Intersection of Closed Sets is Closed/Closure Operator
Let $S$ be a set.
Let $f: \powerset S \to \powerset S$ be a closure operator on $S$.
Let $\AA \subseteq \CC$.
Then $\bigcap \AA \in \CC$.
Let $Q = \bigcap \AA$.
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 $f$.