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

because the complete graph on one node has no edges.

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.