Union of open sets is open

Statement

Let $(X, d)$ be a metric space and let $(U_i)_{i \in I}$ be a family of open subsets of $X$, indexed by any set $I$. Then the union $\bigcup_{i \in I} U_i$ is open in $X$.

Remarks

Proof: let $x$ be in the union. Then $x \in U_j$ for some $j \in I$, and there is $\varepsilon > 0$ with $B(x, \varepsilon) \subseteq U_j$. This ball is also in the union. Then use open sets of a metric space. The family can be infinite, and even uncountable. In a general topological space, this property is one of the axioms.

Explanations

L1 human

Each point of the union lies in one of the open sets, and that set already gives a small ball about the point. For example, the union of the intervals $(1/n, 1)$ for $n \ge 2$ is the open interval $(0, 1)$.