----- > [!proposition] Proposition. ([[compactness characterization for subspaces]]) > Let $X$ be a [[topological space]] and $Y$ a [[subspace topology|subspace]] of $X$. Then $Y$ is [[compact]] if and only if every [[cover|covering]] of $Y$ by sets open *in $X$* contains a finite subcollection covering $Y$. ^9fd510 > [!proof]- Proof. ([[compactness characterization for subspaces]]) > > $\to.$ Suppose $Y$ is [[compact]]. Let $\mathscr{A}$ be a [[cover|covering]] of $Y$ by sets open in $X$. Then the collection $\{ Y \cap A : A \in \mathscr{A} \}$ is an open cover of $Y$ by sets open in $Y$. By compactness of $Y$ it contains a finite subcollection of open sets in $Y$ covering $Y$, $Y = \{ Y \cap A_{i} \}_{i=1}^{n}.$ > Now it is clear that $\{ A_{i} \}_{i=1}^{n} \subset \mathscr{A}$ is a finite subcollection of sets *in $X$* that covers $Y$. > > $\leftarrow.$ Conversely, suppose the other condition holds; we wish to prove $Y$ compact. Take an open cover $\mathscr{A}'$ of $Y$ (by sets open in $Y$). Each member has the form $A \cap Y$ for some open subset $A$ of $X$. The collection $\mathscr{A}$ of all such $A \subset X$ covers $Y$, hence we can obtain a finite subcollection $\{ A_{i} \}_{i=1}^{n}$ covering $Y$. Now $\{ Y \cap A_{i} \}_{i=1}^{n} \subset \mathscr{A}'$ > is the required finite subcover that witnesses $Y$ [[compact]]. ----- #### ---- #### 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 > ```