The uniform metric on product spaces

Jordan Bell

April 3, 2014

1 Metric topology

If $(X,d)$ is a metric space, $a\in X$ , and $r>0$ , then the open ball with center $a$ and radius $r$ is

B_{r}^{d}(a)=\{x\in X:d(x,a)<r\}.

The set of all open balls is a basis for the metric topology induced by $d$ .

If $(X,d)$ is a metric space, define

\overline{d}(a,b)=d(a,b)\wedge 1,\qquad a,b\in X,

where $x\wedge y=\min\{x,y\}$ . It is straightforward to check that $\overline{d}$ is a metric on $X$ , and one proves that $d$ and $\overline{d}$ induce the same metric topologies.¹¹ 1 James Munkres, Topology, second ed., p. 121, Theorem 20.1. The diameter of a subset $S$ of a metric space $(X,d)$ is

\mathrm{diam}(S,d)=\sup_{a,b\in S}d(a,b).

The subset $S$ is said to be bounded if its diameter is finite. The metric space $(X,d)$ might be unbounded, but the diameter of the metric space $(X,\overline{d})$ is

\mathrm{diam}(X,\overline{d})=\sup_{a,b\in X}\overline{d}(a,b)=\mathrm{diam}(X% ,d)\wedge 1,

and thus the metric space $(X,\overline{d})$ is bounded.

2 Product topology

If $J$ is a set and $X_{j}$ are topological spaces for each $j\in J$ , let $X=\prod_{j\in J}X_{j}$ and let $\pi_{j}:X\to X_{j}$ be the projection maps. A basis for the product topology on $X$ are those sets of the form $\bigcap_{j\in J_{0}}\pi_{j}^{-1}(U_{j})$ , where $J_{0}$ is a finite subset of $J$ and $U_{j}$ is an open subset of $X_{j}$ , $j\in J_{0}$ . Equivalently, the product topology is the initial topology for the projection maps $\pi_{j}:X\to X_{j}$ , $j\in J$ , i.e. the coarsest topology on $X$ such that each projection map is continuous. Each of the projection maps is open.²² 2 John L. Kelley, General Topology, p. 90, Theorem 2. The following theorem characterizes convergent nets in the product topology.³³ 3 John L. Kelley, General Topology, p. 91, Theorem 4.

Theorem 1.

Let $J$ be a set and for each $j\in J$ let $X_{j}$ be a topological space. If $X=\prod_{j\in J}X_{j}$ has the product topology and $(x_{\alpha})_{\alpha\in I}$ is a net in $X$ , then $x_{\alpha}\to x$ if and only if $\pi_{j}(x_{\alpha})\to\pi_{j}(x)$ for each $j\in J$ .

Proof.

Let $(x_{\alpha})_{\alpha\in I}$ be a net that converges to $x\in X$ . Because each projection map is continuous, if $j\in J$ then $\pi_{j}(x_{\alpha})\to\pi_{j}(x)$ . On the other hand, suppose that $(x_{\alpha})_{\alpha\in I}$ is a net, that $x\in X$ , and that $\pi_{j}(x_{\alpha})\to\pi_{j}(x)$ for each $j\in J$ . Let $\mathscr{O}_{j}$ be the set of open neighborhoods of $\pi_{j}(x)\in X_{j}$ . For $j\in J$ and $U\in\mathscr{O}_{j}$ , because $\pi_{j}(x_{\alpha})\to\pi_{j}(x)$ we have that $\pi_{j}(x_{\alpha})$ is eventually in $U$ . It follows that if $j\in J$ and $U\in\mathscr{O}_{j}$ then $x_{\alpha}$ is eventually in $\pi_{j}^{-1}(U)$ . Therefore, if $J_{0}$ is a finite subset of $J$ and $U_{j}\in\mathscr{O}_{j}$ for each $j\in J_{0}$ , then $x_{\alpha}$ is eventually in $\bigcap_{j\in J_{0}}\pi_{j}^{-1}(U_{j})$ . This means that the net $(x_{\alpha})_{\alpha\in I}$ is eventually in every basic open neighborhood of $x$ , which implies that $x_{\alpha}\to x$ . ∎

The following theorem states that if $J$ is a countable set and $(X,d)$ is a metric space, then the product topology on $X^{J}$ is metrizable.⁴⁴ 4 James Munkres, Topology, second ed., p. 125, Theorem 20.5.

Theorem 2.

If $J$ is a countable set and $(X,d)$ is a metric space, then

\rho(x,y)=\sup_{j\in J}\frac{\overline{d}(x_{j},y_{j})}{j}=\sup_{j\in J}\frac{% d(x_{j},y_{j})\wedge 1}{j}

is a metric on $X^{J}$ that induces the product topology.

A topological space is first-countable if every point has a countable local basis; a local basis at a point $p$ is a set $\mathscr{B}$ of open sets each of which contains $p$ such that each open set containing $p$ contains an element of $\mathscr{B}$ . It is a fact that a metrizable topological space is first-countable. In the following theorem we prove that the product topology on an uncountable product of Hausdorff spaces each of which has at least two points is not first-countable.⁵⁵ 5 cf. John L. Kelley, General Topology, p. 92, Theorem 6. From this it follows that if $(X,d)$ is a metric space with at least two points and $J$ is an uncountable set, then the product topology on $X^{J}$ is not metrizable.

Theorem 3.

If $J$ is an uncountable set and for each $j\in J$ we have that $X_{j}$ is a Hausdorff space with at least two points, then the product topology on $\prod_{j\in J}X_{j}$ is not first-countable.

Proof.

Write $X=\prod_{j\in J}X_{j}$ , and suppose that $x\in X$ and that $U_{n},n\in\mathbb{N}$ , are open subsets of $X$ containing $x$ . Since $U_{n}$ is an open subset of $X$ containing $x$ , there is a basic open set $B_{n}$ satisfying $x\in B_{n}\subseteq U_{n}$ : by saying that $B_{n}$ is a basic open set we mean that there is a finite subset $F_{n}$ of $J$ and open subsets $U_{n,j}$ of $X_{j}$ , $j\in F_{n}$ , such that

B_{n}=\bigcap_{j\in F_{n}}\pi_{j}^{-1}(U_{n,j}).

Let $F=\bigcup_{n\in\mathbb{N}}F_{n}$ , and because $J$ is uncountable there is some $k\in J\setminus F$ ; this is the only place in the proof at which we use that $J$ is uncountable. As $X_{k}$ has at least two points and $x(k)\in X_{k}$ , there is some $a\in X_{k}$ with $x(k)\neq a$ . Since $X_{k}$ is a Hausdorff space, there are disjoint open subsets $N_{1},N_{2}$ of $X_{k}$ with $x(k)\in N_{1}$ and $a\in N_{2}$ . Define

V_{j}=\begin{cases}N_{1}&j=k\\ X_{j}&j\neq k\end{cases}

and let $V=\prod_{j\in J}V_{j}$ . We have $x\in V$ . But for each $n\in\mathbb{N}$ , there is some $y_{n}\in B_{n}$ with $y_{n}(k)=a\in N_{2}$ , hence $y_{n}(k)\not\in N_{1}$ and so $y_{n}\not\in V$ . Thus none of the sets $B_{n}$ is contained in $V$ , and hence none of the sets $U_{n}$ is contained in $V$ . Therefore $\{U_{n}:n\in\mathbb{N}\}$ is not a local basis at $x$ , and as this was an arbitrary countable set of open sets containing $x$ , there is no countable local basis at $x$ , showing that $X$ is not first-countable. (In fact, we have proved there is no countable local basis at any point in $X$ ; not to be first-countable merely requires that there be at least one point at which there is no countable local basis.) ∎

3 Uniform metric

If $J$ is a set and $(X,d)$ is a metric space, we define the uniform metric on $X^{J}$ by

d_{J}(x,y)=\sup_{j\in J}\overline{d}(x_{j},y_{j})=\sup_{j\in J}d(x_{j},y_{j})% \wedge 1.

It is apparent that $d_{J}(x,y)=0$ if and only if $x=y$ and that $d_{J}(x,y)=d_{J}(y,x)$ . If $x,y,z\in X$ then,

$\displaystyle d_{J}(x,z)$	$\displaystyle=$	$\displaystyle\sup_{j\in J}\overline{d}(x_{j},z_{j})$
	$\displaystyle\leq$	$\displaystyle\sup_{j\in J}\overline{d}(x_{j},y_{j})+\overline{d}(y_{j},z_{j})$
	$\displaystyle\leq$	$\displaystyle\sup_{j\in J}\overline{d}(x_{j},y_{j})+\sup_{j\in J}\overline{d}(% y_{j},z_{j})$
	$\displaystyle=$	$\displaystyle d_{J}(x,y)+d_{J}(y,z),$

showing that $d_{J}$ satisfies the triangle inequality and thus that it is indeed a metric on $X^{J}$ . The uniform topology on $X^{J}$ is the metric topology induced by the uniform metric.

If $(X,d)$ is a metric space, then $X$ is a topological space with the metric topology, and thus $X^{J}=\prod_{j\in J}X$ is a topological space with the product topology. The following theorem shows that the uniform topology on $X^{J}$ is finer than the product topology on $X^{J}$ .⁶⁶ 6 James Munkres, Topology, second ed., p. 124, Theorem 20.4.

Theorem 4.

If $J$ is a set and $(X,d)$ is a metric space, then the uniform topology on $X^{J}$ is finer than the product topology on $X^{J}$ .

Proof.

If $x\in X^{J}$ , let $U=\prod_{j\in J}U_{j}$ be a basic open set in the product topology with $x\in U$ . Thus, there is a finite subset $J_{0}$ of $J$ such that if $j\in J\setminus J_{0}$ then $U_{j}=X$ . If $j\in J_{0}$ , then because $U_{j}$ is an open subset of $(X,d)$ with the metric topology and $x_{j}\in U_{j}$ , there is some $0<\epsilon_{j}<1$ such that $B^{d}_{\epsilon_{j}}(x_{j})\subseteq U_{j}$ . Let $\epsilon=\min_{j\in J_{0}}\epsilon_{j}$ . If $d_{J}(x,y)<\epsilon$ then $d(x_{j},y_{j})<\epsilon$ for all $j\in J$ and hence $d(x_{j},y_{j})<\epsilon_{j}$ for all $j\in J_{0}$ , which implies that $y_{j}\in B^{d}_{\epsilon_{j}}(x_{j})\subseteq U_{j}$ for all $j\in J_{0}$ . If $j\in J\setminus J_{0}$ then $U_{j}=X$ and of course $y_{j}\in U_{j}$ . Therefore, if $y\in B_{\epsilon}^{d_{J}}(x)$ then $y\in U$ , i.e. $B_{\epsilon}^{d_{J}}(x)\subseteq U$ . It follows that the uniform topology on $X^{J}$ is finer than the product topology on $X^{J}$ . ∎

The following theorem shows that if we take the product of a complete metric space with itself, then the uniform metric on this product space is complete.⁷⁷ 7 James Munkres, Topology, second ed., p. 267, Theorem 43.5.

Theorem 5.

If $J$ is a set and $(X,d)$ is a complete metric space, then $X^{J}$ with the uniform metric is a complete metric space.

Proof.

It is straightforward to check that $(X,d)$ being a complete metric space implies that $(X,\overline{d})$ is a complete metric space. Let $f_{n}$ be a Cauchy sequence in $(X^{J},d_{J})$ : if $\epsilon>0$ then there is some $N$ such that $n,m\geq N$ implies that

d_{J}(f_{n},f_{m})<\epsilon.

Thus, if $\epsilon>0$ , then there is some $N$ such that $n,m\geq N$ and $j\in J$ implies that $\overline{d}(f_{n}(j),f_{m}(j))\leq d_{J}(f_{n},f_{m})<\epsilon$ . Thus, if $j\in J$ then $f_{n}(j)$ is a Cauchy sequence in $(X,\overline{d})$ , which therefore converges to some $f(j)\in X$ , and thus $f\in X^{J}$ . If $n,m\geq N$ and $j\in J$ , then

	$\displaystyle\overline{d}(f_{n}(j),f(j))$	$\displaystyle\leq\overline{d}(f_{n}(j),f_{m}(j))+\overline{d}(f_{m}(j),f(j))$
		$\displaystyle\leq d_{J}(f_{n},f_{m})+\overline{d}(f_{m}(j),f(j))$
		$\displaystyle<\epsilon+\overline{d}(f_{m}(j),f(j)).$

As the left-hand side does not depend on $m$ and $\overline{d}(f_{m}(j),f(j))\to 0$ , we get that if $n\geq N$ and $j\in J$ then

\overline{d}(f_{n}(j),f(j))\leq\epsilon.

Therefore, if $n\geq N$ then

d_{J}(f_{n},f)\leq\epsilon.

This means that $f_{n}$ converges to $f$ in the uniform metric, showing that $(X^{J},d_{J})$ is a complete metric space. ∎

4 Bounded functions and continuous functions

If $J$ is a set and $(X,d)$ is a metric space, a function $f:J\to X$ is said to be bounded if its image is a bounded subset of $X$ , i.e. $f(J)$ has a finite diameter. Let $B(J,X)$ be the set of bounded functions $J\to(X,d)$ ; $B(J,X)$ is a subset of $X^{J}$ . Since the diameter of $(X,\overline{d})$ is $\leq 1$ , any function $J\to(X,\overline{d})$ is bounded, but there might be unbounded functions $J\to(X,d)$ . We prove in the following theorem that $B(J,X)$ is a closed subset of $X^{J}$ with the uniform topology.⁸⁸ 8 James Munkres, Topology, second ed., p. 267, Theorem 43.6.

Theorem 6.

If $J$ is a set and $(X,d)$ is a metric space, then $B(J,X)$ is a closed subset of $X^{J}$ with the uniform topology.

Proof.

If $f_{n}\in B(J,Y)$ and $f_{n}$ converges to $f\in X^{J}$ in the uniform topology, then there is some $N$ such that $d_{J}(f_{N},f)<\frac{1}{2}$ . Thus, for all $j\in J$ we have $\overline{d}(f_{N}(j),f(j))<\frac{1}{2}$ , which implies that

d(f_{N}(j),f(j))=\overline{d}(f_{N}(j),f(j))<\frac{1}{2}.

If $i,j\in J$ , then

	$\displaystyle d(f(i),f(j))$	$\displaystyle\leq d(f(i),f_{N}(i))+d(f_{N}(i),f_{N}(j))+d(f_{N}(j),f(j))$
		$\displaystyle\leq\frac{1}{2}+\mathrm{diam}(f_{N}(J),d)+\frac{1}{2}.$

$f_{N}\in B(J,X)$ means that $\mathrm{diam}(f_{N}(J),d)<\infty$ , and it follows that $\mathrm{diam}(f(J),d)\leq\mathrm{diam}(f_{N}(J),d)+1<\infty$ , showing that $f\in B(J,X)$ . Therefore if a sequence of elements in $B(J,X)$ converges to an element of $X^{J}$ , that limit is contained in $B(J,X)$ . This implies that $B(J,X)$ is a closed subset of $X^{J}$ in the uniform topology, as in a metrizable space the closure of a set is the set of limits of sequences of points in the set. ∎

If $J$ is a set and $Y$ is a complete metric space, we have shown in Theorem 5 that $Y^{J}$ is a complete metric space with the uniform metric. If $X$ and $Y$ are topological spaces, we denote by $C(X,Y)$ the set of continuous functions $X\to Y$ . $C(X,Y)$ is a subset of $Y^{X}$ , and we show in the following theorem that if $Y$ is a metric space then $C(X,Y)$ is a closed subset of $Y^{X}$ in the uniform topology.⁹⁹ 9 James Munkres, Topology, second ed., p. 267, Theorem 43.6. Thus, if $Y$ is a complete metric space then $C(X,Y)$ is a closed subset of the complete metric space $Y^{X}$ , and is therefore itself a complete metric space with the uniform metric.

Theorem 7.

If $X$ is a topological space and let $(Y,d)$ is a metric space, then $C(X,Y)$ is a closed subset of $Y^{X}$ with the uniform topology.

Proof.

Suppose that $f_{n}\in C(X,Y)$ and $f_{n}\to f\in Y^{X}$ in the uniform topology. Thus, if $\epsilon>0$ then there is some $N$ such that $n\geq N$ implies that $d_{J}(f_{n},f)<\epsilon$ , and so if $n\geq N$ and $x\in X$ then

\overline{d}(f_{n}(x),f(x))\leq d_{J}(f_{n},f)<\epsilon.

This means that the sequence $f_{n}$ converges uniformly in $X$ to $f$ in the uniform metric, and as each $f_{n}$ is continuous this implies that $f$ is continuous.¹⁰¹⁰ 10 See James Munkres, Topology, second ed., p. 132, Theorem 21.6. We have shown that if $f_{n}\in C(X,Y)$ and $f_{n}\to f\in Y^{X}$ in the uniform topology then $f\in C(X,Y)$ , and therefore $C(X,Y)$ is a closed subset of $Y^{X}$ in the uniform topology. ∎

5 Topology of compact convergence

Let $X$ be a topological space and $(Y,d)$ be a metric space. If $f\in Y^{X}$ , $C$ is a compact subset of $X$ , and $\epsilon>0$ , we denote by $B_{C}(f,\epsilon)$ the set of those $g\in Y^{X}$ such that

\sup\{d(f(x),g(x)):x\in C\}<\epsilon.

A basis for the topology of compact convergence on $Y^{X}$ are those sets of the form $B_{C}(f,\epsilon)$ , $f\in Y^{X}$ , $C$ a compact subset of $X$ , and $\epsilon>0$ . It can be proved that the uniform topology on $Y^{X}$ is finer than the topology of compact convergence on $Y^{X}$ , and that the topology of compact convergence on $Y^{X}$ is finer than the product topology on $Y^{X}$ .¹¹¹¹ 11 James Munkres, Topology, second ed., p. 285, Theorem 46.7. Indeed, we have already shown in Theorem 4 that the uniform topology on $Y^{X}$ is finer than the product topology on $Y^{X}$ . The significance of the topology of compact convergence on $Y^{X}$ is that a sequence of functions $f_{n}:X\to Y$ converges in the topology of compact convergence to a function $f:X\to Y$ if and only if for each compact subset $C$ of $X$ the sequence of functions $f_{n}|C:C\to Y$ converges uniformly in $C$ to the function $f|C:C\to Y$ .