Continuous iff Mapping at Element is Supremum of Compact Elements

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

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

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

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:
 * $\forall x \in S: \map f x = \sup \set {\map f w: w \in S \land w \ll x}$

Let $x \in S$.

By definitions of image of set and compact closure:
 * $\leftset {\map f w: w \in S \land w \preceq_1 x \land w}$ is compact$\rightset {} = f \sqbrk {x^{\mathrm{compact} } }$

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

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

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