Supremum of Empty Set is Smallest Element

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

Suppose that $\map \sup \O$, the supremum of the empty set, exists.

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

That is, $\map \sup \O$ is the smallest element of $S$.

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

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

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

Also see

 * Infimum of Empty Set is Greatest Element