Uniform Spaces

Gustave Jean Jacquet (1846-1909) was a French painter.

We have sections

Uniformities and the Uniform Topology
Uniform Continuity
Metrization
Completeness
Compact spaces

The property of being a Cauchy sequence is not a topological invariant. For example, the map $f$ defined by $f(x)=1/x$ is a homeomorphism of the space of positive real numbers onto itself which carries the Cauchy sequence $\{1/(n+1):n\in\omega\}$ into the non-Cauchy sequence $\{n+1:n\in\omega\}$. However, it is possible to derive the topological results from the statements about Cauchy sequences. For example, a subset $A$ of $\mathbb{R}$ is closed if and only if each Cauchy sequence in $A$ converges to some point of $A$. The reverse sort of implication may also occur; thus, each continuous function on a compact metric space is uniformly continuous. In this case we deduce from a topological premise (that the space is compact) a non-topological conclusion (that a function is uniformly continuous).

A sequence $\{x_{n}\}_{n=1}^{\infty}$ in a pseudo-metric space $(X,d)$ is called a Cauchy sequence if and only if $d(x_{m},x_{n})$ converges to zero as $m$ and $n$ become large. This notion is not meaningful in an arbitrary topological space. In order to define a Cauchy sequence, it is necessary to know, in some sense, for what pairs the distance $d(x,y)$ is small. This statement may be made precise in the following way. If

\[V_{d,r}=\{(x,y):d(x,y)<r\},\]

then $\{x_{n}\}_{n=1}^{\infty}$ is a Cauchy sequence if and only if, for each positive $r$, it is true that $(x_{m},x_{n})$ is a member of $V_{d,r}$ for the large $m$ and $n$. The notion of uniform continuity can also be formulated in terms of the family of all sets of the form $V_{d,r}$. This suggests consideration of a set $X$ and a special family of subsets of $X\times X$.

\begin{equation}{\label{a}}\tag{A}\mbox{}\end{equation}

Uniformities and the Uniform Topology.

Let $X$ be a universal set and consider the Cartesian product $X\times X$. A relation $U$ is a subset of $X\times X$. If $U$ is a relation, the inverse relation $U^{-1}$ is the set of all pairs $(x,y)$ satisfying $(y,x)\in U$. The set of all pairs $(x,x)$ for $x\in X$ is called the identity relation, or the diagonal, and is denoted by $\nabla (X)$ or simply $\nabla$. We also see that $(U^{-1})^{-1}$ is always $U$. If $U=U^{-1}$, then $U$ is called symmetric. If $U$ and $V$ are relations, then the composition $U\circ V$ is the set of all pairs $(x,z)$ satisfying $(x,y)\in V$ and $(y,z)\in U$ for some $y\in X$. It is clear to see
\[U\circ (V\circ W)=(U\circ V)\circ W\]

and

\[(U\circ V)^{-1}=V^{-1}\circ U^{-1}.\]

For each subset $A$ of $X$, we define the set $U(A)$ as
\[U(A)=\left\{y:(x,y)\in U\mbox{ for some $x$ in $A$}\right\}.\]
For $x\in X$, we simply write $U(x)$ as $U(\{x\})$. Given two relations $U$ and $V$, for each subset $A$ of $X$, it is clear to see
\[(U\circ V)(A)=U(V(A)).\]

Proposition. Let $V$ be a symmetric relation. Then, we have
\[V\circ U\circ V=\bigcup_{(x,y)\in U}V(x)\times V(y).\]

\begin{equation}{\label{top221}}\tag{1}\mbox{}\end{equation}
Definition \ref{top221}. Let $X$ be a universal set. A uniformity for $X$ is a nonempty family ${\cal U}$ of subsets of $X\times X$ such that the following conditions are satisfied:

each member of ${\cal U}$ contains the diagonal $\nabla$;
if $U\in {\cal U}$, then $U^{-1}\in {\cal U}$;
if $U\in {\cal U}$, then $V\circ V\subseteq U$ for some $V\in {\cal U}$;
if $U,V\in {\cal U}$, then $U\cap V\in {\cal U}$;
if $U\in {\cal U}$ and $U\subseteq V$, then $V\in {\cal U}$.

The pair $(X,{\cal U})$ is called a uniform space. $\sharp$

In general, it is not necessarily true that the union or the intersection of two uniformities for $X$ is a uniformity. The metric space $(X,d)$ has the similar properties as given in Definition \ref{top221}. The first axiom is derived from the condition that $d(x,x)=0$, and the second axiom derives from the symmetry condition $d(x,y)=d(y,x)$. The third axiom is like a form of the triangle inequality, which roughly says that, given $r$-spheres, there are $(r/2)$-spheres. The fourth and fifth axioms resemble the neighborhood system of a point. There may be many different uniformities for a set $X$. The largest of
these is the family of all those subsets of $X\times X$ which contain $\nabla$ and the smallest is the family whose only member is $X\times X$.

Example. The usual uniformity for $\mathbb{R}$ is the family ${\cal U}$ of all subsets $U$ of $\mathbb{R}\times\mathbb{R}$ satisfying $\{(x,y):|x-y|<\epsilon\}\subseteq U$ for some positive number $\epsilon$. Each member of ${\cal U}$ is a neighborhood of the diagonal $\nabla$ (the line with equation $y=x$). However, we have to emphasize that not every neighborhood of the diagonal is a member of ${\cal U}$. For example, the set $\{(x,y):|x-y|<1/(1+|y|)\}$ is a neighborhood of $\nabla$ but not a member of ${\cal U}$. $\sharp$

Definition. A subfamily ${\cal B}$ of a uniformity ${\cal U}$ is a base for ${\cal U}$ when each member of ${\cal U}$ contains a member of ${\cal B}$. A subfamily ${\cal S}$ is a subbase for ${\cal U}$ when the family of finite intersections of members of ${\cal S}$ is a base for ${\cal U}$. $\sharp$

If ${\cal B}$ is a base for ${\cal U}$, then ${\cal B}$ determines ${\cal U}$ entirely in the sense that a subset of $X\times X$ belongs to ${\cal U}$ if and only if $U$ contains a member of ${\cal B}$. The above definitions are entirely analogous to the definitions of base and subbase for a topology.

\begin{equation}{\label{top222}}\tag{2}\mbox{}\end{equation}
Proposition \ref{top222}. A nonempty family ${\cal B}$ of subsets of $X\times X$ is a base for some uniformity for $X$ if and only if the following conditions are satisfied:

each member of ${\cal B}$ contains the diagonal $\nabla$;
if $U\in {\cal B}$, then $U^{-1}$ contains a member of ${\cal B}$;
if $U\in {\cal B}$, then $V\circ V\subseteq U$ for some $V$ in ${\cal B}$;
the intersection of the members of ${\cal B}$ contains a member of ${\cal B}$. $\sharp$

\begin{equation}{\label{top223}}\tag{3}\mbox{}\end{equation}
Proposition \ref{top223}. A family ${\cal S}$ of subsets of $X\times X$ is a subbase for some uniformity for $X$ if and only if the following conditions are satisfied:

each member of ${\cal S}$ contains the diagonal $\nabla$;
for each $U\in {\cal S}$, the set $U^{-1}$ contains a member of ${\cal S}$;
for each $U\in {\cal S}$, there exists $V\in {\cal S}$ such that $V\circ V\subseteq U$.

In particular, the union of any collection of uniformities for $X$ is the subbase for some uniformity for $X$. $\sharp$

Proposition. Let $(X,{\cal U})$ be a uniform space, and let $\tau$ be the family of all subsets $O$ of $X$ such that, for each $x\in O$, there exists $U\in {\cal U}$ such that $U(x)\subseteq O$. Then $\tau$ is a topology for $X$. $\sharp$

Definition. Let $(X,{\cal U})$ be a uniform space. The topology $\tau$ of the uniformity ${\cal U}$, or the uniform topology, is the family of all subsets $O$ of $X$ such that, for each $x\in O$, there exists $U\in {\cal U}$ such that $U(x)\subseteq O$. $\sharp$

Let us recall that the metric topology is the family of all sets which contain a ball about each point. We see that the uniform topology is the generalization of the metric topology.

Proposition. The interior of a subset $A$ of $X$ with respect to the uniform topology is the set of all points $x$ such that $U(x)\subseteq A$ for some $U$ in ${\cal U}$. $\sharp$

It follows immediately that $U(x)$ is a neighborhood of $x$ for each $U\in {\cal U}$. Consequently, the family of all sets $U(x)$ for $U\in {\cal U}$ is a base for the neighborhood system of $x$.

Proposition. Let ${\cal B}$ be a base (resp. subbase) for the uniformity ${\cal U}$. Then, for each $x$, the family of sets $U(x)$ for $U\in {\cal B}$ is a base (resp. subbase) for the neighborhood system of $x$. $\sharp$

Proposition. For $U\in {\cal U}$, the interior of $U$ is also a member in ${\cal U}$. Moreover, the family of all open symmetric members of ${\cal U}$ is a base for ${\cal U}$. $\sharp$

In view of the foregoing proposition, every member of a uniformity is a neighborhood of the diagonal.

Proposition. Let $A$ be a subset of $X$. The closure of $A$ with respect to the uniform topology is given by
\[\mbox{cl}(A)=\bigcap_{U\in {\cal U}}U(A).\]
The closure of a subset $M$ of $X\times X$ is given by
\[\mbox{cl}(M)=\bigcap_{U\in {\cal U}}U\circ M\circ U.\]

Proposition. The family of closed symmetric members of a uniformity ${\cal U}$ is a base for ${\cal U}$. $\sharp$

Let $(X,{\cal U})$ be a uniform space endowed with a uniform topology $\tau_{X}$. We have the following observations.

The topological space $(X,\tau_{X})$ is completely regular.
The topological space $(X,\tau_{X})$ is Hausdorff if and only if each set consisting of a single point is closed.
Since the closure of the set $\{x\}$ is $\bigcap_{U\in {\cal U}}U(x)$, the topological space $(X,\tau_{X})$ is Hausdorff if and only if $\bigcap_{U\in {\cal U}}U$ is the diagonal $\nabla$.

\begin{equation}{\label{b}}\tag{B}\mbox{}\end{equation}

Uniform Continuity.

Definition. Let $f$ be a function from a uniform space $(X,{\cal U})$ into another uniform space $(Y,{\cal V})$. We say that $f$ is uniformly continuous with respect to ${\cal U}$ and ${\cal V}$ when
\[\{(x,y):(f(x),f(y))\in V\}\in {\cal U}\mbox{ for each }V\in {\cal V}.\]

Let $f$ be a function from $X$ into $Y$, and let $f_{\bullet}$ be the induced function from $X\times X$ into $Y\times Y$ defined by
\[f_{\bullet}(x,y)=(f(x),f(y)).\]
The uniform continuity may also be realized in the following ways.

The function $f$ is uniformly continuous if and only if $f_{\bullet}^{-1}(V)\in {\cal U}$ for each $V\in {\cal V}$.
The function $f$ is uniformly continuous if and only if, given any $V\in {\cal V}$, there exists $U\in {\cal U}$ such that $f_{\bullet}(U)\subseteq V$.
Let ${\cal S}$ be a subbase for ${\cal V}$. Since $f_{\bullet}^{-1}$ preserves unions and intersections, it follows that $f$ is uniformly continuous if and only if $f_{\bullet}^{-1}(V)\in {\cal U}$ for each $V\in {\cal S}$.
Let $Y=\mathbb{R}$, and let ${\cal V}$ be the usual uniformity for $\mathbb{R}$. Then $f$ is uniformly continuous if and only if, given any $\epsilon >0$, there exists $U\in {\cal U}$ such that $|f(x)-f(y)|<\epsilon$ for $(x,y)\in U$.
Let $X=Y=\mathbb{R}$ endowed with the usual uniformities. Then $f$ is uniformly continuous if and only if, given any $\epsilon >0$, there exists $\delta >0$ satisfying $|f(x)-f(y)|<\epsilon$ for $|x-y|<\delta$.

If $f$ is defined on $X$ into $Y$ and $g$ is a function defined on $Y$ into $Z$, then
\[(g\circ f)_{\bullet}=g_{\bullet}\circ f_{\bullet}.\]
Therefore the composition of two uniformly continuous functions is again uniformly continuous.

Definition. Let $f$ be a bijective function from $(X,{\cal U})$ onto $(Y,{\cal V})$. If both $f$ and $f^{-1}$ are uniformly continuous, then $f$ is a uniform isomorphism, and the spaces $(X,{\cal U})$ and $(Y,{\cal V})$ are said to be uniformly equivalent. $\sharp$

The composition of two uniform isomorphisms, the inverse of a uniform isomorphism, and the identity map of a space onto itself are all uniform isomorphisms. Consequently, the collection of all uniform spaces are divided into equivalence classes which are consisting of uniformly equivalent spaces. A property which when possessed by one uniform space is also possessed by every uniformly isomorphic space is a uniform invariance.

Proposition. Each uniformly continuous function is continuous with respect to the uniform topology. In other words, each uniform isomorphism is a homeomorphism. $\sharp$

Let $f$ be a function from a set $X$ into a uniform space $(Y,{\cal V})$. In general, it is not necessarily true that the family of all sets $f_{\bullet}^{-1}(V)$ for $V\in {\cal V}$ is a uniformity for $X$. The problem is that there may be a subset of $X\times X$ which contains some set $f_{\bullet}^{-1}(V)$ that is not the inverse image of any subset of $Y\times Y$. However we have the following interesting result.

Proposition. Let $f$ be a function from a set $X$ into a uniform space $(Y,{\cal V})$. The family of all $f_{\bullet}^{-1}(V)$ is a base for some uniformity ${\cal U}$ for $X$. Moreover $f$ is uniformly continuous with respect to ${\cal U}$ and ${\cal V}$, and ${\cal U}$ is coarser than every other uniformity for which $f$ is uniformly continuous.

Proof. We shall apply Proposition \ref{top222}. It is clear that $f_{\bullet}^{-1}$ preserves inclusions, intersections and inverse, i.e.,
\[f_{\bullet}^{-1}(V^{-1})=(f_{\bullet}^{-1}(V))^{-1}.\]
Therefore, we remain to show that for each $U\in {\cal V}$, there exists $V\in {\cal V}$ satisfying
\[f_{\bullet}^{-1}(V)\circ f_{\bullet}^{-1}(V)\subseteq f_{\bullet}^{-1}(U).\]
For $(x,y)$ and $(y,z)$ belong to $f_{\bullet}^{-1}(V)$, we have $(f(x),f(y))$ and $(f(y),f(z))$ belong to $V$, which implies $(f(x),f(z))\in V\circ V$. Since $V\circ V\subseteq U$ by Proposition \ref{top222}, it follows $(f(x),f(z))\in U$, i.e., $(x,z)\in f_{\bullet}^{-1}(U)$. This completes the proof. $\blacksquare$

Let $(X,{\cal U})$ be a uniform space, and let $Y$ be a subset of $X$. The preceding discussion says that there exists a coarsest uniformity ${\cal V}$ such that the identity map of $(Y,{\cal V})$ into $(X,{\cal U})$ is uniformly continuous. It is clear that if $V\in {\cal V}$ then
\[V=U\cap (Y\times Y)\mbox{ for some }U\in {\cal U}\]
The uniformity ${\cal V}$ is called the relativization of ${\cal U}$ to $Y$, or the relative uniformity for $Y$. In this case, $(Y,{\cal V})$ is called a uniform subspace of the space $(X,{\cal U})$. The topology of the relative uniformity ${\cal V}$ is the relativized topology of ${\cal U}$.

We have seen that there is always a unique coarsest uniformity which makes that a function from a set $X$ into a uniform space is uniformly continuous. This proposition may be extended to a family ${\cal F}$ of functions such that each element $f$ of ${\cal F}$ maps $X$ into a uniform space $(Y_{f},{\cal U}_{f})$. The family of all sets of the form
\[f_{\bullet}^{-1}(U)=\left\{(x,y):(f(x),f(y))\in U_{f}\right\}\mbox{ for }f\in {\cal F}\mbox{ and }U_{f}\in {\cal U}_{f}\]
is a subbase for some uniformity ${\cal U}$ for $X$, and ${\cal U}$ is the coarsest uniformity such that each $f$ is uniformly continuous by referring to Proposition \ref{top223}. Inspired by this way, the product uniformity is defined as follows.

Definition. Let $\{(X_{\alpha},{\cal U}_{\alpha})\}_{\alpha\in\Lambda}$ be a family of uniform spaces The product uniformity for $\prod_{\alpha\in\Lambda}X_{\alpha}$ is the coarsest uniformity such that each projection is uniformly continuous. $\sharp$

The family of all sets of the form
\[\left\{(x,y):(x_{\alpha},y_{\alpha})\in U_{\alpha}\right\}\mbox{ for }\alpha\in\Lambda\mbox{ and }U_{\alpha}\in {\cal U}_{\alpha}\]
is a subbase for the product uniformity. Let $x$ be an element of the product space. Then a subbase for the neighborhood system of $x$ with respect to the uniform topology can be constructed from the subbase for the product uniformity. The family of all sets of the form
\[\{y:(x_{\alpha},y_{\alpha})\in U_{\alpha}\}\]
is a subbase for the neighborhood system of $x$. It follows that a base for the neighborhood system of $x$ with respect to the topology of the product uniformity is the family of finite intersections of sets of the form
\[\left\{y:y_{\alpha}\in U_{\alpha}(x_{\alpha})\right\}\mbox{ for }\alpha\in\Lambda\mbox{ and } U_{\alpha}\in {\cal U}_{\alpha}.\]
The same family is also a base for the neighborhood system of $x$ with respect to the product topology. Therefore the product topology is the topology of the product uniformity.

\begin{equation}{\label{top225}}\tag{4}\mbox{}\end{equation}
Proposition \ref{top225}. The topology of the product uniformity is the product topology. A function $f$ from a uniform space into a product of uniform spaces is uniformly continuous if and only if the composition of $f$ with each projection is uniformly continuous. $\sharp$

\begin{equation}{\label{top226}}\tag{5}\mbox{}\end{equation}
Proposition \ref{top226}. Let $(X,{\cal U})$ be a uniform space, and let $(X,d)$ be a pseudo-metric space. Then $d$ is uniformly continuous on $X\times X$ with respect to the product uniformity if and only if the set $\{(x,y):d(x,y)<r\}$ is a member of ${\cal U}$ for each positive number $r$. $\sharp$

\begin{equation}{\label{c}}\tag{C}\mbox{}\end{equation}

Metrization.

Each pseudo-metric space $(X,d)$ generates a uniformity in the following way. For each positive number $r$, let
\[V_{r}=\{(x,y):d(x,y)<r\}.\]
Let $t=\min\{r,s\}$. It is clear to see
\[(V_{r})^{-1}=V_{r},\quad V_{r}\cap V_{s}=V_{t}\mbox{ and }V_{r}\circ V_{r}\subseteq V_{2r}.\]
It follows that the family of all sets of the form $V_{r}$ is a base for some uniformity for $X$. This uniformity is called the pseudo-metric uniformity, or the uniformity generated by $d$. A uniform space $(X,{\cal U})$ is said to be pseudo-metrizable (resp. metrizable) when there is a pseudo-metric (resp. metric) $d$ such that ${\cal U}$ is the uniformity generated by $d$. The uniformity generated by a pseudo-metric $d$ can be described in another way. According to Proposition \ref{top226}, a pseudo-metric $d$ is uniformly continuous with respect to a uniformity ${\cal V}$ (more precisely, with respect
to the product uniformity constructed from ${\cal V}$) if and only if $V_{r}\in {\cal V}$ for each positive $r$. The uniformity ${\cal U}$ derived from $d$ can be characterized as the coarsest uniformity which makes $d$ to be uniformly continuous on $X\times X$. It should be noticed that the pseudo-metric topology is identical with the uniform topology of ${\cal U}$, since $V_{r}(x)$ is the open $r$-sphere at $x$, and the family of sets of this form is a base for the neighborhood system of $x$ with respect to both topologies.

\begin{equation}{\label{top227}}\tag{6}\mbox{}\end{equation}
Lemma \ref{top227}. (Metrization Lemma). Let $\{U_{n}\}_{n=1}^{\infty}$ be a sequence of subsets of $X\times X$ satisfying $U_{0}=X\times X$, each $U_{n}$ contains the diagonal, and
\[U_{n+1}\circ U_{n+1}\circ U_{n+1}\subseteq U_{n}\]
for each $n$. Then, there exists a nonnegative real-valued function $d$ on $X\times X$ satisfying
\[d(x,y)+d(y,z)\geq d(x,z)\mbox{ for all }x,y,z\in X\]
and
\begin{equation}{\label{top224}}\tag{7}
U_{n}\subseteq\{(x,y):d(x,y)<2^{-n}\}\subseteq U_{n-1}\mbox{ for each }n=1,2,\cdots .
\end{equation}
If we assume further that each $U_{n}$ is symmetric, then there exists a pseudo-metric $d$ satisfying $(\ref{top224})$. $\sharp$

Suppose that a uniformity ${\cal U}$ for $X$ has a countable base $\{V_{n}\}_{n=1}^{\infty}$, Then, by induction, it is possible to construct a family $\{U_{n}\}_{n=1}^{\infty}$ such that each $U_{n}$ is symmetric, $U_{n}\subseteq V_{n}$ and
\[U_{n}\circ U_{n}\circ U_{n}\subseteq U_{n-1}\]
for each $n=1,2,\cdots$. Then, the family $\{U_{n}\}_{n=1}^{\infty}$ is a base for ${\cal U}$. Applying the Metrization Lemma \ref{top227}, it follows that the uniform space $(X,{\cal U})$ is pseudo-metrizable.

Theorem. (Metrization Theorem). A uniform space is pseudo-metrizable if and only if its uniformity has a countable base. $\sharp$

The above theorem clearly says that a uniform space is pseudo-metriczable if and only if it is Hausdorff and its uniformity has a countable base. A uniformity for a set $X$ may be derived from a family ${\cal P}$ of pseudo-metric in the following fashion. Let
\[V_{p,r}=\left\{(x,y):p(x,y)<r\right\}.\]
The family of all sets of the form $V_{p,r}$ for $p\in {\cal P}$ and $r>0$ is a subbase for some uniformity ${\cal U}$ for $X$. This uniformity ${\cal U}$ is defined to be the uniformity generated by ${\cal P}$. The uniformity may be described in many ways.

According to Proposition \ref{top226}, a pseudo-metric $p$ is uniformly continuous on $X\times X$ with respect to the product uniformity derived from ${\cal V}$ if and only if $V_{p,r}\in {\cal U}$ for each positive $r$. Therefore the uniformity generated by ${\cal P}$ is the coarsest uniformity which makes each member $p$ of ${\cal P}$ to be uniformly continuous on $X\times X$.
For a fixed member $p$ of ${\cal P}$, the family of all sets $V_{p,r}$ for $r>0$ is a base for the uniformity of the pseudo-metric space $(X,p)$. If ${\cal U}$ is a uniformity for $X$, then the identity map of $(X,{\cal U})$ into $(X,p)$ is uniformly continuous if and only if $V_{p,r}\in {\cal U}$ for each $r>0$. It follows that the uniformity ${\cal U}$ is the coarsest such that for each $p$ in ${\cal P}$ the identity map of $X$ into $(X,p)$ is uniformly continuous.
Let $Z$ be the product $\prod_{p\in {\cal P}}X$, and let $f$ be the function from $X$ into $Z$ defined by $(f(x))_{p}=x$ for $x\in X$ and $p\in {\cal P}$. Suppose that the $p$-th coordinate space of the product space is assigned the uniformity of the pseudo-metric $p$, and that $Z$ is endowed with the product uniformity. The projection from $Z$ into the $p$-th coordinate space is the identity function from $X$ onto the pseudo-metric space $(X,p)$. Therefore, from Proposition \ref{top225}, the uniformity generated by ${\cal P}$ is the coarsest having the property that the function from $X$ into $Z$ is uniformly continuous. Since $f$ is one-to-one, it follows that $f$ is a uniform isomorphism of $X$ onto a subspace of the product of pseudo-metric spaces.

\begin{equation}{\label{top228}}\tag{8}\mbox{}\end{equation}
Proposition \ref{top228}. Each uniformity for $X$ is generated by the family of all pseudo-metrics which are uniformly continuous on $X\times X$.

Proof. Let $(X,{\cal U})$ be a uniform space, and let ${\cal P}$ be the family of all pseudo-metric spaces for $X$ which are uniformly continuous on $X\times X$. The uniformity generated by ${\cal P}$ is coarser than ${\cal U}$ in view of Proposition \ref{top226}. But the metrization lemma shows that for each member $U$ of ${\cal U}$ there is a member $p$ of ${\cal P}$ such that $\{(x,y):p(x,y)<1/4\}$ is contained in $U$, which says that ${\cal U}$ is coarser than the uniformity generated by ${\cal P}$. $\blacksquare$

There is an interesting corollary to the Proposition \ref{top228}. It has already been observed that, if a uniformity ${\cal U}$ for $X$ is generated by a family ${\cal P}$ of pseudo-metrics, then the space is uniformly isomorphic to a subspace of a product of pseudo-metric spaces, and it is possible to sharpen this result if $(X,{\cal U})$ is Huasdorff. The uniformity ${\cal U}$ is the smallest which makes the identity map of $X$ into the pseudo-metric space $(X,p)$ uniformly continuous for each $p$ in ${\cal P}$. The space $(X,p)$ is isomorphic under a map $h_{p}$ to a metric space $(X_{p},p^{*})$, by Proposition \ref{kelp21}, and it follows that ${\cal U}$ is the smallest uniformity making each of the maps $h_{p}$ uniformly continuous. If a map $h$ of $X$ into $\times\{X_{p}:p\in {\cal P}\}$ is defined by letting $(h(x))_{p}=h_{p}(x)$, then by Proposition \ref{top225} the uniformity ${\cal U}$ is the smallest such that $h$ is uniformly continuous. If $(X,{\cal U})$ is Hausdorff, then $h$ must be one-to-one, and in this case $h$ is uniform isomorphism.

\begin{equation}{\label{top229}}\tag{9}\mbox{}\end{equation}
Proposition \ref{top229}. We have the following properties.

(i) Each uniform space is uniformly isomorphic to a subspace of the product of pseudo-metric spaces.

(ii) Each uniform Hausdorff space is uniformly isomorphic to a subspace of the product of metric spaces. $\sharp$

Corollary. A topology $\tau$ for a set $X$ is the uniform topology for some uniformity of $X$ if and only if the topological space $(X,\tau )$ is completely regular. $\sharp$

Definition. A family ${\cal P}$ of pseudo-metrics for a set $X$ is said to be a gage when there is a uniformity ${\cal U}$ for $X$ such that each member of ${\cal P}$ is uniformly continuous on $X\times X$ with respect to the product uniformity derived from ${\cal U}$. The family ${\cal P}$ is called the gage of the uniformity ${\cal U}$, and ${\cal U}$ is the uniformity of ${\cal P}$ (${\cal U}$ is generated by Proposition \ref{top228}). $\sharp$

Every family of pseudo-metrics generates a uniformity; it will also be said to generate the gage of this uniformity. A direct description of the gage generated by a family ${\cal P}$ of pseudo-metric is possible. The family of all sets of the form $V_{p,r}$ for $p\in {\cal P}$ and $r>0$ is a subbase for the uniformity of the gage. Therefore a pseudo-metric $q$ is uniformly continuous on the product if and only if, given any $s>0$, the set $V_{q,s}$ contains some finite intersection of sets $V_{p,r}$ for $p\in {\cal P}$.

Proposition. Let ${\cal P}$ be a family of pseudo-metrics for a set $X$, and let ${\cal Q}$ be the gage generated by ${\cal P}$. Then a pseudo-metric $q$ belongs to ${\cal Q}$ if and only if, given any $s>0$, there exist $r>0$ and a finite subfamily $p_{1},\cdots p_{n}$ of ${\cal P}$ satisfying
\[\bigcap_{i=1}^{n}V_{p_{i},r}\subseteq V_{q,s}.\]

Each concept which is based on the notion of a uniformity can be described in terms of gage, since each uniformity is completely determined by its gage. Recall that the $p$-distance from a point $x$ to a set $A$ is given by
\[p\mbox{-dist}(x,A)=\inf_{y\in A}p(x,y).\]

Proposition. Let $(X,{\cal U})$ be a uniform space, let ${\cal P}$ be the gage of ${\cal U}$, and let $A$ be a subset of $X$. Then, we have the following properties.

(i) The family of all sets $V_{p,r}$ for $p\in {\cal P}$ and $r>0$ is a base for the uniformity ${\cal U}$.

(ii) The closure of $A$ with respect to the uniform topology is the set of all $x$ satisfying $p\mbox{\em -dist}(x,A)=0$ for each $p\in {\cal P}$.

(iii) The interior of $A$ with respect to the uniform topology is the set of all points satisfying $V_{p,r}(x)\subseteq A$ for some $p\in {\cal P}$ and some $r>0$.

(iv) Suppose ${\cal P}’$ is a subfamily of ${\cal P}$ which generates ${\cal P}$. A net $\{x_{\alpha}\}_{\alpha\in\Lambda}$ in $X$ converges to $x$ if and only if $\{p(x_{\alpha},x)\}_{\alpha\in\Lambda}$ converges to zero for each $p\in {\cal P}’$.

(v) Recall that $f_{\bullet}(x,y)=(f(x),f(y))$. A function $f$ from $X$ into a uniform space $(Y,{\cal V})$ is uniformly continuous if and only if $q\circ f_{\bullet}\in {\cal P}$ for each member $q$ of the gage ${\cal Q}$ of ${\cal V}$. Equivalently, the function $f$ is uniformly continuous if and only if, given any $q\in {\cal Q}$ and $s>0$, there exist $p\in {\cal P}$ and $r>0$ positive such that $p(x,y)<r$ implies $q(f(x),f(y))<s$.

(vi) Let $\{(X_{\alpha},{\cal U}_{\alpha})\}_{\alpha\in\Lambda}$ be a family of uniform spaces, and let ${\cal P}_{\alpha}$ be the gage of ${\cal U}_{\alpha}$ for each $\alpha\in\Lambda$. Then, the gage of the product uniformity for $\prod_{\alpha\in\Lambda}X_{\alpha}$ is generated by all pseudo-metrics of the form $q(x,y)=p_{\alpha}(x_{\alpha},y_{\alpha})$ for $\alpha\in\Lambda$ and $p_{\alpha}\in {\cal P}_{\alpha}$. $\sharp$

\begin{equation}{\label{d}}\tag{D}\mbox{}\end{equation}

Completeness.

Let $(X,{\cal U})$ be a uniform space, and let ${\cal P}$ be the gage of ${\cal U}$; that is, ${\cal P}$ is the family of all pseudo-metrics for $X$ which are uniformly continuous on $X\times X$.

Definition. A net $\{x_{\alpha}\}_{\alpha\in\Lambda}$ in the uniform space $(X,{\cal U})$ is a Cauchy net when, given any $U\in {\cal U}$, there exists $\alpha_{0}\in\Lambda$ satisfying $(x_{\alpha},x_{\beta})\in U$ for $\alpha\succ\alpha_{0}$ and $\beta\succ\alpha_{0}$. $\sharp$

We have the following observations.

Let $\Lambda\times\Lambda$ be the product directed set. Then $\{x_{\alpha}\}_{\alpha\in\Lambda}$ is a Cauchy net if and only if the net $\{(x_{\alpha},x_{\beta})\}_{(\alpha ,\beta )\in\Lambda\times\Lambda}$ is eventually in each member of ${\cal U}$.
Recall that
\[V_{p,r}=\{(x,y):p(x,y)<r\}.\]
Since the family of all sets of the form $V_{p,r}$ for $p\in {\cal P}$ and $r>0$ is a base for the uniformity ${\cal U}$, it follows that $\{x_{\alpha}\}_{\alpha\in\Lambda}$ is a Cauchy net if and only if the net $\{(x_{\alpha},x_{\beta})\}_{(\alpha ,\beta )\in\Lambda\times\Lambda}$ is eventually in each set of the form $V_{p,r}$. In other words, $\{x_{\alpha}\}_{\alpha\in\Lambda}$ is a Cauchy net if and only if $\{p(x_{\alpha},x_{\beta})\}_{(\alpha ,\beta )\in\Lambda\times\Lambda}$ converges to zero for each pseudo-metric $p\in {\cal P}$.

Proposition. A net $\{x_{\alpha}\}_{\alpha\in\Lambda}$ in a uniform space $(X,{\cal U})$ is a Cauchy net if and only if either one of the following statements is true.

(a) The net $\{(x_{\alpha},x_{\beta})\}_{(\alpha ,\beta )\in\Lambda\times\Lambda}$ is eventually in each member of some subbase for the uniformity ${\cal U}$.

(b) The net $\{p(x_{\alpha},x_{\beta})\}_{(\alpha ,\beta )\in\Lambda\times\Lambda}$ converges to zero for each $p$ in some family of pseudo-metrics which generates the gage ${\cal P}$. $\sharp$

\begin{equation}{\label{top230}}\tag{10}\mbox{}\end{equation}
Proposition \ref{top230}. We have the following properties..

(i) The net which is convergent with respect to the uniform topology is a Cauchy net.

(ii) A Cauchy net converges to each of its cluster points. $\sharp$

Definition. A uniform space is complete when every Cauchy net in the space is convergent. $\sharp$

Proposition. We have the following properties.

(i) A closed subspace of a complete space is complete.

(ii) A complete subspace of a Hausdorff uniform space is closed. $\sharp$

We have the following observations.

If the uniformity ${\cal U}$ is the finest possible uniformity for $X$ (i.e., consists of all subsets of $X\times X$ which contain the diagonal), then $(X,{\cal U})$ is complete.
The coarsest uniformity for $X$ also yields a complete space.
If a uniform space $(X,{\cal U})$ is compact with respect to the uniform topology, then it is complete. Indeed, the compactness says that each net has a cluster point. Proposition \ref{top230} says that each Cauchy net converges to some point.
The space of real numbers is complete with respect to the usual uniformity. This ca be realized by verifying that each Cauchy net is eventually in some bounded subset $A$ of the space of real numbers and is therefore eventually in the compact set $\mbox{cl}(A)$.

Recall that a family of sets has the finite intersection property if and only if no finite intersection of members of the family is empty, and a topological space is compact if and only if the intersection of the members of each family of closed sets with the finite intersection property is nonempty. A family ${\cal A}$ of subsets of a uniform space $(X,{\cal U})$ contains {\bf small sets} if and only if for each $U$ in ${\cal U}$ there is a member $A$ of ${\cal A}$ such that $A$ is a subset of $U(x)$ for some point $x$. Another formulation is: for each $U$ in ${\cal U}$ there is $A$ in ${\cal A}$ such that $A\times A\subseteq U$. In terms of the gage ${\cal P}$ of the uniform space, a family ${\cal A}$ contains small sets if and only if for each positive $r$ and each $d$ in ${\cal P}$ there is $A$ in ${\cal A}$ such that the $d$-diameter of $A$ is less than $r$.

Proposition. A uniform space is complete if and only if each family of closed sets which has the finite intersection property and contains small sets has
a nonempty intersection. $\sharp$

Proposition. A pseudo-metrizable uniform space is complete if and only if every Cauchy sequence in the space converges to a point. $\sharp$

Proposition. The product of uniform spaces is complete if and only if each coordinate space is complete. A net in the product is a Cauchy net if and only if its projection into each coordinate space is a Cauchy net. $\sharp$

A function $f$ is uniformly continuous on a subset $A$ of a uniform space $(X,{\cal U})$ when its restriction to $A$, $f|A$, is uniformly continuous with respect to the relativized uniformity. If the range space is complete and Hausdorff and $f$ is uniformly continuous on its domain $A$, then there is a unique uniformly continuous extension whose domain is the closure of $A$.

\begin{equation}{\label{top231}}\tag{11}\mbox{}\end{equation}
Proposition \ref{top231}. Let $f$ be a function whose domain is a subset $A$ of a uniform space $(X,{\cal U})$ and whose values lie in a complete Hausdorff uniform space $(Y,{\cal V})$. If $f$ is uniformly continuous on $A$, then there is a unique uniformly continuous extension $\bar{f}$ of $f$ whose domain is the closure of $A$. $\sharp$

For a metric space $X$ it is possible to find a complete metric space $X^{*}$ such that $X$ is isomorphic to a dense subspace of $X^{*}$ (not just uniformly isomorphic).

Proposition. Each metric (or pseudo-metric) space can be mapped by a one-to-one isometry onto a dense subset of a complete metric (pseudo-metric) space. $\sharp$

Each uniform space is uniformly isomorphic to a subspace of a product of pseudo-metric spaces, and each Hausdorff uniform space is uniformly isomorphic to a product of metric spaces, by Proposition \ref{top229}. The preceding theorem implies that a metrc or pseudo-metric space is uniformly isomorphic to a subspace of a complete space of the same sort.

Proposition. Each uniform space is uniformly isomorphic to a dense subspace of a complete uniform space. Each Hausdorff uniform space is uniformly
isomorphic to a dense subspace of a complete Hausdorff uniform space. $\sharp$

Definition. A completion of a uniform space $(X,{\cal U})$ is a pair, $(f,(X^{*},{\cal U}^{*}))$ where $(X^{*},{\cal U}^{*})$ is a complete uniform space and $f$ is a uniform isomorphism of $X$ into a dense subspace of $X^{*}$. $\sharp$

The completion is Hausdorff if and only if $(X^{*},{\cal U}^{*})$ is a Hausdorff uniform space. The foregoing theorem can then be stated: Each (Hausdorff) uniform space has a (Hausdorff) completion. There is a uniqueness property for Hasudorff completions. If $f$ and $g$ are uniform isomorphisms of $X$ onto dense subspaces of complete Hausdorff uniform spaces $X^{*}$ and $X^{**}$, then both $g\circ f^{-1}$ and $f\circ g^{-1}$ have uniformly continuous extensions to all of $X^{*}$ and $X^{**}$ respectively, by Proposition \ref{top231}. It follows that the extension of $g\circ f^{-1}$ is a uniform isomorphism of $X^{*}$ onto $X^{**}$. Stated roughly: the Hausdorff completion of a Hausdorff uniform space is unique to a uniform isomorphism.

\begin{equation}{\label{e}}\tag{E}\mbox{}\end{equation}

Compact Spaces.

Each completely regular topology $\tau$ for a set $X$ is the uniform topology for some uniformity ${\cal U}$, but the uniformity is usually not unique. If $(X,\tau )$ is compact and regular, then it turns out that there is precisely one uniformity whose topology is $\tau$. In this casem the topology determines the uniformity, topological invariance are uniform invariance.

Proposition. If $(X,{\cal U})$ is a compact uniform space, then every neighborhood of the diagonal $\nabla$ in $X\times X$ is a member of ${\cal U}$ and every pseudo-metric which is continuous on $X\times X$ is a member of the gage of ${\cal U}$. $\sharp$

Each compact regular topological space is completely regular and its topology is therefore the uniform topology for some uniformity.

Corollary. If $(X,\tau )$ is a compact regular topological space, then the family of all neighborhoods of the diagonal $\nabla$ is a uniformity for $X$ and $\tau$ is the uniform topology. $\sharp$

Proposition. Each continuous function on a compact uniform space to a uniform space is uniformly continuous. $\sharp$

Each compact uniform space $(X,{\cal U})$ can be written as the union of a finite number of small sets, in the sense that for each pseudo-metric $d$ belonging to the gage of ${\cal U}$ and each positive $r$ there is a finite cover of $X$ by sets of $d$-diameter less than $r$. This is a direct consequence of compactness, since $X$ can be covered by a finite number of $r/3$ spheres about points and each of these is of diameter less than $r$. A uniform space $(X,{\cal U})$ is totally bounded (or precompact) if and only if $X$ is the union of a finite number of sets of $d$-diameter less than $r$ for each pseudo-metric $d$ of the gage of ${\cal U}$ and each positive $r$. In terms of ${\cal U}$ this can be stated: for each $U$ in ${\cal U}$ the set $X$ is the union of a finite number of sets $B$ such that $B\times B\subseteq U$, or equivalently, for each $U$ in ${\cal U}$ there is a finite subset $F$ of $X$ satisfying $U(F)=X$. A subset $Y$ of a uniform space is called totally bounded if and only if, with the relativized uniformity, is totally bounded.

Proposition. A uniform space $(X,{\cal U})$ is totally bounded if and only if each net in $X$ has a Cauchy subnet. Consequently, a uniform space is compact if and only if it is totally bounded and complete. $\sharp$

Definition. A cover of a subset $A$ of a uniform space $(X,{\cal U})$ is a uniform cover when there is a member $U$ of ${\cal U}$ such that the set $U(x)$ is a subset of some member of the cover for every $x$ in $A$ (that is, the family of $U(x)$ for $x$ in $A$ refines the cover). $\sharp$

In terms of the gage of the uniformity ${\cal U}$, a cover of $A$ is uniform if and only if there is a member $d$ of the gage and a positive number $r$ such that the open sphere of $d$-radius $r$ about each point of $A$ is contained in some member of the cover.

Proposition. Each open cover of a compact subset of a uniform space is a uniform cover. In particular, each neighborhood of a compact subset $A$ contains a neighborhood of the form $U(A)$ where $U$ is a member of the uniformity. $\sharp$

Proposition. (Baire). Let $X$ be either a complete pseudo-metric space or a locally compact regular space. Then, the intersection of a countable family of open dense subsets of $X$ is itself dense in $X$. $\sharp$

A subset $A$ of a topological space is {\bf nowhere dense} in $X$ if and only if the interior of the closure of $A$ is empty; otherwise stated, $A$ is nowhere dense in $X$ if and only if the open set $X\setminus \mbox{cl}(A)$ is dense in $X$. It is evident that the finite union of nowhere dense sets is nowhere dense. A subset $A$ of $X$ is meager in $X$ or of the first category in $X$ if and only if $A$ is the union of a countable family of nowhere dense sets. The Baire theorem can then be stated: the complement of a meager subset of a complete metric space is dense. (The complement of a meager set is sometimes called co-meager or residual in $X$). A set $A$ is non-meager or of the second category in $X$ when it is not meager in $X$.

Proposition. Let $A$ be a subset of a topological space $X$ and let $M(A)$ be the union of all open sets $V$ such that $V\cap A$ is meager in $X$. Then $A\cap \mbox{cl}(M(A))$ is meager in $X$. $\sharp$

An important consequence of the preceding proposition is that if a subset $A$ of a topological space is non-meager then there is a nonempty open set $V$ such that the intersection of $A$ with every neighborhood of each point of $\mbox{cl}(V)$ is non-meager. A map of a uniform space $(X,{\cal U})$ into a uniform space $(Y,{\cal V})$ is uniformly open when, for each $U$ in ${\cal U}$, there is a $V$ in ${\cal V}$ such that $V(f(x))\subseteq f(U(x))$ for each $x$ in $X$. Let $U_{r}=\{(x,y):d(x,y)<r\}$, so that $U_{r}(x)$ is simply the $r$-sphere about $x$.

Lemma. Let $R$ be a closed subset of the product of a complete pseudo-metric space $(X,d)$ with the uniform space $(Y,{\cal V})$ and suppose that for each positive $r$ there is $V$ in ${\cal V}$ such that $\mbox{cl}(R(U_{r}(x)))$ contains $V(y)$ for each $(x,y)\in R$. Then for each $r$ and each positive $e$ it is true that $V(y)\subseteq \mbox{cl}(R(U_{r}(x)))\subseteq R(U_{e+r}(x))$. $\sharp$

Suppose now that $f$ is uniformly open and continuous, that $X$ is complete and pseudo-metrizable, that $Y$ is Hausdorff, and that $Y^{*}$ is a Hausdorff completion of $Y$. Then, the graph of $f$ is a subset of $X\times Y^{*}$ which is closed because $f$ is continuous, and satisfies the condition of the preceding lemma because the map of $X$ into $Y$ is uniformly open. Then, the lemma implies that $f$ is a uniformly open map of $X$ into $Y^{*}$. Finally, since $f(X)$ contains $V(f(X))$ for some $V$ in ${\cal V}$, it must be true that $f(X)$ is closed (and open) in $Y^{*}$; hence $f(X)$ is complete.

Corollary. Let $f$ be a continuous uniformly open map of a complete pseudo-metrizable space into a Hausdorff uniform space. Then, the range of the map is complete. $\sharp$