---
tag: WPTE736R97
kind: proof
name: Proof that the closure is closed
msc: [54A05]
proves: 02TZDKG50Q
uses: [0CNQXVBHCX, 0DRKT1QJR7, 0G5AV9Q4RR, 0839XNQXDK, 05GK32H72H, 0B1FSS1VZW, 01ZQQDCERP]
sources:
  - url: https://proofwiki.org/wiki/Intersection_of_Closed_Sets_is_Closed_in_Topological_Space
---

## Proof

1. Let $(X, \tau)$ be a [[topological space|0CNQXVBHCX]] and let $S \subseteq X$.
2. Let $\mathcal{F}$ be the family of all [[closed sets|0G5AV9Q4RR]] $F$ of $X$ with $S \subseteq F$. By the definition of the [[closure|0DRKT1QJR7]], $\overline{S}$ is the [[intersection|0839XNQXDK]] of the members of $\mathcal{F}$.
3. The family $\mathcal{F}$ is nonempty. Indeed $X \setminus X = \varnothing$ is in $\tau$, so $X$ is closed, and $S \subseteq X$.
4. We claim that $X \setminus \overline{S}$ is the [[union|05GK32H72H]] of the [[complements|0B1FSS1VZW]] $X \setminus F$ for $F \in \mathcal{F}$.
5. Let $x \in X$. Then $x \notin \overline{S}$ if and only if $x \notin F$ for some $F \in \mathcal{F}$, that is, $x \in X \setminus F$ for some $F \in \mathcal{F}$. This proves the claim.
6. For each $F \in \mathcal{F}$, the set $X \setminus F$ is [[open|01ZQQDCERP]], because $F$ is closed.
7. By the second axiom of a topology, a union of members of $\tau$ is in $\tau$. By steps 4 and 6, $X \setminus \overline{S}$ is open.
8. By the definition of a closed set, $\overline{S}$ is closed in $X$.

## Idea

The closure is an intersection of closed sets. Its complement is the union of their complements, which are open. A union of open sets is open by the axioms, so the closure is closed.