Re: Review of Mueckenheims book.

On Mar 2, 12:46 pm, mueck...@xxxxxxxxxxxxxxxxx wrote:
On 2 Mrz., 16:45, "William Hughes" <wpihug...@xxxxxxxxxxx> wrote:



A finite number of nodes does not suffer to construct an infinite
number of paths from them. Look here:
from three nodes
ooo
you can construct three paths
o
oo
ooo

Now let n be any finite number of nodes, then you can prove that there
are not more than n paths possible, can't you? We assumed an infinite
union of *finite* trees. The latter condition guarantees a finite
number of nodes for each set of paths to be costructed from.

Recall: each set of paths used to create an infinite path
consists of an infinite number of finite
paths.

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

H: Each infinite path contains an infinite number of node.

M: We construct the infinite paths from finite paths.
Each finite path contains a finite number of nodes.

H: Each infinite path is contstructed from an infinite number
of finite paths. An infinite number of finite paths contains
an infinite number of nodes.

M: We construct the infinite paths from initital segments.
Each initial segmentment is formed from a finite number
of paths. Therefore each initial segment contains
a finite number of nodes.

H: Each infinite path is contructed from an infinite number
of initial segments. An infinite number of initial seqments
contains an infinite number of nodes.

M: We construct the union tree from finite trees. Each finite
tree contains a finite number of nodes.

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

This is a valid approach to construct the union tree.


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

Or isn't it? Why
not? I do not say that your postulate is incorrect. But different
results of different legal approaches show an inconsistency.

Where? Not in the union of finite trees.

At this point we are just determining (as you requested) the
number of unions of finite paths. This we have done.

You have postulated it. I have shown from the finity of every tree
that your postulate leads to another result than the fundamental
approach used by me.

It is assumed, by ***,
that the infinite path p(oo) is the union of
all finite paths p(n) belonging to p(oo), i.e., having only nodes with
value 0.

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

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


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

Recall: There
are infinitely many trees but every one has a finite number of nodes,
i.e., not omega or aleph, because that would mean an infinite number.
(Recall: There is no natural number omega or aleph_0.) And as I showed
you previously (see above for a simple example, which, however,
exemplifies the things best), a finite number of nodes with value zero
cannot yield an infinite number of paths with only nodes with value
zero.

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.




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.

i: each natural number is of finite size

ii: the set of natural numbers consists only of natural numbers

iii: the set of natural numbers is of finite size

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.

You see, the same is valid for the trees. I repeat: There are
infinitely many of he finite trees T(n), but no one has infinitely
many nodes (but only 2*2^n - 1). As each one is the union of all
smaller trees, there cannot be a tree with infinitely many nodes.
Hence there cannot be a tree with infinitely many paths.


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

Let T be the union of all finite trees.
T is a tree.

i: each finite tree has a finite number of nodes

ii: T is contructed from finite trees

iii: T has a finite number of nodes.

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.

- William Hughes

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