Re: Cantor and the binary tree

In article <1120832988.809905.149890@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
> *** 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.

No, simply an observation.

> > > 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.

I have no idea how you reach that conclusion.

> > 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.

So how would you prove that n*(n+1) is even? For every n you investigate
there are infinitely many more which are not investigated. Therefore
the effect 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?

Sorry? The list is nog complete? You state you give me a list. In that
case I may assume the list is complete methinks. If you give me an
incomplete list nothing can be done at all. I think you are again focussed
on the word "list". A list is nothing more, nor less, than a function
f(n) to objects where n ranges through the natural numbers.

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

No. Because that is an iterative process.

> > 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!

As I stated, the ordering is only needed (and used) when you define the
transcendentals. To define the algebraic numbers, ordering is not used,
and indeed it only plays a minor role in algebraic number theory. How
do you distinguish the (imaginary) numbers i and -i?
--
*** t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~***/
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