Open Subset is Lower Section in Lower Topology

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

Let $A \subseteq S$ such that
 * $A$ is open.

Then $A$ is lower.

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

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