A real number \(b\) is an upper bound (resp. a lower bound) of a set \(A\) of real numbers if \(b \ge a\) (resp. \(b \le a\)) for every \(a \in
A\text{.}\) We write \(\ub(A)\) and \(\lb(A)\) for the sets of upper bounds and lower bounds of \(A\text{,}\) respectively. It is straightforward to check that \(-\ub(A) = \lb(-A)\text{.}\) In other words, \(b\) is an upper bound of \(A\) if and only if \(-b\) is a lower bound of \(-A\text{.}\)
A set \(A\subseteq\R\) is bounded above (resp. bounded below) if it has an upper bound (resp. a lower bound). We say \(A\) is bounded if it is both bounded above and bounded below. A real-valued function is bounded (resp. bounded above, bounded below) if its range is bounded (resp. bounded above, bounded below).
An upper bound of a set is also an upper bound of every subset of that set. Therefore, an intersection of sets that are bounded above is again bounded above. If \(a\) and \(a'\) are upper bounds of \(A\) and \(A'\text{,}\) respectively, then \(\max\{a,a'\}\) is an upper bound of \(A \cup A'\text{.}\) Thus, a finite union of sets bounded above is bounded above. Likewise, the family of subsets of \(\R\) that are bounded below (resp. bounded) is closed under arbitrary intersections and finite unions.
One implication is immediate: if \(A \subseteq [-M,M]\text{,}\) then \(A\) is bounded. For the other implication, suppose \(A\) is bounded. Then so is \(-A\text{.}\) Hence \(A \cup -A\) is bounded above, say by \(M\text{.}\) Then \(|a| = \max\{a,-a\} \le M\) for all \(a\in A\text{,}\) so \(A \subseteq [-M,M]\text{.}\)
A supremum of a set \(A\subseteq\R\text{,}\) denoted by \(\sup A\text{,}\) is the least upper bound of \(A\text{.}\) In other words, \(u=\sup A\) if \(u\) is an upper bound of \(A\) and \(u\le b\) for every upper bound \(b\) of \(A\text{.}\) Dually, an infimum of \(A\text{,}\) denoted by \(\inf A\text{,}\) is the greatest lower bound of \(A\text{.}\) In other words, \(\ell=\inf A\) if \(\ell\) is a lower bound of \(A\) and every lower bound of \(A\) is at most \(\ell\text{.}\) It is clear that \(\sup A\) (likewise \(\inf A\)), if it exists, is unique.
A useful reformulation is: \(u=\sup A\) if and only if \(u\) is an upper bound of \(A\) and for every \(\varepsilon>0\) there exists \(a\in A\) with \(u-\varepsilon<a\le u\text{.}\)
The set of upper bounds and the set of lower bounds of \((0,1)\) are \([1,\infty)\) and \((-\infty,0]\text{,}\) respectively. Hence \(\sup(0,1) = 1\) and \(\inf(0,1)
= 0\text{.}\)
The set of upper bounds and the set of lower bounds of \(H:=\{1/x \colon x \lt 0\}\) are \([0,\infty)\) and the empty set, respectively. Hence \(\sup H = 0\) and \(\inf H\) does not exist.
The set of upper bounds and the set of lower bounds of \(\emptyset\) are both \(\R\text{.}\) Conventionally, we write \(\sup \emptyset = -\infty\) and \(\inf \emptyset = +\infty\text{.}\) Note that neither \(\sup \emptyset\) nor \(\inf \emptyset\) exists in \(\R\text{.}\)
