# 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)

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 $f|C:C\to Y$.