Re: Review of Mueckenheims book.

On 5 Mrz., 02:50, "*** T. Winter" <***.Win...@xxxxxx> wrote:
In article <1173027087.277516.285...@xxxxxxxxxxxxxxxxxxxxxxxxxx> mueck...@xxxxxxxxxxxxxxxxx writes:


I did not know that the definition of N implied limits. It is the axiom
of infinity that grants its existence,

But it couldnot be realized without stating:
lim{n -> oo} {1, 2, 3, ..., n} = N

The axiom of parallels grants the existence of only one to a given
line through a given point. But the lines are not drawn by the axiom,
they are drawn by hand or mind.

Further the axiom states that with n also n+1 is in the set. This
would be useless if we stopped at a given n. Therefore the only
question can be: Is N = lim{n -> oo} {1, 2, 3, ..., n}, or is N
something more (it cannot be less).

> > (2) It is not yet proven that the limit of the cardinality of a sequence
> > of sets is the cardinality of the limit of the sequence of sets, even
> > when both limits are defined.
>
> Therefore I use both limits independent from another.

Yes, and because both do not exist, you are wrong.

Then the axiom of infinity supplies an undefined set.

> > > I fully agree with you. Alas, the limit omega is used in set theory
> > > with improper definitions.
> >
> > In set theory it is *not* a limit.
>
> It is. Omega is a limit ordinal number.

It is a limit ordinal. That does *not* mean it is a limit. The definition
of "limit ordinal" is an ordinal that has no immediate predecessor, that is
all. No limit involved.

Why is it called "limit"? Of course you can state such things like
"proofs by contradiction are not proofs by contradiction" and "limits
are not limits" and "Umformungen are not Umformungen", without being
afraid on any consequences, but this makes the discussion getting
boring.

> > So {{1}, {1, 2}, {1, 2, 3}} can be seen as either a set of sets of naturals
> > as as a set of naturals by taking the union which gives {1, 2, 3}?
> > In what world do you live? A path-bundle is either a set of paths or
> > a set of nodes, it can not be both at the same time.
>
> But at different times. The easiest way to convert one into the other
> is taking the union.

It can also not be both at different times. It is either one or the other.

The set of all nodes of the tree and the set of all nodes of the union
of all paths are identical. That is enough.


> The same is true for the tree. It is defined as a set of nodes. If we
> mean the paths of the tree we have to express this. Important for the
> proof is alone that there are fixed "quantities" of nodes in sets of
> trees and paths.

No. Important for the proof is *how* you express them.

The set of all nodes of the tree and the set of all nodes of the union
of all paths are identical. No atter how this set of nodes is
exressed.

What is half a quark? And again, what is half a point?

That what yields one quark when taking two of them.

> > > > Pray, again, *define* that cross section. If you do not define it, I
> > > > can not consider it.
> > >
> > > The cross section of a tree is the umber of nodes in the last level of
> > > the tree.
> >
> > But there is *no* last level in the union. So how do you define it?
>
> C(oo) is the cardinal number of the union of all levels.

Makes no sense. So now suddenly C(oo) is a cardinal number. So the
cross section of T(oo) is a cardinal number. But the cross sections
of any finite T(n) is a set of nodes. Indeed, makes no sense at all.

You are wrong. I defined the cross section of a tree always as the
number of nodes of the last level. So it is a cardinal number.

C(oo) = Card(U(L(n))) where L(n) = {1,2,3,...,2^n}


> > > The union of finite trees can be identified with the union of finite
> > > numbers. That means:
> > >
> > > 1) There are infinitely many trees.
> > > 2) The number of nodes in the union of finite trees is finite. (This
> > > is true because the sizes of all numbers in the union N are finite.
> > > Recall, the union of finite trees is the largest one of the trees in
> > > the union.)
> >
> > Wrong. The "because" is a non-sequitur. The "numbers of the nodes" are
> > finite, the "number of nodes" is not finite.
>
> So the union of finite tree contains an infinite tree? Or how do you
> get there? Is it a limit process in set theory?

It is *not* a limit process. Unions of arbitrary collections are defined
without limits.

A union of finite numbers cannot contain an infinite number. In the
EIT

1
11
111
....

we have seen that there is no line with infinitely many units. How
could the projection of the lines onto the bottom yield a line longer
than any?

Regards, WM

.

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