Application of the pigeonhole principle

Work in $d$ -dimensions. Let $Q$ be the $d$ -dimensional volume of a closed box, $B\subseteq\mathbb{R}^{d}$ . Suppose there are $N$ distinct points, $p_{0},p_{1},\ldots,p_{N-1}$ in the box. Let $r:=\min_{i\neq j}d(p_{i},p_{j})>0$ . Then $\mathcal{B}(p_{i};r)$ are pairwise disjoint balls,( $\color{#D61F06}{\dagger}$ ) and for each one, at least a fraction $\frac{1}{2^{d}}$ of it must be contained in the closed Box, $B$ . Appealing to these considerations one has

$\begin{array}{c}[mc]{rcl} Q &= &\mathcal{L}_{d}(B)\\ &\geq &\frac{1}{2^{d}}\mathcal{L}_{d}(\bigcup_{i<N}\mathcal{B}(p_{i};r))\\ &= &\frac{1}{2^{d}}\sum_{i<N}\mathcal{L}_{d}(\mathcal{B}(p_{i};r))\\ &= &\frac{1}{2^{d}}\sum_{i<N}V_{d}r^{d}\\ &= &\frac{NV_{d}r^{d}}{2^{d}}\\ \end{array}$

where $\mathcal{L}_{d}(\cdot)$ is the $d$ -dimensional Lebesgue-measure and $V_{d}=$ the volume of a unit $d$ -ball. Hence $N\leq\frac{2^{d}Q}{V_{d}r^{d}}$ . It follows that there can be at most $\frac{2^{d}Q}{V_{d}r^{d}}$ points in the box.

Applying this to our situation, we have the upperbound

$N_{\max}\leq \frac{2^{2}\cdot 3\times 3}{\pi\cdot \sqrt{2}^{2}}=\frac{36}{2\pi}\approx 5.7$

Thus there can be no more than $5$ points in the square. (In fact this is the exact upper bound: placing points on the corner and on in the centre yields $5$ points a distance of at minimum $1.5$ units apart.)

Edit. This approach is overkill. The $r$ -balls are not disjoint. The only property one can derive is that each centre is contained in exactly one ball. This collapses the argument at ( $\color{#D61F06}{\dagger}$ ). To run this approach correctly, it is rather that the balls of radius $r/2$ be disjoint, and filling out the box-like region (expanding both ends of all sides by $r/2$ ), one has

$(s+r)^{d}\geq NV_{d}(r/2)^{d}$

where $s$ is the side-length, so that $N\leq \frac{2^{d}(s+r)^{d}}{V_{d}r^{d}}=\frac{2^{2}\cdot(3+\sqrt{2})^{2}}{\pi\cdot\sqrt{2}^{2}}\approx 12.4$ . Thus upper bound is correct, but tells us at most that $N\geq 13$ is impossible, and nothing about whether $N=10$ is possible. The approach by [ David Ingerman ] is more compact and yields the desired information in this case.

Imagine splitting the 3 by 3 square into nine identical 1 by 1 squares. By pigeonhole principle, there's a pair of points in some small square that would be at most the diagonal length apart, which is $\sqrt{2}$ .