Composition of Compatible Closure Operators
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Theorem
Let $S$ be a set.
Let $f, g: \powerset S \to \powerset S$ be closure operators on $S$.
Let $\CC_f$ and $\CC_g$ 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 $\map f T$ is closed with respect to $g$.
- That is, if $T \in \CC_g$ then $\map f T \in \CC_g$.
- $(2): \quad$ If $T$ is closed with respect to $f$, then $\map g T$ is closed with respect to $f$.
- That is, if $T \in \CC_f$ then $\map g T \in \CC_f$.
Let $\CC_h = \CC_f \cap \CC_g$.
Then:
- $\CC_h$ 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 $\CC_h$ induces a closure operator on $S$.
Let $\AA \subseteq \CC_h$.
By Intersection is Largest Subset:
- $\AA \subseteq \CC_f$
and:
- $\AA \subseteq \CC_g$
Thus by Intersection of Closed Sets is Closed/Closure Operator:
- $\ds \bigcap \AA \in \CC_f$
and
- $\ds \bigcap \AA \in \CC_g$
Thus by the definition of set intersection:
- $\ds \bigcap \AA \in \CC_h$
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:
- $\map f {\map g T} \in \CC_f$
- $\map g T \in \CC_g$
By the premise:
- $\map f {\map g T} \in \CC_g$
Thus by definition of set intersection:
- $\map f {\map g T} \in \CC_f \cap \CC_g = C_h$
So $\map f {\map g T}$ is closed with respect to $h$.
By Set Closure is Smallest Closed Set/Closure Operator:
- $\map h T \subseteq f {\map g T}$
By definition of closed set:
- $\map h T \in C_h$
Thus by definition of set intersection:
- $\map h T \in C_f$
and
- $\map h T \in C_g$
By Set Closure is Smallest Closed Set/Closure Operator:
- $\map g T \subseteq \map h T$
By Closure Operator: Axiom $(2)$ $f$ is order-preserving:
- $\map f {\map g T} \subseteq \map f {\map h T}$
Recall that $\map h T \in C_f$.
By definition of closed set:
- $\map f {\map h T} = \map h T$
Thus:
- $\map f {\map g T} \subseteq \map h T$
We have that:
- $\map h T \subseteq \map f {\map g T}$
So by definition of set equality:
- $\map h T = \map f {\map g T}$
$\blacksquare$