Ordering on Closure Operators iff Images are Including

Theorem
Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a complete lattice.

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

Then $f \preceq g$ $g\left[{S}\right] \subseteq f\left[{S}\right]$

where
 * $\preceq$ denotes the ordering on mappings,
 * $f\left[{S}\right]$ denotes the image of $f$.

Sufficient Condition
Let $f \preceq g$.

Let $x \in g\left[{S}\right]$

By definition of image of mapping:
 * $\exists y \in S: g\left({y}\right) = x$

By definition of closure operator:
 * $g\left({g\left({y}\right)}\right) = g\left({y}\right)$

By definition of ordering on mappings:
 * $f\left({g\left({y}\right)}\right) \preceq g\left({g\left({y}\right)}\right)$

By definition of closure operator:
 * $g\left({y}\right) \preceq f\left({g\left({y}\right)}\right)$

By definition of antisymmetry:
 * $f\left({x}\right) = x$

Thus by definition of image of mapping:
 * $x \in f\left[{S}\right]$

Necessary Condition
Let $g\left[{S}\right] \subseteq f\left[{S}\right]$.

Let $x \in S$.

By definition of image of mapping:
 * $g\left({x}\right) \in g\left[{S}\right]$

By definition of subset:
 * $g\left({x}\right) \in f\left[{S}\right]$

By definition of image of mapping:
 * $\exists a \in S: f\left({a}\right) = g\left({x}\right)$

By definition of closure operator:
 * $f\left({g\left({x}\right)}\right) = g\left({x}\right)$

By definition of closure operator:
 * $x \preceq g\left({x}\right)$

Thus by definition of ordering on mappings:
 * $f\left({x}\right) \preceq g\left({x}\right)$