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applicability: $\mathbb {R}$
Definition: Supremum, Least Upper Bound
Let \(D\) be a nonempty 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 \(y > \sup(D)  \epsilon\).
Notes
 Not all sets have the property that their nonempty 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 $\le \sqrt{2}.$
Examples
 The real number $b$ in the semiopen 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 $D=\{x\in\mathbb R\mid x < \sqrt{2}\}.$
Mentioned in:
Definitions: 1 2 3
Lemmas: 4
Proofs: 5 6 7 8 9 10 11
Propositions: 12 13
Theorems: 14
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References
Bibliography
 Heuser Harro: "Lehrbuch der Analysis, Teil 1", B.G. Teubner Stuttgart, 1994, 11th Edition