Intersection of Closed Sets is Closed/Closure Operator
Let $S$ be a set.
Let $f: \mathcal P(S) \to \mathcal P(S)$ be a closure operator on $S$.
Let $\mathcal A \subseteq \mathcal C$.
Then $\bigcap \mathcal A \in \mathcal C$.
Let $Q = \bigcap \mathcal A$.
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 $f$.