> [!NOTE] Lemma (The finite intersection of open sets is open). > If $U_{1},U_{2},\ldots,U_{m}$ are all [[Open subsets of metric spaces|open]], then the [[Finite Intersection|finite intersection]] $\bigcap_j=1^{m}U_{j}$ is also open. ###### Proof If $p\in\bigcap_{j=1}^{m}U_{j}$ then $\begin{gathered} \exists \varepsilon_{1}>0 \text{ such that } \mathbb{B}_{p}(\varepsilon_{1}) \subset U_{1} \\ \vdots \\ \exists \varepsilon_{m}>0 \text{ such that } \mathbb{B}_{p}(\varepsilon_{m}) \subset U_{m} \end{gathered}$Set $\varepsilon:=\operatorname*{min}\{\varepsilon_{1},\ldots,\varepsilon_{m}\}>0.$ Then $\mathbb{B}(p,\varepsilon)\subset\bigcap_{j=1}^{m}U_{j}$,i.e., $\bigcap_{j=1}^{m}U_{j}$ is open.