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

## Theorem

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

Let $X$ be a closed under directed suprema subset of $S$.

Then $\complement_S\left({X}\right)$ is inaccessible by directed suprema.

## Proof

Let $D$ be a directed subset of $S$ such that

$\sup D \in \complement_S\left({X}\right)$

By definition of relative complement:

$\sup D \notin X$

By definition of closed under directed suprema:

$D \nsubseteq X$
$D \nsubseteq \complement_S\left({\complement_S\left({X}\right)}\right)$
$D \cap \complement_S\left({X}\right) \ne \varnothing$

$\blacksquare$