Helly's theorem is a basic result in discrete geometry on the intersection of convex sets. It was discovered by Eduard Helly in 1913, but not published by him until 1923, by which time alternative proofs by and had already appeared. Helly's theorem gave rise to the notion of a Helly family. Statement Let be a finite collection of convex subsets of , with . If the intersection of every of these sets is nonempty, then the whole collection has a nonempty intersection; that is, For infinite collections one has to assume compactness: Let be a collection of compact convex subsets of , such that every subcollection of cardinality at most has nonempty intersection. Then the whole collection has nonempty intersection. Proof We prove the finite version, using Radon's theorem as in the proof by . 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). The proof is by induction: Base case: Let . By our assumptions, for every there is a point that is in the common intersection of all with the possible exception of . Now we apply Radon's theorem to the set which furnishes us with disjoint subsets of 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 We shall prove that 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 . Inductive Step: Suppose and that the statement is true for . The argument above shows that any subcollection of sets will