Intersection of Closed Sets is Closed/Closure Operator

Theorem

Let $S$ be a set.

Let $f: \powerset S \to \powerset S$ be a closure operator on $S$.

Let $\CC$ be the set of all subsets of $S$ that are closed with respect to $f$.

Let $\AA \subseteq \CC$.

Then $\bigcap \AA \in \CC$.

Proof

Let $Q = \bigcap \AA$.

By the definition of closure operator, $f$ is inflationary, order-preserving, and idempotent.

Let $A \in \AA$.

By Intersection is Largest Subset, $Q \subseteq A$.

Since $f$ is order-preserving, $\map f Q \subseteq \map f A$.

By the definition of closed set, $\map f A = A$

Thus $\map f Q \subseteq A$.

This holds for all $A \in \AA$.

Thus by Intersection is Largest Subset:

$\map f Q \subseteq \bigcap \AA$

Since $\bigcap \AA = Q$:

$\map f Q \subseteq Q$

Since $f$ is inflationary:

$Q \subseteq \map f Q$

Thus by definition of set equality:

$Q = \map f Q$

Therefore $Q$ is closed with respect to $f$.

$\blacksquare$