Contents

# Contents

## Statement

###### Proposition

Let $(X,\tau)$ be a compact topological space, and let $Y \subset X$ be a closed topological subspace. Then also $Y$ is compact.

###### Proof

Let $\{V_i \subset Y\}_{i \in I}$ be an open cover of $Y$. We need to show that this has a finite sub-cover.

By definition of the subspace topology, there exist open subsets $U_i$ of $X$ with

$V_i = U_i \cap Y \,.$

By the assumption that $Y$ is closed, the complement $X \setminus Y$ is an open subset of $X$, and therefore

$\{ X \setminus Y \subset X\} \cup \{ U_i \subset X \}_{i \in I}$

is an open cover of $X$. Now by the assumption that $X$ is compact, this latter cover has a finite subcover, hence there exists a finite subset $J \subset I$ such that

$\{ X \setminus Y \subset X\} \cup \{ U_i \subset X \}_{i \in J \subset I}$

is still an open cover of $X$, hence in particular intersects to a finite open cover of $Y$. But since $Y \cap ( X \setminus Y ) = \empty$, it follows that indeed

$\{V_i \subset Y\}_{i \in J \subset I}$

is a cover of $Y$, and is indeed a finite subcover of the original one.

Last revised on November 24, 2022 at 11:24:59. See the history of this page for a list of all contributions to it.