Re: Review of Mueckenheims book.

In article <1172674257.469668.135280@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
On 26 Feb., 04:14, "*** T. Winter" <***.Win...@xxxxxx> wrote:
> Definition: The cross-section C(n) of a finite tree T(n) is the number
> 2^n of nodes of its last leve L(n).

Obfuscations abound. So C(n) is the number of nodes that are at
distance n from the root.

Correct, namely the maximum of paths which can be separated in this
tree T(n).

Yes, as each path terminates at such a node. You again add obfuscation.

> The cross section C(oo) of the union of finite trees U(T(n)) is C(oo)
> = |2^omega| = aleph_0.

Proof, please. And how do you *define* C(oo)?

I do not define C(oo). We are talking about the union of finite trees
It does not contain a Level n = oo or tree T(oo).

How can you state things about things you do not define? You state C(oo)
as being something, but you do not define C(oo)?

You defined C(n) as the
number in the last level L(n) of T(n). What is the last level of T(oo)?
So what is C(oo)?

If you look at the proof of the harmonic series, there are aleph_0
pairs of parentheses.

What are you babbling about. You used C(oo) and I ask what it is. But
you refrain to answer. I will skip the following, as it is irrelevant
as long as you refuse to answer my question.

Then you see that the cross section of the
union tree U(T(n)) = T(oo) is C(oo) = aleph_0.

What can I see if you even do not give a definition of your terminology?

> Proof: Left as an exercise to the reader. [Hint: Consider the proof of
> the countability of the set of all unit fractions required to prove
> the divergence of the harmonic series:
> 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) ...
> or consider the cardinal number of nodes of the union of finite
> trees.]

Is this a proof? For the divergence of the harmonic series, countability
plays no role. It is just a lack of proof.

You misunderstood. There is the saying that aleph_0 parentheses result
in aleph_0 terms in parentheses although for every finite segment of n
pairs of parentheses we have 2^n terms in the last pair of
parentheses.

I can say nothing about such a saying. This is beyond what is understandable.
--
*** t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~***/
.