K. Jänich, Topology. Chapter I: Fundamental Concepts. 2: Metric Spaces

 Definition: Metric Space Let \(X\) be a set. A metric on \(X\) is a function \(d : X^2 \rightarrow \mathbb{R} \) such that:

• M1. \(d(x,y) \geq 0\) for all \(x,y \in X\), and \(d(x,y) = 0\) if and only if \(x = y\).
• M2. \(d(x,y) = d(y,x)\) for all \(x,y \in X\).
• M3. Triangle Inequality. \(d(x,z) \leq d(x,y) + d(y,z) \) for all \(x,y,z \in X\).

We call \(d(x,y)\) the distance between \(x\) and \(y\).

A metric space is a pair \((X,d)\) such that \(X\) is a set and \(d\) is a metric on \(X\). We usually talk about 'the metric space \(X\)' for 'the metric space \((X,d)\)'. 

Definition. Let \(d\) be a metric on \(X\). Let \(a \in X\) and \(\epsilon > 0\). The open ball with centre \(a\) and radius \(\epsilon\) is \(B_\epsilon(a) = \{ x \in X : d(x,a) < \epsilon \}\).

Definition. Let \(d\) be a metric on \(X\). We say that a set \(U \subseteq X\) is open iff, for all \(a \in U\), there exists \(\epsilon > 0\) such that \(B_\epsilon(a) \subseteq U\). The topology induced by the metric \(d\), denoted \(\mathcal{O}(d)\), is the set of all open sets.

Proposition 1. This is a topology on \(X\).

Proof.

1. Let \(\mathcal{U}\) be a set of open sets; we prove \(\bigcup \mathcal{U}\) is open. Let \(a \in \bigcup \mathcal{U}\). Pick \(U \in \mathcal{U}\) such that \(a \in U\). Pick \(\epsilon > 0\) such that \(B_\epsilon(a) \subseteq U\). Then \(B_\epsilon(a) \subseteq \bigcup \mathcal{U}\) as required.
2. Let \(U\) and \(V\) be open; we prove \(U \cap V\) is open. Let \(a \in U \cap V\). Pick \(\delta, \epsilon > 0\) such that \(B_\delta(a) \subseteq U\) and \(B_\epsilon(a) \subseteq V\). Then \(B_{\min(\delta,\epsilon)}(a) \subseteq U \cap V\) as required.
3. For any \(a \in X\) we have \(B_1(a) \subseteq X\). Thus \(X\) is open.

\(\Box\) 

Proposition 2. Every open ball is open.

Proof. Let \(a \in X\) and \(\epsilon > 0\); we prove \(B_\epsilon(a)\) is open.

Let \(b \in B_\epsilon(a)\). We prove there exists \(\delta > 0\) such that \(B_\delta(b) \subseteq B_\epsilon(a)\).

We have \(d(b,a) < \epsilon\). Let \(\delta = \epsilon - d(b,a) > 0\). We prove that \(B_\delta(b) \subseteq B_\epsilon(a)\).

Let \(x \in B_\delta(b)\). We must prove that \(x \in B_\epsilon(a)\), that is, \(d(x,a) < \epsilon\). Well,

\[ \begin{align} d(x,a) & \leq d(x,b) + d(b,a) & (\text{Triangle Inequality}) \\ & < \delta + d(b,a) & (x \in B_\delta(b)) \\ & = \epsilon & \Box \end{align} \]

Proposition 3. A set \(N\) is a neighbourhood of \(x\) iff it includes an open ball with centre \(x\).

Proof. Assume \(N\) is a neighbourhood of \(x\). Pick an open set \(U\) such that \(x \in U \subseteq N\). Pick \(\epsilon > 0\) such that \(B_\epsilon(x) \subseteq U\). Then \(B_\epsilon(x) \subseteq N\).

Conversely, assume \(B_\epsilon(x) \subseteq N\). Then \(B_\epsilon(x)\) is an open set such that \(x \in B_\epsilon(x) \subseteq N\) and so \(N\) is a neighbourhood of \(x\). \(\Box\)

Proposition 4. Let \(d\) and \(d'\) be two metrics on a set \(X\). Suppose that, for all \(x \in X\), every open \(d\)-ball with centre \(x\) includes an open \(d'\)-ball with centre \(x\). Then \(\mathcal{O}(d) \subseteq \mathcal{O}(d')\).

Proof. Let \(U \in \mathcal{O}(d)\). Let \(x \in U\). Then \(U\) includes an open \(d\)-ball with centre \(x\), which includes an open \(d'\)-ball with centre \(x\). Thus \(U \in \mathcal{O}(d')\). \(\Box\)

Corollary 5. Let \(d\) and \(d'\) be two metrics on a set \(X\). Suppose that, for all \(x \in X\), every open \(d\)-ball with centre \(x\) includes an open \(d'\)-ball with centre \(x\), and every open \(d'\)-ball with centre \(x\) includes an open \(d\)-ball with centre \(x\). Then \(\mathcal{O}(d) =\mathcal{O}(d')\).

Example 6. Let \(n \geq 0 \). Define \(d, d' : \mathbb{R}^n \rightarrow \mathbb{R}\) by

\[ d((x_1, \ldots, x_n),(y_1, \ldots, y_n)) = \sqrt{(y_1 - x_1)^2 + \cdots + (y_n - x_n)^2} \]

\[ d'((x_1,\ldots, x_n),(y_1,\ldots,y_n)) = \max(|y_1 - x_1|, \ldots, |y_n - x_n|) \]

Then \(d\) and \(d'\) are metrics on \(\mathbb{R}^n\) that induce the same topology. We call this topology the standard topology on \(\mathbb{R}^n\).

To prove this, we need some results from analysis:

Cauchy-Schwarz Inequality For any real numbers \(x_1\), ..., \(x_n\), \(y_1\), ..., \(y_n\), we have \[ |x_1 y_1 + \cdots + x_n y_n| \leq \sqrt{x_1^2 + \cdots + x_n^2} \sqrt{y_1^2 + \cdots + y_n^2} \]

Proof Let \(\alpha = \sqrt{x_1^2 + \cdots + x_n^2}\) and \( \beta = \sqrt{y_1^2 + \cdots + y_n^2} \). We may assume \(\alpha\) and \(\beta\) are non-zero. (If \(\alpha = 0\) then \(x_1 = \cdots = x_n = 0\) and both sides of the Cauchy-Schwarz inequality are 0. Likewise if \(\beta = 0\).)

We may also assume \(x_1\), ..., \(x_n\), \(y_1\), ..., \(y_n\) are non-negative. (If any is negative, replace it with its negation.)

Then we have \[ \begin{align} (\beta x_1 - \alpha y_1)^2 + \cdots + (\beta x_n - \alpha y_n)^2 & \geq 0 \\ \therefore \beta^2 (x_1^2 + \cdots + x_n^2) + \alpha^2 (y_1^2 + \cdots + y_n^2) - 2 \alpha \beta (x_1 y_1 + \cdots + x_n y_n) & \geq 0 \\ \therefore 2 \alpha^2 \beta^2 & \geq 2 \alpha \beta (x_1 y_1 + \cdots + x_n y_n) \\ \therefore \alpha \beta & \geq x_1 y_1 + \cdots + x_n y_n \end{align} \] and this is the desired inequality. \(\Box\)

Lemma: Minkowski Inequality For any real numbers \(x_1\), ..., \(x_n\), \(y_1\), ..., \(y_n\), we have \[ \sqrt{(x_1 + y_1)^2 + \cdots + (x_n + y_n)^2} \leq \sqrt{x_1^2 + y_1^2} + \cdots + \sqrt{x_n^2 + y_n^2} \]

Proof \[ \begin{align} & (x_1 + y_1)^2 + \cdots + (x_n + y_n)^2 \\ & = x_1^2 + \cdots + x_n^2 + y_1^2 + \cdots + y_n^2 + 2 (x_1 y_1 + \cdots + x_n y_n) \\ & \geq x_1^2 + \cdots + x_n^2 + y_1^2 + \cdots + y_n^2 + 2 \sqrt{x_1^2 + \cdots + x_n^2} \sqrt{y_1^2 + \cdots + y_n^2} & (\text{Cauchy-Schwarz}) \\ & = (\sqrt{x_1^2 + \cdots + x_n^2} + \sqrt{y_1^2 + \cdots + y_n^2})^2 \end{align} \] The result follows by taking the square root of both sides.

Proof of Example 6.

We first prove \(d\) is a metric.

1. \(d((x_1,y_1),(x_2,y_2)) \geq 0\) immediately from the definition, and \[ \begin{align} d((x_1,\ldots, x_n),(y_1,\ldots,y_n)) = 0 & \Leftrightarrow \sqrt{(x_1 - y_1)^2 + \cdots + (x_n - y_n)^2} = 0 \\ & \Leftrightarrow (x_1 - y_1)^2 + \cdots + (x_n - y_n)^2 = 0 \\ & \Leftrightarrow x_1 - y_1 = 0 \text{ and } \cdots \text{ and } x_n - y_n = 0 \\ & \Leftrightarrow (x_1,\ldots,x_n) = (y_1, \ldots, y_n) \end{align} \]
2. \( \)\[ \begin{align} d((x_1,\ldots, x_n),(y_1, \ldots, y_n)) & = \sqrt{(x_1 - y_1)^2 + \cdots + (x_n - y_n)^2} \\ & = \sqrt{(y_1 - x_1)^2 + \cdots + (y_n - x_n)^2} \\ & = d((y_1,\ldots,y_n),(x_1,\ldots,x_n)) \end{align} \]
3. \( \) \[ \begin{align} & d((x_1,\ldots,x_n),(y_1,\ldots,y_n)) + d((y_1,\ldots,y_n),(z_1,\ldots,z_n)) \\ & = \sqrt{(x_1 - y_1)^2 + \cdots + (x_n - y_n)^2} + \sqrt{(y_1 - z_1)^2 + \cdots + (y_n - z_n)^2} \\ & \geq \sqrt{((x_1 - y_1) + (y_1 - z_1))^2 + \cdots + ((x_n - y_n) + (y_n - z_n))^2} & (\text{Minkowski Inequality}) \\ & = \sqrt{(x_1 - z_1)^2 + \cdots + (x_n - z_n)^2} \\ & = d((x_1,\ldots,x_n),(z_1,\ldots,z_n)) \end{align} \]

We now prove \(d'\) is a metric.

1. \(d'((x_1,\ldots,x_n),(y_1,\ldots,y_n)) \geq 0\) immediately from the definition, and \[ \begin{align} d'((x_1,\ldots,x_n),(y_1,\ldots,y_n)) = 0 & \Leftrightarrow \max(|x_1 - y_1|,\ldots,|x_n - y_n|) = 0 \\ & \Leftrightarrow |x_1 - y_1| = 0 \text{ and } \cdots \text{ and } |x_n - y_n| = 0 \\ & \Leftrightarrow (x_1,\ldots,x_n) = (y_1,\ldots,y_n) \end{align} \]
2. \[ \begin{align} d'((x_1,\ldots,x_n),(y_1,\ldots,y_n)) & = \max(|x_1-y_1|,\ldots,|x_n-y_n|) \\ & = \max(|y_1 - x_1|,\ldots,|y_n-x_n|) \\ & = d'((y_1, \ldots, y_n),(x_1, \ldots, x_n)) \end{align} \]
3. Let \(D = d'((x_1,\ldots,x_n),(y_1,\ldots,y_n)) + d'((y_1,\ldots,y_n),(z_1,\ldots,z_n)) \) \(= \max(|x_1-y_1|,\ldots,|x_n-y_n|) + \max(|y_1-z_1|,\ldots,|y_n-z_n|)\). Then we have \( D \geq |x_i - y_i| + |y_i - z_i| \geq |x_i - z_i| \) for each \(i\). Hence \(D \geq \max(|x_1-z_1|,\ldots,|x_n - z_n|) = d'((x_1,\ldots,x_n),(z_1,\ldots,z_n)) \) as required.

Now, we prove that every open \(d\)-ball with centre \((x_1,\ldots,x_n)\) includes an open \(d'\)-ball with centre \((x_1,\ldots,x_n)\). Let \(\epsilon > 0\) and let \(\delta = \epsilon / \sqrt{n}\). We will prove that \(B^{d'}_\delta((x_1,\ldots,x_n)) \subseteq B^d_\epsilon((x_1,\ldots,x_n))\). Let \((y_1,\ldots,y_n) \in B^{d'}_\delta((x_1,\ldots,x_n)) \). Then we have \(d'((x_1,\ldots,x_n),(y_1,\ldots,y_n)) < \delta \) and so \( |x_i-y_i| < \delta \) for each \(i\). Hence \[ \begin{align} d((x_1,\ldots,x_n),(y_1,\ldots,y_n)) & = \sqrt{(x_1 - y_1)^2 + \cdots + (x_n - y_n)^2} \\ & < \sqrt{\delta^2 + \cdots + \delta^2} \\ & = \delta \sqrt{n} \\ & = \epsilon \end{align} \]

Conversely, we prove that every open \(d'\)-ball with centre \((x_1,\ldots,x_n)\) includes an open \(d\)-ball with centre \((x_1,\ldots,x_n)\). Let \(\epsilon > 0\). We will prove that \(B^d_\epsilon((x_1,\ldots,x_n)) \subseteq B^{d'}_\epsilon((x_1,\ldots,x_n)) \). Let \((y_1,\ldots,y_n) \in B^d_\epsilon((x_1,\ldots,x_n))\). Then we have \( (x_i-y_i)^2 \leq (x_1-y_1)^2 + \cdots + (x_n-y_n)^2 = d((x_1,\ldots,x_n),(y_1,\ldots,y_n))^2 < \epsilon^2 \) for each \(i\). Hence \( d'((x_1,\ldots,x_n),(y_1,\ldots,y_n)) = \max(|x_1-y_1|,\ldots,|x_n-y_n|) < \epsilon \) as required. \(\Box\)

Proposition 7 Let \(d\) be a metric on \(X\). Let \(d'(x,y) = d(x,y)/(1+d(x,y))\). Then \(d'\) is a metric on \(X\) that induces the same topology as \(d\).

Proof We first prove \(d'\) is a metric on \(X\).

1. We have \(d'(x,y) \geq 0\) immediately from the definition and \(d'(x,y) = 0\) iff \(d(x,y) = 0\) iff \(x=y\).
2. \(d'(x,y) = d(x,y)/(1+d(x,y)) = d(y,x)/(1+d(y,x)) = d'(y,x) \)
3. \( \) \[ \begin{align} d(x,z) & \leq d(x,y) + d(y,z) \\ & \therefore d(x,z) + d(x,z)d(y,z) + d(x,z)d(x,y) + d(x,z)d(x,y)d(y,z) \\ & \leq d(x,y) + d(y,z) + d(x,y)d(y,z) + d(x,z)d(x,y) + d(x,z)d(x,y)d(y,z) \\ & \leq d(x,y) + d(x,z)d(y,z) + d(x,z)d(x,y) + 2 d(x,y)d(y,z) + 2 d(x,y)d(y,z)d(x,z) \\ \therefore d(x,z)(1 + d(x,y))(1 + d(y,z)) & \leq d(x,y)(1 + d(y,z))(1+d(x,z)) + d(y,z)(1 + d(x,z))(1 + d(x,y)) \\ \therefore d'(x,z) & \leq d'(x,y) + d'(y,z) \end{align} \]

Now, we prove every \(d\)-ball with centre \(x\) includes a \(d'\)-ball with centre \(x\). Let \(\epsilon > 0\). Let \(\delta = \epsilon / (1 + \epsilon) \). We will prove \(B^{d'}_\delta(x) \subseteq B^d_\epsilon(x)\).

Let \(y \in B^{d'}_\delta(x)\). Then \[ \begin{align} d'(x,y) & < \delta \\ \therefore d(x,y) / (1 + d(x,y)) & < \delta \\ \therefore 1 - 1 / (1 + d(x,y)) & < \delta \\ \therefore 1 / (1 + d(x,y)) & > 1 - \delta \\ \therefore 1 + d(x,y) & < 1 / (1 - \delta) \\ & = 1 + \epsilon \\ \therefore d(x,y) & < \epsilon \end{align} \] as required.

Conversely, we prove that every \(d'\)-ball with centre \(x\) includes a \(d\)-ball with centre \(x\). Let \(\epsilon > 0\). Let \(\delta = \epsilon/(1-\epsilon)\) if \(\epsilon < 1\), or \(\delta = 1\) otherwise. In either case we have \(1/(1+\delta) \geq 1 - \epsilon \) and so, if \(d(x,y) < \delta\), then \[ \begin{align} d'(x,y) & = 1 - 1/(1 + d(x,y)) \\ & < 1 - 1/(1 + \delta) \\ & \leq \epsilon \end{align} \] Thus \( B^d_\delta(x) \subseteq B^{d'}_\epsilon(x) \) as required. \(\Box\)

Definition 8 A topological space is metrizable iff there exists a metric that induces its topology.

Corollary 9 For any metrizable space, there exists a bounded metric that induces its topology (i.e. a metric \(d'\) such that there exists \(B\) such that \(d'(x,y) < B\) for all \(x\), \(y\)).

Proof Take \(d'\) as in Proposition 7 and observe that \(d'(x,y) < 1\) for all \(x\), \(y\). \(\Box\)