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

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

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

Then $f$ is continuous
 * $\forall x \in S, y \in T: y \ll \map f x \iff \exists w \in S: w \ll x \land y \ll \map f w$

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

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

Necessary Condition
Assume that
 * $\forall x \in S, y \in T: y \ll \map f x \iff \exists w \in S: w \ll x \land y \ll \map f w$

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

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

By definition of interior:
 * $\paren {f^{-1} \sqbrk U}^\circ \subseteq f^{-1} \sqbrk U$

We will prove that
 * $f^{-1} \sqbrk U \subseteq \paren {f^{-1} \sqbrk U}^\circ$

Let $x \in f^{-1} \sqbrk U$

By definition of preimage of set:
 * $p := \map f x \in U$

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

By assumption:
 * $\exists w \in S: w \ll x \land u \ll \map f w$

We will prove that
 * $f \sqbrk {w^\gg} \subseteq U$

Let $h \in f \sqbrk {w^\gg}$

By definition of image of set:
 * $\exists z \in w^\gg: h = \map f z$

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 section:
 * $h \in U$

By Preimage of Subset is Subset of Preimage:
 * $f^{-1} \sqbrk {f \sqbrk {w^\gg} } \subseteq f^{-1} \sqbrk U$

By Preimage of Image under Left-Total Relation is Superset:
 * $w^\gg \subseteq \paren {f^{-1} \circ f} \sqbrk {w^\gg}$

By definition of composition of relations:
 * $w^\gg \subseteq f^{-1} \sqbrk {f \sqbrk {w^\gg} }$

By Subset Relation is Transitive:
 * $w^\gg \subseteq f^{-1} \sqbrk U$

By Interior is Union of Way Above Closures:
 * $\paren {f^{-1} \sqbrk U}^\circ = \bigcup \set {g^\gg: g \in S \land g^\gg \subseteq f^{-1} \sqbrk U}$

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

We have:
 * $w^\gg \in \set {g^\gg: g \in S \land g^\gg \subseteq f^{-1} \sqbrk U}$

Thus by definition of union:
 * $x \in \paren {f^{-1} \sqbrk U}^\circ$

By definition of set equality:
 * $\paren {f^{-1} \sqbrk U}^\circ = f^{-1} \sqbrk U$

Thus by definition of interior:
 * $f^{-1} \sqbrk U$ is open.

Hence $f$ is continuous.