Is this bounded Let $d_{k}$ be supremum of the minimum of the pairwise distances between any
$k$ points in the unit square. Is $kd_{k}$ bounded as $k\rightarrow\infty$ ?
 A: The centered hexagonal numbers have the form $3k(k+1) + 1$. This allows us to place $3k(k+1) + 1$ points whose minimum distance apart is $\frac{1}{k-1}$. 
Thus, for $n = 3k(k+1) + 1$, $d_n \geq \frac{1}{k-1}, $ which gives $ n d_n \geq \frac{ 3k(k+1)+1}{k-1} > 3k $. Hence, the sequence is unbounded.
In fact, this shows that for $\alpha > \frac{1}{2}$, $ n^ \alpha d_k$ is unbounded.

This was scratch work to motivate that the sequence is unbounded. It calculates an upper bound for $d_n$.
Fix $n$. we want to calculate an $r$ such that we will find 2 points distance at most $r$ apart. 
Consider circles of radius $\frac{r}{2}$ about each point. These fall within a circle of radius $1+\frac{r}{2}$ If any of the small circles overlap, we have our 2 points. Hence if
$$ n r^2 < (2 +r)^2 \Rightarrow (n-1)r^2 - 4r - 4 < 0,$$
then we can find 2 points that are within $r$ of each other. We can solve this for $r$ to get $ r = \frac {4 + \sqrt{ 16+4(n-1)}} {2(n-1)} $
This strongly suggests that $nr$ is unbounded, and in fact grows like $\sqrt n$. At this point, I'd suggest finding constructions that look like honeycomb packing. 
