Supremum of Empty Set is Smallest Element

Theorem

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

Then:

the supremum of the empty set exists if and only if 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:

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

$\Box$

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 supremum:

$t = \map \sup \O$

$\blacksquare$

Also see

• Infimum of Empty Set is Greatest Element

Sources

• 1982: Peter T. Johnstone: Stone Spaces ... (previous) ... (next): Chapter $\text I$: Preliminaries, Definition $1.2$