Re: Review of Mueckenheims book.

On 2 Mrz., 19:31, "William Hughes" <wpihug...@xxxxxxxxxxx> wrote:

Recall: We construct the union tree from finite trees. Finite trees
have finitely many nodes. Finitely many nodes are not capable of
forming infinitely many paths.

M: There are only a countable number of nodes.
How can we form an uncountable number of infinite paths

Can't you understand or won't you understand?

A finite number of nodes cannot be used to construct an infinite
number of different paths.

Yes O No O

All finite trees have finite numbers of nodes which for the union of
finite trees do not add but are the number of the larger tree.

Yes O No O

Therefore the number of nodes is finite even for the infinite union of
finite trees.

Yes O No O

This is because the natural numbers are finite, even if we consider
infinitely many of them.

Yes O No O

What is difficult about that?

H: The union tree is constructed from an infinite number of
finite trees. The union tree contains an infinite number
of nodes.

The complete tree may contain an infinite number of nodes. The union
tree cannot contain an infinite number of nodes, because this would
mean the existence of an infinite natural number.

Do not mistake: There are infinitely many trees in the union, but
every tree has a finite number of nodes. And in the union of two or
more trees there is not the sum of nodes of the trees but only the
number of nodes of the largest tree of the union.

This is a valid approach to construct the union tree.

Yes, and you end up with an infinite number of nodes.

Not in the union of finite trees. You end up with an infinite number
of trees. But the number of nodes in that union is finite.

OK the infinite path p(oo) consists of the union of
all finite paths having only nodes with value zero.

So it is said. As the nodes of the finite paths must be taken from
finite trees, their number is finite though not bounded.

i: each finite path has a number of nodes which
is finite but not bounded

For each finite path he number of nodes is finite and bounded.

For the union of finite paths the number of nodes is finite but not
bounded.

ii: p(oo) is the union of finite paths

iii: p(oo) has a number of nodes which is finite but not bounded.

That has to be figured out.


Since no one claimed this there is no problem. What was claimed
is that an infinite number of paths, each of which has a finite number
of nodes with value zero, can yield a single infinite path with an
infinite number of nodes with value zero.

Maybe. But there is no infinite number of finite paths.

I hope you will not switch to the claim now that there were natural
numbers of infinite size because there are infinitely many of them?

No. But there are infinitely many of them, so the *set* of
natural numbers is of infinite size.

Correct. But completely irrelevant in this context.

iii does not follow from i and ii. Therefore saying that the *set* of
natural
numbers is of infinite size does not mean that there is a natural
number of infinite size.

Correct. But a finite number of nodes cannot supply an infinite number
of paths.

Your claim is that *any* union of trees is a tree.

Yes, but all are finite.

Let T be the union of all finite trees.

iii does not follow from i and ii. T is the union of an
infinite number of finite trees and therefore has an infinite number
of nodes.

Wrong. This would mean the infinite union of natural numbers contains
an infinite number.

Remember: The union of finite trees is a finite tree, namely the
largest one in that union.

Regards, WM

.

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