Balthasar van der Ast (1593-1657) was a Dutch painter who specialized in still life of flowers and fruit.
Let \(S\) be a set of real numbers. The upper and lower bounds of \(S\) are defined below.
- We say that \(u\) is an upper bound of \(S\) when there exists a real number \(u\in\mathbb{R}\) satisfying \(x\leq u\) for every \(x\in S\). In this case, we also say that \(S\) is bounded above by \(u\).
- We say that \(l\) is a lower bound of \(S\) when there exists a real number \(l\in\mathbb{R}\) satisfying \(x\geq l\) for every \(x\in S\). In this case, we also say that \(S\) is bounded below by \(l\).
The set \(S\) is said to be unbounded above when the set \(S\) has no upper bound. The set \(S\) is said to be unbounded below when the set \(S\) has no lower bound. The maximal and minimal elements of \(S\) are defined below.
- We say that \(u^{*}\) is a maximal element or largest number of \(S\) when there exists a real number \(l^{*}\in S\) satisfying \(x\geq l^{*}\) for every \(x\in S\). In this case, we write \(l^{*}=\min S\).
- We say that \(l^{*}\) is a minimal element or smallest number of \(S\) when there exists a real number \(l^{*}\in S\) satisfying \(x\geq l^{*}\) for every \(x\in S\). In this case, we write \(l^{*}=\min S\).
Example. We provide some concrete examples.
- The set \(\mathbb{R}^{+}=(0,+\infty )\) is unbounded above. It has no upper bounds and no maximal element. It is bounded below by 0, but it has no minimal element.
- The closed interval \(S=[0,1]\) is bounded above by 1 and is bounded below by 0. We also have \(\max S=1\) and \(\max S=0\).
- The half-open interval \(S=[0,1)\) is bounded above by 1, but it has no maximal element. However, we have \(\min S=0\).$\sharp$
Although the set \(S=[0,1)\) is bounded above by 1, it has no maximal element. Therefore, we are going to introduce the concepts of supremum and infinum.
Definition. Let \(S\) be a set of real numbers.
(a) Suppose that \(S\) is bounded above. A real number \(\bar{u}\in\mathbb{R}\) is called a least upper bound or supremum of \(S\) when the following conditions are satisfied.
- \(\bar{u}\) is an upper bound of \(S\).
- If \(u\) is any upper bound of \(S\), then \(u\geq\bar{u}\).
In this case, we write \(\bar{u}=\sup S\). We say that the supremum \(\sup S\) is attained
when \(\bar{u}\in S\).
(b) Suppose that \(S\) is bounded below. A real number \(\bar{l}\in\mathbb{R}\) is called a greatest lower bound or infimum of \(S\) when the following conditions are satisfied.
- \(\bar{l}\) is a lower bound of \(S\).
- If \(l\) is any lower bound of \(S\), then \(l\leq\bar{l}\).
In this case, we write \(\bar{l}=\inf S\). We say that the infimum \(\inf S\) is attained
when \(\bar{l}\in S\). \(\sharp\)
It is clear to see that if the supremum \(\sup S\) is attained then \(\max S=\sup S\).
Similarly, if the infimum \(\inf S\) is attained then \(\min S=\inf S\).
Example. Let \(S=[0,1]\). Then, we have \(\max S=\sup S=1\) and \(\inf S=\min S=0.\) If \(S=[0,1)\), then \(\max S\) does not exists. However, we have \(\sup S=1\).$\sharp$
Proposition. Let \(S\) be a nonempty set of real numbers with \(\bar{u}=\sup S\). Then, given any \(s<\bar{u}\), there exists \(t\in S\) satisfying \(s<t\leq\bar{u}\).
Proof. We are going to prove it by contradiction. Suppose that we have \(t\leq s\) for all \(t\in S\). Then \(s\) is an upper bound of \(S\). According to the definition of supremum, we also have \(s\geq\bar{u}\). This contradiction says that there exists \(t\in S\) satisfying \(s<t\). This completes the proof. \(\blacksquare\)
Proposition. Given any two nonempty subsets \(A\) and \(B\) of \(\mathbb{R}\), we define \(C=A+B\) by
\[C=\left\{x+y:x\in A\mbox{ and }y\in B\right\}.\]
Suppose that the supremum \(\sup A\) and \(\sup B\) are attained. Then, the supremum \(\sup C\) is attained, and we have
\[\sup C=\sup A+\sup B.\]
Proof. We first have \(\sup A=\max A\) and \(\sup B=\max B\). We write \(a=\sup A\) and \(b=\sup B\). Given any \(z\in C\), there exist \(x\in A\) and \(y\in B\) satisfying \(z=x+y\). Since \(x\leq a\) and \(y\leq b\), we have \(z=x+y\leq a+b\), which says that \(a+b\) is an upper bound of \(C\). Therefore, by definition, \(c=\sup C\) satisfies \(c\leq a+b\). Next, we want to show that \(a+b\leq c\). Given any \(\epsilon >0\), Proposition 4 says that there exist \(x\in A\) and \(y\in B\) satisfying \(a-\epsilon <x\) and \(b-\epsilon <y\). We also see that \(x+y\leq c\). Adding these inequalities, we obtain
\[a+b-2\epsilon <x+y\leq c,\]
which says that \(a+b<c+2\epsilon\). Since \(\epsilon\) can be any positive real number, we must have \(a+b\leq c\). This completes the proof.$\blacksquare$
Proposition. Let \(A\) and \(B\) be any two nonempty subsets of \(\mathbb{R}\) satisfying \(a\leq b\) for any \(a\in A\) and \(b\in B\). Then, we have \(\sup A\leq\sup B\).
Proof. Given any fixed \(b^{*}\in B\), we have \(a\leq b^{*}\) for all \(a\in A\), which says that \(b^{*}\) is an upper bound of \(A\). The definition says that
\[\sup A\leq b^{*}\leq\sup B.\]
This completes the proof.\(\blacksquare\)
Let \(f:X\rightarrow\mathbb{R}\) be a real-valued function defined on a normed space \(X\), and let \(K\) be a subset of \(X\). The supremum of function \(f\) on \(K\) is denoted by
\[\sup_{x\in K}f(x)=\sup S\]
with \(S=\{f(x):x\in K\}\). We say that the supremum \(\sup_{x\in K}f(x)\) is attained when there exists \(x^{*}\in K\) satisfying
\[\sup_{x\in K}f(x)=f(x^{*}).\]
The maximum \(\max_{x\in K}f(x)\) means that there exists \(\bar{x}\in K\) satisfying \(f(x)\leq f(\bar{x})\) for all \(x\in K\). However, this situation cannot always be guaranteed. In other words, the maximum \(\max_{x\in K}f(x)\) may not exist. The existence of supremum \(\sup_{x\in K}f(x)\) is guaranteed. When the function \(f\) is upper semi-continuous on the compact set \(K\), the maximum \(\max_{x\in K}f(x)\) is guaranteed, which also says
\[\sup_{x\in K}f(x)=\max_{x\in K}f(x).\]
Proposition. Let \(A\) be a subset of \(\mathbb{R}\), and let \(f\) be a continuous function defined on \(\mbox{cl}(A)\). Then
\[\sup_{a\in A}f(a)=\sup_{a\in{\scriptsize \mbox{cl}(A)}}f(a)\]
and
\[\inf_{a\in A}f(a)=\inf_{a\in{\scriptsize \mbox{cl}(A)}}f(a).\]
Proof. It suffices to prove the case of supremum, since
\[\inf_{a\in A}f(a)=-\sup_{a\in A}[-f(a)].\]
It is clear to see
\[\sup_{a\in A}f(a)\leq\sup_{a\in{\scriptsize \mbox{cl}(A)}}f(a).\]
Given any \(\epsilon >0\), according to the concept of supremum, there exists \(a^{*}\in\mbox{cl}(A)\) satisfying
\[\sup_{a\in{\scriptsize \mbox{cl}(A)}}f(a)-\epsilon\leq f\left (a^{*}\right ).\]
We also see that there exists a sequence \(\{a_{n}\}_{n=1}^{\infty}\) in \(A\) satisfying \(a_{n}\rightarrow a^{*}\). Since \(f\) is continuous on \(\mbox{cl}(A)\), we also have \(f(a_{n})\rightarrow f(a^{*})\) as \(n\rightarrow\infty\). Therefore, we obtain
\begin{align*} \sup_{a\in{\scriptsize \mbox{cl}(A)}}f(a)-\epsilon\leq f\left (a^{*}\right ) & =\lim_{n\rightarrow\infty}f\left (a_{n}\right )\\ & \leq\lim_{n\rightarrow\infty}\left [\sup_{a\in A}f(a)\right ]\\ & =\sup_{a\in A}f(a).\end{align*}
Since \(\epsilon\) is any positive number, it follows
\[\sup_{a\in{\scriptsize \mbox{cl}(A)}}f(a)\leq\sup_{a\in A}f(a).\]
This completes the proof. \(\blacksquare\)
Let \(S\) be a subset of \(\mathbb{R}\). For \(\alpha\in S\) and a sequence \(\{\alpha_{n}\}_{n=1}^{\infty}\) in \(\mathbb{R}\), we write \(\alpha_{n}\uparrow\alpha\) to mean that the sequence \(\{\alpha_{n}\}_{n=1}^{\infty}\) is increasing and converges to \(\alpha\). We also write \(\alpha_{n}\downarrow\alpha\) to mean that the sequence \(\{\alpha_{n}\}_{n=1}^{\infty}\) is decreasing and converges to \(\alpha\).
Proposition. Let \(A\) be a subset of \(\mathbb{R}\). The following statements hold true.
(i) Let \(f\) be a right-continuous function defined on \(\mbox{cl}(A)\). Given any fixed \(r\in\mathbb{R}\), suppose that there exists a sequence \(\{a_{n}\}_{n=1}^{\infty}\) in \(A\) satisfying \(a_{n}\downarrow r\) as \(n\rightarrow\infty\) and \(a_{n}>r\) for all \(n\). Then, we have
\[\sup_{\{a\in A:a>r\}}f(a)=\sup_{\{a\in A:a\geq r\}}f(a)\]
and
\[\inf_{\{a\in A:a>r\}}f(a)=\inf_{\{a\in A:a\geq r\}}f(a).\]
(ii) Let \(f\) be a continuous function defined on \(\mbox{cl}(A)\). Given any fixed \(r\in\mathbb{R}\), suppose that there exists a sequence \(\{a_{n}\}_{n=1}^{\infty}\) in \(A\) satisfying \(a_{n}\rightarrow r\) and \(a_{n}>r\) for all \(n\). Then, we have
\[\sup_{\{a\in A:a>r\}}f(a)=\sup_{\{a\in A:a\geq r\}}f(a)\]
and
\[\inf_{\{a\in A:a>r\}}f(a)=\inf_{\{a\in A:a\geq r\}}f(a).\]
In particular, we can assume \(r\in\mbox{cl}(\{a\in A:a>r\})\).
Proof. It suffices to prove the case of supremum. It is clear to see
\[\sup_{\{a\in A:a>r\}}f(a)\leq\sup_{\{a\in A:a\geq r\}}f(a).\]
To prove part (i), given any \(\epsilon >0\), according to the concept of supremum \(\sup_{\{a\in A:a\geq r\}}f(a)\), there exists \(a^{*}\in A\) with \(a^{*}\geq r\) satisfying
\[\sup_{\{a\in A:a\geq r\}}f(a)-\epsilon\leq f\left (a^{*}\right ).\]
We consider the following two cases.
- Suppose that \(a^{*}>r\). Then, we have \[\sup_{\{a\in A:a\geq r\}}f(a)-\epsilon\leq f\left (a^{*}\right )\leq\sup_{\{a\in A:a>r\}}f(a).\]
- Suppose that \(a^{*}=r\). The assumption says that there exists a sequence \(\{a_{n}\}_{n=1}^{\infty}\) in \(A\) satisfying \(a_{n}\downarrow a^{*}\) as \(n\rightarrow\infty\) and \(a_{n}>r\) for all \(n\). Since \(f\) is right-continuous and \(a^{*}\in\mbox{cl}(A)\), we also have \(f(a_{n})\rightarrow f(a^{*})\) as \(n\rightarrow\infty\). Therefore, we obtain \begin{align*}\sup_{\{a\in A:a\geq r\}}f(a)-\epsilon & \leq f\left (a^{*}\right )=\lim_{n\rightarrow\infty}f\left (a_{n}\right )\\ & \leq\lim_{n\rightarrow\infty}\left [\sup_{\{a\in A:a>r\}}f(a)\right ]\\ & =\sup_{\{a\in A:a>r\}}f(a).\end{align*}
Since \(\epsilon\) is any positive number, it follows
\[\sup_{\{a\in A:a\geq r\}}f(a)\leq\sup_{\{a\in A:a>r\}}f(a).\]
Part (ii) can be similarly obtained, and the proof is complete. \(\blacksquare\)
Proposition. Let \(A\) and \(B\) be any two nonempty subsets of a metric space \((X,d)\). Suppose that, for each fixed \(a\in A\), the infimum \(\inf_{b\in B}d(a,b)\) is attained. Then, we have
\[\sup_{a\in A}\inf_{b\in B}d(a,b)=\inf_{b\in B}\sup_{a\in A}d(a,b).\]
Proof. For any fixed \(a\in A\), we have
\[\inf_{b\in B}d(a,b)\leq d(a,b),\]
which implies
\[\sup_{a\in A}\inf_{b\in B}d(a,b)\leq\sup_{a\in A}d(a,b).\]
By taking infimum on both sides, we obtain
\[\sup_{a\in A}\inf_{b\in B}d(a,b)\leq\inf_{b\in B}\sup_{a\in A}d(a,b).\]
On the other hand, since the infimum \(\inf_{b\in B}d(a,b)\) is attained for each \(a\in A\), we define a function \(b^{*}\) from \(A\) into \(B\) satisfying
\[d(a,b^{}(a))=\inf_{b\in B}d(a,b)=\min_{b\in B}d(a,b)\]
for \(b^{}(a)\in B\). Since
\[\inf_{b\in B}\sup_{a\in A}d(a,b)\leq\sup_{a\in A}d(a,b)\]
for any \(b\in B\), we have
\[\inf_{b\in B}\sup_{a\in A}d(a,b)\leq\sup_{a\in A}d(a,b^{*}(a))=\sup_{a\in A}\inf_{b\in B}d(a,b).\]
This completes the proof. \(\blacksquare\)
