Open balls form a basis of the metric topology

Statement

Let $(X, d)$ be a metric space. Then the set of all open balls $B(x, r)$, with $x \in X$ and real $r > 0$, is a basis of the metric topology of $X$. Also, the balls $B(x, 1/n)$ with $x \in X$ and integer $n \ge 1$ form a basis.

Remarks

Proof: each ball is open, by open balls are open. If $U$ is open and $x \in U$, then some ball $B(x, r)$ lies in $U$, and $B(x, 1/n) \subseteq B(x, r)$ for $n > 1/r$. For each point $x$, the balls $B(x, 1/n)$ form a countable neighbourhood basis at $x$. Thus each metric space is first-countable. In $\mathbb{R}$, the open balls are the bounded open intervals.

Explanations

L1 human

Each open set of a metric space is a union of open balls. So, to check a property for all open sets, it is often enough to check it for balls.