# Composition of Compatible Closure Operators

## Theorem

Let $S$ be a set.

Let $f, g: \mathcal P \left({S}\right) \to \mathcal P \left({S}\right)$ 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.

For each subset $T$ of $S$, let the following hold:

- $(1): \quad$ If $T$ is closed with respect to $g$, then $f \left({T}\right)$ is closed with respect to $g$.
- That is, if $T \in \mathcal C_g$ then $f \left({T}\right) \in \mathcal C_g$.

- $(2): \quad$ If $T$ is closed with respect to $f$, then $g \left({T}\right)$ is closed with respect to $f$.
- That is, if $T \in \mathcal C_f$ then $g \left({T}\right) \in \mathcal C_f$.

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 of mappings.

## Proof

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

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

By Intersection is Largest Subset:

- $\mathcal A \subseteq \mathcal C_f$

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 $f \circ g = h$.

Having established that, it can be seen that $g \circ f = h$ will hold by reversing the variable names.

Let $T \subseteq S$.

By definition of closed set:

- $f \left({g \left({T}\right)}\right) \in \mathcal C_f$
- $g \left({T}\right) \in \mathcal C_g$

By the premise:

- $f \left({g \left({T}\right)}\right) \in \mathcal C_g$

Thus by definition of set intersection:

- $f \left({g \left({T}\right)}\right) \in \mathcal C_f \cap \mathcal C_g = C_h$

So $f \left({g \left({T}\right)}\right)$ is closed with respect to $h$.

By Set Closure is Smallest Closed Set/Closure Operator:

- $h \left({T}\right) \subseteq f \left({g \left({T}\right)}\right)$

By definition of closed set:

- $h \left({T}\right) \in C_h$

Thus by the definition of set intersection:

- $h \left({T}\right) \in C_f$

and

- $h \left({T}\right) \in C_g$

By Set Closure is Smallest Closed Set/Closure Operator:

- $g \left({T}\right) \subseteq h \left({T}\right)$

By Closure Operator: Axiom $(2)$ $f$ is order-preserving:

- $f \left({g \left({T}\right)}\right) \subseteq f \left({h \left({T}\right)}\right)$

Recall that $h \left({T}\right) \in C_f$.

By definition of closed set:

- $f \left({h \left({T}\right)}\right) = h \left({T}\right)$

Thus:

- $f \left({g \left({T}\right)}\right) \subseteq h \left({T}\right)$

We have that:

- $h \left({T}\right) \subseteq f \left({g \left({T}\right)}\right)$

So by definition of set equality:

- $h \left({T}\right) = f \left({g \left({T}\right)}\right)$

$\blacksquare$