Re: Cantor Confusion


*** T. Winter schrieb:

> > >
> > > 0.
> > > / \
> > > 0 1
> > > / \ / \
> > > 0 1 0 1
> > > | | | |
> > > 0 1 0 1
> > > | | | |
> > > 0 1 0 1
> > > ...........................
> This finite tree I call: Trauerweide

Ok. But this does not help either. The complete tree contains the
path 0.0101010101...; none of the 'finite' trees contains that path.
So that path is *not* in the union of the sets of paths of the finite
trees, while it is in the union of trees.

So you say 0.0101010101... is in the union of finite trees while ohers
say it is not?
It cannot be in the union of the finite trees, they say, because it is
not in one of the finite trees. But *all* paths are in the union of the
finite trees, because it is simply silly to suspect different sets of
paths in a tree with all nodes. Therefore the only conclusion is that
the path 0.010101... does not exist.

>
> The answer is: We can use sets of levels, sets of edges, sets of nodes
> or sets of paths. Look at the "Trauerweide".

Yes. But still the union of the sets of paths in the finite trees is not
the set of paths in the complete tree. There are paths in the complete
tree that are in *none* of the finite trees and so not in the union of those
sets of paths.

So you say. If you were right, there should be some indication for the
difference.
What makes the path 0.010101... be in a tree with all nodes but not in
a tree with all nodes?

> > > Which digit is missing?
> >
> > None. But all sets of diagonals of finite squares contain only finite
> > diagonals, so there can not be an infinite diagonal in their union
> > (which is a set of infinite size).
>
> So there cannot be an infinite diagonal at all?

I do not state that. I state that in the union of the sets of diagonals
of the finite squares only finite diagonals do exist (infinitely many
of them). This does *not* prevent an infinite diagonal to exist.

What can we do to switch its existence on and off?

> The Equilateral Infinite Triangle (EIT) contains all lines which can be
> enumerated by natural numbers:
>
> 1 0.1
> 2 0.11
> 3 0.111
> ...........
>
> In addition it contains the complete diagonal 0.111... (Otherwise,
> Cantor's diagonal proof would fail.)

Right. But the complete diagonal is not in the union of the sets of
diagonals of the finite triangles.

Where is it? Or is it nowhere?

> The union of all finite binary trees contains all levels which can be
> enumerated by natural numbers:
>
> 0 0.
> / \
> 1 0 1
> / \ / \
> 2 0 1 0 1
> ...............................
>
> But it does not contain the complete path 0.111... (Otherwise Cantor's
> proof of the uncountability of the real numbers was contradicted).

It does contain that path (according to your definition above).

Yes, that is obvious. But, for a moment, switch to the cut trees,
please. Does the union of all cut trees contain the paths 0.111...? You
would say no.

But
the union of the sets of paths in the finite trees is not the set of
paths in the infinite trees. The first does not contain the path
0.01010101..., but it is in the second.

If this path exists, then it must be constructible from all the nodes
present. There is a bit, 0 or 1, for every index n in N. That is not
sufficient?

> > > There are more limits than sequences?
> >
> > Eh? Why do you think so? Each sequence has only a single limit (when the
> > limit is properly defined and when it exists).
>
> Correct.

So why your question?

There are only countable many sequences.

> > > What we need is: All paths
> > > in the union tree are countable. All possible nodes are in the union
> > > tree.
> >
> > Yes, you desperately need it. I do not need it, but you do.
>
> The countablity of all paths in the union of finite trees is a theorem
> of set theory.

No, it is not. Prove it. You use the *false* assumption that the
union of the sets of paths in the finite trees is the set of paths
in the complete tree. It is not because the first does not contain
the path 0.01010101... .

I use the fact that the union of all nodes contains the nion of all
paths.

> > > A path can only then differ from all paths of the union tree if
> > > there is another node (or if it is running in other directions --- try
> > > that argument?).
> >
> > What is the argument here? I think you are back to the old argument, that
> > is not valid either.
> > (1) For each pair of paths there is an edge were they differ.
>
> in the union of all finite trees. Accepted.
>
> > (2) For each path there is no edge where it differs from all other paths.
>
> in the union of all finite trees. Accepted.

But you state above "A path can only then differ from all paths of the
union tree if there is another node". What do you mean with that?

I mean that in an infinite tree (= union of all finite trees) with all
nodes, there is a set of paths. Paths which are not therein, must
indicate why they are not therein - not be a spell but by a bit (or
node).


(1) and (2) together state that while all paths are different, there
is *no* single edge that makes a single path stand out from all other
paths.

But paths which are not determined by the set of all nodes must stand
out by some indication, or they are determined by the set of all nodes
- or they are existing in imagination only.

Regards, WM

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