Re: Review of Mueckenheims book.

On 7 Mrz., 18:39, "William Hughes" <wpihug...@xxxxxxxxxxx> wrote:
On Mar 7, 11:37 am, mueck...@xxxxxxxxxxxxxxxxx wrote:



Note: There are an infinite number of lines in the EIF.

I pointed jus that fact out to you for many times. There is no chance
to have infinitely many increases by one without getting into the
infinite by size too.

Your usual confusion between "infinite number of lines"
and "infinite line". The statement "there are an infinite number
of lines in the EIF" is true. This does not mean and I did not say
"there is an infinite line in the EIF". The statement "There is no
chance
to have infinitely many increases by one without getting into the
infinite by size too." is false.

Fine. And I accept this position (though it is false). But below you
seem to support just this arguing that the WM(EIT) depends on the
number of lines while it depends *only* on the sizes of the numbers in
the lines (cp. the examples I thought up for you).

But you always refused.

The Waft Maximum of the lines is <oo

Which part of "the fact that the Waft Maximum of the projection is
less than oo does
not mean that the projection is finite" do you fail to understand?

What do you mean by "the projection is finite"?

That the projection contains finitely many filled positions.

Fine. That is what I wanted to hear. (You call it "projection". The
correct name is Waft Maximum.)


The projection
projects infinitely many numbers. It yields one WM.

And as the Waft Maximum is completely irrelevent, who cares?

The Waft Maximum is defined as that what the projection does - above
you say "contains". That is wrong, according to the defintion of WM
given its the inventor. The projection is an application. The WM is
the result. However, in principle, you have absorbed the the correct
idea.


How do you define
"the projection is finite"?

How many I's in the projection of the EIT?
One for every column.
How many columns in the EIT?
One for every line.
How many lines in the EIT?
An infinite number.

Nice, but without any relevance.

The question is: how many I's are in the projection of the
EIT?

What you call the "projection of the EIT" is defined as the WM(EIT) by
its inventor.

How can the answer to this not be relevant?

I understand what you mean. That question is highly relevant. But you
seem to become tangled:
You stated always that the number of lines is irrelevant for the sizes
of numbers in the lines.
Now you seem to *claim* that the number of lines is relevant for the
WM(EIT) which is solely determined by the sizes of the numbers in the
lines. There is some inconsistency, isn't it?

What part of "the Waft Maximum is completely irrelevant" do you
fail to understand?

See above.

Regards, WM

.

Relevant Pages

  • Re: Review of Mueckenheims book.
    ... Do you agree that the triangle, EIT, ... of nodes in the infinite tree are exactly the same] ... i' The number of nodes in the union of all finite trees up to ... the projection contains an infinite number of units. ...
    (sci.math)
  • Re: Review of Mueckenheims book.
    ... infinite by size too. ... Which immediately leads to the Waft Maximum of the ... not mean that the projection is finite" do you fail to understand? ... Let F be a sequence. ...
    (sci.math)
  • Re: Review of Mueckenheims book.
    ... Do you agree that the triangle, EIT, ... of nodes in the infinite tree are exactly the same] ... i' The number of nodes in the union of all finite trees up to ... projection of all lines onto the bottom contain. ...
    (sci.math)
  • Re: Review of Mueckenheims book.
    ... Maximum of a set or about "what the projection does". ... Let P_I be the set of all infinite paths. ... its Waft Maximum is 1. ... Waft Maximum of the set of all finite paths. ...
    (sci.math)
  • Re: Review of Mueckenheims book.
    ... projection of all finite lines is infinite. ... F be a sequence. ... Why then do you think the projection leads to WM = oo in example 7? ...
    (sci.math)
Quantcast