Open sets of a metric space

Statement

Let $(X, d)$ be a metric space and let $U \subseteq X$. Then $U$ is open in the metric topology if and only if for each $x \in U$ there is a real $\varepsilon > 0$ with $B(x, \varepsilon) \subseteq U$. Here $B(x, \varepsilon)$ is the open ball with centre $x$ and radius $\varepsilon$.

Remarks

The number $\varepsilon$ can depend on $x$. In $\mathbb{R}$, the set $(0, \infty)$ is open: for $x > 0$, take $\varepsilon = x$. The set $\mathbb{Q}$ is not open in $\mathbb{R}$, because each ball about a rational number contains an irrational number.

Explanations

L1 human

This is the test for open sets that you use in practice. You pick any point of the set, and you find a small ball about it inside the set.