Re: Cantor and the binary tree



*** T. Winter wrote:

> > *Any* tranposition to be performed is determined from the beginning
> > (given a certain initial well-ordering of the rationals of (0,1)). You
> > can say, for *any* transposition, when it will have to occur and what
> > will be the result. It is not an iterative process. It is equivalent to
> > Cantor's a_nn replaced by b_n.
>
> It is not. But you are too dense to notice.

Interesting argument.
>
> > It is impossible, however, to show the completed result, namely the
> > ordered set of rationals.
>
> There is no completed result.

But there is completed infinity? Otherwise we could not reach beyond
it.
>
> > It is equally impossible to show the
> > completed antidiagonal. The only difference is, that the latter is not
> > so obvious. Terefore some people believe in finished infinity, as yet.
>
> There is no completed antidiagonal, that is you can not give all decimal
> digits. On the other hand it is sufficient to show the existence of a
> real that is different from all the reals on your list.

It would be sufficient. But how can you show it, if there are always
infinitely many remaining, how many and how large n you may ever check?
No. You must rely on the principle. It says that for every n you
investigate, there is a difference. But for every n you investigate
there are infinitely many more which are not investigated. Therefore
the effect of the list is null and void.

Obviously the antidiagonal it is not different from all reals on the
list. Its digits are not complete. The list is not. What does it say
then?

And the above principle, if valid, would also yield the complete set
of ordered rationals.

> The only place where, in the definitions, ordering plays a role is when
> you want to introduce transcendental numbers. For algebraic numbers
> there is no need.

There are three axioms of order for the reals. They are there and they
are valid in mathematics. That is enough!

Regards, WM

.