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  \textbf{Claim 1:} $R_{\eps, \frac{\eps}{2}}$ is triangle-free.

  If there exist $V_i, V_j, V_k$ distinct with
  \[
    \density(V_i, V_j), \density(V_j, V_k), \density(V_i, V_k)
    \ge \eps
  \]
  and are pairwise \epsuniform{\frac{\eps}{2}}.
  Then consider
  \[
    V_i'
    = \{x \in V_i : |N(x) \cap V_j| \ge \eps|V_j|, |N(x) \cap V_k| \ge
    \frac{\eps}{2} |V_k|\}
  .\]
  We know
  \[
    |V_i'|
    \ge (1 - 2\eps)|V_i|
    > 0
  ,\]
  so let $x \in V_i'$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/64e6ad8dbb9c4dc7.png}
  \end{center}
  Now let
  \[
    \{y \in V_j \cap N(x) : |(y) \cap N(x) \cap V_k| \ge \eps|N(x) \cap V_k|\}
  .\]
  Again this set is non-empty, so choose $y$ in it.

  So now note,
  \[
    |N(x) \cap N(y) \cap V_k|
    \ge \eps|N(x) \cap V_k|
    \ge \eps^2|V_k|
    \ge \eps^2 \left( \frac{n}{L} \right)
    > \delta
  .\]
  So there exist $u, v \in N(x) \cap N(y) \cap V_k$ such that $uv \in G$.
  But then $x, y, u, v$ form a $K_4$, contradiction.

  \textbf{Claim 2:} $\density(V_i, V_j) \le \half + 2\eps$, for all $i < j$.

  Throw away $\eps|V_i|$ vertices in $V_i$ with degree $< \half + \eps$ to
  $V_j$.
  There are at least $(1 - 2\eps) \frac{n}{L} > \delta n$ vertices that remain
  in $V_i$.
  So there exist $u, v \in V_i$, $uv \in G$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/80e5690d556d42bf.png}
  \end{center}
  Now note
  \[
    |N(u) \cap N(v) \cap V_j|
    \ge \eps|V_j|
  .\]
  Now use independence number property to find $x, y \in N(u) \cap N(v) \cap
  V_i$ such that $xy \in G$.
  Then $x, y, u, v$ is a $K_4$, contradiction.

  We now estimate the number of edges in $G$ using the claims.
  \begin{align*}
    e(G)
    &\le e(R) \left( \frac{n}{k} \right)^2 \left( \half + 2\eps \right)
    + {k \choose 2} \eps \left( \frac{n}{k} \right)^2
    + \eps{k \choose 2} \left( \frac{n}{k} \right)^2 \\
    &\le \frac{n^2}{8} + 4\eps n^2
    \qedhere
  \end{align*}
\end{proof}

\begin{fcthm}[Bollobás--Erdős, 1970s]
  There exists a graph $G$ on $n$ vertices with $G \not\supset K_4$,
  $\indnum(G) = o(n)$ and $e(G) = \frac{n^2}{8} + o(n^2)$.
\end{fcthm}

\begin{proof}[Sketch]
  Let $d \to \infty$ slowly as $n \to \infty$.
  We consider the sphere $S^d$ in $d + 1$ dimensions.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/783d53274dca4931.png}
  \end{center}
  ``join a red point to a blue point if they are \emph{reasonably} close'' (say
  having inner product bigger than some small quantity).
  ``join red to red and join blue to blue if they have distance $\ge 2 -
  \frac{\eps}{\sqrt{d}}$.
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/9b5f8ae0f3a84e1b.png}
  \end{center}
  
  A little more formally:

  Let $d \to \infty$ slowly, and let $\eps \to 0$ slowly.
  \begin{itemize}
    \item Partition $S^d$ into $\frac{n}{2}$ regions of equal measure and with
      diameter $\le \frac{1}{2\sqrt{d}}$.
    \item Let $R = \{x_1, \ldots, x_{\frac{n}{2}}\}$, $B = \{y_1, \ldots,
      y_{\frac{n}{2}}\}$ be a set of points, with one from each of these
      regions.
  \end{itemize}
  We define $G$ by
  \begin{itemize}
    \item $x \in R$, $y \in B$: $x \sim y$ when $|x - y| < \sqrt{2} -
      \frac{\eps}{\sqrt{d}}$.
    \item $x, x' \in R$: $x \sim x'$ when $|x - y| \ge 2 -
      \frac{\eps}{\sqrt{d}}$.
    \item Same for $y, y' \in B$
  \end{itemize}
  Properties:
  \begin{itemize}
    \item $e(G) = \frac{n^2}{8} + o(n^2)$.
      \begin{center}
        \includegraphics[width=0.6\linewidth]{images/f9458bef49c940e2.png}
      \end{center}
      $x \sim y$ iff $\langle x, y \rangle \ge \frac{c\eps}{\sqrt{d}}$.
      Let's imagine $x = \left( \frac{1}{\sqrt{d}}, \ldots, \frac{1}{\sqrt{d}}
      \right)$.
      \[
        \Pbb \left( \frac{1}{d} \sum_{i = 1}^d y_i \ge \frac{c\eps}{\sqrt{d}}
        \right)
        \label{eq:highdimprob}
        \tag{$*$}
      .\]
      $\Var y_i \sim \frac{1}{d}$, $\Var \sum_{i = 1}^d y_i \sim 1$.
      So \eqref{eq:highdimprob} is $\approx \Pbb(Z \ge c\eps) = \half -
      \Theta(\eps)$, where $Z \sim N(0, 1)$.
      So every vertex has degree $\approx \frac{n}{4}$, so $e(G) \approx
      \frac{n^2}{8}$.
    \item Now we check it is $K_4$-free

      First note $G[R], G[B]$ are triangle-free.
      Let $x, y, z \in R$ and assume they form a $K_3$.
      Then
      \begin{align*}
        0
        &\le \|x + y + z\|^2 \\
        &= \langle x + y + z, x + y + z \rangle \\
        &= 3 + 2(\langle x, y \rangle + \langle y, z \rangle + \langle x, z
        \rangle) \\
        &= 9 - (\|x - y\|^2 + \|y - z\|^2 + \|x - z\|^2) \\
        &= 9 - 3 \left( 2 - \frac{\eps}{\sqrt{d}} \right)^2 \\
        &< 0
      \end{align*}
      Contradiction.

      $G$ is $K_4$-free: consider $\|x + x' - y - y'\|$ where $x, x' \in R$ and
      $y, y' \in B$.
    \item Finally, we need to check $\indnum(G) = o(n)$.
      Let $I \subset V(G)$ be independent.
      Let $I' \cap R$, and without loss of generality say $|I'| \ge
      \frac{|I|}{2}$.
      Let
      \[
        S
        = \bigcup_{x \in I'} R_x
      \]
      where $R_x$ are the regions that we broke the sphere into at the start.
      We have $\mu(S) = \frac{2|I'|}{n}$ (where $\mu$ is such that $\mu(S^d) =
      1$).
      Note $S \subset S^d$ with diameter
      \[
        < 2 - \frac{c}{\sqrt{d}}
      .\]
      Fact from life: this implies $\mu(S) \to 0$ as $d \to \infty$.

      Hence $\frac{|I|}{n} \to 0$, i.e. $|I| = o(n)$.
  \end{itemize}
\end{proof}
