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Um, what? There are two well-known arguments from Cantor: http://en.wikipedia.org/wiki/Cantor's_first_uncountability_proof http://en.wikipedia.org/wiki/Cantor

Um, what? There are two well-known arguments from Cantor: http://en.wikipedia.org/wiki/Cantor's_first_uncountability_proof http://en.wikipedia.org/wiki/Cantor's_first_uncountability_p... http://en.wikipedia.org/wiki/Cantor's_diagonal_argument http://en.wikipedia.org/wiki/Cantor's_diagonal_argument 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. 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. 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. So it's like a mixture of two badly remembered proofs :-)

backlava 13y ago
The proof constructs a monotone bounded sequence. It thus converges. Dividing into three intervals is to rule out the case where both (closed) intervals contain the point.
- Tloewald 13y ago
  Right. There's no reason for the intervals to be closed that I can see.
  - ColinWright 13y ago
    You're right, there isn't, but when you use open intervals you'd be surprised how many people get worked up over the fact that some of the points have been omitted. It's easier to avoid the question entirely. It tends only to be people who already know a lot about this that come up with the objections or comments you're making, and usually they realise that actually, yes, it all works.
ColinWright 13y ago
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.
zodiac 13y ago
> 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. No, the argument is that there is a sequence that converges to some real number, but that real number is not algebraic, hence there exist non-algebraic real numbers. I do agree that the article doesn't clearly describe how to construct said sequence - perhaps the second page of [1] will be clearer. Basically we always have an interval (a_n, b_n) where the a_n and b_n are algebraic, and we keep narrowing the interval by going down the list of algebraic numbers until we find two numbers x and y such that a_n < x < y < b_n; then (x, y) is the new interval. The sequence of the a_n converge but not to an algebraic number, because if it did, that number would eventually become an endpoint of the sequence, which is a contradiction. [1]: http://www.cwu.edu/~harperj/472%20Winter%202012/Cantor-rev2.pdf http://www.cwu.edu/~harperj/472%20Winter%202012/Cantor-rev2....
- ColinWright 13y ago
  > ... the article doesn't clearly describe > how to construct said sequence ... I thought the construction was clear. In particular, this argument does not use Cantor's construction wherein the interval end-points are always algebraics. Instead, it ... * Finds the next listed algebraic a in the current interval, * divides the current interval into three, and * chooses the left-most sub-interval that neither contains nor touches a. The sequence is then the left-most endpoints of the intervals. In particular, the sequence is composed entirely of rationals. More, the argument works even if the set being avoided (which must be countable, but need not be the algebraics) is not dense.

backlava 13y ago

The proof constructs a monotone bounded sequence. It thus converges. Dividing into three intervals is to rule out the case where both (closed) intervals contain the point.

Tloewald 13y ago
Right. There's no reason for the intervals to be closed that I can see.
- ColinWright 13y ago
  You're right, there isn't, but when you use open intervals you'd be surprised how many people get worked up over the fact that some of the points have been omitted. It's easier to avoid the question entirely. It tends only to be people who already know a lot about this that come up with the objections or comments you're making, and usually they realise that actually, yes, it all works.

ColinWright 13y ago

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.

zodiac 13y ago

> 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. No, the argument is that there is a sequence that converges to some real number, but that real number is not algebraic, hence there exist non-algebraic real numbers. I do agree that the article doesn't clearly describe how to construct said sequence - perhaps the second page of [1] will be clearer. Basically we always have an interval (a_n, b_n) where the a_n and b_n are algebraic, and we keep narrowing the interval by going down the list of algebraic numbers until we find two numbers x and y such that a_n < x < y < b_n; then (x, y) is the new interval. The sequence of the a_n converge but not to an algebraic number, because if it did, that number would eventually become an endpoint of the sequence, which is a contradiction. [1]: http://www.cwu.edu/~harperj/472%20Winter%202012/Cantor-rev2.pdf http://www.cwu.edu/~harperj/472%20Winter%202012/Cantor-rev2....

ColinWright 13y ago
> ... the article doesn't clearly describe > how to construct said sequence ... I thought the construction was clear. In particular, this argument does not use Cantor's construction wherein the interval end-points are always algebraics. Instead, it ... * Finds the next listed algebraic a in the current interval, * divides the current interval into three, and * chooses the left-most sub-interval that neither contains nor touches a. The sequence is then the left-most endpoints of the intervals. In particular, the sequence is composed entirely of rationals. More, the argument works even if the set being avoided (which must be countable, but need not be the algebraics) is not dense.