---
tag: JN5P11E6X6
kind: proof
name: Proof that compactness of a subset can be tested with open sets of the ambient space
msc: [54D30]
proves: 0P6PMJB4MA
uses: [0CNQXVBHCX, 00R0X1WGPX, 0QPHW703PV, 0D5YHQGDJY, 0NVPNC12JY, 01ZQQDCERP]
sources:
  - url: https://en.wikipedia.org/wiki/Compact_space
    locator: Compactness of subsets
---

## Proof

We translate covers in each direction with the definition of the subspace topology.

1. Let $X$ be a [[topological space|0CNQXVBHCX]] and let $K \subseteq X$. Give $K$ the [[subspace topology|0D5YHQGDJY]], whose open sets are the sets $K \cap U$ with $U$ [[open|01ZQQDCERP]] in $X$.
2. By definition, $K$ is [[compact|00R0X1WGPX]] when $K$ with this topology is a [[compact space|0QPHW703PV]]. That is, each [[open cover|0NVPNC12JY]] of $K$ by open sets of $K$ has a finite subcover.
3. Assume $K$ is compact. Let $(U_i)_{i \in I}$ be open sets of $X$ with $K \subseteq \bigcup_{i \in I} U_i$.
4. For each $i$, the set $V_i = K \cap U_i$ is open in $K$. Each $x \in K$ lies in some $U_i$, hence in $V_i$. So $(V_i)_{i \in I}$ is an open cover of the space $K$.
5. By compactness, there are $i_1, \ldots, i_n \in I$ with $K = V_{i_1} \cup \cdots \cup V_{i_n}$.
6. Since $V_{i_j} \subseteq U_{i_j}$, we get $K \subseteq U_{i_1} \cup \cdots \cup U_{i_n}$. This proves the first implication.
7. Now assume the cover condition for open sets of $X$. Let $(V_i)_{i \in I}$ be an open cover of the space $K$ by open sets of $K$.
8. For each $i$, choose an open set $U_i$ of $X$ with $V_i = K \cap U_i$. This is possible by step 1.
9. Then $K = \bigcup_i V_i \subseteq \bigcup_i U_i$. By the assumption, there are $i_1, \ldots, i_n \in I$ with $K \subseteq U_{i_1} \cup \cdots \cup U_{i_n}$.
10. Let $x \in K$. Then $x \in U_{i_j}$ for some $j$, and $x \in K$, so $x \in K \cap U_{i_j} = V_{i_j}$.
11. So $V_{i_1}, \ldots, V_{i_n}$ is a finite subcover of $(V_i)_{i \in I}$. Thus $K$ is a compact space, and $K$ is compact.

## Idea

An open set of the subspace $K$ is just an open set of $X$ cut down to $K$.
So a cover of $K$ by open sets of $X$ and a cover by open sets of $K$ carry the same information, and finite subcovers match up.