Re: An uncountable countable set


*** T. Winter schrieb:


> I did already tell you which edges belong to the path 1/3. The nodes
> can be enumerated like the edges.

Yes. But you still only have terminating paths. At each level all paths
are terminated by a node. And so 1/3 is not in the tree because there
is no terminating edge and node.

Then such numbers like 1/3 do not exist (in that representation). But
the same holds for Cantor's list.

Oh, well. As in mathematics the reals require a construction process, so
also your tree requires a construction process. In mathematics the reals
are constructed from the rationals (and I know at least four methods to
do that, that can be shown to give equivalent results). And the rationals
are constructed from the integers. I asked you before, but you never did
reply. Do you know how the rationals are constructed from the integers in
mathematics? More basic, do you know how arithmetic on naturals is defined,
based on the Peano axioms?

More to the topic: Do you know how Cantor's diagonal is constructed
from a list of reals? And how this list is constructed?

> > and there is an
> > edge that terminates at a node terminating the path. I meant the paths
> > during the intermediate process of construction.
>
> There is no process of construction. If you have difficulties to
> comprehend that: it is the same as with Cantor's list. The tree is
> defined once and for all. That's it.

Wrong. But I am not going to explain that again.

What is the difference between my tree and Cantor's list. There was no
explanation and there is none, because there is no difference.

> All real numbers of the interval [0, 1] are there, some of them even
> twice.

Many of them are not. Your tree only contain numbers with a finite binary
expansion. 1/3, 1/5 and 1/7 are not among them.

But Cantor's list only contains numbers with an infinite binary
expansion? In particular his diagonal construction.

> > > How many paths are required to obtain an
> > > uncountable set of edges, according to your opinion?
> >
> You seem to misunderstand the tree. If the diagonal is in Cantor's
> list, then 1/3 is in my tree. Can you give a reason why it should not
> (other than that then set theory is inconsistent)?

I have explained already many times, but you are not willing to listen.

Please give a reference, or better: copy and paste your explanation.

> You recently mentioned an interesting aspect: The algebraic numbers,
> i.e., the polynoms are countable, because they are finite. And,
> therefore, you wanted only to count the finite segments of paths in my
> tree. I oppose, unless you agree that the 1's in 0.111... also are
> uncountable. You see, the problem is the same: We can count finite
> segment but not infinity.

You are using a pretty strange terminology. You can count the algebraic
numbers just because the polynomials remain finite (but there are
infinitely many of them). You can count the finite paths because they
are finite (but there are infinitely many of them). You can count
the digits because each digit position is finite (but there are infinitely
many of them). What inconsistency?

You can count the levels of my tree just because they remain finite
(but there are infinitely many of them).
The number 1/3 has only finite digit positions at finite levels of my
tree. Where should any uncountable edge appear?


> Either you agree that all the edges of my tree are countable or you
> agree that Cantor's diagonal is uncountable. Or you state at least how
> many infinite path in my tree you would consider to have completely
> countable edges.

This question makes not sense to me.

The problem is the following: You assert that the digit positions of
1/3 are countable as well as the levels of my tree (all of them), but
you deny that the edges at these levels are not all countable. That is
wrong because for every n-th level you count, the number 2^n of edges
can also be counted. There is no limit.

Regards, WM

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