> [!NOTE] Lemma (An arbitrary union of open sets is open). > > If $U_\lambda$ is [[Open subsets of metric spaces|open]] for all $\lambda\in\Lambda$, where $\Lambda$ is an > indexing set (which could be uncountable), then [[Union of sets|union]] $\bigcup_{\lambda\in\Lambda} U_\lambda$ is open. ###### Proof If $p\in\bigcup_{\lambda\in\Lambda}U_\lambda$ then $\exists\lambda^*\in\Lambda$ such that $p\in U_{\lambda^*}.$ But $U_{\lambda^*}$ is open and therefore, $\exists\varepsilon>0$ such that $\mathbb{B}(p,\varepsilon)\subset U_\lambda*.$ In particular, $\mathbb{B}(p,\varepsilon)\subset\bigcup_\lambda\in\Lambda U_\lambda$, i.e., $\bigcup_\lambda\in\Lambda U_\lambda$ is open.