Closure Operator from Closed Sets

Theorem
Let $S$ be a set.

Let $\mathcal C$ be a set of subsets of $S$.

Let $\mathcal C$ be closed under arbitrary intersections:
 * $\forall \mathcal K \in \mathcal P \left({\mathcal C}\right): \bigcap \mathcal K \in \mathcal C$

where $\bigcap \varnothing$ is taken to be $S$.

Define $\operatorname{cl}: \mathcal P(S) \to \mathcal C$ by letting:
 * $\operatorname{cl} \left({T}\right) = \bigcap \left\{{C \in \mathcal C: T \subseteq C}\right\}$

Then $\operatorname{cl}$ is a closure operator whose closed sets are the elements of $\mathcal C$.

Proof
First we will show that $\operatorname{cl}$ is a closure operator.

Inflationary
Let $T \subseteq S$.

By Set Intersection Preserves Subsets/General Result/Corollary, $T \subseteq \operatorname{cl} \left({T}\right)$.

Since this holds for all such $T$, $\operatorname{cl}$ is inflationary.

Increasing
Let $T \subseteq U \subseteq S$.

Let $\mathcal T$ and $\mathcal U$ be the sets of elements of $\mathcal C$ containing $T$ and $U$, respectively.

Since Subset Relation is Transitive, every set containing $U$ contains $T$, so $\mathcal U \subseteq \mathcal T$.

By Intersection is Decreasing, $\bigcap \mathcal T \subseteq \bigcap \mathcal U$.

Thus $\operatorname{cl} \left({T}\right) \subseteq \operatorname{cl} \left({U}\right)$.

Idempotent
Let $T \subseteq S$.

By the premise, the intersection of a subset of $\mathcal C$ is in $\mathcal C$.

Thus in particular $\operatorname{cl} \left({T}\right) \in \mathcal C$.

Therefore:
 * $\operatorname{cl} \left({\operatorname{cl} \left({T}\right)}\right) \subseteq \operatorname{cl} \left({T}\right)$

Since $\operatorname{cl}$ is inflationary:
 * $\operatorname{cl} \left({T}\right) \subseteq \operatorname{cl} \left({ \operatorname{cl} \left({T}\right)}\right)$

By definition of set equality:
 * $\operatorname{cl} \left({\operatorname{cl} \left({T}\right)}\right) = \operatorname{cl} \left({T}\right)$

Since this holds for all $T \subseteq S$, $\operatorname{cl}$ is idempotent.

Finally, we need to show that the elements of $\mathcal C$ are the closed sets with respect to $\operatorname {cl}$.

If $C \in \mathcal C$, then since $\operatorname{cl}$ is inflationary:


 * $C \subseteq \operatorname{cl} \left({C}\right)$

But since $C \subseteq C$, $\operatorname{cl} \left({C}\right) \subseteq C$.

Thus by definition of set equality:
 * $\operatorname{cl} \left({C}\right) = C$

so $C$ is closed with respect to $\operatorname{cl}$.

Suppose instead that $C$ is closed with respect to $\operatorname{cl}$.

Then $\operatorname{cl} \left({C}\right) = C$.

Since $\mathcal C$ is closed under intersections, $C \in \mathcal C$.