To take your points:

> There are two well-known arguments from Cantor:
> http://en.wikipedia.org/wiki/Cantor's_first_uncountability_p...
> http://en.wikipedia.org/wiki/Cantor's_diagonal_argument

Yes. And in fact the first uncountability proof comes in two flavours. This is explicitly a re-casting of those two versions of Cantor's first proof.

More explicitly, one version of Cantor's first proof has you list the algebraics, take an interval, find the first two algebraics a and b in the interval, reduce the interval to [a,b], and continue. This has various technical issues. What if there aren't two algebraics in your interval (although there always will be, because they're dense, but the argument goes through for any list, not just algebraics, so we need to consider this possible eventuality if we're using the full generality.) What if the interval doesn't converge to be of length zero? (Which again, for the algebraics, it always will. And again, if we use the argument in general we need to allow for this.) And so on.

The second version of Cantor's first proof avoids some of this by saying: find just the first algebraic in the interval, then choose a smaller sub-interval that avoid it, any carry on. Again, for algebraics the interval will go to length zero, but for a general list that won't necessarily be the case.

> Describing the sequence at the end as something that seems like
> it should converge is a bit unconvincing -- it's a sequence
> explicitly defined not to land on an algebraic number. So the
> argument is that if you assume you can always pick an interval
> that doesn't contain a specified algebraic you won't land on an
> algebraic. This is hardly surprising and kind of presupposes
> the result.

I think you've mis-read the argument. We produce a succession of nested intervals. For a given algebraic, eventually every interval will not include it. That means that no algebraic is in more than finitely many intervals.

The we consider the left end-points. That's an increasing sequence that's bounded above. The reals are complete, so that sequence must converge. But by construction the limit can't be an algebraic.

The comment that the sequence "seems like it should converge" is pointing out that completeness is one of the things we expect of the number line, and it's a way into giving that concept a technical meaning.

> And why divide the interval into thirds? This seems to be
> a result of confusing a different Cantor proof (that the
> measure of the rationals/algebraics in [0,1] is zero, which
> involves dividing things into thirds) with the first proof.

No, it's a way of being explicit in the choice of the sub-interval. I sometimes use fifths. In that way I can discard the end sub-intervals, discard any sub-interval that touches or contains the given algebraic, and thus the end-points are strictly increasing, and not just increasing. I didn't do that here, and you make me wonder if it's worth changing it, so people don't (incorrectly) make the connection with Cantor dust.

But no, this isn't confusing the construction of Cantor dust.

> If we accept the argument at all I don't see how it requires
> three intervals at each step. At most one interval can contain
> the posited nth algebraic, so you've always got the other interval.

The algebraic may be on the boundary, rather than in an interior.

> So it's like a mixture of two badly remembered proofs :-)

With respect, your comments seem like they are generated from an insufficiently careful reading, filtered through an understanding of the existing explanations. It feels like you know about this material, and were assuming this is a simple re-hash.

But I could be wrong. Your comments might have substance that I haven't recognised. If that's the case I'd be delighted if you could expand more on why you think the article is wrong or misleading.

Thanks.