Continuous iff Mapping at Limit Inferior Precedes Limit Inferior of Composition of Mapping and Sequence

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

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

Then $f$ is continuous
 * for all directed set $\left({D, \precsim}\right)$ and Moore-Smith sequence $N:D \to S$ in $S$: $f\left({\liminf N}\right) \preceq_2 \liminf\left({f \circ N}\right)$

Sufficient Condition
Assume that
 * $f$ is continuous.

Let $\left({D, \precsim}\right)$ be a directed set.

Let $N:D \to S$ be a Moore-Smith sequence in $S$.

Aiming for a contradiction suppose that
 * $f\left({\liminf N}\right) \npreceq_2 \liminf\left({f \circ N}\right)$

By definition of lower closure of element:
 * $f\left({\liminf N}\right) \notin \left({ \liminf\left({f \circ N}\right) }\right)^{\preceq_2}$

By definition of relative complement:
 * $f\left({\liminf N}\right) \in \complement_T\left({\left({ \liminf\left({f \circ N}\right) }\right)^{\preceq_2} }\right)$

By Lower Closure of Element is Topologically Closed in Scott Topological Ordered Set:
 * $\left({ \liminf\left({f \circ N}\right) }\right)^{\preceq_2}$ is closed.

By definition of closed set:
 * $\complement_T\left({\left({ \liminf\left({f \circ N}\right) }\right)^{\preceq_2} }\right)$ is open.

By definition of continuous:
 * $f^{-1}\left[{\complement_T\left({\left({ \liminf\left({f \circ N}\right) }\right)^{\preceq_2} }\right)}\right]$ is open.

By Open iff Upper and with Property (S) in Scott Topological Lattice:
 * $f^{-1}\left[{\complement_T\left({\left({ \liminf\left({f \circ N}\right) }\right)^{\preceq_2} }\right)}\right]$ has property (S).

By Set of Infima is Directed:
 * $X := \left\{ {\inf \left({N\left[{\precsim\left({j}\right)}\right]}\right): j \in D}\right\}$ is directed.

By definition of limit inferior:
 * $\liminf N = \sup X$

By definition of preimage of set:
 * $\liminf N \in f^{-1}\left[{\complement_T\left({\left({ \liminf\left({f \circ N}\right) }\right)^{\preceq_2} }\right)}\right]$

By definition of property (S):
 * $\exists y \in X:\forall x \in X: y \preceq_1 x \implies x \in f^{-1}\left[{\complement_T\left({\left({ \liminf\left({f \circ N}\right) }\right)^{\preceq_2} }\right)}\right]$

We have:
 * $\exists j \in D: y = \inf \left({N\left[{\precsim\left({j}\right)}\right]}\right)$

By definition of reflexivity:
 * $y \preceq_1 y$

Then:
 * $y \in f^{-1}\left[{\complement_T\left({\left({ \liminf\left({f \circ N}\right) }\right)^{\preceq_2} }\right)}\right]$

Define $y' := \inf \left({\left({f \circ N}\right)\left[{\precsim\left({j}\right)}\right]}\right)$

Define $X' := \left\{ {\inf \left({\left({f \circ N}\right)\left[{\precsim\left({j}\right)}\right]}\right): j \in D}\right\}$

Necessary Condition
Assume that
 * for all directed set $\left({D, \precsim}\right)$ and Moore-Smith sequence $N:D \to S$ in $S$: $f\left({\liminf N}\right) \preceq_2 \liminf\left({f \circ N}\right)$

By Mapping at Limit Inferior Precedes Limit Inferior of Composition Mapping and Sequence implies Mapping Preserves Directed Suprema:
 * $f$ preserves directed suprema.

Thus by Mapping Preserves Directed Suprema implies Mapping is Continuous:
 * $f$ is continuous.