Lower Closure of Element is Topologically Closed in Scott Topological Ordered Set

Theorem
Let $\struct {S, \preceq}$ be an up-complete ordered set.

Let $T = \struct {S, \preceq, \tau}$ be a relational structure with the Scott topology.

Let $x \in S$.

Let $x^\preceq$ denote the lower closure of $x$.

Then $x^\preceq$ is topologically closed.

Proof
By Lower Closure of Element is Closed under Directed Suprema:
 * $x^\preceq$ is closed under directed suprema.

By Lower Closure of Singleton:
 * $\set x^\preceq = x^\preceq$

By Lower Closure is Lower Set:
 * $x^\preceq$ is a lower set.

Thus by Closed Set iff Lower and Closed under Directed Suprema in Scott Topological Ordered Set:
 * $x^\preceq$ is closed.