Supremum (least upper bound)

Let $(X,\le)$ be a partially ordered set and let $S\subseteq X$. An element $s^\ast\in X$ is the supremum of $S$, written $s^\ast=\sup S$, if:

• $s^\ast$ is an upper bound of $S$, i.e. $\forall s\in S,\ s\le s^\ast$, and
• $s^\ast$ is the least such upper bound: for every upper bound $u$ of $S$, one has $s^\ast\le u$.

Suprema are “best possible” upper bounds and are the key completeness feature of $\mathbb{R}$. In general ordered sets, $\sup S$ need not exist.

Examples:

• In $\mathbb{R}$, $\sup(0,1)=1$ (even though $1\notin(0,1)$).
• In $\mathbb{R}$, if $S=\{x\in\mathbb{R}: x^2<2\}$, then $\sup S=\sqrt{2}$.
• In $\mathbb{Q}$ with its usual order, the set $S=\{q\in\mathbb{Q}: q^2<2\}$ has no supremum in $\mathbb{Q}$ (its least upper bound in $\mathbb{R}$ is $\sqrt{2}\notin\mathbb{Q}$).