> > > How then can it be contained in a set which contains only sets of
> > > finite paths?
> >
> > It is *not* an element of a set which contains only sets of finite
> > paths.
> > Pray get your statements correct. A set that contains only sets of
> > finite
> > paths has as elements sets of finite paths. Not paths.
>
> We can form the union of the sets. Then we get a set of finite paths.

Yes, we can do that. And then we get a set of finite paths that does *not*
contain a non-terminating path as an element, and also not the path-length
of such a path as an element. So what now?

This set of paths forms a tree which does not differ by any node from
the complete tree T(oo).

A finite chain of nodes can, by definition, only be a path in a tree in
which it is of maximal length, so no finite chain of nodes can be a path
in a CIBT.

Thus a CIBT either has all paths endless or no paths at all.

Again "contain". But, so what? I fail to see the relevance to the
path-length of non-terminating paths.

To spell it out clearly: If you take the infinite union U(T(n)) of all
finite trees T(n), then you do not have any infinite path.

The you also do not have any CIBT, and in particular, you do not have
T(oo).

> > > why then do you believe that an infinite path (number) is contained
> > > in
> > > the union of sets of finite paths (numbers).
> >
> > It is not an element of the union of sets of finite paths. Why do you
> > think that I do think so? The union of sets of finite paths is a set
> > of finite paths, so there is no infinite path in it as element.
>
> Therefore, after having the union of all paths which are elements of
> finite trees, there is no infinite path.

Wrong. In the first place, paths are *not* elements of trees according to
your definitions. Paths are *subsets* of trees according to them. Pray
use the correct wording.

OK. Subsets.

Second, the union of all paths etc., is *not* a
path, it is just an infinite set of nodes, the same as the union of all
trees. The paths are in it as subsets, as are the infinite paths.

The union of all *finite* paths gives the same infinite set of nodes
as the union of all finite trees. But there is no infinite path
involved / required.

Then there is no infinite tree required.

If a set of nodes forms a tree at all, then every maximal chain of
parent-to-child-edges nodes in /that/ set of nodes is a path in /that/
tree.

The union of all finite paths is an infinite union of finite paths,
but not an infinite path.

The union of all finite paths is the node set of the entire tree.
To get only its paths, one needs to consider special sets of paths.

If one has a family of finite paths in which each path, as a set of
nodes, is a subset of every longer path in the family, then the union of
that family, as a set of nodes, is an infinite path in the tree formed
by the "union" of those finite trees, as sets of nodes.




Pray define "isolated".

Two path are isolated at level k if they have digits m and n with m =/
= n at a level i <= k.

In other words, the paths have the same nodes up to some level i prior
to level k, and at that level i they branch in different directions.

For any finite tree, any path is isolated from /all/ others only at its
leaf (last) node level.

For a CIBT, there are no leaf nodes, so being isolated from all other
paths at any node is no longer possible.

But that is no more marvelous that the density of the rationals or
reals, neither of which allows one member to be "isolated" from all
others of its type.
.

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