What is the maximum number of closed intervals that can be specified on the number line such that any three of them intersect, but the intersection of any four is empty.
Let $J_{1}$ be the smallest right endpoint of the intervals given in accordance with the condition; we discard from the intervals - according to the condition, at most three - those that contain $J_{1}$. None of the remaining intervals contain $J_{1}$, and repeating the above procedure on them, we obtain the endpoint $J_{2}$, and we can again discard at most three more intervals. However, no further interval can remain, since any such interval could not have a common point with either a $J_{1}$ or a $J_{2}$ endpoint, and these two do not have a common point with each other either.

We have thus found that at most 6 intervals can be given in the prescribed manner. This is possible, however, if we "nest" three of them inside each other.

$1988-11-389-1 . e p s$

Remark. In the course of the solution, we essentially showed that if in a system of intervals any three have two intersecting, then there exist at most two points - these were $J_{1}$ and $J_{2}$ - that "capture" all the intervals, meaning that each interval contains at least one of these points.

This is generally true. In any system of intervals, the minimum number of points that can capture the interval system is equal to the maximum number of intervals that can be given without any pair having a common point. Fewer points are clearly insufficient to capture all intervals, since different points are needed to capture disjoint intervals. And that it is possible to capture all intervals with this many points can be proven with the reasoning of the solution, since the selected $J_{1}, J_{2}, \ldots J_{k}$ endpoints are the endpoints of a system of disjoint intervals.