Supremum is Dual to Infimum

Theorem
Let $\left({S, \preceq}\right)$ 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 a infimum for $T$

Proof
By definition, $a$ is a supremum for $T$ iff:


 * $a$ 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 a infimum for $T$.

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

Also see

 * Duality Principle (Order Theory)