Re: An uncountable countable set


*** T. Winter schrieb:


> Then such numbers like 1/3 do not exist (in that representation).

Indeed, in your tree with terminating edges, such numbers do not exist.

> But
> the same holds for Cantor's list.

I do not see how you can jump to that conclusion.

Cantor's list has as many lines as my tree. The diagonal has as many
digits as each path of my tree. The only difference is that the paths
split while the diagonal does not.

> > 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?

Again, the second proof was *not* about the reals.

Please spare your nonsense. Cantor did not consider anything else than
the reals. If today the proof concerns some wider range then this is
not due to Cantor.

But I know how the
diagonal is constructed. I have no idea how the list is constructed.
But again (it appears that I have to repeat myself a lot in this
discussion), there is *no* construction of the list given. The proof
is along the lines of: "given *any* list, I can show such". So the
proof is valid without regards to the actual construction of the list.

The same is true for my tree.

> > > 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.

You give a construction process for your tree. In the case of the list
there is no construction process involved. The proof is about "given
*any* tree".

Forget the construction process of the tree. Take any tree which
contains all reals. As you say: My proof is about "given *any*
infinite tree".

The list is not a list that is constructed, but a list that is given, or
assumed. No construction involved. Whether the list contains finite
binary expansions or not is completely irrelevant. The process with Cantors
proof is more like the following:
give me a list of infinite sequences of 0's and 1's, and I show that there
is a sequence that is not in the list.
Do you have problems with proofs by contradiction?

Give my a tree of infinite paths consisting of 0's and 1's, and I show
that there
are not less edges than paths.

> > > 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.

I have repeated my argument above. But I think you will not read it.
1/3 is not in your tree because all edges (and hence paths) are terminating.
The diagonal is not in the list because it is defined in such a way that it
can not be in the list. There is no relation between the two.

The list, if existing, contains a diagonal (before anything is
exchanged). Doesn't it?

> The number 1/3 has only finite digit positions at finite levels of my
> tree. Where should any uncountable edge appear?

Depends on how you count. If you count only the edges leading to 1/3 you
can do that (but never reach 1/3).

That is self-evident for all counting processes. Why do you stress it?
Because you try to argue the counting of all edges would fail for that
reason? That is wrong.

If you want to do the same with 1/5
the case is similar. The problem appears when you want to count *all*
edges. At each level, the number of edges is 2^n - 1 or somesuch. But
also at each level we still have paths that need to be distinguished.

You know that sets of order 2^omega are countable?

> > > 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.

As I have stated again and again, if your edges *do* terminate, the number
of edges is countable, but 1/3 is not in your tree. There is no terminating
edge that leads to 1/3.

"Countable" means that you can count on and on ad that for any element
a natural number is reserved which is mapped on that element.
"Countable" does ot mean that we reach an end.

Give up your arguing. All set theorists know that the edges of my tree
are countable. Even Virgil knows it.

Regards, WM

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