CronoDAS comments on The Lifespan Dilemma - Less Wrong
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You are not the first person to try to explain this to me, but it doesn't seem "surprising", it seems like everybody is cooperating at pulling my leg. Since I'm aware that such a conspiracy would be impractical and that I am genuinely terrible at math, I don't think that's actually happening, but the fact remains that I just do not get this (and, at this point, no longer seriously entertain the hope of learning to do so). It is only slightly less obvious to me that there are more numbers between 0 and 2 than 0 and 1, than it is that one and one are two.
To put it a little differently, while I can understand the proofs that show how you may line up all the rationals in a sensible order and thereby assign an integer to each, it's not obvious to me that that is the way you should count them, given that I can easily think of other ways to count them where the integers will be used up first. Nothing seems to recommend the one strategy over the other except the consensus of people who don't seem to share my intuitions anyway.
Well, that was the short explanation. The long one makes a little more sense. (By the way, the technical term for the number of members in a set is the cardinality of a set.)
Let's try this from a different angle.
If you have two sets, X and Y, and you can map X to a subset of Y and still have some members of Y left over, then X can't be a bigger set than Y is. In other words, a set can't "fit inside" a set that's smaller than itself. For example, {1,2,3} can fit inside {a,b,c,d}, because you can map 1 to a, 2 to b, 3 to c, and still have "d" left over. This means that {1,2,3} can't be bigger than {a,b,c,d}. It shouldn't matter how you do the mapping, because we only care about whether or not the whole thing fits. Am I making sense here?
Now, because you can map the positive even integers to a subset of the positive integers (for example, by mapping each positive integer to itself) and still have positive integers left over (all the odd ones), the set of positive even integers fits inside the set of positive integers, and so it can't be bigger than the set of positive integers.
On the other hand, the positive integers can also fit inside the positive even integers. Just map every positive integer n to the positive even integer 2*(n+1). You get the list (1,4), (2,6), (3,8), and so on. You've used up every positive integer, but you still have a positive even integer - 2 - left over. So, because the positive integers fit inside the positive even integers, so they're not bigger, either.
If the positive even integers aren't bigger than the positive integers, and the positive integers aren't bigger than the positive even integers, then the only way that could happen is if they are both exactly the same size. (Which, indeed, they are.)
So in fact, we count them both ways, get both answers, and conclude that since each answer says that it is not the case that the one set is bigger than the other, they must be the same size?
Congratulations! I think I have, if not a perfect understanding of this, at least more of one than I had yesterday! Thanks :)
You're welcome. I like to think that I'm good at explaining this kind of thing. ;) To give credit where credit is due, it was the long comment thread with DanArmak that helped me see what the source of your confusion was. And, indeed, all the ways of counting them matter. Mathematicians really, really hate it when you can do the same thing two different ways and get two different answers.
I learned about all this from a very interesting book I once read, which has a section on Georg Cantor, who was the one who thought up these ways of comparing the sizes of different infinite sets in the first place.