Complement of Inaccessible by Directed Suprema Subset is Closed under Directed Suprema

Theorem
Let $L = \left({S, \preceq}\right)$ be an up-complete ordered set.

Let $X$ be an inaccessible by directed suprema subset of $S$.

Then $\complement_S\left({X}\right)$ is closed under directed suprema.

Proof
Let $D$ be a directed subset of $S$ such that
 * $D \subseteq \complement_S\left({X}\right)$

By Empty Intersection iff Subset of Relative Complement:
 * $D \cap X = \varnothing$

By definition of inaccessible by directed suprema:
 * $\sup D \notin X$

Thus by definition of relative complement:
 * $\sup D \in \complement_S\left({X}\right)$

Also See

 * Complement of Closed under Directed Suprema Subset is Inaccessible by Directed Suprema