applicability: $\mathbb {R}$

Definition: Supremum, Least Upper Bound

Let $D$ be a non-empty subset of real numbers. The real number $\sup(D)$ is called the supremum (or the least upper bound) of $D$, if

$\sup(D)$ is an upper bound of $D$ and
it is less than or equal to every upper bound of $D$, i.e. if $B$ is any upper bound of $D$ then $\sup(D)\le B$.

Equivalently, for every $\epsilon > 0$ there exists an $y\in D$ with $y > \sup(D) - \epsilon$.

Not all sets have the property that their non-empty subsets have a supremum. In fact, this definition is valid only for complete ordered fields, since in them, the existence of a supremum is ensured.
For instance, the set of all rational numbers $0 \le x \le \sqrt{2}$ does not have a supremum since we cannot identify the least rational upper bound that is still $\le \sqrt{2},$ however, the set of all real numbers $0 \le x \le \sqrt{2}$ has the supremum $\sqrt{2},$ since it is the least real upper bound $\le \sqrt{2}.$

The real number $b$ in the semi-open interval $[a,b)$ is the supremum of this interval because for every $\epsilon > 0$ (no matter how small) there is an $y\in[a,b)$ lying between the numbers $b-\epsilon$ and $b.$
$\sqrt{2}\not\in\mathbb Q$ is the supremum of the set or real numbers whose square is less or equal $2$, i.e. the set $A:=\{x\in\mathbb Q\mid x^2\le 2\}$, because for every $\epsilon > 0$ there is a real number $d\in A$ with $d > \sqrt{2}-\epsilon.$
Please note that the $\sup(D)$ does not have to be an element of $D$, like $b\not\in[a,b).$ However, $\sup(D)$ must be in the same domain as $D$, like $\sqrt{2}$ is not a supremum of $D=\{x\in\mathbb Q\mid x < \sqrt{2}\},$ but it is a supremum of $D=\{x\in\mathbb R\mid x < \sqrt{2}\}.$

Definitions: 1 2 3
Lemmas: 4
Proofs: 5 6 7 8 9 10 11
Propositions: 12 13
Theorems: 14

Thank you to the contributors under CC BY-SA 4.0!

Heuser Harro: "Lehrbuch der Analysis, Teil 1", B.G. Teubner Stuttgart, 1994, 11th Edition