Complement of Upper Closure of Element is Open in Lower Topology

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

Let $x \in S$.

Then $\complement_S\left({x^\succeq}\right)$ is open and $x^\succeq$ is closed.

Proof
Define $B := \left\{ {\complement_S\left({y^\succeq}\right): y \in S}\right\}$

By definition of lower topology:
 * $B$ is sub-basis of $T$.

By definition of sub-basis:
 * $B \subseteq \tau$

By definition of $B$:
 * $\complement_S\left({x^\succeq}\right) \in B$

Thus by definition of subset:
 * $\complement_S\left({x^\succeq}\right) \in \tau$

Thus by definition:
 * $x^\succeq$ is closed.