Finite subcover lemma - Imported Knowlpedia Content

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}.

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.

Remarks

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.