# Intersection of Closed Sets is Closed/Closure Operator

## Theorem

Let $S$ be a set.

Let $f: \mathcal P(S) \to \mathcal P(S)$ be a closure operator on $S$.

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

Let $\mathcal A \subseteq \mathcal C$.

Then $\bigcap \mathcal A \in \mathcal C$.

## Proof

Let $Q = \bigcap \mathcal A$.

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

Let $A \in \mathcal A$.

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

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

By the definition of closed set, $f(A) = A$

Thus $f(Q) \subseteq A$.

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

Thus by Intersection is Largest Subset:

- $f(Q) \subseteq \bigcap \mathcal A$

Since $\bigcap \mathcal A = Q$:

- $f(Q) \subseteq Q$

Since $f$ is inflationary:

- $Q \subseteq f(Q)$

Thus by definition of set equality:

- $Q = f(Q)$

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

$\blacksquare$