Closed Subset is Upper Section 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 closed.

Then $A$ is an upper section of $S$.

Proof
By definition of closed set:
 * $S \setminus A$ is open.

By Open Subset is Lower Section in Lower Topology:
 * $S \setminus A$ is a lower section.

Thus by Complement of Lower Section is Upper Section and Relative Complement of Relative Complement:
 * $A$ is an upper section of $S$.