Prove the inequality for the given condition Suppose that the boxes $B_1, . . . , B_n ⊂ R^d$ pairwise touch each other, that is, any
two of them are separated and have a common boundary point. Prove that $n ≤ 2^d$
I am trying to use Brunn-Minkowski inequality for brick set and the concept of volume to solve this. Like I am taking a hyperplane H that exactly divide it into two equal parts so that $B_1,. .,B_\frac{n}{2}$ lies in one plane i.e. $H^+$ and $B_\frac{n}{2},..,B_n$  lies in $H^-$ plane. But I am unable to proceed from here. Is there any other method to do so? Any help is appreciated.
Thanks in advance.
 A: I'm gonna keep this concise, so as it might be easy to verify it's correctness.
Theorem 1. Let $\mathcal{B}$ be any finite set of boxes in $\mathbb{R}^d$ which satisfy the property that any pair of distinct boxes are separated and share boundary points. Then $ \bigcap_{B \in \mathcal{B}} \partial B \neq \varnothing$.
Proof. Use strong induction on $d$.
Induction Step: Let $s \geq 1$ . Given that any set of boxes $\mathcal{B}$ from $\mathbb{R}^t  (1 \le t \le s)$ which are pairwise separated and pairwise share boundary points satisfy $ \bigcap_{B \in \mathcal{B}} \partial B \neq \varnothing$, we need to show that any set of boxes $\mathcal{B}'$ from $\mathbb{R}^{s+1} $ which are pairwise separated and pairwise share boundary points satisfy $ \bigcap_{B \in \mathcal{B}'} \partial B \neq \varnothing$
Let $n := |\mathcal{B}'|$. To prove this for $n \geq 2$, we use strong induction on $n$.
Induction Step: Let $2 \le k \le n$. Given that any $k$ set of boxes of $\mathbb{R}^{s+1}$ (call it $\mathcal{B}$) which are pairwise separated and pairwise share boundary points satisfy $ \bigcap_{B \in \mathcal{B}} \partial B \neq \varnothing$, we need to show that any $k+1$ set of boxes of $\mathbb{R}^{s+1}$ which are pairwise separated and share boundary points have a non-empty intersection of their boundaries.
Let $\mathcal{B}'$ be any $k+1$ set of boxes of $\mathbb{R}^{s+1}$ which are pairwise separated and pairwise share boundary points. Let $B_0 \in \mathcal{B}'$. Consider all $k$ sized subsets of $\mathcal{B}'$ which contains $B_0$. There are $k$ such subsets, denote them by $\mathcal{B}_1$, $\mathcal{B}_2$, .. $\mathcal{B}_k$. By the inner induction hypothesis then there exists $k$ faces $F_1$, $F_2$, ... $F_k$ of $B_0^c$ which are the least dimensional faces of $B_0^c$ containing the intersection of the boxes in $\mathcal{B}_1$, $\mathcal{B}_2$, .. $\mathcal{B}_k$ respectively.
Claim $\bigcap_{i = 1}^k F_i \neq \varnothing$
Lemma 1. For any $i \neq j$, $F_i \cap F_j \neq \varnothing$.
Proof. There exists boxes $B_1, B_2$ s.t. $B_1 \in \mathcal{B}_i \setminus \mathcal{B}_j$ and $B_2 \in \mathcal{B}_j \setminus \mathcal{B}_i$. If $F_i \cap F_j = \varnothing$, then $B_1$ and $B_2$ cannot share a boundary point (because if two boxes share a face, then there doesn't exist any other box, which shares two disjoint faces with each of these boxes i.e. $B_0$ can't exist). This is a contradiction.
Lemma 2. Let $k_0 = \min_{k} (dim F_k)$. There exists $A \subseteq \{ F_1, F_2 , ... F_k \}$ such that $F \in A \Rightarrow \dim F = k_0$ and $\bigcap_{F \in A} = \bigcap_{i=1}^k F_i$
Proof. As $F_i \cap F_j \neq \varnothing$, if $F_i$ has dimension less than $F_j$, then $F_i \subset F_j$. Hence we can remove $F_j$ from the set.
Now if $\bigcap_{F \in A} = \varnothing$, then all faces in $A$ are separable. In this case, we may use the induction hypothesis of outermost induction, to get $\bigcap_{F \in A} \neq \varnothing$, which is a contradiction. This proves the claim.
And the claim implies that induction hypothesis. Qed.
Theorem 2. Let $\mathcal{B}$ be a set of boxes in $\mathbb{R}^d$ which are separated and satisfy $ \bigcap_{B \in \mathcal{B}} \partial B \neq \varnothing$. Let $B_0 \in \mathcal{B}$ such that $n$ ($0 \le n < d$) is the minimum dimension of a face of $B_0$ which contains  $ \bigcap_{B \in \mathcal{B}} \partial B$. Then
$$|\mathcal{B}| \le  2^{d-n}$$
Proof (simple counting argument on the $\mathbb{R}$ intervals of the boxes, given they have a common $n$ dimensional face). Qed.
Final Proof.
Use Theorem 1 and Theorem 2 to show that $\mathbb{R}^n$ cannot have $> 2^n$ boxes which are pairwise separable and pairwise share boundary points (use contradiction).
