Proof
We prove the finite version, using Radon's theorem as in the proof by Radon (1921). The infinite version then follows by the finite intersection property characterization of compactness: a collection of closed subsets of a compact space has a non-empty intersection if and only if every finite subcollection has a non-empty intersection (once you fix a single set, the intersection of all others with it are closed subsets of a fixed compact space).
Suppose first that . By our assumptions, there is a point that is in the common intersection of
Likewise, for every
there is a point that is in the common intersection of all with the possible exception of . We now apply Radon's theorem to the set
Radon's theorem tells us that there are disjoint subsets such that the convex hull of intersects the convex hull of . Suppose that is a point in the intersection of these two convex hulls. We claim that
Indeed, consider any . Note that the only element of that may not be in is . If, then, and therefore . Since is convex, it then also contains the convex hull of and therefore also . Likewise, if, then, and by the same reasoning . Since is in every, it must also be in the intersection.
Above, we have assumed that the points are all distinct. If this is not the case, say for some, then is in every one of the sets, and again we conclude that the intersection is nonempty. This completes the proof in the case .
Now suppose that and that the statement is true for, by induction. The above argument shows that any subcollection of sets will have nonempty intersection. We may then consider the collection where we replace the two sets and with the single set
In this new collection, every subcollection of sets will have nonempty intersection. The inductive hypothesis therefore applies, and shows that this new collection has nonempty intersection. This implies the same for the original collection, and completes the proof.
Read more about this topic: Helly's Theorem
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