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If you start with an infinite set, you can have a sequence of nested sets which converge to a single point. (ie Intersection of $\left(\large\frac{-1}{n}, \frac{1}{n}\right)$ as $n\to \infty$)

However, at no time during the sequence is there a first element with only one point. In fact, for any finite (but unbounded) $n$, the number of points in the interval is uncountably large. So, how do you understand the idea that this infinite intersection contains just one point? How do you make sense of it?

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"In mathematics you don't understand things. You just get used to them." - John von Neumann. – Asaf Karagila Sep 30 '10 at 14:44
I don't think it is helpful to think of a sequence of sets as "converging". Here we are just considering intersections and as others have pointed out the question is: which elements are in all sets in the sequence? Of course one can find a nested chain of uncountable subsets of $\mathbb{R}$ whose intersection is empty. – Robin Chapman Sep 30 '10 at 18:03
@Robin: I disagree. The intersection of a decreasing sequence of sets, as in this case, is a reasonable notion of convergence, especially in measure theory, where (finite) measures are continuous with respect to this sort of convergence. Indeed, you will often see limit notation used explicitly for such set convergence. – Nate Eldredge Oct 1 '10 at 12:41
up vote 3 down vote accepted

A point is in the intersection if it's in every set of the sequence. Focus on that. No matter how close to $0$ a point is (excluding $0$ itself), there will be a sufficiently large $n$ such that the set $(-1/n,1/n)$ does not contain it, and therefore that point will not be in the (infinite) intersection, because, again, a point is in the intersection iff and it is in every set specified by the intersection.

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I think the source of the confusion is this: taking the cardinality of a set seems like such a simple operation that we subconsciously expect it to be continuous. We might think that simple limiting operations like countable nested intersections should give convergent cardinalities. But the simple fact is that they don't; our subconscious guess is just wrong, and this example proves it. Cardinality is not continuous in this sense.

This is related to the standard "paradox" of the supertask where at time $1-2^{-n}$ you add to a box two balls numbered $2n-1$ and $2n$, and remove the one numbered $n$. Despite the fact that the cardinality of the balls in the box is always increasing, in the limit at time 1 the box is empty. Again, the paradox arises only if you feel that cardinality should be continuous. If you realize that it isn't, then you no longer see such examples as paradoxical.

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In short: if the situation doesn't fit your intuition, it may be time to refit your intuition to the situation. In this case, cardinality just isn't that pertinent a concept, while in the original problem, a different concept --- measure --- is. – Niel de Beaudrap Oct 1 '10 at 6:53
Thanks for introducting me to the notion of a supertask. Very intersting... – user2003 Oct 1 '10 at 9:33

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