Operator Generated by Closure System Preserves Directed Suprema iff Closure System Inherits Directed Suprema

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

Let $S = \left({Y, \precsim}\right)$ be a closure system on $L$.

Then $\operatorname{operator}\left({S}\right)$ preserves directed suprema $S$ inherits directed suprema.

where $\operatorname{operator}\left({S}\right)$ denotes the operator generated by $S$.

Sufficient Condition
Assume that
 * $\operatorname{operator}\left({S}\right)$ preserves directed suprema.

Let $Z$ be directed subset of $Y$ such that
 * $Z$ admits a supremum in $L$.

By Image of Operator Generated by Closure System is Set of Closure System:
 * $\operatorname{operator}\left({S}\right)\left[{X}\right] = Y$

By Operator Generated by Closure System is Closure Operator:
 * $\operatorname{operator}\left({S}\right)$ is closure operator.

By definition of closure operator/idempotent:
 * $\operatorname{operator}\left({S}\right)\left[{Z}\right] = Z$

By definitions of Definition:Mapping Preserves Supremum/Directed and Definition:Mapping Preserves Supremum/Subset:
 * $\sup_L Z = \operatorname{operator}\left({S}\right)\left({\sup_L Z}\right)$

Thus by definition of image of mapping:
 * $\sup_L Z \in Y$

Necessary Condition
Assume that
 * $S$ inherits directed suprema.

Define $f := \operatorname{operator}\left({S}\right)$.

By Image of Operator Generated by Closure System is Set of Closure System:
 * $f\left[{X}\right] = Y$

Let $A$ be a directed subset of $X$ such that
 * $A$ admits a supremum in $L$.

Thus by definition of complete lattice:
 * $f\left[{A}\right]$ admits a supremum in $L$.

We will prove that
 * $f\left({\sup_L A}\right)$ is upper bound for $f\left[{A}\right]$

Let $x \in f\left[{A}\right]$.

By definition of image of subset:
 * $\exists y \in A: x = f\left({y}\right)$

By definitions of supremum and upper bound:
 * $y \preceq \sup_L A$

Thus by definition of closure operator/increasing:
 * $x \preceq f\left({\sup_L A}\right)$

By definition of supremum:
 * $\sup_L\left({f\left[{A}\right]}\right) \preceq f\left({\sup_L A}\right)$

We will prove that
 * $\sup_L\left({f\left[{A}\right]}\right)$ is upper bound for $A$

Let $x \in A$.

By definition of image of set:
 * $f \left({x}\right) \in f \left[{A}\right]$

By definitions of supremum and upper bound:
 * $f \left({x}\right) \preceq \sup_L \left({ f \left[{A}\right]}\right)$

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

Thus by definition of transitivity:
 * $x \preceq \sup_L \left({ f \left[{A}\right]}\right)$

By definition of supremum:
 * $\sup_L A \preceq \sup_L \left({ f \left[{A}\right]}\right)$

By Image of Directed Subset under Increasing Mapping is Directed:
 * $f \left[{A}\right]$ is directed in $L$.

By definition of ordered subset:
 * $f \left[{A}\right]$ is directed in $S$.

By definition of directed suprema inheriting:
 * $\sup_L \left({ f \left[{A}\right]}\right) \in Y$

By definition of closure operator/idempotent:
 * $f \left({\sup_L \left({ f \left[{A}\right]}\right)}\right) = \sup_L \left({ f \left[{A}\right]}\right)$

By definition of closure operator/increasing:
 * $f \left({\sup_L A}\right) \preceq \sup_L \left({ f \left[{A}\right]}\right)$

Thus by definition of antisymmetry:
 * $f \left({\sup_L A}\right) = \sup_L \left({ f \left[{A}\right]}\right)$