Closed balls are closed

Statement

Let $(X, d)$ be a metric space, let $x \in X$ and let $r \ge 0$. Then the closed ball $\bar{B}(x, r)$ is a closed set of $X$.

Remarks

Proof: let $y \notin \bar{B}(x, r)$ and let $\varepsilon = d(x, y) - r > 0$. If $d(y, z) < \varepsilon$, then $d(x, z) \ge d(x, y) - d(y, z) > r$ by the triangle inequality. Thus the complement of $\bar{B}(x, r)$ is open. For $r = 0$, this shows that each one-point set $\{x\}$ is closed in a metric space.

Explanations

L1 human

A point outside the closed ball is at some positive distance beyond its edge. A small ball about that point stays outside, so the outside is open.