# 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

$d(x, y) \leq M \Forall x, y \in A.$

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

$d(x, y) \leq M \Forall x, y \in X.$

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:

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

$$A$$ 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 $$\diam X$$ is finite.

Proposition 3.14

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

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

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

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

$\diam B(x,r) = \sup d(y, z) \leq 2 r.$

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

Proposition 3.15

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

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

For the converse, we proceed as follows:

1. Let $$\diam A = 0$$.

2. Assume $$A$$ is not a singleton.

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

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

5. But then, $$\diam A \geq d(x,y) > 0$$.

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

Proposition 3.16

If $$A \subseteq B$$, then $$\diam A \leq \diam B$$.

Proof. We proceed as follows:

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

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

3. Thus, $$d(x,y) \leq \diam B$$ (by definition).

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

Proposition 3.17

Let $$x \in A$$ and $$y \in B$$. Then $$d(x,y) \leq \diam (A \cup B)$$.

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

$d(x,y) \leq \diam (A \cup B).$

Proposition 3.18

If $$A \cap B \neq \EmptySet$$, then

$\diam (A \cup B) \leq \diam A + \diam B.$

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

1. If both $$x,y \in A$$, then $$d(x,y) \leq \diam A$$.

2. If both $$x,y \in B$$, then $$d(x,y) \leq \diam B$$.

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

4. Since $$A \cap B \neq \EmptySet$$, we can pick $$z \in A \cap B$$.

5. Then, by triangle inequality:

$d(x, y) \leq d(x, z) + d(y, z).$
6. Since $$x, z \in A$$, hence $$d(x, z) \leq \diam A$$.

7. Since $$y, z \in B$$, hence $$d(y, z) \leq \diam B$$.

8. Combining $$d(x, y) \leq \diam A + \diam B$$.

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

$\diam (A \cup B) \leq \diam A + \diam B.$

## 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 $$r = \diam A$$.

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 $$d(a,x) \leq r$$.

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

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

$d(x, y) \leq d(a, x) + d(a, y) < r + r = 2 r.$
3. Taking supremum on the L.H.S. over all $$x,y \in A$$, we get $$\diam A \leq 2 r$$.

4. Thus, $$A$$ is bounded.