Composition of Compatible Closure Operators

Theorem
Let $S$ be a set.

Let $f, g: \mathcal P(S) \to \mathcal P(S)$ be closure operators on $S$.

Let $\mathcal C_f$ and $\mathcal C_g$ be the sets of closed sets of $S$ with respect to $f$ and $g$, respectively.

Suppose that for each subset $T$ of $S$:


 * If $T$ is closed with respect to $g$, then $f(T)$ is closed with respect to $g$. That is, if $T \in \mathcal C_g$ then $f(T) \in \mathcal C_g$.


 * If $T$ is closed with respect to $f$, then $g(T)$ is closed with respect to $f$. That is, if $T \in \mathcal C_f$ then $g(T) \in \mathcal C_f$.

Let $\mathcal C_h = \mathcal C_f \cap \mathcal C_g$.

Then:
 * $\mathcal C_h$ induces a closure operator $h$ on $S$
 * $f \circ g = g \circ f = h$, where $\circ$ represents composition.

Proof
First, we show that $\mathcal C_h$ induces a closure operator on $S$.

Let $\mathcal A \subseteq \mathcal C_h$.

By Intersection Largest, $\mathcal A \subseteq \mathcal C_f$ and $\mathcal A \subseteq \mathcal C_g$.

Thus by Intersection of Closed Sets is Closed/Closure Operator, $\bigcap \mathcal A \in \mathcal C_f$ and $\bigcap \mathcal A \in \mathcal C_g$.

Thus by the definition of set intersection, $\bigcap \mathcal A \in \mathcal C_h$.

Thus by Closure Operator from Closed Sets, $C_h$ induces a closure operator $h$ on $S$.

Now we will show that $f \circ g = h$. That $g \circ f = h$ will hold by reversing the variable names.

Let $T \subseteq S$.

$f(g(T)) \in \mathcal C_f$ because Closure is Closed/Power Set.

$g(T) \in \mathcal C_g$ because Closure is Closed/Power Set.

By the premise, then, $f(g(T)) \in \mathcal C_g$.

Thus $f(g(T)) \in \mathcal C_f \cap \mathcal C_g = C_h$ by the definition of intersection.

So $f(g(T))$ is closed with respect to $h$.

By Set Closure is Smallest Closed Set/Closure Operator, $h(T) \subseteq f(g(T))$.

By Closure is Closed/Power Set, $h(T) \in C_h$.

Thus by the definition of intersection, $h(T) \in C_f$ and $h(T) \in C_g$.

By Set Closure is Smallest Closed Set/Closure Operator, $g(T) \subseteq h(T)$.

Because $f$ is order-preserving, $f(g(T)) \subseteq f(h(T))$.

Recall that $h(T) \in C_f$. By Closure is Closed/Power Set, $f(h(T)) = h(T)$.

Thus $f(g(T)) \subseteq h(T)$.

We have that:
 * $h(T) \subseteq f(g(T))$

So by definition of set equality:
 * $h(T) = f(g(T))$