Lower Section with no Maximal Element

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $L \subseteq S$.


Then:

$L$ is a lower section in $S$ with no maximal element

if and only if:

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

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


Proof

By Dual Pairs (Order Theory):

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

$\blacksquare$


Also see