Re: Attempts to Refute Cantor's Uncountability Proof?

Dave Seaman wrote:

On Tue, 11 Jul 2006 06:09:21 -0400, Hatto von Aquitanien wrote:

Originally my (incorrect) understanding was that Cantor first put all the
rational numbers into a counted list, and then proved he hadn't counted
everything. I now realize that was incorrect. However, /suppose/ we
/did/ claim that we have a counted list of rational numbers (1-1
correspondence
with N+), but don't explain how we came by it. Would Cantor's diagonal
insertion method not indicate there are always more rational numbers?

No, not at all. The if all the rationals are in the list, then the
diagonal method must necessarily produce an irrational number.

What does that mean? It /seems/ to me that would require an infinite number
of steps. That doesn't mean it's an invalid argument, but it does make me
suspicious.

And we don't need to "suppose" we have a counted list of rational numbers.
It's been done. See, for example,
<http://www.lacim.uqam.ca:16080/~plouffe/OEIS/citations/recounting.pdf>.

No, you have an algorithm. I can produce an algorithm for generating "all"
the "reals" iteratively. In a finite number of steps, it will only produce
rationals, but I can show that it approaches the reals with every
iteration.

This question may in fact be the origin of my previous confusion. I may
have entertained this same question years ago, and never followed up on
it.

It has nothing to do with how "long" it takes to make the list or to
construct the diagonal value. As soon as you postulate the
existence of the list, the diagonal value exists as well, and that
value is always missing from the list.

I trust this has been suggested before, but suppose we are talking about
a two-valued underlying domain, and use Cantor's diagonal method to
enumerate the sequences.

A two-valued underlying domain? Isn't that basically happens with the
Cantor
proof that |X| < |P(X)|? A subset of X is identified by its
characteristic function, which is a mapping f: X -> {0,1}.

Perhaps so. I will have to think about it.

<http://www.mathacademy.com/pr/prime/articles/cantor_theorem/>
Nice site. Thanks.

--
Nil conscire sibi
.

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