MAT 307 : Combinatorics Lecture 14 : Topological methods

1 The Borsuk-Ulam theorem We have seen how combinatorics borrows from probability theory. Another area which has been very beneficial to combinatorics, perhaps even more surprisingly, is topology. We have already seen Brouwer's fixed point theorem and its combinatorial proof. Theorem 1 (Brouwer). For any continuous function f : B n → B n , there is a point x ∈ B n such that f (x) = x. A more powerful topological tool which seems to stand at the root of most combinatorial applications is a somewhat related result which can be stated as follows. Here, S n denotes the n-dimensional sphere, i.e. the surface of the (n + 1)-dimensional ball B n+1. such that f (x) = f (−x). There are many different proofs of this theorem, some of them elementary and some of them using a certain amount of the machinery of algebraic topology. All the proofs are, however, more involved than the proof of Brouwer's theorem. We will not give the proof here. In the following, we use a corollary (in fact an equivalent re-statement of the Borsuk-Ulam theorem). Theorem 3. For any covering of S n by n + 1 open or closed sets A 0 ,. .. , A n , there is a set A i which contains two antipodal points x, −x. Let's just give some intuition how this is related to Theorem 2. For now, let us assume that all the sets A i are closed. (The extension to open sets is a technicality but the idea is the same.) We define a continuous function f : S n → R n , f (x) = (dist(x, A 1), dist(x, A 2),. .. , dist(x, A n)) where dist(x, A) = inf y∈A ||x − y|| is the distance of x from A. By Theorem 2, there is a point x ∈ S n such that f (x) = f (−x). This means that dist(x, A i) = dist(−x, A i) for 1 ≤ i ≤ n. If dist(x, A i) = 0 for some i, then we are done. If dist(x, A i) = dist(−x, A i) = 0 for all i ∈ {1,. .. , n}, it means that x, −x / ∈ A 1 ∪. .. ∪ A n. But then x, −x ∈ A 0. Similarly to the previous sections, Kneser graphs are derived from the intersection …