Re: Cantor Confusion


*** T. Winter schrieb:


> > It has been
> > proven that *every* Cauchy sequence yields a real number. And it is
> > easy to see that *every* decimal expansion is in fact a Cauchy sequence.
> > 10^-n is *never* zero. You need limits.
>
> Yes. And they say: As long as 10^-k is not zero, the sum {n=1 to k} a_n
> + 10^-n is not an irrational number.

Right. And as 10^-k is never zero, that sum is never an irrational number.
You need limits to get irrational numbers.

The infinite binary tree contains this limit. That's why he tree is
called infinite.

> > > There is no edge in the *finite* domain which is "passed in full". In
> > > the infinite series there is an edge passed in full.
> >
> > Which edge is passed in full to the path that goes at each step to
> > the left?
>
> I don't claim the existence of this representation of 0.

Do you claim that is an answer to my question?

Yes. There is no acually infinite sequence of digits (or anything else)
which can be distinguished from any other infinite sequence.

Again, re-read my question.
I am not talking about representations, I am talking about paths. You were
talking about paths. Infinite paths. Hence my question.

A path is a representaton as well as any decimal or n-ary number.

> > > Otherwise the path
> > > does not exist at all.
> >
> > That is what you are trying to prove. But you can not just state it. You
> > hav to *prove* it.
>
> I need not prove that two entities which have no different property,
> cannot be distinguished.

Your statement was that if there is no edge in the infinite series that is
passed in full (meaning that the edge does belong only to that path), the
path does not exist. That is what you have to prove. You again fail to
do so.

A path is an (ordered) set. If there is no element distinguishing it
from every other set, then it is not different from every other set.

> > > The infinite sum 1 + 1/2 + 1/4 + ... yields 2, although there is no
> > > smallest term 2^-n.
> >
> > Indeed.
> >
> > > It can be applied to calculate the edges per path. You should know that
> > > absolutely converging series can be reordered.
> >
> > You should know that you can *not* reverse absolutely converging series.
>
> Proof?

The infinite series has no last term, so the reversal does not have a first
term, and so is not an infinite series.

But it has the same value.

> > And that is what you are doing. Because that would result in a series
> > without start.
>
> No reordering of an absolutely converging series changes its sum.

But reversal makes it something quite different. What is the sum of the
first 10 terms of the reversal of the above series? What is the sum
of the first 20 terms?

Not defined. What is defined is the sum of all terms. That's enough.


> > > > But in the complete tree
> > > >the parts of edges that are assigned to an infinite path are not 1,
> > > >1/2, 1/4 etc. There is *no* edge that is completely assigned to a
> > > > path,
> > >
> > > Then the tree is incomplete. You fail to understand infiniteness
> > > already on this low level?
> >
> > Circular reasoning abounds here. Care to give a *proof* that the tree
> > is not complete in that case? This statement is completely similar to
> > the statement that the completed set of natural numbers should have a
> > last element.
>
> There is no largest element, but the infinite path is nothing than the
> union of all finite paths. Therefore we can calculate the union of all
> finite sequences 1 + 1/2 + 1/4 + ...+1/2^n.

Interesing. How do you calculate the union of sequences?

By taking the limit n --> oo. (The union of sequences is only a
*slightly* sloppy expression.)

But again my
question: care to give a "proof" that the tree is not complete if there
is no edge that is completely assigned to a path?

The tree is as complete as a tree an be. The lacking personal edge
shows the non-existence of irrational numbers.

> > > > Note that in each finite tree, the complete edge assigned to a path is
> > > > the last edge on the path. As there is no last edge in an infinite
> > > > path, this does not hold in the infinite tree.
> > >
> > > So there are different paths without different edges?
> >
> > No. I never did state that.
>
> So at least one path has its own edge?

No, I never did state that. And it is false.

Correct. Even in an infinite tree this is wrong, because there are no
irrational numbers. Irrational. In its second meaning we see: Nomen est
omen.

Regards, WM

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