[proofplan] We argue by induction on $n$. Fix the coordinate $i$ and decompose the closed neighbourhood $N(A)$ into its contributions from the two layers of $Q_n$ determined by membership in $\{i\}$: the "$-$ layer" (vertices not containing $i$) and the "$+$ layer" (vertices containing $i$). Each layer's contribution is a union of two families in $Q_{n-1}$ — the closed neighbourhood of one $i$-section together with the other $i$-section. After $i$-compression, both sections become initial segments of the simplicial order in $\mathcal{P}(X \setminus \{i\})$, and since initial segments are nested (a smaller one is always contained in a larger one), each union collapses to a maximum. The general set-theoretic bound $|S \cup T| \geq \max(|S|, |T|)$ then shows each original union term is at least as large as the corresponding maximum for the compressed set, completing the comparison. [/proofplan]

[step:Establish the base case $n = 1$] When $n = 1$, the cube $Q_1$ has two vertices $\varnothing$ and $\{1\}$ joined by a single edge. Every non-empty family $A \subseteq Q_1$ satisfies $N(A) = Q_1$, and $C_1(A)$ is the initial segment of $\mathcal{P}(\{1\})$ of size $|A|$, which is either $\varnothing$ (if $|A| = 0$) or a non-empty family. In either case $|N(C_1(A))| = |N(A)|$, so the inequality holds. [/step]

[step:Decompose $|N(A)|$ into contributions from the two layers determined by coordinate $i$]Fix $n \geq 2$ and assume the result holds for $Q_{n-1}$. Let $A \subseteq Q_n$, fix $1 \leq i \leq n$, and set $B = C_i(A)$. Write $X = \{1, \ldots, n\}$ and $X' = X \setminus \{i\}$. Recall the $i$-sections of $A$, viewed as subsets of $\mathcal{P}(X') \cong Q_{n-1}$: \begin{align*} A_-^{(i)} &= \{ x \in A : i \notin x \}, \\ A_+^{(i)} &= \{ x \setminus \{i\} : x \in A,\; i \in x \}. \end{align*} The vertex set of $Q_n$ partitions into the **$-$ layer** $\{ v \in \mathcal{P}(X) : i \notin v \}$ and the **$+$ layer** $\{ v \in \mathcal{P}(X) : i \in v \}$, each isomorphic to $Q_{n-1}$ via the map that either includes or excludes $\{i\}$. A vertex $v$ in the $-$ layer lies in $N(A)$ if and only if at least one of the following holds: (a) $v \in A_-^{(i)}$; (b) $v$ is adjacent within $Q_{n-1}$ to some element of $A_-^{(i)}$; or (c) the vertex $v \cup \{i\}$ in the $+$ layer belongs to $A$, i.e., $v \in A_+^{(i)}$. Conditions (a) and (b) together say $v \in N_{Q_{n-1}}(A_-^{(i)})$. So the $-$ layer contribution to $N(A)$ is $N_{Q_{n-1}}(A_-^{(i)}) \cup A_+^{(i)}$. By the same reasoning with the layers interchanged, the $+$ layer contribution (re-indexed via $v \mapsto v \setminus \{i\}$ to a subset of $Q_{n-1}$) is $N_{Q_{n-1}}(A_+^{(i)}) \cup A_-^{(i)}$. Since the two layers partition $V(Q_n)$ and each is identified with $Q_{n-1}$: \begin{align*} |N(A)| = \bigl|N_{Q_{n-1}}(A_-^{(i)}) \cup A_+^{(i)}\bigr| + \bigl|N_{Q_{n-1}}(A_+^{(i)}) \cup A_-^{(i)}\bigr|. \end{align*} The identical decomposition holds for $B$: \begin{align*} |N(B)| = \bigl|N_{Q_{n-1}}(B_-^{(i)}) \cup B_+^{(i)}\bigr| + \bigl|N_{Q_{n-1}}(B_+^{(i)}) \cup B_-^{(i)}\bigr|. \end{align*}[/step]

[guided]Why does this decomposition work? In $Q_n$, every vertex $v$ has exactly one neighbour obtained by toggling membership of $i$: if $i \notin v$, that neighbour is $v \cup \{i\}$; if $i \in v$, it is $v \setminus \{i\}$. All other neighbours of $v$ differ from $v$ in some coordinate $j \neq i$ and therefore lie in the same layer as $v$. So a vertex $v$ in the $-$ layer belongs to $N(A)$ precisely when either $v$ has a neighbour in $A$ within the $-$ layer (meaning $v \in N_{Q_{n-1}}(A_-^{(i)})$), or the unique cross-layer neighbour $v \cup \{i\}$ belongs to $A$ (meaning $v \in A_+^{(i)}$). The $-$ layer's contribution is therefore $N_{Q_{n-1}}(A_-^{(i)}) \cup A_+^{(i)}$. The argument for the $+$ layer is symmetric: a vertex $w$ with $i \in w$ lies in $N(A)$ when either $w \setminus \{i\} \in N_{Q_{n-1}}(A_+^{(i)})$ (adjacency within the $+$ layer or membership in $A_+^{(i)}$), or $w \setminus \{i\} \in A_-^{(i)}$ (the cross-layer neighbour $w \setminus \{i\}$ is in $A$). Re-indexing the $+$ layer by $w \mapsto w \setminus \{i\}$, the contribution is $N_{Q_{n-1}}(A_+^{(i)}) \cup A_-^{(i)}$. Since every vertex of $Q_n$ is in exactly one layer, the total neighbourhood size is the sum of the two contributions. No vertex is double-counted: the two sets live in disjoint layers of $Q_n$ (or, equivalently, the two copies of $Q_{n-1}$ after re-indexing are summed, not unioned). This decomposition relies critically on the product structure of $Q_n$ — each vertex has exactly one cross-$i$ neighbour.[/guided]

[step:Show that the compressed sections are initial segments whose neighbourhoods are also initial segments, making all relevant sets pairwise nested]By definition of $i$-compression, $B_+^{(i)}$ is the initial segment of $\mathcal{P}(X')$ in the simplicial order of size $|A_+^{(i)}|$, and $B_-^{(i)}$ is the initial segment of size $|A_-^{(i)}|$. Initial segments of the simplicial order are nested: if $|S_1| \leq |S_2|$, then the initial segment of size $|S_1|$ is contained in the initial segment of size $|S_2|$. [claim:Neighbourhood of an initial segment is an initial segment] If $I \subseteq \mathcal{P}(X')$ is an initial segment of the simplicial order, then $N_{Q_{n-1}}(I)$ is also an initial segment of the simplicial order. [/claim] [proof] The simplicial order on $\mathcal{P}(X')$ lists elements by non-decreasing size, with ties broken by the lex order within each level. Write $I = X'^{(\leq r)} \cup L$ where $r \geq 0$ and $L$ is a (possibly empty) lex initial segment of $X'^{(r+1)}$. The closed neighbourhood of $I$ in $Q_{n-1}$ consists of all elements of $I$ together with every vertex adjacent to some element of $I$. Every element of $X'^{(\leq r)}$ has all its neighbours in $X'^{(\leq r+1)} \subseteq X'^{(\leq r)} \cup X'^{(r+1)}$, and every element of $L \subseteq X'^{(r+1)}$ has neighbours in $X'^{(r)} \cup X'^{(r+2)}$ together with the upper shadow of $L$ in $X'^{(r+2)}$. Hence \begin{align*} N_{Q_{n-1}}(I) = X'^{(\leq r+1)} \cup \partial^+ L, \end{align*} where $\partial^+ L = \{ y \in X'^{(r+2)} : y \supset x \text{ for some } x \in L \}$ is the upper shadow of $L$. When $L = \varnothing$, we get $N_{Q_{n-1}}(I) = X'^{(\leq r+1)}$, which is an initial segment. When $L \neq \varnothing$, the upper shadow $\partial^+ L$ of a lex initial segment of $X'^{(r+1)}$ is itself a lex initial segment of $X'^{(r+2)}$ (this is a direct consequence of the lex order structure: the upper shadow respects the lex ordering because the elements of $L$, being the first $|L|$ sets of size $r+1$ in lex order, produce upper shadow elements that appear earliest in lex order at level $r+2$). Therefore $X'^{(\leq r+1)} \cup \partial^+ L$ is an initial segment of the simplicial order. [/proof] Since $B_+^{(i)}$, $B_-^{(i)}$, $N_{Q_{n-1}}(B_+^{(i)})$, and $N_{Q_{n-1}}(B_-^{(i)})$ are all initial segments of the simplicial order on $\mathcal{P}(X')$, any two of them are comparable by containment: the one with fewer elements is a subset of the one with more. In particular, for each pair appearing in the decomposition of $|N(B)|$, the union of two nested sets equals the larger: \begin{align*} \bigl|N_{Q_{n-1}}(B_-^{(i)}) \cup B_+^{(i)}\bigr| &= \max\bigl\{|N_{Q_{n-1}}(B_-^{(i)})|,\; |B_+^{(i)}|\bigr\}, \\ \bigl|N_{Q_{n-1}}(B_+^{(i)}) \cup B_-^{(i)}\bigr| &= \max\bigl\{|N_{Q_{n-1}}(B_+^{(i)})|,\; |B_-^{(i)}|\bigr\}. \end{align*} Therefore \begin{align*} |N(B)| = \max\bigl\{|N_{Q_{n-1}}(B_-^{(i)})|,\; |B_+^{(i)}|\bigr\} + \max\bigl\{|N_{Q_{n-1}}(B_+^{(i)})|,\; |B_-^{(i)}|\bigr\}. \end{align*}[/step]

[guided]The crucial property at work is **nesting**. For the original family $A$, the sections $A_+^{(i)}$ and $A_-^{(i)}$ are arbitrary subsets of $Q_{n-1}$, so the sets $N_{Q_{n-1}}(A_-^{(i)})$ and $A_+^{(i)}$ may overlap partially — neither need contain the other. Their union can therefore be strictly larger than the maximum of their sizes: \begin{align*} \bigl|N_{Q_{n-1}}(A_-^{(i)}) \cup A_+^{(i)}\bigr| = |N_{Q_{n-1}}(A_-^{(i)})| + |A_+^{(i)}| - |N_{Q_{n-1}}(A_-^{(i)}) \cap A_+^{(i)}|, \end{align*} and the intersection $|N_{Q_{n-1}}(A_-^{(i)}) \cap A_+^{(i)}|$ could be much smaller than $\min(|N_{Q_{n-1}}(A_-^{(i)})|, |A_+^{(i)}|)$, making the union large. After compression, everything is an initial segment: $B_-^{(i)}$, $B_+^{(i)}$, and their neighbourhoods $N_{Q_{n-1}}(B_-^{(i)})$, $N_{Q_{n-1}}(B_+^{(i)})$ are all initial segments (the last two by the claim just proved). Initial segments of the same ground set are nested — the smaller one is always a subset of the larger — so every pair satisfies the containment hypothesis of $|S \cup T| = \max(|S|, |T|)$. This is the tightest possible: the union is as small as it can be given the sizes of the two sets. The "spreading" that inflated the unions for $A$ is eliminated by compression.[/guided]

[step:Establish by induction that initial segments have the smallest closed neighbourhood at each size][claim:Initial segments minimise the closed neighbourhood in $Q_m$] For every $m \geq 1$ and every family $S \subseteq Q_m$, the initial segment $\hat{S}$ of the simplicial order with $|\hat{S}| = |S|$ satisfies $|N_{Q_m}(\hat{S})| \leq |N_{Q_m}(S)|$. [/claim] [proof] We prove this by induction on $m$. When $m = 1$, the cube $Q_1$ has two vertices and the claim is immediate: any non-empty family has $N_{Q_1}(S) = Q_1$, and the initial segment of the same size has the same neighbourhood. For $m \geq 2$, fix any coordinate $j$ with $1 \leq j \leq m$ and apply the $i$-section decomposition to $S$ with respect to $j$ (working in $Q_m$ exactly as in the main proof): \begin{align*} |N_{Q_m}(S)| = \bigl|N_{Q_{m-1}}(S_-^{(j)}) \cup S_+^{(j)}\bigr| + \bigl|N_{Q_{m-1}}(S_+^{(j)}) \cup S_-^{(j)}\bigr|. \end{align*} Let $\hat{S}_+$ and $\hat{S}_-$ be the initial segments of $\mathcal{P}(\{1,\ldots,m\} \setminus \{j\})$ of sizes $|S_+^{(j)}|$ and $|S_-^{(j)}|$ respectively. By the inductive hypothesis applied in $Q_{m-1}$: \begin{align*} |N_{Q_{m-1}}(\hat{S}_+)| \leq |N_{Q_{m-1}}(S_+^{(j)})|, \qquad |N_{Q_{m-1}}(\hat{S}_-)| \leq |N_{Q_{m-1}}(S_-^{(j)})|. \end{align*} Using the general bound $|P \cup Q| \geq \max(|P|, |Q|)$ for the original terms, and the nesting identity $|P \cup Q| = \max(|P|, |Q|)$ when both $P$ and $Q$ are initial segments (since one contains the other): \begin{align*} |N_{Q_m}(S)| &\geq \max\bigl\{|N_{Q_{m-1}}(S_-^{(j)})|,\; |S_+^{(j)}|\bigr\} + \max\bigl\{|N_{Q_{m-1}}(S_+^{(j)})|,\; |S_-^{(j)}|\bigr\} \\ &\geq \max\bigl\{|N_{Q_{m-1}}(\hat{S}_-)|,\; |\hat{S}_+|\bigr\} + \max\bigl\{|N_{Q_{m-1}}(\hat{S}_+)|,\; |\hat{S}_-|\bigr\} \\ &= \bigl|N_{Q_{m-1}}(\hat{S}_-) \cup \hat{S}_+\bigr| + \bigl|N_{Q_{m-1}}(\hat{S}_+) \cup \hat{S}_-\bigr| \\ &= |N_{Q_m}(\hat{S})|, \end{align*} where the last line uses the fact that the family $\hat{S} \subseteq Q_m$ whose $j$-sections are the initial segments $\hat{S}_+$ and $\hat{S}_-$ is itself the initial segment of $\mathcal{P}(\{1,\ldots,m\})$ of size $|\hat{S}_+| + |\hat{S}_-| = |S|$ (because initial segments of the simplicial order decompose over any coordinate into a pair of initial segments), and the layer decomposition of its neighbourhood gives the displayed expression. [/proof][/step]

[guided]Why is this induction well-founded and non-circular? At each stage, the claim for $Q_m$ is reduced to the same claim for $Q_{m-1}$: we decompose the neighbourhood via $j$-sections, bound the original union terms below by maxima, bound the neighbourhood sizes of the sections using the inductive hypothesis, and use nesting to convert the resulting maxima back into exact union sizes for initial segments. The base case $m = 1$ is trivial. At no point do we invoke any external isoperimetric result — the entire argument rests on two elementary facts: (i) $|S \cup T| \geq \max(|S|, |T|)$ for any finite sets $S, T$; and (ii) initial segments of the simplicial order are pairwise nested, so their unions attain this lower bound with equality. The claim that $\hat{S}$'s $j$-sections are themselves initial segments deserves verification. The simplicial order on $\mathcal{P}(\{1,\ldots,m\})$ lists elements by size, with lex tiebreaking within each level. For any coordinate $j$, the $j$-sections of the initial segment of size $k$ are: $\hat{S}_- = \{ x \in \hat{S} : j \notin x \}$ (an initial segment of $\mathcal{P}(\{1,\ldots,m\} \setminus \{j\})$ because removing elements not containing $j$ from a lex-by-size initial segment preserves the initial segment property) and $\hat{S}_+ = \{ x \setminus \{j\} : x \in \hat{S},\; j \in x \}$ (also an initial segment, by the analogous argument). The sizes satisfy $|\hat{S}_+| + |\hat{S}_-| = |\hat{S}| = |S|$ and are uniquely determined by $|S|$ and $m$.[/guided]

[step:Compare term by term to conclude $|N(B)| \leq |N(A)|$]Applying the claim from the previous step to the sections $A_+^{(i)}$ and $A_-^{(i)}$ in $Q_{n-1}$, with initial segments $B_+^{(i)}$ and $B_-^{(i)}$ of the same sizes: \begin{align*} |N_{Q_{n-1}}(B_+^{(i)})| \leq |N_{Q_{n-1}}(A_+^{(i)})|, \qquad |N_{Q_{n-1}}(B_-^{(i)})| \leq |N_{Q_{n-1}}(A_-^{(i)})|. \end{align*} Since $|B_+^{(i)}| = |A_+^{(i)}|$ and $|B_-^{(i)}| = |A_-^{(i)}|$ by construction, the max terms satisfy: \begin{align*} \max\bigl\{|N_{Q_{n-1}}(B_-^{(i)})|,\; |B_+^{(i)}|\bigr\} &\leq \max\bigl\{|N_{Q_{n-1}}(A_-^{(i)})|,\; |A_+^{(i)}|\bigr\}, \\ \max\bigl\{|N_{Q_{n-1}}(B_+^{(i)})|,\; |B_-^{(i)}|\bigr\} &\leq \max\bigl\{|N_{Q_{n-1}}(A_+^{(i)})|,\; |A_-^{(i)}|\bigr\}. \end{align*} Combining the layer decomposition of $|N(A)|$ with the general bound $|P \cup Q| \geq \max(|P|, |Q|)$: \begin{align*} |N(A)| &= \bigl|N_{Q_{n-1}}(A_-^{(i)}) \cup A_+^{(i)}\bigr| + \bigl|N_{Q_{n-1}}(A_+^{(i)}) \cup A_-^{(i)}\bigr| \\ &\geq \max\bigl\{|N_{Q_{n-1}}(A_-^{(i)})|,\; |A_+^{(i)}|\bigr\} + \max\bigl\{|N_{Q_{n-1}}(A_+^{(i)})|,\; |A_-^{(i)}|\bigr\} \\ &\geq \max\bigl\{|N_{Q_{n-1}}(B_-^{(i)})|,\; |B_+^{(i)}|\bigr\} + \max\bigl\{|N_{Q_{n-1}}(B_+^{(i)})|,\; |B_-^{(i)}|\bigr\} \\ &= |N(B)|. \end{align*} The first inequality is $|S \cup T| \geq \max(|S|, |T|)$ applied to each term. The second inequality uses the neighbourhood-minimisation property of initial segments (the claim from the previous step) together with $|B_\pm^{(i)}| = |A_\pm^{(i)}|$ and the monotonicity of $\max$. The final equality is the nesting identity from the third step. This completes the inductive step and the proof that $|N(C_i(A))| \leq |N(A)|$.[/step]

[guided]Let us trace the full chain of reasoning to confirm there is no gap. We established four facts: 1. **Layer decomposition** (second step): $|N(A)| = |N_{Q_{n-1}}(A_-^{(i)}) \cup A_+^{(i)}| + |N_{Q_{n-1}}(A_+^{(i)}) \cup A_-^{(i)}|$, and the same formula holds for $B$. 2. **Nesting identity** (third step): since $B_\pm^{(i)}$ and $N_{Q_{n-1}}(B_\pm^{(i)})$ are all initial segments and therefore pairwise nested, $|N(B)| = \max\{|N_{Q_{n-1}}(B_-^{(i)})|, |B_+^{(i)}|\} + \max\{|N_{Q_{n-1}}(B_+^{(i)})|, |B_-^{(i)}|\}$. 3. **Neighbourhood minimisation** (fourth step): initial segments of $Q_{n-1}$ have the smallest closed neighbourhood among all families of the same size, i.e., $|N_{Q_{n-1}}(B_\pm^{(i)})| \leq |N_{Q_{n-1}}(A_\pm^{(i)})|$. 4. **Size preservation**: $|B_\pm^{(i)}| = |A_\pm^{(i)}|$ by the definition of $i$-compression. From (1), applying the elementary bound $|S \cup T| \geq \max(|S|, |T|)$ to each union term in $|N(A)|$: \begin{align*} |N(A)| \geq \max\bigl\{|N_{Q_{n-1}}(A_-^{(i)})|,\; |A_+^{(i)}|\bigr\} + \max\bigl\{|N_{Q_{n-1}}(A_+^{(i)})|,\; |A_-^{(i)}|\bigr\}. \end{align*} From (3) and (4), each max term for $B$ is at most the corresponding max term for $A$, because $\max(a, c) \leq \max(b, c)$ whenever $a \leq b$: \begin{align*} \max\bigl\{|N_{Q_{n-1}}(B_-^{(i)})|,\; |B_+^{(i)}|\bigr\} &\leq \max\bigl\{|N_{Q_{n-1}}(A_-^{(i)})|,\; |A_+^{(i)}|\bigr\}, \\ \max\bigl\{|N_{Q_{n-1}}(B_+^{(i)})|,\; |B_-^{(i)}|\bigr\} &\leq \max\bigl\{|N_{Q_{n-1}}(A_+^{(i)})|,\; |A_-^{(i)}|\bigr\}. \end{align*} Summing and chaining: $|N(B)| \overset{(2)}{=} \text{LHS sum} \leq \text{RHS sum} \leq |N(A)|$. What makes this argument self-contained is that the neighbourhood-minimisation property (fact 3) is itself proved by the same layer-decomposition-plus-nesting strategy, applied inductively in $Q_{n-1}$. The induction bottoms out at $Q_1$ where the claim is immediate. No external isoperimetric theorem is invoked — the entire proof rests on two elementary set-theoretic ingredients: $|S \cup T| \geq \max(|S|, |T|)$ for arbitrary sets, and $|S \cup T| = \max(|S|, |T|)$ when one set contains the other.[/guided]