Maximum number of $\varepsilon$-separated points in unit ball. We place ourselves in $\Bbb{R}^d$, for $d\geq 1$. For a given $\varepsilon>0$, how many points $x_i$ can we place on the unit ball so that $\lVert x_i-x_j\lVert>\varepsilon$ for $i\neq j$?
I found the following bound: the unit ball is contained in the unit cube of side $1$, so we construct an $\frac{\varepsilon}{\sqrt{d}}$-grid on the unit cube, that divides in in "little cubes" of diameter $\varepsilon$ (the diameter of each little cube is the lenght of its big diagonal, which is equal to $\varepsilon$). Then we can only have at most only one $x_i$ in any little cube, and the number of those little cubes (contained in the unit square) being roughly $\left(\frac{\sqrt{d}}{\varepsilon}\right)^d$, we find a bound. 
But the the thing is, apparently I can find a bound in $\mathcal{O}\left(\frac{1}{\varepsilon}\right)^{d-1}$. It makes sense because the unit ball is smaller than the unit square, but I don't know how to show it.
 A: Asymptotically, $O(\varepsilon^{-d})$ is the right bound. Another way to prove it is to let $B_i$ be the ball of radius $\varepsilon/2$ centered at $x_i$; then the balls $B_i$ are disjoint and contained in the ball of radius $(1+\varepsilon/2)$. Hence, the sum of volumes of $B_i$ is at most the volume of the ball of radius $(1+\varepsilon/2)$, which yields
$$N \le \frac{(1+\varepsilon/2)^{d}}{(\varepsilon/2)^d} = O(\varepsilon^{-d})\quad \text{ as } \varepsilon\to 0$$
The exponent $-d$ cannot be improved; in particular there is no bound of the form $O(\varepsilon^{1-d})$. Indeed, the unit ball contains a cube with sidelength $2/\sqrt{d}$. Divide it into cubes of sidelength slightly greater than $\varepsilon$; there are about $\left(\frac{2}{\varepsilon\sqrt{d}}\right)^d$ such cubes, which is $\varepsilon^{-d}$ up to a constant. The centers of these cubes are separated from each other by distance $>\varepsilon$. 
I would not expect there to be an explicit formula for the maximum number of $\varepsilon$-separated points.
