Re: Cantor Confusion


*** T. Winter schrieb:

In article <1169547646.733664.94830@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
> *** T. Winter schrieb:
> >
> > > If the union of singletons {1} U {2} U {3} U... U {n} U ... is an
> > > infinite number omega, then the union of domains of sequences {1} U {1,
> > > 2} U {1, 2, 3} U... U {1, 2, 3, ..., n} U ... is the domain of an
> > > infinite seqeunce {1, 2, 3, ...}.
> >
> > Pray explain what you understand under "domain of", and what you understand
> > under "union of domains". That is complete non-standard terminology.
>
> The domain of a sequence is a natural number. The domain of an infinite
> sequence is omega.

That is not an explication. That is obfuscation. That is, it is still
clear as mud.

It is from a book on set theory. The domain of a sequence is also
called its pre-image, in German Definitionsbereich. I don't know f
other Enlish words for it.
> > But
> > whatever, whenever in a collection of sets A_k, none of the sets A_k
> > contains a particular element E, it is also not in the union.
>
> Therefore there is no uncuntable set of path in the union tree. But
> there are all paths in the union trees.

Conclusion is false.

Summary

1) Every complete infinite binary tree T (containing all nodes and
edges) contains all paths.
2) The union tree T(oo) of all finite trees is well defined (as I have
shown elsewhere) and yields the complete infinite binary tree
containing all nodes and edges: T = T(oo).
3) The union of all finite trees includes the union of all nodes and,
with it, the union of all such subsets which are paths (because every
path is a well defined subset of the set of nodes if the structure of
the tree is well defined).
4) The set of paths in T(oo) is a subset of the countable set of finite
sets of all paths in the finite trees.
5) A countable union of countable sets is a countable set (according to
ZF with AC).
==> The set of all path is countable. (==> The real numbers are
countable.)

Going on, we can say:

6) T(oo) = T contains only finite paths.
7) T(oo) = T contains all paths including all infinite paths.
==> There are no infinite paths. (There are no irrational numbers.)



> Right. It is the union of all nodes (and edges and levels and paths).

Wrong. If you state that T1 is the countable union of all finite sets of
finite paths, it is a set of paths, and so does not have nodes, edges or
levels as element.

> > > And T1 = T2.
> >
> > No. T1 is a set of paths, it is not a tree. T1 does not contains nodes
> > or edges as elements. Only paths.
>
> The nodes can be enumerated in various ways.

That does not matter. T1 as you defined above is a set of paths.

No. I *define* T(oo) is an ordered set of nodes, the order being
expressed by the edges.

*Not*
a tree. And a set of paths does not have nodes or edges as elements, it
has paths as elements.

The union of paths is a set of nodes as a path is set of nodes.

And if you consider a path as a set of node, then
T1 *still* does not have nodes as elements but sets of nodes.

T(oo) is the union of the T(n) which are sets of nodes. Therefo T(oo)
is a set of nodes.

> The union of trees is a union of their nodes. The paths are merely
> subsets which are defined by the nodes and the special kind of tree.

But you have to be careful when uniting sets of paths, that is uniting
sets of sets of nodes. That union contains all paths from both tree,
you can only omit duplicates from that union, so only paths that are
*identical*, i.e. consist of the same set of nodes.

The union of trees is defined by nodes! Paths are maximal subsets
following some prescription. Therefore in long trees there are no short
paths, but there are long paths only.

Suppose {a, b, c}
is a path in tree T1, and {a, b, c, d, e} is a path in tree T2, than
the union of the sets of paths of T1 and T2 contains *both* paths.

But not the union of the elements (nodes a, b, c, d).


>
> The domain of the path 1,0,1 is 1,2,3.

Still not clear. You can not define things by example.

Here is a quote from Hrabacek and Jech, p. 55: We begin with some new
terminology. A sequence is a function whose domain is either a natural
number or N. A sequence whose domain is some natural number ne N is
called a finite sequence of length n and is denoted. (By the way,
remembering an old discussion of ours, they speak of "adic" numbers,
not "ary".)

> A path is a sequence. A sequence
> is defined as a mapping f from a natural number or omega into the
> reals. The set of reals (here only 0 and 1) is called the range.
> Abbreviations are usual: dom f and ran f.

Finally now you define a the domain of a sequence. So it is now also
clear that a path is an ordered set of nodes. And the domain is the
initial subset of the natural numbers that terminates at the cardinality
of the path (as set of nodes). Still if T1 contains the path {a, b, c} and
T2 contains the path {a, b, c, d, e}, the union of the sets of paths contains
both.

But not the unin of the trees T1 and T2. Tghe union contains only the
path {a, b, c, d, e}.

> > > The tree is an (ordered) set of nodes. The nodes can be used to form
> > > paths, i.e., subsets of the tree.
> >
> > As a tree is a set of nodes, any subset of it is a set of nodes.
>
> Of course, therefore the subsets which are paths must exist in the
> tree if the subsets exist there.

Ok, if a path is defined as a set of nodes. But still see above about
the union of sets of paths.

> >Not
> > every subset of it is a path. But whatever, the number of subsets of
> > the final tree is not countable. The union of the sets of paths of
> > the finite trees is not the set of paths of the complete tree.
>
> If all nodes and edges are in the union, then all subsets are in the
> union too.

You misread again. The set of subsets of the complete tree is *not*
equal to the union of the set of subsets of the finite trees.

No. The set of subsets (which are maximal, i.e., which are paths) of
the complete tree is only a subset of the union of the set of subsets
of the finite trees

The
reason is clear: that union does not contain an infinite subset, while
the complete tree *does* contain infinite subsets.

The union contains all we have. There is nothing remaining.

Regards, WM

.

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