Finite subcover lemma

Finite subcover lemma: Let $(X,d)$ be a metric space and let $K\subseteq X$ be compact . If $\{U_\alpha\}_{\alpha\in A}$ is an open cover of $K$, meaning $K\subseteq \bigcup_{\alpha\in A} U_\alpha,$ then there exist $\alpha_1,\dots,\alpha_N\in A$ such that $K\subseteq U_{\alpha_1}\cup\cdots\cup U_{\alpha_N}.$

This is the defining operational feature of compactness and is used as a “black box” step in many arguments: convert infinitely many local pieces into finitely many.

Examples:

• The interval $[0,1]$ is compact, so any open cover of $[0,1]$ contains a finite subcover.
• The open interval $(0,1)$ is not compact: the cover $\{(1/n,1):n\in\mathbb{N}\}$ has no finite subcover.