Closed Subset is Upper 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 closed.

Then $A$ is upper.

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

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

Thus by Complement of Lower Set is Upper Set and Relative Complement of Relative Complement:
 * $A$ is upper.