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  Let's see that $G - T$ is triangle-free.
  If not, then $x, y, z \in V(G)$ with $x \in V_i$, $y \in V_j$, $z \in V_r$.
  So we must have $\density(V_i, V_j) \ge \frac{\eps}{2}$ and $(V_i, V_j)$ is
  \epsuniform{\frac{\eps}{4}}.

  So we can apply the ``counting lemma'' for triangles to find that the number
  of triangles is at least
  \[
    (1 - 2\eps) \eps^3 \left( \frac{n}{k} \right)^3
    \ge \left( \half \frac{\eps^2}{L^3} \right) n^3
    > \delta n^3
  ,\]
  which contradicts the initial assumption.
\end{proof}

\begin{remark*}
  \leavevmode
  \begin{itemize}
    \item This proof gives tower-type dependence
      \[
        \delta = \frac{1}{\ub{2^{2^{2^{\iddots^2}}}}_{\frac{1}{\eps^c}}}
      \]
      between $\delta$ and $\eps$.
    \item It's a big question to improve these bounds.
    \item Best known is due to Fox, who proved that
      \[
        \delta
        < \frac{1}{\ub{2^{2^{\iddots^2}}}_{\left( \log \frac{1}{\eps} \right)^C}}
      \]
      is sufficient.
  \end{itemize}
\end{remark*}

\begin{fcthm}[Roth]
  For $\eps > 0$ there exists $n(\eps) \in \Nbb$ such that if $n \ge n(\eps)$
  and $A \subset [n]$ with $|A| \ge \eps n$, then $A$ contains $a, a + d, a +
  2d$ where $d \neq 0$.
\end{fcthm}

\begin{center}
  \includegraphics[width=0.6\linewidth]{images/7b1d367a044f4394.png}
\end{center}

\begin{proof}
  For $\eps > 0$ given, let $\delta = \delta(\eps)$ be the $\delta$ from
  \nameref{thm:TRL}.
  Let $n(\eps) \ge \frac{1}{\delta(\eps)}$.
  Now given a set $A \subset [n]$, we define a graph $G$ on $V_1 \cup V_2 \cup
  V_3$ where $V_i$ are disjoint copies of $[3n]$ and:
  \begin{align*}
    E(V_1, V_2)
    &= \{\{x, x + a\} : x \in V_1, x + a \in V_2, a \in A\} \\
    E(V_2, V_3)
    &= \{\{y, y + a\} : y \in V_2, y + a \in V_3, a \in A\} \\
    E(V_1, V_3)
    &= \{\{x, x + 2a\} : x \in V_1, x + 3a \in V_3, a \in A\} \\
  \end{align*}
  Say then $x \in V_1$, $y \in V_2$, $z \in V_3$ form a triangle in $G$.
  So $y = x + a$, $z = y + b$, $z = x + 2c$ for some $a, b, c \in A$.
  Then
  \[
    y + z + (x + 2c)
    = (x + a) + (y + b) + z
  ,\]
  hence $a + b = 2c$.
  If $A$ is no $3$-term arithmetic progressions, then $G$ only contains
  triangles of the form $\{x, x + a, x + 2a\}$, where $a \in A$.
  So $G$ contains $\le 3n|A| \le 3n^2$ triangles.

  So apply the \nameref{thm:TRL} to find $T \subset E(G)$ with $|T| \le \eps n^2$
  such that $G - T$ is triangle-free.
  This is possible since $n > \frac{10}{\delta}$.
  But now notice that 
  \[
    \{\{x, x + a, x + 2a\} : x \in [n], a \in A\}
  \]
  is a set of edge-disjoint triangles of size $n|A|$ and hence
  \[
    n|A|
    \le |T|
    \le \eps n^2
  ,\]
  so $|A| \le \eps n$.
\end{proof}

\begin{remark*}
  \leavevmode
  \begin{itemize}
    \item Again this gives bad bounds on our minimum density $\eps$.
    \item Roth showed that if $A \subset [n]$, $|A| \ge c\frac{n}{\log\log n}$,
      then $A$ contains a $3$-term arithmetic progression.
    \item Kelley--Meka showed that the same holds for $|A| \ge n e^{-c (\log
      n)^{\frac{1}{12}}}$.

      There exist $A \subset [n]$ with no $3$-term arithmetic progression but
      \[
        |A|
        \ge n e^{-c\sqrt{\log n}}
      .\]
  \end{itemize}
\end{remark*}

\newpage
\section{Ramsey--Turán}

\begin{fcthm}[Turán]
  Assuming:
  - $G$ is an $n$-vertex graph with $G \not\supset K_{r + 1}$
  Then:
  \[
    e(G)
    \le \left( 1 - \frac{1}{r} \right) \frac{n^2}{2}
  .\]
  In other words,
  \[
    \ex(n, K_{r + 1})
    = \left( 1 - \frac{1}{r} \right) \frac{n^2}{2} + o(n^2)
  .\]
\end{fcthm}

Example:
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/b5d47238524b46c1.png}
\end{center}

\begin{fcdefn}[Ramsey--Turán number]
  \glssymboldefn{RT}%
  Let $H$ be a graph.
  Then we define
  \[
    \RTinternal(n, H, k)
    = \max \{e(G) : \text{$G$ is on $n$ vertices, $G \not\supset H$, $\indnum(G)
    \le k$}\}
  .\]
\end{fcdefn}

\begin{example*}
  $\RT(n, K_3, k) \le \frac{(k - 1) n}{2}$, since the neighbourhood of every
  vertex must be an independent set (using the fact that we are triangle-free).
\end{example*}

\begin{fcthm}[Szemerédi]
  $\RT(n, K_4, o(n)) \le \frac{n^2}{8} + o(n^2)$.

  This was shown to be sharp by Bollobás and Erdős.
\end{fcthm}

\begin{fcthm}[]
  For every $\eps > 0$, there exists $\delta > 0$ such that if $\indnum(G) \le
  \delta n$ and $G$ is $K_4$-free, we have
  \[
    e(G)
    \le \frac{n^2}{8} + \eps n^2
  .\]
\end{fcthm}

\begin{center}
  \includegraphics[width=0.6\linewidth]{images/a924680239b54ca2.png}
\end{center}

\begin{proof}
  Given $\eps > 0$, let $L = L \left( \frac{10}{\eps}, \frac{\eps}{2} \right)$
  and $\delta = \frac{\eps^2}{2L}$ and apply \nameref{thm:regularity} with
  parameters $\frac{\eps}{2}$ and $l = \frac{10}{\eps}$.
  We define the graph on vertex set $\{V_1, \ldots, V_k\}$.
  Define the reduced graph $R_{\eps, \frac{\eps}{2}}$ on $V(R_{\alpha, \eps}) =
  \{V_1, \ldots, V_k\}$ and $V_i \sim V_j$ if $i \neq j$,
  \epsuniform{\frac{\eps}{2}} and $\density(V_i, V_j) \ge \eps$.