Complement of Lower Closure of Element is Open 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 $\complement_S\left({x^\preceq}\right)$ is topologically open,

where
 * $x^\preceq$ denotes the lower closure of $x$,
 * $\complement_S\left({x^\preceq}\right)$ denotes the relative complement of $x^\preceq$.

Proof
By Lower Closure of Element is Topologically Closed in Scott Topological Ordered Set:
 * $x^\preceq$ is closed.

By definition of closed set:
 * $\complement_S\left({x^\preceq}\right) \in \tau$

Thus by definition:
 * $\complement_S\left({x^\preceq}\right)$ is a open set.