Ramsey's Theorem

Theorem
In any coloring of the edges of a sufficiently large complete graph (that is, a simple graph in which an edge connects every pair of vertices), one will find monochromatic complete subgraphs. For 2 colors, Ramsey's theorem states that for any pair of positive integers (r,s), there exists a least positive integer R(r,s) such that for any complete graph on R(r,s) vertices, whose edges are colored red or blue, there exists either a complete subgraph on r vertices which is entirely red, or a complete subgraph on s vertices which is entirely blue. Here R(r,s) signifies an integer that depends on both r and s. It is understood to represent the smallest integer for which the theorem holds.

Proof
First we prove the theorem for the 2-color 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.

Claim: R(r, s) ≤ R(r − 1, s) + R(r, s − 1): Consider a complete graph on R(r − 1, s) + R(r, s − 1) vertices. 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 $$ M \cup \{v\} $$ 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. We now prove the result for the general case of c colors. The proof is again by induction, this time on the number of colors c. We have the result for c = 1 (trivially) and for c = 2 (above). Now let c &gt; 2.

Claim: R(n1, ..., nc) ≤ R(n1, ..., nc−2, R(nc−1, nc))

Note, that the right hand side only contains Ramsey numbers for c − 1 colors and 2 colors, and therefore exists and is the finite number t, by the inductive hypothesis. Thus, proving the claim will prove the theorem.

Proof of claim: Consider a graph on t vertices and color its edges with c colours. Now 'go color-blind' and pretend that c − 1 and c are the same color. Thus the graph is now (c − 1)-colored. By the inductive hypothesis, it contains either a Kni  monochromatically coloured with color i for some 1 ≤ i ≤ (c − 2) or a KR(nc−1,nc) -colored in the 'blurred color'. In the former case we are finished. In the latter case, we recover our sight again and see from the definition of R(nc−1, nc) we must have either a (c − 1)-monochrome Knc−1 or a c-monochrome Knc. In either case the proof is complete.