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Alternative version:

\begin{fclemma}[Shearer, expectation version]
  \label{lemma:4.2}
  Assuming:
  - $X = (X_1, \ldots, X_n)$ a random variable
  - $A \subset [n]$ a randomly chosen subset of
  $[n]$, according to some probability
  distribution (don't need any independence
  conditions!)
  - for each $i \in [n]$, $\Pbb[i \in A] \ge \mu$
  Then:
  \[
    \ent X
    \le \mu^{-1} \Ebb_A \ent{X_A}
  .\]
\end{fclemma}

\begin{proof}
  As before,
  \[
    \ent{X_A}
    \ge \sum_{a \in A} \cent{X_a}{X_{< a}}
  .\]
  So
  \begin{align*}
    \Ebb_A \ent{X_A}
    &\ge \Ebb_A \sum_{a \in A} \cent{X_a}{X_{< a}}
    \\
    &\ge \mu \sum_{a = 1}^n \cent{X_a}{X_{<a}} \\
    &= \mu \ent X
    \qedhere
  \end{align*}
\end{proof}

\begin{fcdefnstar}[$P_A$]
  Let $E \subset \Zbb^n$ and let $A \subset [n]$.
  Then we write $P_A E$ for the set of all $u \in
  \Zbb^A$ such that there exists $v \in \Zbb^{[n]
  \setminus A}$ such that $[u, v] \in E$, where
  $[u, v]$ is $u$ suitably intertwined with $v$
  (i.e. $u \cup v$ as functions).
\end{fcdefnstar}

\begin{fccoro}[]
  \label{coro:4.3}
  Assuming:
  - $E \subset \Zbb^n$
  - $\mathcal{A}$ a family of subsets of $[n]$
  such that every $i \in [n]$ is contained at
  least $r$ sets $A \in \mathcal{A}$
  Then:
  \[
    |E|
    \le \prod_{A \in \mathcal{A}} |P_A
    E|^{\frac{1}{r}}
  .\]
\end{fccoro}

\begin{proof}
  Let $X$ be a uniform random element of $E$. Then
  by \nameref{lemma:4.1},
  \[
    \ent X
    \le \frac{1}{r} \sum_{A \in \mathcal{A}}
    \ent{X_A}
  .\]
  But $X_A$ tkaes values in $P_A E$, so
  \[
    \ent{X_A}
    \le \log|P_A X
  ,\]
  so
  \[
    \log|E|
    \le \frac{1}{r} \sum_A \log|P_A E|
    .
    \qedhere
  \]
\end{proof}

If $\mathcal{A} = \{[n] \setminus \{i\} : i = 1,
\ldots, n\}$ we get
\[
  |E|
  \le \prod_{i = 1}^{n} |P_{[n] \setminus \{i\}}
  E|^{\frac{1}{n - 1}}
.\]
\glsnoundefn{thm:dlw}{discrete Loomis-Whitney}{NA}
This case is the discrete Loomis-Whitney theorem.

\begin{fcthm}[]
  \label{thm:4.4}
  Assuming:
  - $G$ a graph with $m$ edges
  Then:
  $G$ has at most $\frac{(2m)^{\frac{3}{2}}}{6}$
  triangles.
\end{fcthm}

Is this bound natural? Yes: if $m = {n \choose
2}$, and we consider a complete graph on $n$
vertices, then we get approximately
$\frac{(2m)^{\frac{2}{3}}}{6}$ triangles.

\begin{proof}
  Let $(X_1, X_2, X_3)$ be a random ordered
  triangle (without loss of generality
  $G$ has a triangle so that this is possible).

  Let $t$ be the number of triangles in $G$. By
  \nameref{lemma:4.1},
  \[
    \log(6t)
    = \ent{X_1, X_2, X_3}
    \le \half (\ent{X_1, X_2} + \ent{X_1, X_3} +
    \ent{X_2, X_3})
  .\]
  Each edge $\ent{X_i, X_j}$ is supported in the
  set of edges $G$, given a direction, i.e.
  \[
    \half (\ent{X_1, X_2} + \ent{X_1, X_3} +
    \ent{X_2, X_3})
    \le \frac{3}{2} \cdot \log(2m)
    .
    \qedhere
  \]
\end{proof}

\begin{fcdefnstar}[]
  \glsadjdefn{tint}{$\Delta$-intersecting}{family}%
  Let $X$ be a set of size $n$ and let
  $\mathcal{G}$ be a set of graphs with vertex set
  $X$. Then $\mathcal{G}$ is
  \emph{$\Delta$-intersecting} (read as
  ``triangle-intersecting'') if for all $G_1, G_2
  \in \mathcal{G}$, $G_1 \cap G_2$ contains a
  triangle.
\end{fcdefnstar}

\begin{fcthm}[]
  \label{thm:4.5}
  Assuming:
  - $|V| = n$
  - $\mathcal{G}$ a \gls{tint} family with vertex
  set $V$
  Then:
  $\mathcal{G}$ has size at most $2^{{n \choose 2}
  - 2}$.
\end{fcthm}

\begin{proof}
  Let $X$ be chosen uniformly at random from
  $\mathcal{G}$. We write $V^{(2)}$ for the set of
  (unordered) pairs of elements of $V$. Think of
  any $G \in \mathcal{G}$ as a function from
  $V^{(2)}$ to $\{0, 1\}$. So $X = (X_e : e \in
  V^{(2)})$.

  For each $R \subset V$, let $G_R$ be the graph
  $K_R \cup K_{V \setminus R}$
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/ac52c73cf5694a15.png}
  \end{center}
  For each $R$, we shall look at the projection
  $X_{G_R}$, which we can think of as taking
  values in the set $\{G \cap G_R : G \in
  \mathcal{G}\} \eqdef \mathcal{G}_R$.

  Note that if $G_1, G_2 \in \mathcal{G}$, $R
  \subset [n]$, then $G_1 \cap G_2 \cap G_R \neq
  \emptyset$, since $G_1 \cap G_2$ contains a
  triangle, which must intersect $G_R$ by
  Pigeonhole Principle.

  Thus, $\mathcal{G}_R$ is an intersecting family,
  so it has size at most $2^{|E(G_R)| - 1}$. By
  \nameref{lemma:4.2},
  \begin{align*}
    \ent X
    &\le 2 \Ebb_R \ent{X_{G_R}}
    &&(\text{since each $e$ belongs to $G_R$ with
    probability $1/2$}) \\
    &\le 2\Ebb_R (|E(G_R)| - 1) \\
    &= 2 \left( \half {m \choose 2} - 1 \right) \\
    &= {n \choose 2} - 2
    &&\qedhere
  \end{align*}
\end{proof}