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

Theorem
Let $T = \left({S, \preceq, \tau}\right)$ be a relational structure with Scott topology

where $\left({S, \preceq}\right)$ is an up-complete ordered set.

Let $x \in S$.

Then $x^\preceq$ is topologically closed,

where $x^\preceq$ denotes the lower closure of $x$.

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

By Lower Closure of Singleton:
 * $\left\{ {x}\right\}^\preceq = x^\preceq$

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.