Re: Review of Mueckenheims book.

In article <1172771276.240487.107540@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
On 1 Mrz., 02:44, "*** T. Winter" <***.Win...@xxxxxx> wrote:
> The finite paths form a countable set.
> The sets like {p(0), p(1), p(2), ...} of finite paths like p(n) which
> belong to an infinite path like p(oo) form a countable set.
> The unions U{p(0), p(1), p(2), ...} of finite paths like p(n) which
> belong to an infinite path like p(oo) form a countable set.

Proof, please... p(0) is not used in a single path, nor is any of the
p(i).

It is a subset of every p(n).

Do you consider that a proof of your statement?

Do you know how that is *defined*? I think not. That argument would
be sufficient to refute Cantor's diagonal proof. Let me redefine a
few things to make it simpler. We omit "0." from each path, leaving
p(0) = "0" or "1" (depending on the direction you are taking), next we
replace each p(k) by the digit in which it is different from p(k - 1),
So it is either "0" or "1". That does not change anything in the
argument, as the replacements can be undone (there is nothing
irretrievable lost). Now replace "0" by 'm' and "1" by 'w'. And
you are stating:
Therefore the number of sequence {k(0), k(1), k(2), ...} where
each k(i) is either 'm' or 'w' is countable. Because 2^omega
is countable.

Yes that is one inconsistency of set theory.

Not an inconsistency. You fail to see what the definitio of 2^omega
actually is.

(I note that with these replacements it becomes the sequences that
Cantor considers in his diagonal proof.) Now we can state either
w < m or m < w, that does not matter.

In WM we have W < M, so let us choose that.

Ok.

The crucial point is that in
the definition of 2^omega only finitely many elements are *not* mapped
to the smallest of the two. If there are not finitely many elements
mapped to the smallest element, that operation is not defined. It has
to do with well-ordering.

I don't understand what you are trying to say.

No. That appears to be the case all to often. 2^omega is defined as the
order type of the set of mappings from a set with order type omega to the
set {0, 1}. So far so good. But in order to define it, the mappings should
only be those for which only finitely elements are mapped to 1. When that
is not the case the operation is not defined.

> You said it is p(oo) = U{p(0), p(1), p(2), ...}. If so, then the set
> P(oo) of all infinite paths p(oo) is countable. If infinite paths do
> not exist, then the set P is empty.

Not so, see above.

The set P(oo) is he set of infinite paths. If infinite paths do not
exist, then it is empty.

That is trivially true. But there are infinite paths.
--
*** t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~***/
.

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