Helly's theorem

      Helly's theorem for the Euclidean plane: if a family of convex sets has a nonempty intersection for every triple of sets, then the whole family has a nonempty intersection.

      Helly's theorem is a basic result in discrete geometry describing the ways that convex sets may intersect each other. It was discovered by Eduard Helly in 1913,[1] but not published by him until 1923, by which time alternative proofs by Radon (1921) and König (1922) had already appeared. Helly's theorem gave rise to the notion of a Helly family.

      Statement

      Suppose that

      X_1,X_2,\dots,X_n

      is a finite collection of convex subsets of \Bbb{R}^d, where  n > d . If the intersection of every d+1 of these sets is nonempty, then the whole collection has a nonempty intersection; that is,

      \bigcap_{j=1}^n X_j\ne\varnothing.

      For infinite collections one has to assume compactness: If \{X_\alpha\} is a collection of compact convex subsets of R^d and every subcollection of cardinality at most d+1 has nonempty intersection, then the whole collection has nonempty intersection.

      ↑Jump back a section

      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 n=d+2. By our assumptions, there is a point x_1 that is in the common intersection of

      X_2,X_3,\dots,X_n.

      Likewise, for every

      j=1,2,3,\dots,n

      there is a point x_j that is in the common intersection of all X_i with the possible exception of X_j. We now apply Radon's theorem to the set

      A=\{x_1,x_2,\dots,x_n\}.

      Radon's theorem tells us that there are disjoint subsets A_1,A_2\subset A such that the convex hull of A_1 intersects the convex hull of A_2. Suppose that p is a point in the intersection of these two convex hulls. We claim that

      p\in\bigcap_{j=1}^n X_j.

      Indeed, consider any j\in\{1,2,...,n\}. Note that the only element of A that may not be in X_j is x_j. If x_j\in A_1, then x_j\notin A_2, and therefore X_j\supset A_2. Since X_j is convex, it then also contains the convex hull of A_2 and therefore also p\in X_j. Likewise, if x_j\notin A_1, then X_j\supset A_1, and by the same reasoning p\in X_j. Since p is in every X_j, it must also be in the intersection.

      Above, we have assumed that the points x_1,x_2,\dots,x_n are all distinct. If this is not the case, say x_i=x_k for some i\ne k, then x_i is in every one of the sets X_j, and again we conclude that the intersection is nonempty. This completes the proof in the case n=d+2.

      Now suppose that n>d+2 and that the statement is true for n-1, by induction. The above argument shows that any subcollection of d+2 sets will have nonempty intersection. We may then consider the collection where we replace the two sets X_{n-1} and X_n with the single set

      X_{n-1}\cap X_n.

      In this new collection, every subcollection of d+1 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.

      ↑Jump back a section

      References

      ↑Jump back a section

      Read in another language

      This page is available in 4 languages

      Last modified on 28 April 2013, at 01:02