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\section*{Ramsey Theory}

\paragraph{Example:}
\mbox{}\\    Given 6 people,
\\    either there are 3 who all like each other,
\\    or there are 3 no two of whom like each other.

\paragraph{Abstract version:}
\mbox{}\\    $K_n :=$ "complete graph on $n$ vertices"
\mbox{}\\ \mbox{\hspace{2em}}$= n$ points with an edge between each pair.

    Colour the edges of $K_6$ each either red or blue,
\\    then there's a red copy of $K_3$ or there's a blue copy of $K_3$;
\\    i.e. there is a monochromatic triangle.

    Denote this fact
\mbox{}\\ \mbox{\hspace{2em}}$K_6 \rightarrow  K_3,K_3$

\paragraph{Proof:}
\mbox{}\\    Pick a vertex $v_0$.
\\    Consider the 5 edges from it.
\\    3 of them are red or 3 of them are blue, since $5 > (3-1) + (3-1)$.
\\    Say 3 are red, and consider the 3 other vertices of these red edges.
\\    If the edges between them are all blue, they form a blue triangle and
    we're done.
\\    Else, some edge is red; but then it along with the edges from $v_0$ form a
    red triangle, and we're done.

\paragraph{Ramsey's Theorem for 2-coloured graphs:}
\mbox{}\\    Given $n$ and $m$ positive integers,
\\    there exists $r$ such that for any red-blue colouring of the edges of $K_r$,
\\    there are $n$ vertices all edges between which are red
    or there are $m$ vertices all edges between which are blue.
\paragraph{Notation:}
\mbox{}\\    We write
\mbox{}\\ \mbox{\hspace{2em}}$K_r \rightarrow  K_n,K_m$
\\    to mean that $r$ has this property,
\\    and we let $r(m,n)$ ("the (m,n)th Ramsey number") be the least such $r$.
\paragraph{Remarks:}
\mbox{}\\    We saw that $K_6 \rightarrow  K_3,K_3$;
\\    it's easy to see that $K_5 \not\rightarrow  K_3,K_3$,
\\    so $r(3,3) = 6$.

    It has been shown that
\mbox{}\\ \mbox{\hspace{2em}}$r(3,4) = 9$
\mbox{}\\ \mbox{\hspace{2em}}$r(3,5) = 14$
\mbox{}\\ \mbox{\hspace{2em}}$r(4,4) = 18$

    $r(5,5)$ is unknown! All we know is
\mbox{}\\ \mbox{\hspace{2em}}$43 \leq  r(5,5) \leq  49$.

    Erd\"{o}s:
\\	"Suppose aliens invade the earth and threaten to obliterate it in a
	year's time unless human beings can find the Ramsey number for red
	five and blue five. We could marshal the world's best minds and
	fastest computers, and within a year we could probably calculate the
	value. If the aliens demanded the Ramsey number for red six and blue
	six, however, we would have no choice but to launch a preemptive
	attack."

\paragraph{Proof of Theorem:}
\mbox{}\\    Suppose inductively that
\mbox{}\\ \mbox{\hspace{2em}}$K_b \rightarrow  K_{n-1},K_m$
\\    and
\mbox{}\\ \mbox{\hspace{2em}}$K_c \rightarrow  K_n,K_{m-1}$.

    We show that
\mbox{}\\ \mbox{\hspace{2em}}$K_{b+c} \rightarrow  K_n,K_m$.

    So colour $K_{b+c}$, and suppose there's no red $K_n$ and no blue $K_m$.

    Pick a vertex $v_0$; consider the $b+c-1$ edges from it.
\\    Since $b+c-1 > (b-1)+(c-1)$,
\\    $b$ of the edges are red or $c$ of the edges are blue.

    Say $b$ are red.
\\	Consider the $K_b$ formed by the vertices these edges connect to $v_0$.
\\	By the inductive hypothesis, it contains a red $K_{n-1}$ or a blue $K_m$.
\\	If it contains a red $K_{n-1}$, adjoining $v_0$ yields a red $K_n$;
\\	contradiction.
\\	If it contains a blue $K_m$, then so does our original $K_{b+c}$;
\\	contradiction.

    A symmetrical argument applies in the case that $c$ of the edges from
    $v_0$ are blue.

\paragraph{Remark:}
\mbox{}\\    This proof yields a recursive upper bound on the Ramsey numbers:
\mbox{}\\ \mbox{\hspace{2em}}$r(m,n) \leq  r(n-1,m) + r(n,m-1)$
\\    (but this is far from sharp).
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