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\begin{remark*}
  Vital in the proof of \cref{thm_2_8}, and of
  \cref{harper}, that the extremal sets (in
  dimension $n - 1$) were nested.
\end{remark*}

\glssymboldefn{inum}%
The \emph{isoperimetric} number of a graph $G$ is
\[
  i(G) = \min \left\{ \frac{|\bounde A|}{|A|} : A
  \subset G, \frac{|A|}{|G|} \le \half \right\}
.\]
$\frac{|\bounde A|}{|A|}$ is the ``average
out-degree of $|A|$''.

\begin{corollary}
  \label{coro_2_10}
  % Corollary 10
  $\inum(\Q_n) = 1$.
\end{corollary}

\begin{proof}
  Taking $A = \powset{n - 1}$, we show $\inum(\Q_n)
  \le 1$ (as $\frac{|\bounde A|}{|A|} = \frac{2^{n
  - 1}}{2^n} = \half$).

  To show $\inum(\Q_n) \ge \half$, just need to
  show that if $C$ is an initial segment of binary
  with $|C| \le 2^{n - 1}$ then $|\bounde C| \ge
  |C|$.

  But $C \subset \powset{n - 1}$, so certainly
  $|\bounde C| \ge |C|$.
\end{proof}

\subsection{Inequalities in the grid}

\glsnoundefn{grid}{grid}{grids}%
For any $k = 2, 3, \ldots$, the \emph{grid} is the
graph on $[k]^n$ in which $x$ is joined to $y$ if
for some $i$ we have $x_j = y_j$ for all $j \neq
i$ and $|x_i - y_i| = 1$.

``distance is $l_1$-distance''.

\begin{example*}
  $[4]^2$
  \begin{center}
    \includegraphics[width=0.3\linewidth]{images/c76e4d732dec4749.png}
  \end{center}
\end{example*}

Note that for $k = 2$ this is exactly $\Q_n$.

Do we have analogues of \cref{harper} and
\cref{thm_2_8} for the \gls{grid}?

Starting with vertex-isoperimetric: which sets $A
\subset [k]^n$ (of given size) minimise
$|\neigh(A)|$?

\begin{example*}
  In $[k]^2$:
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/61a641e320b94e40.png}
  \end{center}
  For $A$: $|\bound(A)| \sim r \sim \sqrt{2|A|}$.

  For $B$: $|\bound(B)| = 2r = 2\sqrt{|B|}$.
\end{example*}

This suggests we ``go up in levels'' according to
$|x| = \sum_{i = 1}^{n} |x_i|$ -- e.g. we'd take
$\{x \in [k]^n : |x| \le r\}$.

What if $|\{x \in [k]^n : |x| \le r\}| < |A| <
|\{x \in [k]^n : |x| \le r + 1\}$?

Guess: take $A = \{x \in [k]^n : |x| \le r\}$ plus
some points with $|x| = r + 1$, but which points?

\begin{example*}
  In $[k]^3$:
  \begin{center}
    \includegraphics[width=0.3\linewidth]{images/ad5658e028034110.png}
  \end{center}
  so ``keep $x_1$ large''.
\end{example*}

\glsnoundefn{simpordk}{simplicial
order}{simplicial ordering}%
This suggests in the \emph{simplicial order} on
$[k]^n$, we set $x < y$ if either $|x| < |y|$ or
$|x| = |y|$ and $x_i > y_i$, where $i = \min,cbj :
x_j \neq y_j\}$.

\begin{note*}
  Agrees with the previous definition of
  \gls{simpord} when $k = 2$.
\end{note*}

\begin{example*}
  On $[3]^2$: $(1,1)$, $(2,1)$, $(1,2)$, $(1,1)$,
  $(2, 2)$, $(1, 3)$, $(3, 2)$, $(2, 3)$, $(3, 3)$.
  \begin{center}
    \includegraphics[width=0.3\linewidth]{images/9efbb6b7704c4181.png}
  \end{center}
  On $[4]^3$: $(1, 1, 1)$, $(2, 1, 1)$, $(1, 2, 1)$,
  $(1, 1, 2)$, $(3, 1, 1)$, $(2, 2, 1)$, $(2, 1,
  2)$, $(1, 3, 1)$, $(1, 2, 2)$, $(1, 1, 3)$, $(4,
  1, 1)$, $(3, 2, 1)$, \ldots
\end{example*}

\glsnoundefn{isec}%
For $A \subset [k]^n$ ($n \ge 2$), and $1 \le i
\le n$, the \emph{$i$-sections} of $A$ are the
sets $A_1, \ldots, A_k$ (or $A_1^{(i)}, \ldots,
A_k^{(i)}$) as a subset of $[k]^{n - 1}$ defined
by:
\[
  A_t = \{x \in [k]^{n - 1} : (x_1, x_2, \ldots,
  x_{i - 1}, t, x_i, x_{i + 1}, \ldots, x_{n - 1})
  \in A\}
,\]
for each $1 \le t \le k$.
\begin{center}
  \includegraphics[width=0.3\linewidth]{images/50d4e3b8997f4787.png}
\end{center}

\glssymboldefn{isec}%
\glssymboldefn{gridCi}%
The \emph{$i$-compression} of $A$ is $C_i(A)
\subset [k]^n$ is defined by giving its
$i$-sections:
\[
  C_i(A)_{\isec t} = \text{initial segment of $[k]^{n -
  1}$ of size $|A_{\isec t}|$, for each $1 \le t \le k$}
.\]
Thus $|\gridCi(A)| = |A|$.

\glsadjdefn{gridiced}{$i$-compressed}{family}%
Say $A$ is \emph{$i$-compressed} if $\gridCi(A) =
A$.

\begin{fcthm}[Vertex-isoperimetric inequality in
  the grid]
  \label{thm_2_11}
  % Theorem 11
  Assuming:
  - $A \subset [k]^n$
  - $C$ is the initial segment of \gls{simpordk} on $[k]^n$
  with $|C| = |A|$
  Then:
  $|\neigh(C)| \le |\neigh(A)|$. In particular, if
  $|A| \ge |\{x : |x| \le r\}|$ then $|\neigh(A)|
  \ge |\{x : |x| \le r + 1\}|$.
\end{fcthm}

\begin{proof}
  Induction on $n$. For $n = 1$ it is trivial: if
  $A \subset [k]^1 \neq \emptyset, [k]^1$, then
  $|\neigh(A)| \ge |A| + 1 = |\neigh(C)|$.

  Given $n > 1$ and $A \subset [k]^n$: fix $1 \le
  i \le n$.

  \textbf{Claim:} $|\neigh(\gridCi(A)| \le
  |\neigh(A)|$.

  Proof of claim: write $B$ for $\gridCi(A)$. For
  any $1 \le t \le k$, we have
  \[
    \neigh(A)_{\isec t} = \ub{\neigh(A_{\isec
    t})}_{\text{from $x_i = t$}} \cup
    \ub{A_{\isec{t - 1}}}_{\text{from $x_i = t - 1$}}
    \cup \ub{A_{\isec{t + 1}}}_{\text{from $x_i = t +
    1$}}
  .\]
  \begin{center}
    \includegraphics[width=0.3\linewidth]{images/50d4e3b8997f4787.png}
  \end{center}
  (where $A_0, A_{k + 1} = \emptyset$).

  Also,
  \[
    \neigh(B)_{\isec t} = \neigh(B_{\isec t}) \cup
    B_{\isec{t - 1}} \cup B_{\isec{t + 1}}
  .\]
  Now, $|B_{\isec{t - 1}}| = |A_{\isec{t - 1}}|$ and
  $|B_{\isec{t + 1}}| = |A_{\isec{t + 1}}|$, and
  $|\neigh(B_{\isec t})| \le |\neigh(A_{\isec t})|$
  (induction hypothesis). But the sets $B_{\isec{t
  - 1}}$, $B_{\isec{t + 1}}$, $\neigh(B_{\isec t})$ are
  nested (as each is an initial segment of
  \gls{simpordk} on $[k]^{n - 1}$).

  Hence $|\neigh(B)_{\isec t}| \le |\neigh(A)_{\isec
  t}|$ for each $t$. Thus $|\neigh(B)| \le
  |\neigh(A)|$.

  Among all $B \subset [k]^n$ with $|B| = |A|$ and
  $|\neigh(B)| \le |\neigh(A)|$, pick one that
  minimises the quantity
  $\sum_{x \in B} \text{position of $x$ in
  \gls{simpordk}}$.

  Then $B$ is \gls{gridiced} for all $i$. Note
  however, that this time we will make use of this
  minimality property of $B$ for more than just
  deducing that $B$ is \gls{gridiced} for all $i$.

  \textbf{Case 1:} $n = 2$. What we know is
  precisely that $B$ is a down-set ($A \subset
  [k]^n$ is a \emph{down-set} if $x \in A$, $y_i
  \le x_i ~\forall i \implies y \in A$)
  \begin{center}
    \includegraphics[width=0.3\linewidth]{images/7ae72066fa3c4783.png}
  \end{center}