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, \(\diam A \geq d(x,y) > 0\).

  6. A contradiction.

  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.