Lower Section with no Maximal Element

Theorem
Let $(S, \preceq)$ be an ordered set.

Let $L \subseteq S$.

Then $L$ is a lower set in $S$ with no maximal element iff:


 * $L = \bigcup \left\{{ {\dot\downarrow}l: l \in L }\right\}$

where ${\dot\downarrow}l$ is the strict down-set of $l$.

Proof
By Dual Pairs (Order Theory):
 * Lower set is dual to upper set.
 * Maximal element is dual to minimal element.
 * Strict down-set is dual to strict up-set.

Thus the theorem holds by the duality principle.