Closure in a metric space

Statement

Let $(X, d)$ be a metric space, let $S \subseteq X$ and let $x \in X$. Then $x$ is in the closure $\overline{S}$ if and only if for each real $\varepsilon > 0$ there is $s \in S$ with $d(x, s) < \varepsilon$.

Remarks

Equivalently, $x \in \overline{S}$ if and only if each open ball $B(x, \varepsilon)$ meets $S$. For nonempty $S$, this means $d(x, S) = 0$, where $d(x, S)$ is the distance from the point to the set. In $\mathbb{R}$, the point $0$ is in the closure of $\{1/n : n \ge 1\}$, because $1/n < \varepsilon$ for large $n$.

Explanations

L1 human

A point is in the closure of $S$ when there are points of $S$ as close to it as you want. The point does not need to be in $S$ itself.