# Epsilon Nets

#### A common technique in arguments in sketching is to appeal to the concept of an Epsilon Net. This article explains what that is and why it always exists.

We will be working in the column space of a matrix $A \in \mathbb{R}^{n \times d}$. Without loss of generality, we will assume that $A$ has orthonormal columns; if it doesn’t, we can always find an orthonormal basis for its column space by considering its SVD, $A=U\Sigma V^\ast$, and redefining $A:=U$. That is enough because the set $\{ Ax \;|\; x \in \mathbb{R}^d\}$ is the same as the set $\{ Uz \;|\; z \in \mathbb{R}^\rho \}$ where $\rho$ is the rank of $A$.

**Definition**: Define the set $\mathcal{S} = \{ y \;|\; y=Ax, \; \lVert y \rVert_2 = 1 \}$. An $\epsilon$-net is a subset $\mathcal{N} \subset \mathcal{S}$ such that, for every $y \in \mathcal{S}$, there exists a $w \in \mathcal{N}$such that $\lVert y - w \rVert_2 \leq \epsilon$.

### Existence

**Lemma**: For $\epsilon \in (0, 1)$, there exists an $\epsilon$-net $\mathcal{N}$ for the set $\mathcal{S}$ (defined above) whose size is at most $(1 + 2/\epsilon)^d$.

**Proof**: Let’s consider the unit sphere in $\mathbb{R}^d: \mathbb{S}^{d-1}$. Instead of finding a net for $\mathcal{S}$, we will argue by finding a net for $\mathbb{S}^{d - 1}$. That’s because $A$ provides an isometry when operating on the unit sphere, so that a net for $\mathbb{S}^{d - 1}$ gives us a net for the image of $\mathbb{S}^{d - 1}$ under $A$.

Finding a net over the unit sphere is trivial. We can choose a maximal set $\mathcal{N}^\prime$ such that no two points in the set are within distance $\epsilon$ from each other. This means that balls of radius $\epsilon/2$ centered at points in $\mathcal{N}^\prime$ are disjoint. These balls all fit in the ball with radius $(1 + \epsilon/2)$ centered at the origin. The volume of this enlarged ball is $(1 + \epsilon/2)^d / (\epsilon/2)^d$ times larger than the small balls. As such, there can be at most $(1 + 2/\epsilon)^d$ small balls, implying that $\lvert \mathcal{N}^\prime \rvert \leq (1 + 2/\epsilon)^d$. [This is the covering number of the unit sphere, which is a compact set.]

Now let $\mathcal{N} = \{ y \;|\; y = Ax, \; x \in \mathcal{N}^\prime \}$. We can show by contradiction that, for any $y \in \mathcal{S}$ there must exist a point $w$ in $\mathcal{N}$ such that $\lVert y - w \rVert_2 \leq \epsilon$.$\mathbb{}$ Suppose there exists no such $w$ for some $y=Ax$ for some $x \in \mathbb{S}^{d - 1}$. That implies that, there exists no point $v$ in $\mathcal{N}^\prime$ for which $\lVert x - v \rVert_2 \leq \epsilon$. But that contradicts our construction of $\mathcal{N}^\prime$.