Supremum is Dual to Infimum

Theorem

Let $\struct {S, \preceq}$ be an ordered set.

Let $a \in S$ and $T \subseteq S$.

The following are dual statements:

$a$ is a supremum for $T$

$a$ is an infimum for $T$

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 Dual Pairs (Order Theory).

By definition, this means $a$ is an infimum for $T$.

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

$\blacksquare$

Also see

• Duality Principle (Order Theory)