Lower Section with no Maximal Element

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

Let $L \subseteq S$.

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


 * $\ds L = \bigcup \set {l^\prec: l \in L}$
 * $\ds L = \bigcup \set {l^\prec: l \in L}$

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