Bounded above

Let $(X,\le)$ be an ordered set and let $S\subseteq X$. The set $S$ is bounded above if there exists an element $u\in X$ such that

Equivalently, $S$ is bounded above iff $S$ has an upper bound.

Boundedness above is the hypothesis needed to speak meaningfully about $\sup S$ (and in $\mathbb{R}$, completeness asserts existence of $\sup S$ for every nonempty bounded-above set).

Examples:

• $(0,1)\subseteq\mathbb{R}$ is bounded above (e.g. by $u=1$).
• $\{n\in\mathbb{N}\}\subseteq\mathbb{R}$ is not bounded above.
• The set $\{x\in\mathbb{R}: x\le 5\}$ is bounded above (e.g. by $u=5$).