Problem 1876

(a) Let $A$ be an admissible set in $\mathbb{Z}^2$, choose a point $a_0$ of $A$, and, for each positive integer $k$, choose a point $a_k$ of $A$ different from $a_{k-1}$, having the same first coordinate as the latter if $k$ is odd, and the same second coordinate if $k$ is even. Eventually we must choose an $a_n = a_m$, $m < n$. Assume $a_n$ is the first point to duplicate a preceding point. If $m$ and $n$ have like parities, then $a_m, a_{m+1}, \dots, a_{n-1}$ form a null set, and if they have opposite parities, then $a_{m+1}, \dots, a_{n-1}$ do.

(b) We exhibit a minimal admissible set $A$ in $\mathbb{Z}^3$ which is not itself null. Here and hereafter, minimality refers to the fact that no proper subset is admissible. Since every null finite set is admissible, the conclusion follows. The set $A$ is a set of lattice points in the parallelepiped $[0, 3] \times [0, 3] \times [0, 4]$. We describe it by successive horizontal cross-sections:
$$A = A_0 \times 0 \cup A_1 \times 1 \cup A_2 \times 2 \cup A_3 \times 3 \cup A_4 \times 4,$$
where $A_0 = \{0, 3\} \times \{0, 3\}$, $A_1 = \{0, 1\} \times \{2, 3\} \cup \{1, 2\} \times \{0, 1\}$, $A_2 = \{0, 1\} \times \{1, 2\} \cup \{2, 3\} \times \{2, 3\}$, $A_3 = \{1, 2\} \times \{2, 3\} \cup \{2, 3\} \times \{0, 1\}$, and $A_4 = \{0, 1\} \times \{0, 1\} \cup \{2, 3\} \times \{1, 2\}$. Notice that, for $k = 1, 2, 3$, the configuration $A_{k+1}$ is obtained from $A_k$ by a clockwise rotation through $\pi/2$ about the centre of the square $[0, 3] \times [0, 3]$.

The set $A$ is admissible, since each horizontal cross-section $A_k \times k$ is admissible in $\mathbb{Z}^2 \times k$, and the perpendicular in $\mathbb{Z}^3$ to any horizontal cross-section through any one of its points meets at least one other horizontal cross-section.

To prove minimality, we exhibit a connected geometric lattice graph $G$ on $A$ such that the line of support of each edge of $G$ is a line in $\mathbb{Z}^3$ stabbing $A$ at exactly two points, namely, the end points of that edge. The existence of such a graph implies minimality, since removal of any one point in $A$ entails removal of all its neighbours in $G$, and eventually removal of all of $A$.

Begin by noticing that each of the verticals $i \times j \times \mathbb{Z}$ through a point of $A$, where either $i$ or $j$ is in $\{0, 3\}$, stabs exactly two horizontal cross-sections of $A$. Join the corresponding points of $A$ by the vertical segment they determine.

Next, consider the generic planar lattice paths $\alpha_1 = (1 \times 0)(2 \times 0)(2 \times 1)(1 \times 1)$ and $\alpha'_1 = (1 \times 2)(0 \times 2)(0 \times 3)(1 \times 3)$, and, for $k = 1, 2, 3$, let $\alpha_{k+1}$ and $\alpha'_{k+1}$ be obtained from $\alpha_k$ and $\alpha'_k$, respectively, by a clockwise rotation through $\pi/2$ about the centre of the square $[0, 3] \times [0, 3]$. The edges of the lattice paths $\alpha_k \times k$ and $\alpha'_k \times k$, joining points in $A_k \times k$, $k = 1, 2, 3, 4$, along with the vertical edges in the previous paragraph form the desired connected geometric lattice graph $G$ on $A$.

Finally, the set $A$ is not null, for the vertical $1 \times 1 \times \mathbb{Z}$ stabs exactly three horizontal cross-sections of $A$, namely, $A_1 \times 1$, $A_2 \times 2$ and $A_4 \times 4$; in fact, each of the lines $i \times j \times \mathbb{Z}$, $i, j \in \{1, 2\}$, stabs exactly three horizontal cross-sections of $A$.