Supremum of Empty Set is Smallest Element

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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 supremun:

$t = \map \sup \O$

$\blacksquare$


Also see


Sources