Supremum of Empty Set is Smallest Element

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

Then:
 * the supremum of the empty set exists the smallest element of $S$ exists

in which case:
 * $\map \sup \O$ is the smallest element of $S$

Proof
Observe that, vacuously, any $s \in S$ is an upper bound for $\O$.

Necessary Condition
Let $\map \sup \O$ exist.

For any upper bound $s$ of $\O$, $\map \sup \O \preceq s$ by definition of supremum.

Hence:
 * $\forall s \in S: \map \sup \O \preceq s$

Sufficient Condition
Let $t$ be the smallest element of $S$.

Then $t$ is an upper bound of $\O$.

For any upper bound $s$ of $\O$, $t \preceq s$ by definition of the smallest element.

By definition of the supremun:
 * $t = \map \sup \O$

Also see

 * Infimum of Empty Set is Greatest Element