Lower Section with no Maximal Element

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

Let $L \subseteq S$.

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


 * $\displaystyle L = \bigcup \left\{{l^\prec: l \in L }\right\}$
 * $\displaystyle L = \bigcup \left\{{l^\prec: l \in L }\right\}$

where $l^\prec$ is the strict lower closure of $l$.

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

Thus the theorem holds by the duality principle applied to Upper Set with no Minimal Element.

Also see

 * Upper Set with no Minimal Element