Proof of The Theorem
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.
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 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 colours. The proof is again by induction, this time on the number of colours c. We have the result for c = 1 (trivially) and for c = 2 (above). Now let c > 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 colours and 2 colours, and therefore exists and is a 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 colour its edges with c colours. Now 'go colour-blind' and pretend that c − 1 and c are the same colour. Thus the graph is now (c − 1)-coloured. By the inductive hypothesis, it contains either a Kni monochromatically coloured with colour i for some 1 ≤ i ≤ (c − 2) or a KR(nc−1,nc)-coloured in the 'blurred colour'. 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. – In the 2-color case, if R(r − 1, s) and R(r, s − 1) are both even, the induction inequality can be strengthened to R(r, s) ≤ R(r − 1, s) + R(r, s − 1) − 1.
Read more about this topic: Ramsey's Theorem
Famous quotes containing the words proof of the, proof of, proof and/or theorem:
“The fact that several men were able to become infatuated with that latrine is truly the proof of the decline of the men of this century.”
—Charles Baudelaire (18211867)
“Sculpture and painting are very justly called liberal arts; a lively and strong imagination, together with a just observation, being absolutely necessary to excel in either; which, in my opinion, is by no means the case of music, though called a liberal art, and now in Italy placed even above the other twoa proof of the decline of that country.”
—Philip Dormer Stanhope, 4th Earl Chesterfield (16941773)
“The fact that several men were able to become infatuated with that latrine is truly the proof of the decline of the men of this century.”
—Charles Baudelaire (18211867)
“To insure the adoration of a theorem for any length of time, faith is not enough, a police force is needed as well.”
—Albert Camus (19131960)