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\section{The Ramsey numbers $R(3, k)$}

\begin{fcthm}[Erdos-Szekeres, 1935]
  \label{thm:erdosszekeres}
  $R(l, k) \le {k + l - 2 \choose l - 1}$.
\end{fcthm}

\begin{proof}
  Induction.
\end{proof}

\begin{note*}
  $R(k) = R(k, k) \le {2(k - 1) \choose k - 1} \sim c \frac{4^k}{\sqrt{k}}$.
\end{note*}

\begin{fccoro}[]
  \label{coro:trivialr3kupperbound}
  $R(3, k) \le k^2$.
\end{fccoro}

\textbf{Idea 1:} If randomly colour $K_n$, we have $\sim \frac{1}{8} {n \choose
3}$ triangles \intosad

\textbf{Idea 2:} We want to bias our distribution in favour of $K_k$ -- colour
``red''.
Say $G \sim G(n, p)$.
Sad news: if $p \gg \frac{1}{n}$ then $G$ contains a $K_3$ with high
probability. \intosad

\begin{fcthm}[Erdos, 1960s]
  \label{thm:firstr3klowerbound}
  $R(3, k) \ge c \left( \frac{k}{\log k} \right)^{3/2}$ for some $c > 0$.
\end{fcthm}

\textbf{Proof idea:}
Sample $G \sim G(n, p)$, then define $\tilde{G} = G - (\text{a vertex from
each triangle})$.
For this to work we need $\frac{\# \text{vertices}}{2} > \# \text{triangles}$.
So $\frac{n}{2} > p^3 {n \choose 3}$.
So take $p \sim n^{-2/3}$.

\begin{remark*}
  This is called the ``deletion method''.
\end{remark*}

\textbf{Fact} (Markov)\textbf{.} Let $X$ be a non-negative random variable.
Let $t > 0$.
Then
\[
  \Pbb(X \ge t)
  \le \frac{\Ebb X}{t}
.\]
That is
\[
  \Pbb(X \ge s\Ebb X)
  \le \frac{1}{s}
,\]
for any $s > 0$.

\textbf{Fact.} For $1 \le k \le n$ we have
\[
  {n \choose k}
  \le \left( \frac{en}{k} \right)^k
\]
and
\[
  {n \choose k}
  \ge \left( \frac{n}{k} \right)^k
.\]

\begin{fcdefn}[Independence number]
  \glsadjdefn{ind}{independent}{set}%
  \glssymboldefn{indnum}%
  Let $G$ be a graph.
  Then $I \subset V(G)$ is \emph{independent} if $x \not\sim y$ in $G$ for all
  $x, y \in I$.

  Let $\alpha(G) = \text{size of the largest independent set}$.
\end{fcdefn}

\begin{center}
  \includegraphics[width=0.3\linewidth]{images/6e7733c2214b44de.png}
\end{center}

\begin{note*}
  To prove \cref{thm:firstr3klowerbound}, we are trying to construct a triangle
  free graph on $k^{3/2 - o(1)}$ vertices with $\indnum(G) < k$.
\end{note*}

\begin{fclemma}[]
  \label{lemma:randomindnum}
  Assuming:
  - $\frac{1}{n} \ll p \le \half$
  - $G \sim G(n, p)$
  Then:
  \[
    \indnum(G)
    \le (2 + o(1)) \frac{\log np}{p}
  ,\]
  with probability tending to $1$ as $n \to \infty$ (known as ``with high
  probability (w.h.p.)'').
\end{fclemma}

\begin{remark*}
  We actually have $=$ in this lemma.
\end{remark*}

\begin{proof}
  Let $k = (2 + \delta) \frac{\log np}{p}$, for some $\delta > 0$.
  Let $X$ be the random variable that counts the number of independent $k$ sets
  in $G$.
  Using $1 - x \le e^{-x}$,
  \begin{align*}
    \Ebb X
    &= \Ebb \sum_{I \in [n]^{(k)}} \indicator{\text{$I$ is \gls{ind}}} \\
    &= \sum_{I \in [n]^{(k)}} \Pbb(\text{$I$ is \gls{ind}}) \\
    &= {n \choose k} (1 - p)^{{k \choose 2}} \\
    &\le \left( \frac{en}{k} \right)^k e^{-p {k \choose 2}} \\
    &= \left( \frac{en}{k} \cdot e^{- \left( \frac{k - 1}{2} \right) p}
    \right)^k \\
    &\to 0
  \end{align*}
  by plugging in $k$.
  So by Markov,
  \[
    \Pbb(X \ge 1)
    \le \Ebb X
    \to 0
    .
    \qedhere
  \]
\end{proof}

We have a method for bounding the probability that $X$ is large: Markov says
that $\Pbb(X > t) \le \frac{\Ebb X}{t}$.
What about bounding the probability that it is small?

\textbf{Fact:} Let $X$ be a random variable with $\Ebb X$, $\Var X$ finite.
Then for all $t > 0$,
\[
  \Pbb(|X - \Ebb X| \ge t)
  \le \frac{\Var X}{t^2}
.\]

Reminder:
\[
  \Var X
  = \Ebb X^2 - (\Ebb X)^2
  = \Ebb(X - \Ebb X)^2
.\]

\begin{fclemma}[]
  Assuming:
  - $\frac{1}{n} \ll p \le 1$
  - $G \sim G(n, p)$
  Then:
  \[
    \# \text{triangles in $G$}
    = (1 + o(1)) p^3 {n \choose 3}
  \]
  with high probability.
\end{fclemma}

\begin{proof}
  Let $X$ be the random variable counting the number of triangles in $G$.
  Note
  \[
    X
    = \sum_{T \in [n]^{(3)}} \indicator{T^{(2)} \subset G}
  .\]
  Mean is straightforward:
  \[
    \Ebb X
    = p^3 {n \choose 3}
  .\]
  Variance is a little more tricky.
  First:
  \[
    X^2
    = \sum_{S, T \in [n]^{(3)}} \indicator{S^{(2)}, T^{(2)} \subset G}
  .\]