Ordering on Closure Operators iff Images are Including

Theorem
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a complete lattice.

Let $f, g:S \to S$ be closure operators on $L$.

Then $f \preceq g$ $g \sqbrk S \subseteq f \sqbrk S$

where
 * $\preceq$ denotes the ordering on mappings,
 * $f \sqbrk S$ denotes the image of $f$.

Sufficient Condition
Let $f \preceq g$.

Let $x \in g \sqbrk S$

By definition of image of mapping:
 * $\exists y \in S: \map g y = x$

By definition of closure operator/idempotent:
 * $\map g {\map g y} = \map g y$

By definition of ordering on mappings:
 * $\map f {\map g y} \preceq \map g {\map g y}$

By definition of closure operator/inflationary:
 * $\map g y \preceq \map f {\map g y}$

By definition of antisymmetry:
 * $\map f x = x$

Thus by definition of image of mapping:
 * $x \in f \sqbrk S$

Necessary Condition
Let $g \sqbrk S \subseteq f \sqbrk S$.

Let $x \in S$.

By definition of image of mapping:
 * $\map g x \in g \sqbrk S$

By definition of subset:
 * $\map g x \in f \sqbrk S$

By definition of image of mapping:
 * $\exists a \in S: \map f a = \map g x$

By definition of closure operator/idempotent:
 * $\map f {\map g x} = \map g x$

By definition of closure operator/inflationary:
 * $x \preceq \map g x$

Thus by definition of closure operator/increasing:
 * $\map f x \preceq \map g x$