Re: Review of Mueckenheims book.
- From: mueckenh@xxxxxxxxxxxxxxxxx
- Date: 15 Mar 2007 04:05:23 -0700
On 12 Mrz., 20:41, "William Hughes" <wpihug...@xxxxxxxxxxx> wrote:
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.
The have not the same Wat Maximum. Look up the definition.
Now let F1 be: the projection of all paths of theunion 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.
How do you see that? LSD, ethanol? If you continue with such non-
technical nonsense, I will stop this discussion.
Regards, WM
.
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