Compactness via open covers

Statement

Let $X$ be a topological space and let $K \subseteq X$. Then $K$ is compact if and only if each family $(U_i)_{i \in I}$ of open subsets of $X$ with $K \subseteq \bigcup_{i \in I} U_i$ has a finite subfamily $U_{i_1}, \ldots, U_{i_n}$ with $K \subseteq U_{i_1} \cup \cdots \cup U_{i_n}$.

Remarks

Proof: the open sets of the subspace topology on $K$ are the sets $U \cap K$ with $U$ open in $X$. So covers of $K$ by open sets of $K$ and covers by open sets of $X$ match each other. For $K = X$, the statement is the definition of a compact space. Thus compactness of a subset does not depend on the space around it. If $K \subseteq Y \subseteq X$, then $K$ is compact in $Y$ if and only if it is compact in $X$.

Explanations

L1 human

To test if a subset is compact, you can use open sets of the big space. Each time you cover the subset with open sets, finitely many of them must already be enough.