Finite intersection of open sets is open

Statement

Let $(X, d)$ be a metric space, let $n \ge 1$ be an integer, and let $U_1, \ldots, U_n$ be open subsets of $X$. Then the intersection $U_1 \cap \cdots \cap U_n$ is open in $X$.

Remarks

Proof: let $x$ be in the intersection. For each $k$ there is $\varepsilon_k > 0$ with $B(x, \varepsilon_k) \subseteq U_k$. Then $\varepsilon = \min(\varepsilon_1, \ldots, \varepsilon_n) > 0$ and $B(x, \varepsilon)$ lies in each $U_k$. The statement is false for infinite families: in $\mathbb{R}$, the intersection of the open intervals $(-1/n, 1/n)$ for $n \ge 1$ is $\{0\}$, which is not open. In a general topological space, this property is one of the axioms.

Explanations

L1 human

Each open set gives its own small ball about the point, and the smallest of these finitely many balls fits in all the sets. With infinitely many sets, the radii can shrink to zero, and then no ball is left.