Re: Cantor Confusion

In article <1167393772.865867.58640@xxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
*** T. Winter schrieb:
> 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.

Yes, so what?

> > > 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.

Strange. You state that in the infinite tree for each path has an edge
that is passed in full. When I ask you what particular edge is passed in
full by a particular path you claim something about not claiming something
about a representation. I ask you to substantiate your claim that for each
path there is an edge that is passed in full. Especially for the path
that goes to the left at each step. I am not talking about representations.

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.

Does not matter. Your claim was about edges and paths, my question was
about a specific path. But apparently you decline to answer the question.

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.

Each path can be distinguished from each other path by some edge. There
is *no* edge that distinguishes a particular path from all other paths.
Back to that again. Quantifier dyslexia.

Similar to: each natural number written in decimal can be distinguished
from each other natural number by some (actually in most cases many)
decimal digit. For each natural number there is *no* decimal digit that
distinguishes it from all other decimal digits.

> > 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.

Why? Proof?

> 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.

How do you define that? With your own particular definition? Is that
definition consistent?

> 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 that union of sequences does actually not cover the path that leads to
1/3.

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.

And a host of rational numbers. But even of the tree of rational numbers
ending in either 111... or 000... there is no personal edge for any of
those numbers. A lack of personal edges abounds in your tree.

> > 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.

And a host of rational numbers also do not exist.
--
*** t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~***/
.