Re: Review of Mueckenheims book.

On 2 Mrz., 04:20, "*** T. Winter" <***.Win...@xxxxxx> wrote:
In article <1172771927.176281.193...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueck...@xxxxxxxxxxxxxxxxx writes:

> On 1 Mrz., 03:38, "*** T. Winter" <***.Win...@xxxxxx> wrote:
...
> > > > > 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)?
>
> I do not define C(oo) *starting from the complete tree* T(oo) but
> using the union U(T(n)) of the finite trees, (because it is in
> question whether T(oo) = U(T(n))).

So, my question remains, how do you *define* C(oo)? Until now I have not
yet seen a definition.

C(oo) is the cross section of U(T(n)).

But you are now apparently stating that
T(oo) != U(T(n))?
Because I state they are equal.

I think it might be possible that T(oo) != U(T(n)). We will
investigate it.

> > > 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.
>
> The harmonic series as split off in Oresmes proof is isomorphic with
> the binary tree:

I do not ask about that, I ask about the saying, and where it comes from.

The saying that there are aleph_0 parentheses in this proof comes from
the fact that here are infinitely many.
The saying that there are less than 2^aleph_0 unit fractions in these
parentheses comes from the assumption that there are less than
2^aleph_0 natural numbers what implies that there are less than
2^aleph_0 unit fractions, namely exactly aleph_0.

Regards, WM

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