Complement of Lower Section is Upper Section

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

Let $L$ be a lower section.

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


Let $u \in S \setminus L$.

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

Aiming for a contradiction, suppose $s \notin S \setminus L$.

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.


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