[Munkres Topology] 20. The Metric Topology

 This article is one of the posts in the Textbook Commentary Project.

Definition. A metric on a set $X$ is a function $$d:X\times X\rightarrow R$$ having the following properties:

(1) $d(x,y)\ge 0$ for all $x,y \in X$; equality holds if and only if $x=y$.

(2) $d(x,y)=d(y,x)$ for all $x,y\in X$.

(3) (Triangle inequality) $d(x,y)+d(y,z)\ge d(x,z)$, for all $x,y,z\in X$.

Given a metric $d$ on $X$, the number $d(x,y)$ is often called the distance between $x$ and $y$ in the metric $d$. Given $\epsilon>0$, consider the set $$B_d(x,\epsilon)=\{y| d(x,y)<\epsilon\}$$ of all points $y$ whose distance from $x$ is less than $\epsilon$. It is called the $\epsilon$-ball centered at $x$. Sometimes we omit the metric $d$ from the notation and write this ball simply as $B(x,\epsilon)$, when no confusion will arise.

Definition. If $d$ is a metric on the set $X$, then the collection of all $\epsilon$-balls $B_d(x,\epsilon)$, for $x\in X$ and $\epsilon>0$, is a basis for a topology on $X$, called the metric topology induced by $d$.

The first condition for a basis is trivial, since $x\in B(x,\epsilon)$ for any $\epsilon >0$. Before checking the second condition for a basis, we show that if $y$ is a point of the basis element $B(x,\epsilon)$. Define $\delta$ to be the positive number $\epsilon-d(x,y)$. Then $B(y,\delta)\subset B(x,\epsilon)$, for if $z\in B(y,\delta)$, then $d(y,z)<\epsilon-d(x,y)$, from which we conclude that $$d(x,z)\le d(x,y)+d(y,z)<\epsilon.$$ See Figure 20.1.

Now check the second condition for a basis, let $B_1$ and $B_2$ be two basis elements and let $y\in B_1 \cap B_2$. We have just shown that we can choose positive numbers $\delta_1$ and $\delta_2$ so that $B(y,\delta_1)\subset B_1$ and $B(y,\delta_2)\subset B_2$. Letting $\delta$ be the smaller of $\delta_1$ and $\delta_2$, we conclude that $B(y,\delta)\subset B_1\cap B_2$.

Using what we have just proved, we can rephrase the definition of the metric topology as follows:

A set $U$ is open in the metric topology induced by $d$ is and only if for each $y\in U$, there is a $\delta>0$ such that $B_d(y,\delta)\subset U$.

Clearly this condition implies that $U$ is open. Conversely, if $U$ is open, it contains a basis element $B=B_d(x,\epsilon)$ containing $y$, and $B$ in turn contains a basis element $B_d(y,\delta)$ centered at $y$.

Definition. If $X$ is a topological space, $X$ is said to be metrizable if there exists a metric $d$ on the set $X$ that induces the topology of $X$. A metric space is a metrizable space $X$ together with a specific metric $d$ that gives the topology of $X$.

Definition. Let $X$ be a metric space with metric $d$. A subset $A$ of $X$ is said to be bounded if there is some number $M$ such that $$d(a_1,a_2)\le M$$ for every pair $a_1,a_2$ of points of $A$. If $A$ is bounded and nonempty, the diameter of $A$ is defined to be the number $$\mbox{diam }A=\sup \{d(a_1,a_2)|a_1,a_2\in A\}.$$

Boundedness of a set is a topological property, for it depends on the particular metric $d$ that is used for $X$. For instance, if $X$ is a metric space with metric $d$, then there exists a metric $\bar{d}$ that gives the topology of $X$, relative to which every subset of $X$ is bounded. It is defined as follows:

Theorem 20.1. Let $X$ be a metric space with metric $d$. Define $\bar{d}:X\times X\rightarrow \mathbb{R}$ by the equation $$\bar{d}(x,y)=\min \{d(x,y),1\}.$$ Then $\bar{d}$ is a metric that induces the same topology as $d$.

The metric $\bar{d}$ is called the standard bounded metric corresponding to $d$.

Proof. Checking the first two conditions for a metric is trivial. Let us check the triangle inequality: $$\bar{d}(x,z)\le \bar{d}(x,y)+\bar{d}(y,z).$$ Now if either $d(x,y)\le 1$ or $d(y,z)\le 1$, then the right side of this inequality is at least $1$; since the left side is (by definition) at most $1$, the inequality holds. It remains to consider the case in which $d(x,y)<1$ and $d(y,z)<1$. In this case, we have $$d(x,z)\le d(x,y)+d(y,z)=\bar{d}(x,y)+\bar{d}(y,z).$$ Since $\bar{d}(x,z)\le d(x,z)$ by definition, the triangle inequality holds for $\bar{d}$.

Now we note that in any metric space, the collection of $\epsilon$-balls with $\epsilon<1$ forms a basis for the metric topology, for every basis element containing $x$ contains such an $\epsilon$-ball centered at $x$. It follows that $d$ and $\bar{d}$ induced the same topology on $X$, because the collection of $\epsilon$-balls with $\epsilon<1$ under these two metrics are the same collection.

Definition. Given $\mathbf{x}=(x_1,\cdots,x_n)$ in $\mathbb{R}^n$, we define the norm of $\mathbf{x}$ by the equation $$\lVert x\rVert = (x_1^2+\cdots+x_n^2)^{1/2};$$ and we define the euclidean metric $d$ on $\mathbb{R}^n$ by the equation $$d(\mathbf{x},\mathbf{y})=\lVert \mathbf{x}-\mathbf{y}\rVert = [(x_1-y_1)^2+\cdots+(x_n-y_n)^2]^{1/2}.$$ We define the square metric $\rho$ by the equation $$\rho(\mathbf{x},\mathbf{y})=\max \{ \left| x_1-y_1\right| ,\cdots,\left| x_n - y_n\right|\}.$$

The proof that $d$ is a metric requires some work; it is probably already familiar to you.

To show that $\rho$ is a metric is easier. Only the triangle inequality is nontrivial. From the triangle inequality for $\mathbb{R}$ it follows that for each positive integer $i$, $$\left| x_i - z_i\right| \le \left| x_i - y_i\right| + \left| y_i - z_i\right|.$$ The by definition of $\rho$, $$\left| x_i - z_i\right| \le \rho (\mathbf{x},\mathbf{y})+\rho(\mathbf{y},\mathbf{z}).$$ As a resulf $$\rho(\mathbf{x},\mathbf{z})=\max \{\left| x_i -z_i \right|\} \le \rho (\mathbf{x},\mathbf{y})+\rho(\mathbf{y},\mathbf{z}),$$ as desired.

Lemma 20.2. Let $d$ and $d'$ be two metrics on the set $X$; let $\mathcal{T}$ and $\mathcal{T}'$ be the topologies they induce, respectively. Then $\mathcal{T}'$ is finer than $\mathcal{T}$ if and only if for each $x$ in $X$ and each $\epsilon >0$, there exists a $\delta>0$ such that $$B_{d'}(x,\delta)\subset B_d(x,\epsilon).$$

Proof. Suppose that $\mathcal{T}'$ is finer than $\mathcal{T}$. Given the basis element $B_d(x,\epsilon)$ for $\mathcal{T}$, there is by Lemma 13.3 a basis element $B'$ for the topology $\mathcal{T}'$ such that $x\in B'\subset B_d(x,\epsilon)$. Within $B'$ we can find a ball $B_{d'}(x,\delta)$ centered at $x$. 

Conversely, suppose the $\delta-\epsilon$ condition holds. Given a basis element $B$ for $\mathcal{T}$ containing $x$, we can find within $B$ a ball $B_d(x,\epsilon)$ centered at $x$. By the given condition, there is a $\delta$ such that $B_{d'}(x,\delta)\subset B_d(x,\epsilon)$. The Lemma 13.3 applies to show $\mathcal{T}'$ is finer than $\mathcal{T}$.

Theorem 20.3. The topologies on $\mathbb{R}^n$ induced by the euclidean metric $d$ and the square metric $\rho$ are the same as the product topology on $\mathbb{R}^n$.

Proof. Let $\mathbf{x}=(x_1,\cdots,x_n)$ and $\mathbf{y}=(y_1,\cdots,y_n)$ be two points of $\mathbb{R}^n$. It is simple algebra to check that $$\rho(\mathbf{x},\mathbf{y})\le d(\mathbf{x},\mathbf{y})\le \sqrt{n} \rho (\mathbf{x},\mathbf{y}).$$ The first inequality shows that $$B_d(\mathbf{x},\epsilon)\subset B_\rho (\mathbf{x},\epsilon)$$ for all $\mathbf{x}$ and $\epsilon$, since if $d(\mathbf{x},\mathbf{y})<\epsilon$, then $\rho(\mathbf{x},\mathbf{y})<\epsilon$ also. Similarly, the second inequality shows that $$B_\rho (\mathbf{x},\epsilon/\sqrt{n})\subset B_d(\mathbf{x},\epsilon)$$ for all $\mathbf{x}$ and $\epsilon$. It follows from the preceding lemma that the two metric topologies are the same.

Now we show that the product topology is the same as that given by the metric $\rho$. First, let $$B=(a_1,b_1)\times \cdots \times (a_n,b_n)$$ be a basis element for the product topology, and let $\mathbf{x}=(x_1,\cdots,x_n)$ be an element of $B$. For each $i$, there is an $\epsilon_i$ such that $$(x_i-\epsilon_i, x_i+\epsilon_i)\subset (a_i,b_i);$$ choose $\epsilon=\min\{\epsilon_1,\cdots,\epsilon_n\}$. Then $B_\rho(\mathbf{x},\epsilon)\subset B$,a as you can readily check. As a result, the $\rho$-topology is finer than the product topology.

Conversely, let $B_\rho(\mathbf{x},\epsilon)$ be a basis element for the $\rho$-topology. Given the element $\mathbf{y}\in B_\rho (\mathbf{x},\epsilon)$, we need to find a basis element $B$ for the product topology such that $\mathbf{y}\in B \subset B_\rho (\mathbf{x},\epsilon).$$ But this is trivial, for $$B_\rho(\mathbf{x},\epsilon)=(x_1-\epsilon,x_1+\epsilon)\times \cdots \times (x_n-\epsilon,x_n+\epsilon)$$ is itself a basis element for the product topology.

Now we consider the infinite cartesian product $\mathbb{R}^\omega$. It is natural to try to generalize the metrics $d$ and $\rho$ to this space.

Definition. Given an index set $J$, and given points $\mathbf{x}=(x_\alpha)_{\alpha \in J}$ and $\mathbf{y}=(y_\alpha)_{\alpha\in J}$ of $\mathbb{R}^J$, let us define a metric $\bar{\rho}$ on $\mathbb{R}^J$ by the equation $$\bar{\rho}(\mathbf{x},\mathbf{y})=\sup \{ \bar{d}(x_\alpha, y_\alpha) | \alpha \in J\},$$ where $\bar{d}$ is the standard bounded metric on $\mathbb{R}$. It is east to check that $\bar{\rho}$ is indeed a metric; it is called the uniform metric on $\mathbb{R}^J$, and the topology it induces is called the uniform topology.

Theorem 20.4. The uniform topology on $\mathbf{R}^J$ is finer than the product topology and coarser than the box topology; these three topologies are all different if $J$ is infinite.

Proof. 

[Uniform $\supset$ Product] Suppose that we are given a point $\mathbf{x}=(x_\alpha)_{\alpha \in J}$ and a product topology basis element $\prod U_\alpha$ about $\mathbf{x}$. Let $\alpha_1,\cdots,\alpha_n$ be the indices for which $U_\alpha\ne \mathbb{R}$. Then for each $i$, choose $\epsilon_i >0$ so that the $\epsilon_i$-ball centered at $x_{\alpha_i}$ in the $\bar{d}$ metric is contained in $U_{\alpha_i}$: this we can do because $U_{\alpha_i}$ is open in $\mathbb{R}$. Let $\epsilon=\min\{ \epsilon_1,\cdots,\epsilon_n\}$; then the $\epsilon$-ball centered at $\mathbf{x}$ in the $\bar{\rho}$ metric is contained in $\prod U_\alpha$. For if $\mathbf{z}$ is a point of $\mathbb{R}^J$ such that $\bar{\rho}(\mathbf{x},\mathbf{z})<\epsilon$, then $\bar{d}(x_\alpha,z_\alpha)<\epsilon$ for all $\alpha$, so that $\mathbf{z} \in \prod U_\alpha$. It follows that the uniform topology is finer than the product topology.

[Uniform $\subset$ Box] On the other hand, let $B$ be the $\epsilon$-ball centered at $\mathbf{x}$ in the $\bar{\rho}$ metric. Then the box neighborhood $$U=\prod (x_\alpha - \frac{1}{2} \epsilon, x_\alpha+ \frac{1}{2}\epsilon)$$ of $\mathbf{x}$ is contained in $B$. For if $\mathbf{y}\in U$, then $\bar{d}(x_\alpha,y_\alpha)<\frac{1}{2} \epsilon$ for all $\alpha$, so that $\bar{\rho}(\mathbf{x},\mathbf{y})\le \frac{1}{2} \epsilon$.

In the case where $J$ is infinite, we still have not determined whether $\mathbb{R}^J$ is metrizable in either the box or the product topology. It turns out that the only one of these cases where $\mathbb{R}^J$ is metrizable is the case where $J$ is countable and $\mathbf{R}^J$ has the product topology. As we shall see.

Theorem 20.5. Let $\bar{d}(a,b)=\min \{\left| a-b\right|,1\}$ be the standard bounded metric on $\mathbb{R}$. If $\mathbf{x}$ and $\mathbf{y}$ are two points of $\mathbb{R}^\omega$, define $$D(\mathbf{x},\mathbf{y})=\sup \{ \frac{\bar{d}(x_i,y_i)}{i}\}.$$ Then $D$ is a metric that induces the product topology on $\mathbb{R}^\omega$.

Proof. The properties of a metric are satisfied trivially except for the triangle inequality, which is proved by noting that for all $i$, $$\frac{\bar{d}(x_i,z_i)}{i}\le \frac{\bar{d}(x_i,y_i)}{i}+\frac{\bar{d}(y_i,z_i)}{i}\le D(\mathbf{x},\mathbf{y})+D(\mathbf{y},\mathbf{z}).$$ so that $$\sup \{ \frac{\bar{d}(x_i,z_i)}{i}\} \le D(\mathbf{x},\mathbf{y})+D(\mathbf{y},\mathbf{z}).$$

The fact that $D$ gives the product topology requires a little more work. First, let $U$ be open in the metric topology and let $\mathbf{x}\in U$; we find and open set $V$ in the product topology such that $\mathbf{x}\in V\subset U$. Choose an $\epsilon$-ball $B_D(\mathbf{x},\epsilon)$ lying in $U$. Then choose $N$ large enough that $1/N<\epsilon$. Finally, let $V$ be the basis element for the product topology $$V=(x_1-\epsilon,x_1+\epsilon)\times \cdots \times (x_N-\epsilon,x_N+\epsilon)\times \mathbb{R}\times \mathbb{R}\times \cdots.$$ We assert that $V\subset B_D(\mathbf{x},\epsilon)$: Given any $\mathbf{y}$ in $\mathbb{R}^\omega$, $$\frac{\bar{d}(x_i,y_i)}{i}\le \frac{1}{N} \quad \mbox{for } i \ge N.$$ Therefore, $$D(\mathbf{x},\mathbf{y})\le \max \{ \frac{\bar{d}(x_1,y_1)}{1},\cdots,\frac{\bar{d}(x_N,y_N)}{N},\frac{1}{N}\}.$$ If $\mathbf{y}$ is in $V$m this expression is less than $\epsilon$, so that $V\subset B_D(\mathbf{x},\epsilon)$, as desired.

Conversely, consider a basis element $$U=\prod_{i\in \mathbb{Z}_+} U_i$$ for the product topology, where $U_i$ is open in $\mathbb{R}$ for $i=\alpha_1,\cdots,\alpha_n$ and $U_i=\mathbb{R}$ for all other indices $i$. Given $\mathbf{x}\in U$, we find an open set $V$ of the metric topology such that $\mathbf{x}\in V\subset U$. Choose an interval $(x_i-\epsilon_i,x_i + \epsilon_i)$ in $\mathbb{R}$ centered about $x_i$ and lying in $U_i$ for $i=\alpha_1,\cdots,\alpha_n$; choose each $\epsilon_i \le 1$. Then define $$\epsilon = \min \{ \epsilon_i /i | i=\alpha_1,\cdots,\alpha_n\}.$$ We assert that $$\mathbf{x}\in B_D(\mathbf{x},\epsilon)\subset U.$$ Let $\mathbf{y}$ be a point of $B_D(\mathbf{x},\epsilon)$. Then for all $i$, $$\frac{\bar{d}(x_i,y_i)}{i}\le D(\mathbf{x},\mathbf{y})<\epsilon.$$ Now if $i=\alpha_1,\cdots,\alpha_n$, then $\epsilon \le \epsilon_i/i$, so that $\bar{d}(x_i,y_i)<\epsilon_i\le 1$; it follows that $\left| x_i-y_i\right| < \epsilon_i$. Therefore, $\mathbf{y}\in \prod U_i$, as desicred.