Mapping at Element is Supremum of Compact Elements implies Mapping at Element is Supremum that Way Below

Theorem
Let $\left({S, \vee_1, \wedge_1, \preceq_1}\right)$ and $\left({T, \vee_2, \wedge_2, \preceq_2}\right)$ be complete lattices.

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
 * $\forall x \in S: f\left({x}\right) = \sup \left\{ {f\left({w}\right): w \in S \land w \ll x}\right\}$

Proof
Let $x \in S$.

Define $X = \left\{ {f\left({w}\right): w \in S \land w \preceq_1 x \land w}\right.$ is compact$\left.{}\right\}$

Define $Y = \left\{ {f\left({w}\right): w \in S \land w \ll x}\right\}$

We will prove that
 * $X \subseteq Y$

Let $b \in X$.

By definition of $X$:
 * $\exists w \in S: b = f\left({w}\right) \land w \preceq_1 x \land w$ is compact.

By definition of compact element:
 * $w \ll w$

By Preceding and Way Below implies Way Below:
 * $w \ll x$

Thus by definition of $Y$:
 * $b \in Y$

We will prove that
 * $f\left({x}\right)$ is upper bound for $Y$.

Let $b \in Y$.

By definition of $Y$:
 * $\exists w \in S: b = f\left({w}\right) \land w \ll x$

By Way Below implies Preceding:
 * $w \preceq_1 x$

By Mapping at Element is Supremum implies Mapping is Increasing:
 * $f$ is an increasing mapping.

Thus by definition of increasing mapping:
 * $b \preceq_2 f\left({x}\right)$

We will prove that
 * $\forall b \in T: b$ is upper bound for $Y \implies f\left({x}\right) \preceq_2 b$

Let $b \in T$ such that
 * $b$ is upper bound for $Y$.

By Upper Bound for Subset:
 * $b$ is upper bound for $X$.

By assumption:
 * $f\left({x}\right) = \sup X$

Thus by definition of supremum:
 * $f\left({x}\right) \preceq_2 b$

Thus by definition of supremum:
 * $f\left({x}\right) = \sup Y$