Mapping at Element is Supremum of Compact Elements implies Mapping is Increasing

Theorem
Let $\struct{S, \vee_1, \wedge_1, \preceq_1}$ be a lattice.

Let $\struct{T, \vee_2, \wedge_2, \preceq_2}$ be a complete lattice.

Let $f: S \to T$ be a mapping such that
 * $\forall x \in S: \map f x = \sup \leftset{ \map f w : w \in S \land w \preceq_1 x \land w}$ is compact$\rightset{}$

Then $f$ is increasing.

Proof
Let $x, y \in S$ such that
 * $x \preceq_1 y$

By Compact Closure is Increasing:
 * $x^{\mathrm{compact} } \subseteq y^{\mathrm{compact} }$

By Image of Subset under Relation is Subset of Image/Corollary 2:
 * $f \sqbrk{x^{\mathrm{compact} } } \subseteq f\sqbrk{y^{\mathrm{compact} } }$

By assumption:
 * $\map f x = \sup \leftset{\map f w: w \in S \land w \preceq_1 x \land w}$ is compact$\rightset{}$

and
 * $\map f y = \sup \leftset{\map f w: w \in S \land w \preceq_1 y \land w}$ is compact$\rightset{}$

By definitions of image of set and compact closure:
 * $\map f x = \map \sup {f \sqbrk{x^{\mathrm{compact} } } }$

and
 * $\map f y = \map \sup {f \sqbrk{y^{\mathrm{compact} } } }$

Thus by Supremum of Subset and definition of complete lattice:
 * $\map f x \preceq_2 \map f y$