Re: Review of Mueckenheims book.

On Mar 11, 4:08 am, mueck...@xxxxxxxxxxxxxxxxx wrote:
On 10 Mrz., 23:03, "William Hughes" <wpihug...@xxxxxxxxxxx> wrote:



On Mar 10, 10:32 am, mueck...@xxxxxxxxxxxxxxxxx wrote:
The following two statments are true:

The Waft Maximum of the EIT is finite.
The projection of the EIT contains an infinite number of elements.

Congratulations. You got it. (If you only apply this wisdom to the
paths.)

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.


The projection of R contains an infinite number of elements.

Obviously you need some more tutorial. Here it is:

The EIT

1
12
123
1234
...
123456789
...
123.......................n
...

has a WaftMaximum less than omega, as you said.

This projection is equivalent to the projection of all paths of the
union of all finite trees:

OK.

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.

Let F2 = the projection of R.

Note, the elements of F1 and F2 are not projections
of paths, but projections of nodes contained in paths.

Every element of F1 is an element of F2.
Every element of F2 is an element of F1

Therefore F2 = F1

Since F1 contains an infinite number of elements, we conclude that
F2 contains an infinite number of elements.

Therefore R must contain at least one infinite path.


Stop! Don't hurry. We have time. We were talking about the paths of
all finite trees

No, we were talking about paths of the *union* of all finite
trees.

The question is "does the fact that the Waft Maximum of a set of paths
is finite mean that the set of paths does not contain an infinite
path".
The answer is no.

Restoring the question you do not want to answer.

What bit of "the fact that the Waft Maximum is finite does
not mean there is no infinite path" do you fail to understand?

- William Hughes

.

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