Ramsey's Theorem

Theorem
In any coloring of the edges of a sufficiently large complete graph, one will find monochromatic complete subgraphs.

For 2 colors, Ramsey's theorem states that for any pair of positive integers $$\left({r, s}\right)$$, there exists a least positive integer $$R \left({r, s}\right)$$ such that for any complete graph on $$R \left({r, s}\right)$$ 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.

More generally, for any given number of colors $$c$$, and any given integers $$n_1, \ldots, n_c$$, there is a number $$R \left({n_1, \ldots, n_c}\right)$$ such that:
 * if the edges of a complete graph of order $$R \left({n_1, \ldots, n_c}\right)$$ are colored with $$c$$ different colours, then for some $$i$$ between $$1$$ and $$c$$, it must contain a complete subgraph of order $$n_i$$ whose edges are all color $$i$$.

This number $$R \left({n_1, \ldots, n_c}\right)$$ is called the Ramsey number for $$n_1, \ldots, n_c$$.

The special case above has $$c = 2$$ (and $$n_1 = r$$ and $$n_2 = s$$).

Here $$R \left({r, s}\right)$$ 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:
 * $$\forall n \in \N: R \left({n, 1}\right) = R \left({1, n}\right) = 1$$.

This is the base case.

We prove that $$R \left({r, s}\right)$$ exists by finding an explicit bound for it.

By the inductive hypothesis, $$R \left({r - 1, s}\right)$$ and $$R \left({r, s - 1}\right)$$ exist.

Proof for Two Colors
We will show that:
 * $$R \left({r, s}\right) \le R \left({r - 1, s}\right) + R \left({r, s - 1}\right)$$

Consider a complete graph on $$R \left({r - 1, s}\right) + R \left({r, s - 1}\right)$$ vertices.

Pick a vertex $$v$$ from the graph, and partition the remaining vertices into two sets $$M$$ and $$N$$, such that for every vertex $$w$$:
 * $$w \in M$$ if $$\left({v, w}\right)$$ is blue;
 * $$w \in N$$ if $$\left({v, w}\right)$$ is red.

Because the graph has $$R \left({r - 1, s}\right) + R \left({r, s - 1}\right) = \left|{M}\right| + \left|{N}\right| + 1$$ vertices, it follows that either $$\left|{M}\right| \ge R \left({r - 1, s}\right)$$ or $$\left|{N}\right| \ge R \left({r, s - 1}\right)$$.

In the former case, if $$M$$ has a red $$K_s$$ then so does the original graph and we are finished.

Otherwise $$M$$ has a blue $$K_{r−1}$$, and so $$M \cup \left\{{v}\right\}$$ has blue $$K_r$$ 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 $$c > 2$$.

Proof for More than Two Colors
We will show that:
 * $$R \left({n_1, \ldots, n_c}\right) \le R \left({n_1, \ldots, n_{c-2}, R \left({n_{c-1}, n_c}\right)}\right)$$

We note that the RHS only contains only Ramsey numbers for $$c - 1$$ colors and 2 colors, and therefore exists.

Thus it is the finite number $$t$$, by the inductive hypothesis.

So proving this will prove the theorem.

Consider a graph on $$t$$ vertices and color its edges with $$c$$ colors.

Now "go color-blind" and pretend that $$c - 1$$ and $$c$$ are the same color.

Thus the graph is now $$\left({c - 1}\right)$$-colored.

By the inductive hypothesis, it contains either:


 * a complete monochromatic graph $$K_{n_i}$$ with color $$i$$ for some $$1 \le i \le \left({c - 2}\right)$$, or
 * a complete monochromatic graph $$K_{R \left({n_{c - 1}, n_c}\right)}$$-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 $$R \left({n_{c-1}, n_c}\right)$$ we must have either:
 * a complete $$\left({c - 1}\right)$$-monochromatic graph $$K_{n_{c-1}}$$, or
 * a complete $$c$$-monochromatic graph $$K_{n_c}$$.

In either case the proof is complete.