Complement of Lower Section is Upper Section

Theorem
Let $\struct {S, \preceq}$ be an ordered set.

Let $L$ be a lower section.

Then $S \setminus L$ is an upper section.

Proof
Let $u \in S \setminus L$.

Let $s \in S$ such that $u \preceq s$.

$s \notin S \setminus L$.

Then $s \in L$.

By definition of lower section, $u \in L$, a contradiction.

Hence $s \in S \setminus L$.

Since this holds for all such $u$ and $s$, $S \setminus L$ is an upper section.

Also see

 * Complement of Upper Section is Lower Section