3.3. Boundedness

Let $(X,d)$ be a metric space.

Definition 3.29 (Boundedness of a set)

A subset $A\subseteq X$ is called bounded if there exists a number $M>0$ such that

$$d(x,y)\le M\mathrm{\forall }x,y\in A.$$

Definition 3.30 (Boundedness of the metric space)

A metric space $(X,d)$ is called bounded if there exists a number $M>0$ such that

$$d(x,y)\le M\mathrm{\forall }x,y\in X.$$

Even if a metric space $(X,d)$ is unbounded, it is possible to introduce a metric $\rho$ for it which makes the metric space $(X,\rho )$ bounded. See Theorem 3.56 for details.

3.3.1. Diameter

Definition 3.31 (Diameter)

The diameter of a nonempty subset $A$ of $(X,d)$ is defined as:

$$\mathrm{diam}A\triangleq sup\{d(x,y)|x,y\in A\}.$$

$A$ is bounded if its diameter is finite (i.e. the supremum on the R.H.S. exists). Otherwise, it is unbounded.

Remark 3.4

$(X,d)$ is bounded if and only if $\mathrm{diam}X$ is finite.

Proposition 3.14

The diameter of an open ball $B(x,r)$ is bounded by $2r$.

Proof. Let $y,z\in B(x,r)$. Then by triangle inequality:

$$d(y,z)\le d(x,y)+d(x,z)<r+r=2r.$$

Taking supremum on the L.H.S., we get:

$$\mathrm{diam}B(x,r)=supd(y,z)\le 2r.$$

For an example where $B(x,r)<2r$, see Proposition 3.26.

Proposition 3.15

$\mathrm{diam}A=0$ if and only if $A$ is a singleton set.

Proof. Let $A$ be singleton. Then, $A=\{x\}$. Then $\mathrm{diam}A=d(x,x)=0$.

For the converse, we proceed as follows:

1. Let $\mathrm{diam}A=0$.

2. Assume $A$ is not a singleton.

3. Then there exist distinct $x,y\in A$.

4. Since $x\ne y$, hence $d(x,y)>0$.

5. But then, $\mathrm{diam}A\ge d(x,y)>0$.

6. A contradiction.

7. Hence, $A$ must be a singleton.

Proposition 3.16

If $A\subseteq B$, then $\mathrm{diam}A\le \mathrm{diam}B$.

Proof. We proceed as follows:

1. Let $x,y\in A$.

2. Then $x,y\in B$.

3. Thus, $d(x,y)\le \mathrm{diam}B$ (by definition).

4. Taking supremum over all pairs of $x,y\in A$ in the L.H.S., we get: $\mathrm{diam}A\le \mathrm{diam}B$.

Proposition 3.17

Let $x\in A$ and $y\in B$. Then $d(x,y)\le \mathrm{diam}(A\cup B)$.

Proof. Since $x$ and $y$ both belong to $A\cup B$, hence, by definition:

$$d(x,y)\le \mathrm{diam}(A\cup B).$$

Proposition 3.18

If $A\cap B\ne \varnothing$, then

$$\mathrm{diam}(A\cup B)\le \mathrm{diam}A+\mathrm{diam}B.$$

Proof. Let $x,y\in A\cup B$.

1. If both $x,y\in A$, then $d(x,y)\le \mathrm{diam}A$.

2. If both $x,y\in B$, then $d(x,y)\le \mathrm{diam}B$.

3. Now, consider the case when $x\in A$ and $y\in B$.

4. Since $A\cap B\ne \varnothing$, we can pick $z\in A\cap B$.

5. Then, by triangle inequality:

   $$d(x,y)\le d(x,z)+d(y,z).$$

6. Since $x,z\in A$, hence $d(x,z)\le \mathrm{diam}A$.

7. Since $y,z\in B$, hence $d(y,z)\le \mathrm{diam}B$.

8. Combining $d(x,y)\le \mathrm{diam}A+\mathrm{diam}B$.

9. Taking the supremum over all pairs $x,y\in A\cup B$,

   $$\mathrm{diam}(A\cup B)\le \mathrm{diam}A+\mathrm{diam}B.$$

3.3.2. Characterization of Boundedness

Theorem 3.29

A set $A\subseteq X$ is bounded if and only if there exists $a\in X$ and $r>0$ such that

$$A\subseteq B(a,r).$$

In other words, $A$ is bounded if and only if $A$ is contained in an open ball.

Proof. Assume $A$ is bounded.

1. Let $r=\mathrm{diam}A$.

2. Fix some $a\in A$.

3. Consider an open ball $B(a,r+1)$.

4. Consider any $x\in A$.

5. Since $r$ is diameter of $A$ and $a,x\in A$, hence $d(a,x)\le r$.

6. Thus, $d(x,a)\le r<r+1$.

7. Thus, $x\in B(a,r+1)$.

8. Since $x$ was arbitrary, hence $A\subseteq B(a,r+1)$.

Now assume that there is some $a\in X$ and $r>0$ such that $A\subseteq B(a,r)$.

1. Let $x,y\in A$. Then, $x,y\in B(a,r)$.

2. By triangle inequality

   $$d(x,y)\le d(a,x)+d(a,y)<r+r=2r.$$

3. Taking supremum on the L.H.S. over all $x,y\in A$, we get $\mathrm{diam}A\le 2r$.

4. Thus, $A$ is bounded.