Piercing axis-parallel boxes

Abstract

Let $\F$ be a finite family of axis-parallel boxes in ${\mathbb{R}}^d$ such that $\F$ contains no $k+1$ pairwise disjoint boxes. We prove that if $\F$ contains a subfamily $\M$ of $k$ pairwise disjoint boxes with the property that for every $F\in \F$ and $M\in \M$ with $F\cap M\ne \varnothing$, either $F$ contains a corner of $M$ or $M$ contains $2^{d-1}$ corners of $F$, then $\F$ can be pierced by $O(k)$ points. One consequence of this result is that if $d=2$ and the ratio between any of the side lengths of any box is bounded by a constant, then $\F$ can be pierced by $O(k)$ points. We further show that if for each two intersecting boxes in $\F$ a corner of one is contained in the other, then $\F$ can be pierced by at most $O(k\log \log (k))$ points, and in the special case where $\F$ contains only cubes this bound improves to $O(k)$.