Re: An uncountable countable set

In article <1152208684.463235.35190@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> mueckenh@xxxxxxxxxxxxxxxxx writes:
Dik T. Winter schrieb:
> > Moreover,
> > different orderings of the same subset of transpositions will yield
> > different results. In the formula I gave for the diagonal number,
> > calculation of one digit does *not* depend on the calculation of
> > another digit.
>
> And what is the consequence of this?

That the calculation of the digits can be done in parallel?

It is done in zero time. It is determined from the beginning. Why
should something "be done"?

It was you who remarked on the calculations...

> > It is Cauchy's theorem that proves that the limit exists, using the
> > epsilon argument.
>
> But Cantor's arguing is that without epsilon.

I do not know Cantor's argument exactly. I think that he implicitly
uses that result. In my formulation it was abundantly clear that I did
use it.

But he needs to consider every digit with equal weight. That is not a
limit process.

What did he *mean* when he wrote "with equal weight"? For comparison
purposes the digits have equal weight (i.e. it is not the case that some
numbers on the list are less different from the diagonal than others).
However, when you wish to know whether the number constructed is a real
number you need limits.

And again, it maintains the well order as long as you remain in the finite
domain. So up to every finite n, you have ordered the first n elements
of the well-ordered list, retaining a finite list. And back again to
problem 1 with this. How do you define that when n grows without bound?

I remain in omega! So I always stay with finite numbers, they may grow
as far as you like. Wha holds for small natural numbers does equally
well hold for big ones, as far as well-order is concerned.

Yes, as I said, for every finite n your list of transpositions will have
transformed the list (implying well-ordering) of rationals to another list
of rationals. It will not have changed to the rationals under standard
order. You have just sorted the first n rationals using a bubble sort.
There are still infinitely many rationals that are not sorted according
to their standard order.

So you claim that with that process you get a well-ordering of the rationals
with their standard order can only be made when, in some way, you define
what happens when the process continues indefinitely. And for that you
need some kind of limit. And next the question is whether well-ordering
is preserved under taking the limit. And I think that is false.

How do you define the "limit"? And if you define that, is that "limit"
also well-ordered? Those are things you have to prove.

I define limit by: *Using all finite natural numbers* just as like as
Cantor does.

No. You do *not* and he does *not*. In the Cantor diagonal the number
obtained is a real number using limits (in the Cauchy sense) with the
definition of real number by many others (that can be proven to all be
equivalent with each other). The limit used by Cantor is precisely the
epsilon argument, together with majoration and minoration on the ordered
set of rational numbers. By Cauchy any sequence of decimal digits has a
limit, but it is not certain whether that limit is in the defined set of
numbers. By Cantor, Dedekind, Weierstrass and a host of others (using
various formulations), such a limit (when starting with rational numbers)
is defined as a real number. That is the place where Cantor uses the
limit (although he may not have expressed it as such), i.e. showing that
the resulting number is a real.

On the other hand, you have *not* defined what an infinite number of
transpositions means. "Using all finite natural numbers" is not a
definition.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
.

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