Closure Operator from Closed Sets

Theorem
Let $S$ be a set.

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

Suppose that $\forall \mathcal K \in \mathcal P (\mathcal C): \bigcap \mathcal K \in \mathcal C$, where $\bigcap \varnothing$ is taken to be $S$.

That is, suppose that $\mathcal C$ is closed under arbitrary intersections.

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

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, $T \subseteq \operatorname{cl} (T)$.

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

Idempotent
Let $T \subseteq S$.

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

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

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

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

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

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:


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

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

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

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

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

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

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