----- > [!proposition] Proposition. ([[finite union of compact is compact]]) > Let $X$ be a [[topological space]] and $K_{1},\dots,K_{n}$ be [[compact]] [[subspace topology|subspaces]] of $X$. Then $K:=K_{1} \cup \dots \cup K_{n}$ is also a [[compact]] subspace of $X$. ^3a0678 > [!basicnonexample] > The result is generally false for infinite unions. For example, the set $\mathbb{R}= \bigcup_{n \in \mathbb{Z}}^{} [n, n+2] $ is not [[compact]], despite being the union of [[closed interval|closed (hence compact) intervals]]. ^496c6e > [!proof]- Proof. ([[finite union of compact is compact]]) > We will use the [[compactness characterization for subspaces]]. Consider a [[cover|covering]] $\mathscr{A}$ of $K$ by open subsets of $X$. This induces a covering $\mathscr{A}_{i} \subset \mathscr{A}$ of each $K_{i}, i \in [n]$. Because $K_{i}$ is [[compact]], $\mathscr{A}_{i}$ admits a finite subcover $\mathcal{A}_{i}$. Then the finite collection $\{ \mathcal{A}_{i} \}_{i=1}^{n} \subset \mathscr{A}$ is a finite subcollection of open sets in $X$ that covers $K$. Hence, $K$ is [[compact]]. ^f62f88 ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```