---
tag: 066HW6D9A7
kind: statement
role: proposition
name: Closed subsets of compact sets are compact
msc: [54D30]
uses: [0P6PMJB4MA, 0G5AV9Q4RR]
sources:
  - url: https://en.wikipedia.org/wiki/Compact_space
    locator: Properties of compact spaces
---

## Statement

Let $X$ be a [[topological space|0CNQXVBHCX]], let $K \subseteq X$ be [[compact|00R0X1WGPX]] and let $F \subseteq X$ be [[closed|0G5AV9Q4RR]] with $F \subseteq K$.
Then $F$ is compact.

## Remarks

Proof: let $(U_i)$ be open sets of $X$ that cover $F$. Add the open set $X \setminus F$ to get an open cover of $K$. A finite subcover of $K$ gives, after you remove $X \setminus F$, a finite subcover of $F$.
More generally, if $K$ is compact and $F$ is closed, then $K \cap F$ is compact.
For example, the [[Cantor set|0BXNDZFQME]] is compact, because it is a closed subset of $[0, 1]$.

## Explanations

### L1 human

A closed part of a compact set is again compact.
The open complement of the closed part covers everything outside it, so it does the job of all the extra open sets.