Mapping Preserves Directed Suprema implies Mapping is Continuous

Theorem
Let $\struct {S, \preceq_1, \tau_1}$ and $\struct {T, \preceq_2, \tau_2}$ be up-complete ordered sets with Scott topologies.

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

Then $f$ is continuous.

Proof
Let $P$ be a closed subset of $T$.

By Closed Set iff Lower and Closed under Directed Suprema in Scott Topological Ordered Set:
 * $P$ is lower and closed under directed suprema.

We will prove that
 * for all directed subset $D$ of $S$: $D \subseteq f^{-1} \sqbrk P \implies \sup D \in f^{-1} \sqbrk P$

Let $D$ be a directed subset of $S$ such that
 * $D \subseteq f^{-1} \sqbrk P$

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

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

By definition of mapping preserves the supremum:
 * $\map \sup {f \sqbrk D} = \map f {\sup D}$

By Directed Suprema Preserving Mapping is Increasing:
 * $f$ is increasing.

By Image of Directed Subset under Increasing Mapping is Directed:
 * $f \sqbrk D$ is directed.

By Preimage is Subset of Set implies Image of Set is Subset:
 * $f \sqbrk D \subseteq P$

By definition of closed under directed suprema:
 * $\map \sup {f \sqbrk D} \in P$

Thus by definition of preimage of set:
 * $\sup D \in f^{-1} \sqbrk P$

By definition:
 * $f^{-1} \sqbrk P$ is closed under directed suprema.

By Preimage of Lower Section under Increasing Mapping is Lower:
 * $f^{-1} \sqbrk P$ is lower.

Thus by Closed Set iff Lower and Closed under Directed Suprema in Scott Topological Ordered Set:
 * $f^{-1} \sqbrk P$ is closed.