Ramsey’s Theorem on Graphs

Imagine that you have 6 people at a party. We assume that, for every pair of them, either THEY KNOW EACH OTHER or NEITHER OF THEM KNOWS THE OTHER. So we are assuming that if x knows y, then y knows x. Claim: Either there are at least 3 people all of whom know one another, or there are at least 3 people no two of whom know each other (or both). Proof of Claim: Let the people be p1, p2, p3, p4, p5, p6. Now consider p6. Among the other 5 people, either there are at least 3 people that p6 knows, or there are at least 3 people that p6 does not know. Why is this? Well, suppose that, among the other 5 people, there are at most 2 people that p6 knows, and at most 2 people that p6 does not know. Then there are only 4 people other than p6, which contradicts the fact that there are 5 people other than p6. Suppose that p6 knows at least 3 of the others. We consider the case where p6 knows p1, p2, and p3. All the other cases are similar. If p1 knows p2, then p1, p2, and p6 all know one another. HOORAY! If p1 knows p3, then p1, p3, and p6 all know one another. HOORAY! If p2 knows p3, then p2, p3, and p6 all know one another. HOORAY! What if none of these scenarios holds? Then none of these three people (p1, p2, p3) knows either of the other 2. HOORAY! End of Proof of Claim We want to generalize this observation.