Complement of Lower Section is Upper Section

Theorem
Let $\left({S, \preceq}\right)$ be an ordered set.

Let $L$ be a lower set.

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

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

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

Suppose for the sake of contradiction that $s \notin S \setminus L$.

Then $s \in L$.

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

Hence $s \in S \setminus L$.

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

Also see

 * Complement of Upper Set is Lower Set