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  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/031eed4190194a07.png}
  \end{center}
  Note
  \[
    Y
    = \sum_{\substack{\{y_1, \ldots, y_s\} \in (V(G))^{(s)} \\ |N(y_1) \cap
    \cdots \cap N(y_s)| < k}}
    \indicator{y_1, \ldots, y_s \in N(x_1) \cap \cdots \cap N(x_t)}
  ,\]
  so
  \begin{align*}
    \Ebb Y
    &\le \sum_{\substack{\{y_1, \ldots, y_s\} \in (V(G))^{(s)} \\ |N(y_1) \cap
    \cdots \cap N(y_s)| < k}}
    \Pbb(x_1, \ldots, x_t \in N(y_1) \cap \cdots \cap N(y_s)) \\
    &\le {n \choose s} \left( \frac{k}{n} \right)^t
  \end{align*}
  To finish, define $R$ to be $N(x_1) \cap \cdots \cap N(x_t)$ with a vertex
  deleted from each $\{y_1, \ldots, y_s\} \in \left( \bigcap_{i = 1}^t N(x_i)
  \right)^{(s)}$ with $|N(y_1) \cap \cdots \cap N(y_s)| < k$.
  By definition, $R$ is \gls{skrich}, so we now just need to ensure that $|R|
  \ge r$ holds for some choice of $x_1, \ldots, x_t$.
  Indeed,
  \begin{align*}
    |R|
    &\ge |N(x_1) \cap \cdots \cap N(x_t)|
    - Y(x_1, \ldots, x_t) \\
    \Ebb|R|
    &\ge \frac{(2m)^t}{n^{2t - 1}} - {n \choose s} \left( \frac{k}{n} \right)^t
    \\
    &\ge r
  \end{align*}
  The last line follows from the assumption.
\end{proof}

\begin{fcdefn}[Extremal number]
  \glsnoundefn{exnum}{extremal number}{extremal numbers}%
  \glssymboldefn{exnum}%
  Let $H$ be a graph.
  The extremal numbers of $H$ are
  \[
    \exinternal(n, H)
    = \max \{e(G) : G \not\supset H, \text{$G$ on $n$ vertices}\}
  .\]
\end{fcdefn}

\begin{fcthm}[Erdős-Stone]
  \label{thm:erdosstone}
  Assuming:
  - $H$ a graph
  - $\chi(H) = r \ge 3$
  Then:
  \[
    \exnum(n, H)
    = \left( 1 - \frac{1}{r - 1} + o(1) \right) {n \choose 2}
  .\]
\end{fcthm}

\begin{fcdefn}[Complete bipartite graph]
  $K_{s, t}$ is the complete bipartite graph with parts of sizes $s$, $t$
  respectively.
\end{fcdefn}

\begin{center}
  \includegraphics[width=0.3\linewidth]{images/9697a243afff4cd0.png}
\end{center}

If $\chi(H) \ge 3$, then for all $s, t$, we have $K_{s, t} \not\supset H$.
More generally, a complete $(\chi(H) - 1)$-partite graph will not contain a copy
of $H$.
\nameref{thm:erdosstone} says that these constructions are optimal, up to
$o(1){n \choose 2}$ edges.

\nameref{thm:erdosstone} doesn't help us when $\chi(H) = 2$.
In this case, it is true that
\[
  cn^{2 - \frac{1}{2t}}
  \le \exnum(n, K_{t, t})
  \le C_t n^{2 - \frac{1}{t}}
.\]
See Part II Graph Theory for proofs of both bounds.

These bounds are essentially the best known for $t \ge 5$.

We now prove a generalisation of this upper bound that holds for graphs other
than complete bipartite graphs:

\begin{fcthm}[Füredi]
  \label{thm:furedi}
  Assuming:
  - $H$ be a bipartite graph with vertex partition $A \cup B$
  - $\deg(x) \le s$ for all $x \in B$
  Then:
  \[
    \exnum(n, H)
    \le C_H n^{2 - \frac{1}{s}}
  ,\]
  for some $C_H > 0$ ($C_H$ depends only on $H$ and not on $n$).
\end{fcthm}

\begin{fclemma}[]
  \label{lemma:maxdegembed}
  Assuming:
  - $H$ be a bipartite graph with vertex partition $A \cup B$
  - $\deg(x) \le s$ for all $x \in B$
  - $G$ a graph
  - $R \subset V(G)$ a \glsref[skrich]{$(s, |V(H)|)$-rich} set
  - $|R| \ge |A|$
  Then:
  $G \supset H$.
\end{fclemma}

\begin{center}
  \includegraphics[width=0.3\linewidth]{images/a38016f563194897.png}
\end{center}

\begin{proof}
  First inject $A \to R$.
  We now inductively place the vertices $B = \{y_1, \ldots, y_l\}$.
  Assume we have embedded $y_1, \ldots, y_i$ into $G$.
  To embed $y_{i + 1}$, simply consider the $N(y_{i + 1})$.
  These have been embedded inside of $R$ and thus have $\ge |V(H)|$ common
  neighbours.
  So there must be a choice for $y_{i + 1}$ that does not clash with the
  previous choices.
\end{proof}

\begin{proof}[Proof of \nameref{thm:furedi}]
  We are given $G$ with $e(G) \ge Cn^{2 - \frac{1}{s}}$.
  We choose $C = 2|H|$, and will show $G \supset H$.
  By lemma, enough to find a \glsref[skrich]{$(s, |V(H)|)$-rich} set of size
  $\ge |A|$.
  Let $h = |V(H)|$ and $t = s$.
  Then
  \begin{align*}
    \frac{(2Cn^{2 - \frac{1}{s}})^t}{n^{2t - 1}}
    - {n \choose s} \left( \frac{h}{n} \right)^t
    &\ge (2C)^s - n^s \left( \frac{h}{n} \right)^s \\
    &= (4h)^s - h^s \\
    &\ge h \\
    &\ge |A|
  \end{align*}
  as desired.
\end{proof}

\begin{fcdefn}[Hypercube graph]
  \glssymboldefn{cube}%
  $Q_d$ denotes the \emph{hypercube graph}: the graph with $V(Q_d) = \{0, 1\}^d$
  and $x, y \in \{0, 1\}^d$ are adjacent if they differ in exactly one
  coordinate.
\end{fcdefn}

\begin{fcthm}[]
  $R(\cube_d) \le \mathcloze{2^{10d}}$.
\end{fcthm}

\begin{proof}
  Let $n = 2^{10d}$ and let $\chi$ be a colouring of $K_n$.
  Let $G$ be the graph of the majority colour.
  So $e(G) \ge \half {n \choose 2} \ge \frac{n^2}{8}$.
  So we apply \nameref{lemma:drc} to find a \skrich{d}{2^d} set of size $\ge
  2^d$.
  If $t = 2d$, then
  \begin{align*}
    \frac{(n^2 / 4)^t}{n^{2t - 1}} - \left( \frac{en}{d} \right)^d \cdot \left(
    \frac{2^d}{n} \right)^t
    &\ge \frac{n}{4^{2d}} - n^d \frac{2^{d \cdot t}}{n^{2d}} \\
    &\ge \frac{n}{4^{2d}} - \frac{2^{2d^2}}{2^{10d^2}} \\
    &\ge 2^d
  \end{align*}
  Then finish by applying \cref{lemma:maxdegembed}.

  The reasoning for choosing $t = 2d$ is that we wanted to make $\left(
  \frac{2^d}{n} \right)^t$ win against the $\left( \frac{en}{d} \right)^d$.
  Setting $t = d$ makes the powers of $n$ cancel, but then we are still left
  with the big $2^{dt}$ term, so instead we make $t$ a bit bigger than $d$.
\end{proof}

\begin{remark*}
  One can use this proof to get
  \[
    R(\cube_d)
    \le 2^{3d}
  .\]
  Burr-Erd\H{o}s conjectured that $R(\cube_d) \le C 2^d$.
  State of the art: $R(\cube_d) \le 2^{(2 - \eps)d}$ by Tikhomirov 2023.
\end{remark*}