Hyperrectangle that contains most of cube's interior (except its vertices)

Question

Let $n>0$, and let $p,q\in (0,1)$ such that $p<q$.

Is there a hyperrectangle $H$ that satisfies the following:

• $\forall i\in{1,\dots,n}:\\ H\supset \prod_{j=1,\dots,n} \begin{cases} [p,q], &\text{if j=i}\\ [0,1], &\text{else} \end{cases}$
• $\{0,1\}^n\cap H=\emptyset$

I tried many examples by trial and error, so I was able to conclude that there exists such a hyperrectangle (or actually, a square) in 2D, but I couldn't find such a hyperrectangle in 3D or in higher dimensions, or to disprove its existence.

Any help would be appreciated.

Answer 1

In dimension $n\geqslant 3$ it is not possible. Indeed, the hyperrectangle $H$ has $2n$ facets and the cube $\{0,1\}^n$ has $2^n>2n$ vertices. For any vertex $v$ there should exist a facet $\alpha \subset H$ which separates $v$ from $H$. By pigeonhole principle there exist two vertices $v_1,v_2$ with the same $\alpha$. Then $\alpha$ separates the whole segment between $v_1$ and $v_2$. If $v_1$ and $v_2$ differ in $i$'th coordinate, this segment contains a point with $i$-th coordinate belonging to $[p,q]$. This contradicts to your assumption.

Answer 2

Note: Not really an answer to your question per se, but rather a different direction in which to generalise your 2D result.

A really straightforward higher-dimensional generalisation is given by hyperoctahedra, the duals of hypercubes. In 2D they are the squares you found, but in 3D they are octahedra. Their vertices correspond to hyperfaces, and conversely the vertices of the hypercube correspond to the hyperoctahedron's hyperfaces. As you scale the hyperoctahedron up, it will accommodate larger and larger values of $p,q \in (0,1)$, until the point where its faces finally touch the vertices of the hypercube.