Open Subset is Lower in Lower Topology

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

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

Then $A$ is lower.

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

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

By definitions of sub-basis and basis:
 * $\ds \BB = \set {\bigcap \FF: \FF \subseteq B, \FF \text{ is finite} }$ is a basis.

By definition of basis:
 * $\ds \tau \subseteq \set {\bigcup X: X \subseteq \BB}$

Let $x \in A$, $y \in S$ such that:
 * $y \preceq x$

By definition of open set:
 * $A \in \tau$

By definition of subset:
 * $\exists Y \subseteq \BB: A = \bigcup Y$

By definition of union:
 * $\exists Z \in Y: x \in Z$

By definition of subset:
 * $\exists X \subseteq B: Z = \bigcap X \land X$ is finite.

We will prove that:
 * $\forall Q \in X: y \in Q$

Let $Q \in X$.

By definition of subset:
 * $Q \in B$

By definition of $B$:
 * $\exists z \in S: Q = \relcomp S {z^\succeq}$

By definition of intersection:
 * $x \in Q$

By Upper Closure is Upper Set:
 * $z^\succeq$ is an upper set.

By Complement of Upper Set is Lower Set:
 * $Q$ is a lower set.

Thus by definition of lower set:
 * $y \in Q$

By definition of intersection:
 * $y \in Z$

Thus by definition of union:
 * $y \in A$