# The uniform metric on product spaces

## 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

$$ |

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

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

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

where $x\wedge y=\mathrm{min}\{x,y\}$.
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.^{1}^{1}
1
James Munkres, Topology, second ed., p. 121, Theorem 20.1.
The diameter of a subset $S$ of a metric space $(X,d)$ is

$$\mathrm{diam}(S,d)=\underset{a,b\in S}{sup}d(a,b).$$ |

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

$$\mathrm{diam}(X,\overline{d})=\underset{a,b\in X}{sup}\overline{d}(a,b)=\mathrm{diam}(X,d)\wedge 1,$$ |

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.^{2}^{2}
2
John L. Kelley, General Topology,
p. 90, Theorem 2. The following theorem characterizes convergent nets in the product topology.^{3}^{3}
3
John L. Kelley, General Topology, p. 91, Theorem 4.

###### Theorem 1.

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

###### 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 ${\mathcal{O}}_{j}$ be the set of open neighborhoods of ${\pi}_{j}(x)\in {X}_{j}$. For $j\in J$ and $U\in {\mathcal{O}}_{j}$, 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 $U\in {\mathcal{O}}_{j}$ then ${x}_{\alpha}$ is eventually in ${\pi}_{j}^{-1}(U)$. Therefore, if ${J}_{0}$ is a finite subset of $J$ and ${U}_{j}\in {\mathcal{O}}_{j}$ 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.^{4}^{4}
4
James
Munkres, Topology, second ed., p. 125, Theorem 20.5.

###### Theorem 2.

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

$$\rho (x,y)=\underset{j\in J}{sup}\frac{\overline{d}({x}_{j},{y}_{j})}{j}=\underset{j\in J}{sup}\frac{d({x}_{j},{y}_{j})\wedge 1}{j}$$ |

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 $\mathcal{B}$ of open sets each of which contains $p$ such that each open set
containing $p$ contains an element of $\mathcal{B}$.
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.^{5}^{5}
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 $j\mathrm{\in}J$ we have that ${X}_{j}$ is a Hausdorff space with at least two points, then the product topology on ${\mathrm{\prod}}_{j\mathrm{\in}J}{X}_{j}$ 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 $x(k)\ne a$. 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

$${V}_{j}=\{\begin{array}{cc}{N}_{1}\hfill & j=k\hfill \\ {X}_{j}\hfill & j\ne k\hfill \end{array}$$ |

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 ${y}_{n}(k)\notin {N}_{1}$ and so ${y}_{n}\notin V$. 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

$${d}_{J}(x,y)=\underset{j\in J}{sup}\overline{d}({x}_{j},{y}_{j})=\underset{j\in J}{sup}d({x}_{j},{y}_{j})\wedge 1.$$ |

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,

${d}_{J}(x,z)$ | $=$ | $\underset{j\in J}{sup}\overline{d}({x}_{j},{z}_{j})$ | ||

$\le $ | $\underset{j\in J}{sup}\overline{d}({x}_{j},{y}_{j})+\overline{d}({y}_{j},{z}_{j})$ | |||

$\le $ | $\underset{j\in J}{sup}\overline{d}({x}_{j},{y}_{j})+\underset{j\in J}{sup}\overline{d}({y}_{j},{z}_{j})$ | |||

$=$ | ${d}_{J}(x,y)+{d}_{J}(y,z),$ |

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 ${X}^{J}$.^{6}^{6}
6
James
Munkres, Topology, second ed., p. 124, Theorem 20.4.

###### Theorem 4.

If $J$ is a set and $\mathrm{(}X\mathrm{,}d\mathrm{)}$ 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 $$ such that ${B}_{{\u03f5}_{j}}^{d}({x}_{j})\subseteq {U}_{j}$. Let $\u03f5={\mathrm{min}}_{j\in {J}_{0}}{\u03f5}_{j}$. If $$ then $$ for all $j\in J$ and hence $$ for all $j\in {J}_{0}$, which implies that ${y}_{j}\in {B}_{{\u03f5}_{j}}^{d}({x}_{j})\subseteq {U}_{j}$ 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 $y\in {B}_{\u03f5}^{{d}_{J}}(x)$ then $y\in U$, i.e. ${B}_{\u03f5}^{{d}_{J}}(x)\subseteq U$. 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.^{7}^{7}
7
James Munkres, Topology, second ed., p. 267, Theorem 43.5.

###### Theorem 5.

If $J$ is a set and $\mathrm{(}X\mathrm{,}d\mathrm{)}$ 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 $\u03f5>0$ then there is some $N$ such that $n,m\ge N$ implies that

$$ |

Thus, if $\u03f5>0$, then there is some $N$ such that $n,m\ge N$ and $j\in J$ implies that $$. 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 $n,m\ge N$ and $j\in J$, then

$\overline{d}({f}_{n}(j),f(j))$ | $\le \overline{d}({f}_{n}(j),{f}_{m}(j))+\overline{d}({f}_{m}(j),f(j))$ | ||

$\le {d}_{J}({f}_{n},{f}_{m})+\overline{d}({f}_{m}(j),f(j))$ | |||

$$ |

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

$$\overline{d}({f}_{n}(j),f(j))\le \u03f5.$$ |

Therefore, if $n\ge N$ then

$${d}_{J}({f}_{n},f)\le \u03f5.$$ |

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 $\le 1$, 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.^{8}^{8}
8
James Munkres,
Topology, second ed., p. 267, Theorem 43.6.

###### Theorem 6.

If $J$ is a set and $\mathrm{(}X\mathrm{,}d\mathrm{)}$ is a metric space, then $B\mathit{}\mathrm{(}J\mathrm{,}X\mathrm{)}$ 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 $$. Thus, for all $j\in J$ we have $$, which implies that

$$ |

If $i,j\in J$, then

$d(f(i),f(j))$ | $\le d(f(i),{f}_{N}(i))+d({f}_{N}(i),{f}_{N}(j))+d({f}_{N}(j),f(j))$ | ||

$\le {\displaystyle \frac{1}{2}}+\mathrm{diam}({f}_{N}(J),d)+{\displaystyle \frac{1}{2}}.$ |

${f}_{N}\in B(J,X)$ means that $$, and it follows that $$, 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.^{9}^{9}
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 $\mathrm{(}Y\mathrm{,}d\mathrm{)}$ is a metric space, then $C\mathit{}\mathrm{(}X\mathrm{,}Y\mathrm{)}$ 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 $\u03f5>0$ then there is some $N$ such that $n\ge N$ implies that $$, and so if $n\ge N$ and $x\in X$ then

$$ |

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.^{10}^{10}
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 $\u03f5>0$, we denote by ${B}_{C}(f,\u03f5)$ the set of those $g\in {Y}^{X}$ such that

$$ |

A basis for the topology of compact convergence on ${Y}^{X}$ are those sets of the form ${B}_{C}(f,\u03f5)$,
$f\in {Y}^{X}$, $C$ a compact subset of $X$, and $\u03f5>0$. 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 ${Y}^{X}$.^{11}^{11}
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 $f|C:C\to Y$.