Let $(M, d)$ be a metric space and $A \subset M$. The following are equivalent:

1. $A$ is totally bounded. 2. For every $\varepsilon > 0$, there exists a finite subset $F \subset M$ such that $A \subset \overline{B}_\varepsilon(F) := \{x \in M : \operatorname{dist}(x, F) \le \varepsilon\}$. 3. The closure $\overline{A}$ is totally bounded. 4. $A$ has a totally bounded $\varepsilon$-net for every $\varepsilon > 0$ (meaning the $\varepsilon$-net itself can be chosen to be totally bounded --- in fact, finite).