$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

# 3.9: Bounded Sets. Diameters

$$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

I. Geometrically, the diameter of a closed globe in $$E^{n}$$ could be defined as the maximum distance between two of its points. In an open globe in $$E^{n},$$ there is no "maximum" distance (why?), but we still may consider the supremum of all distances inside the globe. Moreover, this makes sense in any set $$A \subseteq(S, \rho) .$$ Thus we accept it as a general definition, for any such set.

Definition

The diameter of a set $$A \neq \emptyset$$ in a metric space $$(S, \rho),$$ denoted $$d A,$$ is the supremum (in $$E^{*}$$) of all distances $$\rho(x, y),$$ with $$x, y \in A ;^{1}$$ in symbols,

$d A=\sup _{x, y \in A} \rho(x, y).$

If $$A=\emptyset,$$ we put $$d A=0 .$$ If $$d A<+\infty, A$$ is said to be bounded $$($$ in $$(S, \rho) ) .$$

Equivalently, we could define a bounded set as in the statement of the following theorem.

Theorem $$\PageIndex{1}$$

$$A$$ set $$A \subseteq(S, \rho)$$ is bounded iff $$A$$ is contained in some globe. If so, the center p of this globe can be chosen at will.

Proof

If $$A=\emptyset,$$ all is trivial.

Thus let $$A \neq \emptyset ;$$ let $$q \in A,$$ and choose any $$p \in S .$$ Now if $$A$$ is bounded, then $$d A<+\infty,$$ so we can choose a real $$\varepsilon>$$ $$\rho(p, q)+d A$$ as a suitable radius for a globe $$G_{p}(\varepsilon) \supseteq A($$ see Figure 11 for motivation$$).$$ Now if $$x \in A,$$ then by the definition of $$d A$$ $$\rho(q, x) \leq d A ;$$ so by the triangle law,

\begin{aligned} \rho(p, x) & \leq \rho(p, q)+\rho(q, x) \\ & \leq \rho(p, q)+d A<\varepsilon; \end{aligned}

i.e., $$x \in G_{p}(\varepsilon) .$$ Thus $$(\forall x \in A) x \in G_{p}(\varepsilon)$$ as required.

Conversely, if $$A \subseteq G_{p}(\varepsilon),$$ then any $$x, y \in A$$ are also in $$G_{p}(\varepsilon) ;$$ so $$\rho(x, p)<\varepsilon$$ and $$\rho(p, y)<\varepsilon,$$ whence

$\rho(x, y) \leq \rho(x, p)+\rho(p, y)<\varepsilon+\varepsilon=2 \varepsilon.$

Thus 2$$\varepsilon$$ is an upper bound of all $$\rho(x, y)$$ with $$x, y \in A .$$ Therefore,

$d A=\sup \rho(x, y) \leq 2 \varepsilon<+\infty;$

i.e., $$A$$ is bounded, and all is proved. $$\square$$ As a special case we obtain the following.

Theorem $$\PageIndex{1}$$

$$A$$ set $$A \subseteq E^{n}$$ is bounded iff there is a real $$K>0$$ such that

$(\forall \overline{x} \in A) \quad|\overline{x}|<K$

(*similarly in $$C^{n}$$ and other normed spaces).

Proof

By Theorem $$1($$ choosing $$\overline{0}$$ for $$p), A$$ is bounded iff $$A$$ is contained in some globe $$G_{\overline{0}}(\varepsilon)$$ about $$\overline{0} .$$ That is,

$(\forall \overline{x} \in A) \quad \overline{x} \in G_{\overline{0}}(\varepsilon)\text{ or } \rho(\overline{x}, \overline{0})=|\overline{x}|<\varepsilon.$

Thus $$\varepsilon$$ is the required $$K$$ .(*The proof for normed spaces is the same.) $$\square$$

Note 1. In $$E^{1},$$ this means that

$(\forall x \in A) \quad-K<x<K;$

i.e., $$A$$ is bounded by $$-K$$ and $$K .$$ This agrees with our former definition, given in Chapter 2, §§8-9.

Caution: Upper and lower bounds are not defined in $$(S, \rho),$$ in general.

Example $$\PageIndex{1}$$

(1) $$\emptyset$$ is bounded, with $$d \emptyset=0,$$ by definition.

(2) Let $$A=[\overline{a}, \overline{b}]$$ in $$E^{n},$$ with $$d=\rho(\overline{a}, \overline{b})$$ its diagonal. By Corollary 1 in §7 $$d$$ is the largest distance in $$A .$$ In nonclosed intervals, we still have

$d=\sup _{x, y \in A} \rho(x, y)=d A<+\infty\text{ (see Problem 10 (ii)).}$

Thus all intervals in $$E^{n}$$ are bounded.

(3) Each globe $$G_{p}(\varepsilon)$$ in $$(S, \rho)$$ is bounded, with $$d G_{p}(\varepsilon) \leq 2 \varepsilon<+\infty,$$ as was shown in the proof of Theorem $$1 .$$ See, however, Problems 5 and 6 below.

(4) All of $$E^{n}$$ is not bounded, under the standard metric, for if $$E^{n}$$ had a finite diameter $$d,$$ no distance in $$E^{n}$$ would exceed $$d ;$$ but $$\rho\left(-d \overline{e}_{1}, d \overline{e}_{1}\right)=2 d$$, a contradiction!

(5) On the other hand, under the discrete metric §11, Example (3)), any set (even the entire space) is contained in $$G_{p}(3)$$ and hence bounded. The same applies to the metric $$\rho^{\prime}$$ defined for $$E^{*}$$ in Problem 5 of §§11, since distances under that metric never exceed $$2,$$ and so $$E^{*} \subseteq G_{p}(3)$$ for any choice of $$p$$.

Note 2. This shows that boundedness depends on the metric $$\rho .$$ A set may be bounded under one metric and not bounded under another. A metric $$\rho$$ is said to be bounded iff all sets are bounded under $$\rho$$ (as in Example (5)).

Problem 9 of §11 shows that any metric $$\rho$$ can be transformed into a bounded one, even preserving all sufficiently small globes; in part (i) of the problem, even the radii remain the same if they are $$\leq 1$$.

Note 3. An idea similar to that of diameter is often used to define distances between sets. If $$A \neq \emptyset$$ and $$B \neq \emptyset$$ in $$(S, \rho),$$ we define $$\rho(A, B)$$ to be the infimum of all distances $$\rho(x, y),$$ with $$x \in A$$ and $$y \in B .$$ In particular, if $$B=\{p\}$$ (a singleton$$),$$ we write $$\rho(A, p)$$ for $$\rho(A, B).$$ Thus

$\rho(A, p)=\inf _{x \in A} \rho(x, p).$

II. The definition of boundedness extends, in a natural manner, to sequences and functions. We briefly write $$\left\{x_{m}\right\} \subseteq(S, \rho)$$ for a sequence of points in $$(S, \rho)$$, and $$f : A \rightarrow(S, \rho)$$ for a mapping of an arbitrary set $$A$$ into the space $$S .$$ Instead of "infinite sequence with general term $$x_{m},$$" we say "the sequence $$x_{m}$$."

Definition

A sequence $$\left\{x_{m}\right\} \subseteq(S, \rho)$$ is said to be bounded iff its range is bounded in $$(S, \rho),$$ i.e., iff all its terms $$x_{m}$$ are contained in some globe in $$(S, \rho).$$

In $$E^{n},$$ this means (by Theorem 2$$)$$ that

$(\forall m) \quad\left|x_{m}\right|<K$

for some fixed $$K \in E^{1}.$$

Definition

A function $$f : A \rightarrow(S, \rho)$$ is said to be bounded on a set $$B \subseteq A$$ iff the image set $$f[B]$$ is bounded in $$(S, \rho) ;$$ i.e. iff all function values $$f(x),$$ with $$x \in B,$$ are in some globe in $$(S, \rho)$$.

In $$E^{n},$$ this means that

$(\forall x \in B) \quad|f(x)|<K$

for some fixed $$K \in E^{1}.$$

If $$B=A,$$ we simply say that $$f$$ is bounded.

Note 4. If $$S=E^{1}$$ or $$S=E^{*},$$ we may also speak of upper and lower bounds. It is customary to call sup $$f[B]$$ also the supremum of $$f$$ on $$B$$ and denote it by symbols like

$\sup _{x \in B} f(x)\text{ or } \sup \{f(x) | x \in B\}.$

In the case of sequences, we often write sup $$_{m} x_{m}$$ or sup $$x_{m}$$ instead; similarly for infima, maxima, and minima.

Example $$\PageIndex{1}$$

(a) The sequence

$x_{m}=\frac{1}{m} \quad\text{ in } E^{1}$

is bounded since all terms $$x_{m}$$ are in the interval $$(0,2)=G_{1}(1) .$$ We have inf $$x_{m}=0$$ and $$\sup x_{m}=\max x_{m}=1.$$

(b) The sequence

$x_{m}=m \quad\text{ in } E^{1}$

is bounded below (by 1$$)$$ but not above. We have inf $$x_{m}=\min x_{m}=1$$ and $$\sup x_{m}=+\infty$$ (in $$E^{*})$$.

(c) Define $$f : E^{1} \rightarrow E^{1}$$ by

$f(x)=2 x.$

This map is bounded on each finite interval $$B=(a, b)$$ since $$f[B]=$$ $$(2 a, 2 b)$$ is itself an interval and hence bounded. However, $$f$$ is not bounded on all of $$E^{1}$$ since $$f\left[E^{1}\right]=E^{1}$$ is not a bounded set.

(d) Under a bounded metric $$\rho,$$ all functions $$f : A \rightarrow(S, \rho)$$ are bounded.

(e) The so-called identity map on $$S, f : S \rightarrow(S, \rho),$$ is defined by

$f(x)=x.$

Clearly, $$f$$ carries each set $$B \subseteq S$$ onto itself; i.e., $$f[B]=B .$$ Thus $$f$$ is bounded on $$B$$ iff $$B$$ is itself a bounded set in $$(S, \rho).$$

(f) Define $$f : E^{1} \rightarrow E^{1}$$ by

$f(x)=\sin x.$

Then $$f\left[E^{1}\right]=[-1,1]$$ is a bounded set in the range space $$E^{1} .$$ Thus $$f$$ is bounded on $$E^{1}$$ (briefly, bounded).