3.9: Bounded Sets. Diameters

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).