Re: Review of Mueckenheims book.

On Mar 12, 1:10 pm, mueck...@xxxxxxxxxxxxxxxxx wrote:
On 11 Mrz., 19:35, "William Hughes" <wpihug...@xxxxxxxxxxx> wrote:

On Mar 11, 4:08 am, mueck...@xxxxxxxxxxxxxxxxx wrote:

Consider the set R consisting of exactly two paths:

R= { (0,0,0,...),(1,1,1,...) }

The projection of R is the same as the projection of all paths
so:

The Waft Maximum of R is finite.

Why?

Because the Waft Maximum of R is the same as the Waft Maximum of the
EIT.

You wrote that R contains an infinite path (0,0,0,...). The EIT does
*not* contain any infinite line.

Indeed, R and the EIT are different. However two different
things can have the same Waft Maximum. Note they have the
same projection. If they have the same projection they
have the same Waft Maximum.

<snip>

Now let F1 be: the projection of all paths of the
union of all finite trees:

We have F1 is equivalent to the projection of the EIT.
Therefore since the projection of the EIT contains
infinitely many elements, so does F1.

Of course F1 contains infinitely many paths.



Let F2 = the projection of R.

You mean the projection of

0.000...
0.111...

This projection yields the waft maximum aleph_0.

No F1 has a finite waft maximum. F2 is equal to F1.
F2 has a finite waft maximum.

Which of the the following statments do you disagree with

-every element of F1 is an element of F2
-every element of F2 is an element of F1
-F2 has the same waft maximum as F1

While you ponder that we can go back.

You now concede

The EIT contains an infinite number of elements
The projection of the EIT contains an infinite number of
elements.

Equivalently we have ( U(T(n)) is the union of all finite trees )

U(T(n)) contains an infinite number of nodes
The projection of U(T(n)) contains an infinite number of elements

[So the statements you made on March 2

[T]he number of nodes is finite even for the infinite union
of
finite trees.

The union tree cannot contain an infinite number of nodes

were simply wrong]

We can now go back to the posts that started all this (Feb 28)

M: Now enumerate the nodes of the union tree
M: U(T(n)) by natural numbers.

H: And as the paths are *subsets* of the set of nodes you have
H: to consider the power set of the set of nodes. The fact that
H: the set of nodes is countable does not mean that the power
H: set of the set of nodes is countable.

Let the set of all infinite paths be P_I.

There are a countable number of nodes.
Each infinite path is the union of an
infinite number of nodes. The infinite subsets
of a countable set are uncountable
P_I is a subset of an uncountable set.

or alternately

There are a countable number of finite paths.
Each infinite path is the union of an
infinite number of finite paths. The infinite subsets
of a countable set are uncountable.
P_I is a subset of an uncountable set.

So at this point we do not know if P_I is countable
or uncountable.

- William Hughes


.

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