This question is a bit abstract, but I found it fun to play around with it. I've broken it up into pieces that hopefully provide a large enough range of difficulties to make it interesting.


Easiest version:
You're given an infinite number of intervals E_n of real numbers of the form [a, b] (that is, [a, b] = the set of points x where a <= x <= b), one for every whole number n. Each interval fits in the unit interval [0, 1], so 0 <= a < b <= 1. Is some point contained in infinitely many of the E_n's?

Answer (highlight): no. Take the intervals [0, 1/2], [1/2, 3/4], [3/4, 7/8], ....


Easy version:
Add the condition that each interval E_n = [a_n, b_n] has a certain minimum length, say D > 0. That is, b_n - a_n >= D. Is there some point contained in infinitely many of the E_n's?

Answer (highlight): yes. There are several ways to see this. My current favorite is to look at the midpoints of the intervals and find a strictly increasing or strictly decreasing sequence of midpoints (one must exist; this is another interesting thing to show). Such a sequence must converge (yet another interesting thing to think about; this can be taken as one of the defining properties of the real numbers, or it can be derived in any particular construction), and the point of convergence is in infinitely many of the intervals we've taken midpoints from. In fact the interval centered on the point of convergence of length D (possibly excluding its left and right end point) has the property that each point is in infinitely many of the above intervals.


Moderate version:
Pick some whole number N. For each interval E_n, let E_n be the union of up to N disjoint intervals, where the sum of the lengths of the intervals is still >= D for each E_n. Is there some point contained in infinitely many of the E_n's?

Answer (highlight): yes. For each E_n, one of the intervals making it up must be of length D/N or greater--for otherwise the length of E_n is less than D. Apply the easy version to these intervals.


Hard version:
For each interval E_n, let E_n be the union of a possibly infinite sequence of disjoint intervals, where the (possibly infinite) sum of the lengths is >= D for each E_n. For each whole number K, is there some point contained in at least K different E_n's? If so, what is the minimum "length" (defined as best you can) of the set of points which are contained in at least K different E_n's?

Fair warning: Most of the properties of "length" you might wish were true can be made true rigorously, though I won't go into details. You can lose essentially nothing by taking the E_n's to be finite unions of disjoint intervals instead, and, for a given K, by considering only some (finite but large) number of the E_n. This should bypass the technicalities associated with the concept of "length" almost entirely.

Answer (highlight): Yes, and D. Look at the "length" of points in 1, 2, 3, 4, etc. or more intervals as you keep adding intervals. Try D=1/2 first. After adding enough intervals, the first "few" lengths we were considering must have grown to, say, larger than D/2. Generalize this to get the result.


Hardest version:
For each interval E_n, instead make it a Lebesgue-measurable set of measure >= D. What is the minimum measure of the set of points contained in infinitely many of the intervals E_n?

Fair warning: I'd be impressed by anyone who can answer this (even letting the E_n be as in the "hard" version) without using results from measure theory. This entire question took me a week or so of pondering, though it shouldn't have taken so long--I had forgotten something I'd recently shown for another problem and so spent quite a while unfruitfully messing around.

Answer (highlight): D. My proof is quite long and would be pretty incomprehensible without some background in measure theory and analysis. The basic idea is to show that for each K, the points in at least K E_n's has measure D, so their intersection has measure at least D. Doing so involves the reasoning I suggested for the "hard" version. This minimum is achieved trivially by E_n = [0, D], so the result is the best possible.