Supremum of Empty Set is Smallest Element

Theorem
Let $\left({S, \preceq}\right)$ be an ordered set.

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

Then $\forall s \in S: \sup \varnothing \preceq s$.

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

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

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

Hence:
 * $\forall s \in S: \sup \varnothing \preceq s$

Also see

 * Infimum of Empty Set is Greatest Element