3.11. Special Topics#

This section covers details on specific functions from the perspective of metric spaces

3.11.1. Distance from a Point#

Theorem 3.122 (Distance function is uniformly continuous)

Let \((X,d)\) be a metric space. Fix some point \(a \in X\). Define a function \(f: X \to \RR\) as:

\[ f(x) = d(x, a) \Forall x \in X. \]

Then, \(f\) is uniformly continuous on \(X\).

Proof. Let \(x, y \in X\). Recall from triangle inequality that:

\[ |d (x, a) - d(y, a)| \leq d(x,y). \]
  1. Let \(\epsilon > 0\).

  2. Choose \(\delta = \epsilon\).

  3. Assume \(d(x, y) < \delta\).

  4. Then

    \[ |f(x) - f(y)| = |d (x, a) - d(y, a)| \leq d(x,y) < \delta = \epsilon. \]
  5. Thus, \(d(x,y) < \delta \implies |f(x) - f(y)| < \epsilon\).

  6. Thus, \(f\) is uniformly continuous.

3.11.2. Distance from a Set#

Recall from Definition 3.5 that the distance between a point \(x \in X\) and a set \(A \subseteq X\) is given by:

\[ d(x, A) = \inf \{ d(x,a) \Forall a \in A \}. \]

Proposition 3.32 (Distance from a singleton set)

For every \(x, y \in X\),

\[ d(x, \{ y \}) = d(x, y). \]

Proof. Let \(A = \{ y \}\) be a singleton set. Then

\[ d(x, A) = d(x, \{ y \}) = \inf\{ d(x, y) \} = d(x, y). \]

Proposition 3.33 (Distance from a subset)

Let \(\EmptySet \neq A \subseteq B \subseteq X\). Then, for any \(x \in X\)

\[ d(x, A) \geq d(x, B). \]

Proof. We note that since \(A \subseteq B\), hence:

\[ \{ d(x,a) \Forall a \in A \} \subseteq \{ d(x,a) \Forall a \in B \}. \]

The infimum over a larger set is smaller. Hence,

\[ d(x, B) \leq d(x, A). \]

Proposition 3.34 (Distance from a containing subset)

Let \(x \in A \subseteq X\). Then

\[ d(x, A) = 0. \]

Proof. We note that:

\[ d(x, x) \in \{ d(x,a) \Forall a \in A \} \]

since \(x \in A\).


\[ d(x,A) = d(x, x) = 0. \]

Theorem 3.123 (Continuity of set distance function)

Let \(A \subseteq X\) be nonempty. Then, the function \(f: X \to \RR\) given by:

\[ f(x) = d(x, A) = \inf \{ d(x,a) \Forall a \in A \} \]

is continuous.

We provided a proof in Example 3.16. We rephrase it here.

Proof. Let \(a \in X\). We shall show that \(f\) is continuous at \(a\).

  1. Let \(r > 0\).

  2. Consider the open ball/interval \(V = B(f(a), r)\) in \(\RR\).

  3. Consider the open ball in \(X\) given by \(U = B(a, r)\).

  4. Let \(x \in U\) and let \(y \in A\).

  5. Then, using triangle inequality:

    \[ d(a, y) + d(x, a) \geq d(x, y) \]


    \[ d(a, x) + d(x, y) \geq d(a, y). \]
  6. Taking the infimum over \(y \in A\) in both the inequalities, we get:

    \[ d(a, A) + d(x, a) \geq d(x, A) \]


    \[ d(x, a) + d(x, A) \geq d(a, A). \]
  7. Rewriting it further, we get:

    \[ d(x, A) \leq d(a, A) + d(x, a) \]


    \[ d(a, A) - d(x, a) \leq d(x, A). \]
  8. Since \(d(a,x) < r\), hence, we obtain:

    \[ d(a, A) - r < d(x, A) < d(a, A) + r. \]
  9. Substituting \(f(x) = d(x, A)\) and \(f(a) = d(a, A)\),

    \[ f(a) - r < f(x) < f(a) + r. \]
  10. Thus, \(f(x) \in B(f(a), r) = V\).

  11. In other words, \(f(U) \subseteq V\).

We have shown that for every \(r > 0\), there exists \(\delta = r\) such that \(x \in B(a, \delta)\) implies \(f(x) \in B(f(a), r)\). Thus, \(f\) is continuous at \(a\).

Theorem 3.124 (Open neighborhoods of a set)

Let \(A \subseteq X\) be nonempty. Let \(\epsilon > 0\) and consider the set \(A_{\epsilon}\) given by

\[ A_{\epsilon} = \{x \in X \ST d(x, A) < \epsilon \}. \]

Then \(A_{\epsilon}\) is open for every \(\epsilon > 0\).

Proof. We established in Theorem 3.123 that \(f(x) = d(x, A)\) is continuous. The set \(A_{\epsilon}\) is the inverse image of the open interval \((-\epsilon, \epsilon)\) in \(\RR\). Since \(f\) is continuous, hence the inverse image of an open interval is an open set.

Following is a direct proof.

  1. Let \(x \in A_{\epsilon}\).

  2. Then, there is \(y \in A\) such that \(d(x, y) < \epsilon\).

  3. Let \(\delta > 0\) be small enough such that \(d(x,y) + \delta < \epsilon\).

  4. Consider the open ball \(B(x, \delta)\).

  5. For all \(z \in B(x, \delta)\), we have:

    \[ d(z, y) \leq d(z, x) + d(x, y) < \delta + d(x,y) < \epsilon. \]
  6. Thus, \(d(z, A) < \epsilon\) since \(y \in A\).

  7. Thus, \(z \in A_{\epsilon}\).

  8. Thus, \(B(x, \delta) \subseteq A_{\epsilon}\).

  9. Thus, \(A_{\epsilon}\) is open.

Theorem 3.125 (Closure and set distance)

Let \(A \subseteq X\) be a nonempty set. Then, for any \(x \in X\), \(d(x, A) = 0\) if and only if \(x \in \closure A\).

In other words, the distance of a point from a set is zero if and only if the point is a closure point of the set.

Proof. Let \(x \in \closure A\).

  1. Then either \(x \in A\) or \(x \in \boundary A\).

  2. If \(x \in A\), then by Proposition 3.34, \(d(x,A) = 0\).

  3. Now, assume \(x \notin A\) and \(x \in \boundary A\).

  4. Let \(r > 0\).

  5. Since \(x \in \boundary A\), hence there exists \(y \in A\) such that \(d(x, y) < r\).

  6. Therefore,

    \[ d(x, A) = \inf \{d (x, y) \ST y \in A \} < r. \]
  7. Since this is true for every \(r > 0\), hence \(d(x, A) = 0\).

Now assume that \(x \notin \closure A\).

  1. Thus, there exists an open ball \(B(x, r)\) for some \(r > 0\) such that \(B(x, r) \cap A = \EmptySet\).

  2. Therefore, for every \(a \in A\), \(d(x, a) \geq r > 0\).

  3. Therefore,

    \[ d(x, A) = \inf \{ d(x, a) \ST a \in A \} \geq r > 0. \]

Corollary 3.4

Let \(A \subseteq X\) be a nonempty closed set. Then, for any \(x \in X\), \(d(x, A) = 0\) if and only if \(x \in A\).

Proof. A closed set contains all its closure points. Thus, \(d(x, A) = 0\) if and only if \(x\) is a closure point of \(A\) if and only if \(x \in A\).

Theorem 3.126 (Distance minimizer in a compact set)

Let \(A\) be a nonempty compact subset of \(X\). Let \(x \in X\). Then, there exists an \(a \in A\) such that

\[ d(x, A) = d(x, a). \]

Proof. For a fixed \(x \in X\), consider the real valued function \(f : X \to \RR\) given by:

\[ f(y) = d(y, x) \Forall y \in X. \]

Then, \(f\) is continuous as it is a distance function from a singleton set \(\{ x \}\).

Hence, \(f(A)\) attains a minimum value at some \(a \in A\). See Theorem 3.85.

Theorem 3.127 (Distance minimizer in a closed set)

Let \(X\) be a Euclidean metric space. Let \(A \subseteq X\) be a nonempty closed set. Let \(x \in X\). Then, there exists an \(a \in A\) such that

\[ d(x, A) = d(x, a). \]

In other words, the infimum of the distance between \(x\) and points of \(A\) is realized at a point in \(A\).

Proof. Since \(A\) is nonempty, we can pick some \(b \in A\) and compute \(r = d(x, b)\). Clearly,

\[ d(x,A) \leq d(x, b) = r. \]

Consider the set

\[ C = \{a \in A \ST d(x, a) \leq d(x, b) = r\} = A \cap B[x,r]. \]

In other words, \(C\) is the intersection of \(A\) and the closed ball \(B[x,r]\) centered at \(x\) and of radius \(r\). Since \(A\) is closed and \(B[x,r]\) is closed, hence \(C\) is closed.

It is easy to see that

\[ d(x,A) = d(x, C). \]

For any points \(u,v \in C\), we have:

\[ d(u, v) \leq d(u, x) + d(x, v) \leq r + r = 2 r. \]

Thus, \(C\) is closed and bounded. By Heine-Borel theorem, \(C\) is compact.

Now, for a fixed \(x \in X\), consider the function \(f : X \to \RR\) given by:

\[ f(y) = d(y, x) \Forall y \in X. \]

Then, \(f\) is continuous.

  1. By Theorem 3.78, \(f(C)\) is compact (continuous images of compact sets are compact).

  2. But \(f(C) \subseteq \RR\).

  3. Hence, \(f(C)\) is closed and bounded as \(\RR\) is Euclidean (Heine-Borel theorem).

  4. Hence, \(f(C)\) attains a minimum value at some \(a \in C\). See Theorem 3.85.

3.11.3. Distance Between Sets#

Recall from Definition 3.6 that the distance between two subsets of a metric space is given by

\[ d(A, B) = \inf \{ d(a,b) \ST a \in A, b \in B \}. \]

Theorem 3.128 (Distance between disjoint compact and closed sets)

Let \((X, d)\) be a metric space. Let \(A, B \subseteq X\) be disjoint (i.e., \(A \cap B = \EmptySet\)). Assume that \(A\) is compact and \(B\) is closed.

Then, there is \(\delta > 0\) such that

\[ |x - y | \geq \delta \Forall x \in A, y \in B. \]

In other words, the distance between \(A\) and \(B\) is nonzero; i.e., \(d(A, B) > 0\).

Proof. We prove this by contradiction.

  1. Assume that there is no such \(\delta > 0\).

  2. Then there exist sequences \(\{x_n \}\) of \(A\) and \(\{ y_n \}\) of \(B\) such that

    \[ \lim_{n \to \infty} | x_n - y_n | = 0. \]
  3. Since \(A\) is compact, hence \(\{x_n \}\) has a convergent subsequence, say \(\{ x_{n_k} \}\) which converges to some \(x \in A\).

  4. Now,

    \[ |x - y_{n_k}| \leq | x - x_{n_k} | + |x_{n_k} - y_{n_k} |. \]
  5. Since \(\{ x_n - y_n \}\) is a convergent sequence, hence \(\{ x_{n_k} - y_{n_k} \}\) also converges to \(0\).

  6. Thus,

    \[ \lim_{n \to \infty} |x - y_{n_k}| \leq \lim_{n \to \infty} | x - x_{n_k} | + \lim_{n \to \infty} |x_{n_k} - y_{n_k} | = 0 + 0 = 0. \]
  7. Thus, \(\{ y_{n_k} \}\) is a convergent sequence of \(B\) converging to \(x\).

  8. Since \(B\) is closed, hence \(x \in B\).

  9. But then, \(x \in A \cap B\).

  10. Thus, \(A \cap B \neq \EmptySet\) which is a contradiction.