Continuous iff Mapping at Element is Supremum of Compact Elements

Theorem
Let $L = \left({S, \preceq_1, \tau_1}\right)$ and $R = \left({T, \preceq_2, \tau_2}\right)$ be complete algebraic topological lattices with Scott topologies.

Let $f: S \to T$ be a mapping.

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

Proof
By Algebraic iff Continuous and For Every Way Below Exists Compact Between:
 * $L$ and $R$ are continuous.

Sufficient Condition
Assume that@
 * $f$ is continuous.

By Continuous iff Mapping at Element is Supremum:
 * $\forall x \in S: f\left({x}\right) = \sup \left\{ {f\left({w}\right): w \in S \land w \ll x}\right\}$

Let $x \in S$.

By definitions of image of set and compact closure:
 * $\left\{ {f\left({w}\right): w \in S \land w \preceq_1 x \land w}\right.$ is compact$\left.{}\right\} = f\left[{x^{\mathrm{compact} } }\right]$

By Compact Closure is Directed:
 * $D := x^{\mathrm{compact} }$ is directed.

By Continuous iff Directed Suprema Preserving:
 * $f$ preserves directed suprema.

By definition of mapping preserves directed suprema:
 * $f$ preserves the supremum of $D$.

By definition of complete lattice:
 * $D$ admits a supremum.

By definition of algebraic:
 * $L$ satisfies axiom of K-approximation.

Thus

Necessary Condition
Assume 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$\left.{}\right\}$

By Mapping at Element is Supremum of Compact Elements implies Mapping at Element is Supremum that Way Below:
 * $\forall x \in S: f \left({x}\right) = \sup \left\{ {f \left({w}\right): w \in S \land w \ll x}\right\}$

Thus by Continuous iff Mapping at Element is Supremum:
 * $f$ is continuous.