Piercing axis-parallel boxes

Let F be a finite family of axis-parallel boxes in 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 ∈ F and M ∈ M with F ∩ M ≠ ∅, 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).