Supremum is Dual to Infimum
Jump to navigation
Jump to search
Theorem
Let $\struct {S, \preceq}$ be an ordered set.
Let $a \in S$ and $T \subseteq S$.
The following are dual statements:
Proof
By definition, $a$ is a supremum for $T$ if and only if:
- $a$ is an upper bound for $T$
- $a \preceq b$ for all upper bounds $b$ of $T$
The dual of this statement is:
- $a$ is a lower bound for $T$
- $b \preceq a$ for all lower bounds $b$ of $T$
By definition, this means $a$ is an infimum for $T$.
The converse follows from Dual of Dual Statement (Order Theory).
$\blacksquare$