>We write $\sup S = l$ if $l$ is the *supremum* of $S$ and $\sup S =\infty$ if $S$ is not bounded above. >We write $\inf S = l$ if $l$ is the *infimum* of $S$ and $\inf S =-\infty$ if $S$ is not bounded below. > [!NOTE] **Definition (Bounded Set of Reals)** > Suppose $S \subset \mathbb{R}$ is a set of [[Real numbers|reals]]. We say that $S$ is *bounded* if there exists $r\in \mathbb{R}$ such that $|s|\leq r$ for all $s\in S$ (i.e. $S$ is *bounded above* by $r$ and *bounded below* by $-r$/ $r$ is an upper bound for $S$ and $-r$ is a lower bound for $S$). > > [!NOTE] **Definition (Bounded above, upper bound)** > Suppose $S \subset \mathbb{R}$. An *upper bound* for $S$ is a real number $r$ such that $s \leq r$ for all $s \in S$. > If a set has an upper bound then it is *bounded above*. ^3387f0 # Properties > [!NOTE] Theorem (Completeness of reals) > Any non-empty set of reals that is bounded above has a *least upper bound*. >See [[Real numbers|Proof]]. > [!NOTE] **Lemma** (Tightness) > Suppose $S \subset R$ is bounded above then $l= \sup S$ iff $l$ is an *upper bound* for $S$ and for every $\varepsilon>0$ there exists $s \in S$ with $s>l-\varepsilon$. > ^fcf26c > *Proof*. BWOC suppose $l= \sup S$, $t<l$ and there is no $s \in S$ with $s>t$. > Then every $s \in S$ must satisfy $s \leq t$ and so $t$ is an upper bound for $S$ that is strictly less that $l$, which contradicts the fact that $l$ is the least upper bound of $S$. > [!NOTE] Lemma (Negation) > If $r$ is an upper bound for $S$, then $-r$ is an lower bound for $-S = \{ -s \mid s\in S \}$ >*Proof*. For all $s\in S$, $s \leq r \implies -s \geq -r$. $\square$ > [!NOTE] Lemma (Upper bound in set) > If $r\in S$ is an upper bound for $S$ then $\sup S = r.$ >*Proof*. We already have that $r$ is an upper bound for $S.$ If $m$ is an upper bound for $S,$ then $r\leq m$ since $r\in s$. $\square$ > [!NOTE] Corollary (Greatest upper bound property) Any non-empty subset of $\mathbb{R}$ that is *bonded below* has a *greatest lower bound.* > *Proof*. Suppose that $S$ is a non-empty that is bounded below by $m$. Then $-S$ is a non-empty set that is bounded above by $-m$, so it has a least upper bound $-l$ by LUBA. > > Now $l$ is a lower bound for $S.$ Also if $-m$ is another upper bound for $-S$, we must have $-l \leq -m$, which shows that $l \geq m$ so that $-l$ is indeed the *greatest lower bound* for $S$. # Applications - [[Bounded Sequence]]. - [[Bounded Real Function]]. - [[Real intervals]].