Let $(M,d)$ be a [metric space](/page/Metric%20Space), and let $\mathbb{N} := \{1,2,3,\dots\}$. For $x \in M$ and $r > 0$, write $B(x,r) := \{y \in M : d(x,y) < r\}$. Assume that for every $\varepsilon > 0$ there exists a finite set $F \subset M$ such that $M \subset \bigcup_{a \in F} B(a,\varepsilon)$. Then $(M,d)$ is separable; that is, there exists a countable subset $D \subset M$ whose closure in $M$ is all of $M$.