Lower Closure is Dual to Upper Closure

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

Let $a, b \in S$.

Let $T \subseteq S$

The following are pairs of dual statements:


 * $b \in a^\preccurlyeq$, the lower closure of $a$
 * $b \in a^\succcurlyeq$, the upper closure of $a$


 * $b \in T^\preccurlyeq$, the lower closure of $T$
 * $b \in T^\succcurlyeq$, the upper closure of $T$

Elements
By definition of lower closure, $b \in a^\preccurlyeq$ :


 * $b \preccurlyeq a$

The dual of this statement is:


 * $a \preccurlyeq b$

by Dual Pairs (Order Theory).

By definition of upper closure, this means $b \in a^\succcurlyeq$.

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

Sets
By the definition of lower closure, $b \in T^\preccurlyeq$ :


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

The dual of this statement is:


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

by Dual Pairs (Order Theory).

By the definition of upper closure, this means $b \in T^\succcurlyeq$.

Also see

 * Duality Principle (Order Theory)