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 $T$

Proof

Elements

By definition of lower closure, $b \in a^\preccurlyeq$ if and only if:

$b \preccurlyeq a$

The dual of this statement is:

$a \preccurlyeq b$

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

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

$\Box$

Sets

By the definition of lower closure, $b \in T^\preccurlyeq$ if and only if:

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

The dual of this statement is:

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

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

$\blacksquare$