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\begin{fcdefnstar}[Edge-boundary]
  \glssymboldefn{bound}%
  Let $G$ be a graph and let $A \subset V(G)$. The
  \emph{edge-boundary} $\partial A$ of $A$ is the
  set of edges $xy$ such that $y \notin A$.

  If $G = \Zbb^n$ or $\{0, 1\}^n$ and $i \in [n]$,
  then the $i$-th boundary $\partial_i A$ is the
  set of edges $xy \in \partial A$ such that $x -
  y = \pm e_i$, i.e. $\partial_i A$ consists of
  deges pointing in direction $i$.
\end{fcdefnstar}

\begin{fcthm}[Edge-isoperimetric inequality in
  $Z^n$]
  \label{thm:4.6}
  Assuming:
  - $A \subset \Zbb^n$ a finite set
  Then:
  $|\ebound A| \ge 2n|A|^{\frac{n - 1}{n}}$.
\end{fcthm}

\begin{proof}
  By the \gls{thm:dlw} inequality,
  \begin{align*}
    |A|
    &\le \prod_{i = 1}^n |P_{[n] \setminus \{i\}}
    A|^{\frac{1}{n - 1}} \\
    &= \left( \prod_{i = 1}^n |P_{[n] \setminus
    \{i\}} A|^{\frac{1}{n}} \right)^{\frac{n}{n -
    1}} \\
    &\le \left( \frac{1}{n} \sum_{i = 1}^n |P_{[n]
    \setminus \{i\}} A| \right)^{\frac{n}{n - 1}}
  \end{align*}
  But $|\ebound_i A| \ge 2|P_{[n] \setminus \{i\}}
  A|$ since each fibre contributes at least 2.

  So
  \begin{align*}
    |A|
    &\le \left( \frac{1}{2n} \sum_{i = 1}^{n}
    |\ebound_i A| \right)^{\frac{n}{n - 1}} \\
    &= \left( \frac{1}{2n} |\ebound A|
    \right)^{\frac{n}{n - 1}}
    \qedhere
  \end{align*}
\end{proof}

\begin{fcthm}[Edge-isoperimetric inequality in the
  cube]
  \label{thm:4.7}
  Assuming:
  - $A \subset \{0, 1\}^n$ (where we take the
  usual graph)
  Then:
  $|\ebound A| \ge |A| (n - \log |A|)$.
\end{fcthm}

\begin{proof}
  Let $X$ be a uniform random element of $A$ and
  write $X = (X_1, \ldots, X_n)$. Write
  $X_{\setminus i}$ for $(X_1, \ldots, X_{i - 1},
  X_{i + 1}, \ldots, X_n)$. By
  \nameref{lemma:4.1},
  \begin{align*}
    \ent X
    &\le \frac{1}{n - 1} \sum_{i = 1}^{r}
    \ent{X_{\setminus i}} \\
    &= \frac{1}{n - 1} \sum_{i = 1}^{n} \ent X -
    \cent{X_i}{X_{\setminus i}}
  \end{align*}
  Hence
  \[
    \sum_{i = 1}^{n}  \cent{X_i}{X_{\setminus i}}
    \le \ent X
  .\]
  Note
  \[
    \cent{X_i}{X_{\setminus i} = u}
    = \begin{cases}
      1 & |P_{[n] \setminus \{i\}}^{-1}(u)| = 2 \\
      0 & |P_{[n] \setminus \{i\}}^{-1}(u)| = 1
    \end{cases}
  \]
  The number of points of the second kind is
  $|\partial_i A|$, so $\cent{X_i}{X_{\setminus
  i}} = 1 - \frac{|\partial_i A|}{|A|}$. So
  \begin{align*}
    \ent X
    &\ge \sum_{i = 1}^{n} \left( 1 -
    \frac{|\ebound_i A|}{|A|} \right) \\
    &= n - \frac{|\ebound A|}{|A|}
  \end{align*}
  Also, $\ent X = \log |A|$. So we are done.
\end{proof}

\begin{fcdefnstar}[Lower shadow]
  \glssymboldefn{shadow}%
  Let $\mathcal{A}$ be a family of sets of size
  $d$. The \emph{lower shadow} $\partial
  \mathcal{A}$ is $\{B : |B| = d - 1, \exists A
  \in \mathcal{A}, B \subset A\}$.
\end{fcdefnstar}

\begin{notation*}
  \glssymboldefn{h}%
  Let $h(x) = x\log \frac{1}{x} + (1 - x) \log
  \frac{1}{1 - x}$ (for $x \in [0, 1]$).
\end{notation*}

\begin{fcthm}[Kruskal-Katona]
  Assuming:
  - $|\mathcal{A}| = {t \choose d} = \frac{t(t -
  1) \cdots (t - d - 1)}{d!}$ for some real
  number $t$
  Then:
  $|\shadow \mathcal{A}| \ge {t \choose d - 1}$.
\end{fcthm}

\begin{proof}
  Let $X = (X_1, \ldots, X_d)$ be a random
  ordering of the elements of a uniformly random
  $A \in \mathcal{A}$. Then
  \[
    \ent X = \log\left(d! {t \choose d}\right)
  .\]
  Note that $(X_1, \ldots, X_{d - 1})$ is an
  ordering of the elements of some $B \in \shadow
  \mathcal{A}$, so
  \[
    \ent{X_1, \ldots, X_{d - 1}}
    \le \log\left((d - 1)! |\shadow \mathcal{A}|\right)
  .\]
  So it's enough to show
  \[
    \ent{X_1, \ldots, X_{d - 1}}
    \ge \log\left((d - 1)! {t \choose d - 1}\right)
  .\]
  Also,
  \[
    \ent X
    = \ent{X_1, \ldots, X_{d - 1}}
    + \cent{X_d}{X_1, \ldots, X_{d - 1}}
  \]
  and
  \[
    \ent X
    = \ent{X_1}
    + \cent{X_2}{X_1}
    + \cdots
    + \cent{X_d}{X_1, \ldots, X_{d - 1}}
  .\]
  We would like an \emph{upper bound} for
  $\ent{X_d}{X_{<d}}$. Our strategy will be to
  obtain a \emph{lower bound} for
  $\cent{X_k}{X_{<k}}$ in terms of $\cent{X_{k +
  1}}{X_{< k + 1}}$. We shall prove that
  \[
    2^{\cent{X_k}{X_{< k}}} \ge 2^{\cent{X_{k +
    1}}{X_{< k + 1}}} + 1 \qquad \forall k
  .\]
  Let $T$ be chosen independently of $X_1, \ldots,
  X_{k - 1}$ with
  \[
    T = \begin{cases}
      0 & \text{probability $p$} \\
      1 & \text{probability $1 - p$}
    \end{cases}
  \]
  ($p$ will be chosen and optimised later).

  Given $X_1, \ldots, X_{k - 1}$, let
  \[
    X^* = \begin{cases}
      X_{k + 1} & T = 0 \\
      X_k & T = 1
    \end{cases}
  \]
  Note that $X_k$ and $X_{k + 1}$ have the same
  distribution (given $X_1, \ldots, X_{k - 1}$),
  so $X^*$ does as well. Then
  \begin{align*}
    \cent{X_k}{X_{<k}}
    &= \cent{X^*}{X_{< k}} \\
    &\ge \cent{X^*}{X_{\le k}}
    &&\text{(\nameref{submod})} \\
    &= \cent{X^*, T}{X_{\le k}}
    &&\text{($X_{\le k}$ and $X^*$ determine $T$)}
    \\
    &= \cent{T}{X_{\le k}} + \cent{X^*}{T, X_{\le
    k}}
    &&\text{(\gls{additivity})}
    \\
    &= \ent T
    + p \cent{X_{k + 1}}{X_1, \ldots, X_k} \\
    &~~~+ (1 - p) \cent{X_k}{X_1, \ldots, X_k} \\
    &= \h(p) + ps
  \end{align*}