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

Theorem
Let $\left({S, \vee_1, \wedge_1, \preceq_1}\right)$ be a lattice.

Let $\left({T, \vee_2, \wedge_2, \preceq_2}\right)$ be a complete lattice.

Let $f: S \to T$ be a mapping such that
 * $\forall x \in S: f\left({x}\right) = \sup \left\{ {f\left({w}\right): w \in S \land w \preceq_1 x \land w}\right.$ is compact$\left.{}\right\}$

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:
 * $f\left[{x^{\mathrm{compact} } }\right] \subseteq f\left[{y^{\mathrm{compact} } }\right]$

By assumption:
 * $f\left({x}\right) = \sup \left\{ {f\left({w}\right): w \in S \land w \preceq_1 x \land w}\right.$ is compact$\left.{}\right\}$

and
 * $f\left({y}\right) = \sup \left\{ {f\left({w}\right): w \in S \land w \preceq_1 y \land w}\right.$ is compact$\left.{}\right\}$

By definitions of image of set and compact closure:
 * $f\left({x}\right) = \sup \left({f\left[{x^{\mathrm{compact} } }\right]}\right)$

and
 * $f\left({y}\right) = \sup \left({f\left[{y^{\mathrm{compact} } }\right]}\right)$

Thus by Supremum of Subset and definition of complete lattice:
 * $f\left({x}\right) \preceq_2 f\left({y}\right)$