Lower Closure is Dual to Upper Closure
Jump to navigation
Jump to search
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 $b \in T^\succcurlyeq$.
$\blacksquare$