Closed Set iff Lower and Closed under Directed Suprema in Scott Topological Ordered Set

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

where $\left({S, \preceq}\right)$ is an up-complete ordered set.

Let $A$ be a subset of $S$.

Then $A$ is closed $A$ is lower and closed under directed suprema.

Sufficient Condition
Assume that
 * $A$ is closed.

By definition of closed set:
 * $\complement_S\left({A}\right) \in \tau$

By definition of Scott topology:
 * $\complement_S\left({A}\right)$ is upper and inaccessible by directed suprema.

By Complement of Complement:
 * $\complement_S\left({\complement_S\left({A}\right)}\right) = A$

Thus by Complement of Upper Set is Lower Set:
 * $A$ is a lower set.

Thus by Complement of Inaccessible by Directed Suprema Subset is Closed under Directed Suprema:
 * $A$ is closed under directed suprema.

Necessary Condition
Assume that $A$ is lower and closed under directed suprema.

By Complement of Lower Set is Upper Set and Complement of Closed under Directed Suprema Subset is Inaccessible by Directed Suprema:
 * $\complement_S\left({A}\right)$ is upper and inaccessible by directed suprema.

By definition of Scott topology:
 * $\complement_S\left({A}\right) \in \tau$

Thus by definition:
 * $A$ is closed set.