Ramsey's theorem

In combinatorial mathematics, Ramsey's theorem states that one will find monochromatic cliques in any edge labelling (with colours) of a sufficiently large complete graph. To demonstrate the theorem for two colours (say, blue and red), let r and s be any two positive integers. Ramsey's theorem states that there exists a least positive integer R(r, s) for which every blue-red edge colouring of the complete graph on R(r, s) vertices contains a blue clique on r vertices or a red clique on s vertices. (Here R(r, s) signifies an integer that depends on both r and s.)Lemma 1. R(r, s) ≤ R(r − 1, s) + R(r, s − 1):Lemma 2. If c>2, then R(n1, …, nc) ≤ R(n1, …, nc−2, R(nc−1, nc)).Theorem. Let X be some infinite set and colour the elements of X(n) (the subsets of X of size n) in c different colours. Then there exists some infinite subset M of X such that the size n subsets of M all have the same colour. In combinatorial mathematics, Ramsey's theorem states that one will find monochromatic cliques in any edge labelling (with colours) of a sufficiently large complete graph. To demonstrate the theorem for two colours (say, blue and red), let r and s be any two positive integers. Ramsey's theorem states that there exists a least positive integer R(r, s) for which every blue-red edge colouring of the complete graph on R(r, s) vertices contains a blue clique on r vertices or a red clique on s vertices. (Here R(r, s) signifies an integer that depends on both r and s.) Ramsey's theorem is a foundational result in combinatorics. The first version of this result was proved by F. P. Ramsey. This initiated the combinatorial theory now called Ramsey theory, that seeks regularity amid disorder: general conditions for the existence of substructures with regular properties. In this application it is a question of the existence of monochromatic subsets, that is, subsets of connected edges of just one colour. An extension of this theorem applies to any finite number of colours, rather than just two. More precisely, the theorem states that for any given number of colours, c, and any given integers n1, …, nc, there is a number, R(n1, …, nc), such that if the edges of a complete graph of order R(n1, ..., nc) are coloured with c different colours, then for some i between 1 and c, it must contain a complete subgraph of order ni whose edges are all colour i. The special case above has c = 2 (and n1 = r and n2 = s). Suppose the edges of a complete graph on 6 vertices are coloured red and blue. Pick a vertex, v. There are 5 edges incident to v and so (by the pigeonhole principle) at least 3 of them must be the same colour. Without loss of generality we can assume at least 3 of these edges, connecting the vertex, v, to vertices, r, s and t, are blue. (If not, exchange red and blue in what follows.) If any of the edges, (r, s), (r, t), (s, t), are also blue then we have an entirely blue triangle. If not, then those three edges are all red and we have an entirely red triangle. Since this argument works for any colouring, any K6 contains a monochromatic K3, and therefore R(3, 3) ≤ 6. The popular version of this is called the theorem on friends and strangers. An alternative proof works by double counting. It goes as follows: Count the number of ordered triples of vertices, x, y, z, such that the edge, (xy), is red and the edge, (yz), is blue. Firstly, any given vertex will be the middle of either 0 × 5 = 0 (all edges from the vertex are the same colour), 1 × 4 = 4 (four are the same colour, one is the other colour), or 2 × 3 = 6 (three are the same colour, two are the other colour) such triples. Therefore, there are at most 6 × 6 = 36 such triples. Secondly, for any non-monochromatic triangle (xyz), there exist precisely two such triples. Therefore, there are at most 18 non-monochromatic triangles. Therefore, at least 2 of the 20 triangles in the K6 are monochromatic. Conversely, it is possible to 2-colour a K5 without creating any monochromatic K3, showing that R(3, 3) > 5. The unique colouring is shown to the right. Thus R(3, 3) = 6. The task of proving that R(3, 3) ≤ 6 was one of the problems of William Lowell Putnam Mathematical Competition in 1953, as well as in the Hungarian Math Olympiad in 1947. First we prove the theorem for the 2-colour case, by induction on r + s. It is clear from the definition that for all n, R(n, 1) = R(1, n) = 1. This starts the induction. We prove that R(r, s) exists by finding an explicit bound for it. By the inductive hypothesis R(r − 1, s) and R(r, s − 1) exist. Proof. Consider a complete graph on R(r − 1, s) + R(r, s − 1) vertices whose edges are coloured with two colours. Pick a vertex v from the graph, and partition the remaining vertices into two sets M and N, such that for every vertex w, w is in M if (v, w) is blue, and w is in N if (v, w) is red. Because the graph has R(r − 1, s) + R(r, s − 1) = |M| + |N| + 1 vertices, it follows that either |M| ≥ R(r − 1, s) or |N| ≥ R(r, s − 1). In the former case, if M has a red Ks then so does the original graph and we are finished. Otherwise M has a blue Kr−1 and so M ∪ {v} has blue Kr by definition of M. The latter case is analogous. Thus the claim is true and we have completed the proof for 2 colours.