Complement of Upper Closure of Element is Open in Lower Topology

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

Let $x \in S$.

Then $\relcomp S {x^\succeq}$ is open and $x^\succeq$ is closed.

Proof
Define $B := \set {\relcomp S {y^\succeq}: y \in S}$

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

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

By definition of $B$:
 * $\relcomp S {x^\succeq} \in B$

Thus by definition of subset:
 * $\relcomp S {x^\succeq} \in \tau$

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