Continuous iff Way Below iff There Exists Element that Way Below and Way Below

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

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

Then $f$ is continuous
 * $\forall x \in S, y \in T: y \ll f\left({x}\right) \iff \exists w \in S: w \ll x \land y \ll f\left({w}\right)$

Sufficient Condition
Assume that
 * $f$ is continuous.

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

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

Thus by Mapping at Element is Supremum implies Way Below iff There Exists Element that Way Below and Way Below:
 * $\forall x \in S, y \in T: y \ll f\left({x}\right) \iff \exists w \in S: w \ll x \land y \ll f\left({w}\right)$

Necessary Condition
Assume that
 * $\forall x \in S, y \in T: y \ll f\left({x}\right) \iff \exists w \in S: w \ll x \land y \ll f\left({w}\right)$

Let $U$ be an open subset of $T$.

By definition of Scott topology:
 * $U$ is upper.

By definition of interior:
 * $\left({f^{-1}\left[{U}\right]}\right)^\circ \subseteq f^{-1}\left[{U}\right]$

We will prove that
 * $f^{-1}\left[{U}\right] \subseteq \left({f^{-1}\left[{U}\right]}\right)^\circ$

Let $x \in f^{-1}\left[{U}\right]$

By definition of preimage of set:
 * $p := f\left({x}\right) \in U$

By Open implies There Exists Way Below Element:
 * $\exists u \in T: u \ll p \land u \in U$

By assumption:
 * $\exists w \in S: w \ll x \land u \ll f\left({w}\right)$

We will prove that
 * $f\left[{w^\gg}\right] \subseteq U$

Let $h \in f\left[{w^\gg}\right]$

By definition of image of set:
 * $\exists z \in w^\gg: h = f\left({z}\right)$

By definition of way above closure:
 * $w \ll z$

By assumption:
 * $u \ll h$

By Way Below implies Preceding:
 * $u \preceq_2 h$

Thus by definition of upper set:
 * $h \in U$

By Image of Subset under Relation is Subset of Image/Corollary 3:
 * $f^{-1}\left[{f\left[{w^\gg}\right]}\right] \subseteq f^{-1}\left[{U}\right]$

By Preimage of Image under Relation is Superset:
 * $w^\gg \subseteq \left({f^{-1} \circ f}\right)\left[{w^\gg}\right]$

By definition of composition of relations:
 * $w^\gg \subseteq f^{-1}\left[{f\left[{w^\gg}\right]}\right]$

By Subset Relation is Transitive:
 * $w^\gg \subseteq f^{-1}\left[{U}\right]$

By Interior is Union of Way Above Closures:
 * $\left({f^{-1}\left[{U}\right]}\right)^\circ = \bigcup \left\{ {g^\gg: g \in S \land g^\gg \subseteq f^{-1}\left[{U}\right]}\right\}$

By definition of way above closure:
 * $x \in w^\gg$

We have:
 * $w^\gg \in \left\{ {g^\gg: g \in S \land g^\gg \subseteq f^{-1}\left[{U}\right]}\right\}$

Thus by definition of union:
 * $x \in \left({f^{-1}\left[{U}\right]}\right)^\circ$

By definition of set equality:
 * $\left({f^{-1}\left[{U}\right]}\right)^\circ = f^{-1}\left[{U}\right]$

Thus by definition of interior:
 * $f^{-1}\left[{U}\right]$ is open.

Hence $f$ is continuous.