Constructing an $\epsilon$-net of $l_2$ unit ball

I am interested in probabilistic or explicit ways to construct an $\epsilon$-net of the $l_2$ unit ball in $\mathbb{R}^{d}$.

I know that, for every $\epsilon > 0$, there exists an $\epsilon$-net $\mathcal{N}_{\epsilon}$ for the unit sphere in $d$ dimensions such that $$ M\triangleq\left|\mathcal{N}_{\epsilon}\right| \le \left( 1+\frac{2}{\epsilon}\right)^{d}. $$ (Lemma 5.2 in https://arxiv.org/abs/1011.3027) To my understanding, the aforementioned bound holds for an $\epsilon$-net of the entire ball, not only the sphere.

In the case of the sphere, we can construct an $\epsilon$-net with high probability, by drawing a sufficient number ($O(M\log{M})$) of independent random vectors according to a Gaussian distribution $N(\mathbf{0}, \mathbf{I})$, and normalizing the length to $1$. I believe that one way to get an $\epsilon$-net for the ball, would be to repeat the above procedure $O(1/\epsilon)$ times, for all spheres of radii $\epsilon, 2\epsilon,3\epsilon, \dots, 1$. The union of the $\epsilon$-nets, should be able to cover the ball. However, it would require $\tilde{O}\left((1+2/{\epsilon})^{d+1}\right)$ points (ignoring the logarithmic factor).

• Is there a simple way to construct an $\epsilon$-net for the unit ball directly, $\textit{i.e.}$, without constructing nets for multiple spheres?
• Is there way to achieve the bound on $\left|\mathcal{N}_{\epsilon}\right|$ (possibly up to logarithmic factors)?

I would appreciate any pointers to either probabilistic or explicit methods.

We can use surface area instead of volume.

Let $N_\epsilon$ be a subset of $S^{n-1}$ (embedded in $\mathbb{R}^n$) such that any two points in $N_\epsilon$ have distance strictly greater than $\epsilon$ and $N_\epsilon$ is maixmal. This means that the balls of radius $\epsilon/2$ centered at points in $N_\epsilon$ are disjoint and are contained in $B(0,1+\epsilon/2)$ and are outside the ball $B(0,1-\epsilon/2)$. Let $V(r)$ be the volume of the $n$ dimensional ball of radius $r$ and let $S(r)$ be the surafce area of the ball of radius $r$. Then \begin{align*} |N_\epsilon| V(\epsilon/2) &\le V(1+\epsilon/2) - V(1-\epsilon/2) \\ &= \int_{1-\epsilon/2}^{1+\epsilon/2} S(r) dr \\ &\le \int_{1-\epsilon/2}^{1+\epsilon/2} S(1+\epsilon/2) dr \\ &= \epsilon S(1+\epsilon/2) \\ &= \epsilon (1+\epsilon/2)^{n-1} S(1) \end{align*}

Then using the relation between the volume and the surface area, we get that $$ |N_\epsilon| \le \epsilon (1+\epsilon/2)^{n-1} \frac{2^n}{\epsilon^n} \frac{S(1)}{V(1)} = 2n \left(1+\frac{2}{\epsilon}\right)^{n-1}. $$

Here we gained a factor of $n$ in front, but we reduced the power from $n$ to $n-1$. Hence when $\epsilon$ is small, this is a better bound than the volumetric bound.