Lower Closure is Dual to Upper Closure

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

Let $a, b \in S$.

Let $T \subseteq S$

The following are pairs of dual statements:


 * $b \in {\bar\downarrow} a$, the lower closure of $a$
 * $b \in {\bar\uparrow} a$, the upper closure of $a$


 * $b \in {\downarrow} T$, the lower closure of $T$
 * $b \in {\uparrow} T$, the upper closure of $T$

Elements
By definition of lower closure, $b \in {\bar\downarrow} a$ iff:


 * $b \preceq a$

The dual of this statement is:


 * $a \preceq b$

by Dual Pairs (Order Theory).

By definition of upper closure, this means $b \in {\bar\uparrow} a$.

The converse follows from Dual of Dual Statement (Order Theory).

Sets
By the definition of lower closure, $b \in {\downarrow} T$ iff:


 * $\exists a \in T: b \preceq a$

The dual of this statement is:


 * $\exists a \in T: a \preceq b$

by Dual Pairs (Order Theory).

By the definition of upper closure, this means $b \in {\uparrow} T$.

Also see

 * Duality Principle (Order Theory)