Re: Cantor Confusion

In article <1168005688.719735.164370@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:

Virgil schrieb:

In article <1167860527.440727.143820@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
mueckenh@xxxxxxxxxxxxxxxxx wrote:

*** T. Winter schrieb:


> But you did not say which edges or nodes distinguish the complete
> tree
> from the union of all rational trees.
> Do you believe that there are such distinguishing edges but that one
> cannot name them?
> Or do you think that there are same edges in both trees but that
> some
> paths can form in one tree which cannot from in the other tree?

The last one. In the union there is *no* infinite path.

> Wouldn't both answers point to some matheology?

No, they are based on the definition of union and some elementary
logic.
(1) Each node is in one of the finite trees, so it is also in the
union.

So there is an infinite number of nodes in the union of finite numbers
of nodes.

Only if WM can prove that one can generate the necessary infinite number
of trees from a merely finite number of nodes. Which I doubt.

So the union of all finite trees has ony a finite number of nodes?

(2) Each edge is in one of the finite trees, so it is also in the
union.

So there is an infinite number of edges in the union of finite numbers
of nodes.

Only if WM can prove that one can generate the necessary infinite number
of trees from a merely finite number of edges. Which I doubt.

So the union of all finite trees has ony a finite number of edges?

WM may believe that, but that does not follow from anything that I said
or from anything logical.

The union of all finite trees does not contain any infinite path but it
does contain infinitely many finite paths, as well as infinitely many
nodes and infinitely many edges,.

That is wrong. If an infinite number of nodes is in any path of the
union, then the path is infinite too.

But it is impossible for any finite path in any finite tree to contain
more than finitely many nodes, or edges.

Fine. That is my arguing with respect to the natural numbers. Their
number is not finite.

As no path in any set of the
union has infinitely many nodes, and the union can contain only those
paths, from which finite tree is WM getting his infinite path?

Fine. Infinity does not exist. (Because the complete tree cannot
contain more than the union of all finite trees.)

That infinite union of finite trees contains infinitely many nodes and
infinitely many edges in infinitely many finite paths, but no infinite
paths.


If EVERY edge of a path is shared by another path, then both paths
cannot be distinguished.

There is no "both paths" until one has chosen both paths to be
compared, at which point for those two one edge in each not in the other
is easy to find, the two edges branching from last node they have in
common.

There is a both paths, if you claim that no edge is unique.

Every edge is, in some sense, unique, but I have no idea in what respect
WM is requiring some edge to be unique.

In the sense that if two paths are distingishable there is some edge in
one but not in the other, fine. For any two infinite paths there are
infinitely many such "unique" edges, in one but not the other.
.

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