Uniform space

In mathematics, and more specifically in topology, the notions of a uniform structure and a uniform space generalize the notions of a metric (distance function) and a metric space respectively. As a human activity, the theory of uniform spaces is a chapter of general topology. From the formal point of view, the notion of a uniform space is a sibling of the notion of a topological space. While uniform spaces are significant for mathematical analysis, the notion seems less fundamental than that of a topological space. The notion of uniformity is auxiliary rather than an object to be studied for its own sake (specialists on uniform spaces may disagree though).

For two points of a metric space, their distance is given, and it is a measure of how close each of the given two points is to another. The notion of uniformity catches the idea of two points being near one another in a more general way, without assigning a numerical value to their distance. Instead, given a subset $W\subseteq X\times X\$, we may say that two points  $x,y\in X\$ are W-near one to another, when $\ (x,y)\in W$; certain such sets $W\subseteq X\times X\$ are called entourages (see below), and then the mathematician Roman Sikorski would write suggestively:

$d(x,y)<W\$

meaning that this whole mathematical phrase stands for: $W\$ is an entourage, and $\ (x,y)\in W$.  Thus we see that in the general case of uniform spaces, the distance between two points is (not measured but) estimated by the entourages to which the ordered pair of the given two points belongs.

Historical remarks

The uniform ideas, in the context of finite dimensional real linear spaces (Euclidean spaces), appeared already in the work of the pioneers of the precision in mathematical analysis (A.-L. Cauchy, E. Heine). Next, George Cantor constructed the real line by metrically completing the field of rational numbers, while Frechet introduced metric spaces. Then Felix Hausdorff extended the Cantor's completion construction onto arbitrary metric spaces. General uniform spaces were introduced by Andre Weil in a 1937 publication.

The uniform ideas may be expressed equivalently in terms of coverings. The basic idea of an abstract triangle inequality in terms of coverings has appeared already in the proof of the metrization Aleksandrov-Urysohn theorem (1923).

A different but equivalent approach was introduced by V.A. Efremovich, and developed by Y.M.Smirnov. Efremovich axiomatized the notion of two sets approaching one another (infinitely closely, possibly overlapping). In terms of entourages, two sets approach one another if for every entourage $W\$ there is an ordered pair of points $\ (x,y)\in W$, one from each of the given two sets, i.e. for which the Sikorski's inequality holds:

$d(x,y)<W\$

According to P.S.Aleksandrov, this kind of approach to uniformity, in the language of nearness, goes back to Riesz (perhaps F.Riesz).

Topological prerequisites

This article assumes that the reader is familiar with certain elementary, basic notions of topology, namely:

• topology (as a family of open sets), topological space;
• neighborhoods (of points and sets), bases of neighborhoods;
• separation axioms:
  • $T_{0}\$  (Kolmogorov axiom);
  • $T_{1}\$
  • $T_{2}\$  (Hausdorff axiom);
  • regularity axiom and  $T_{3}\$
  • complete regularity (Tichonov axiom) and  $\ T_{3{\frac {1}{2}}}$;
  • normal spaces and  $T_{4}\$;
• continuous functions (maps, mappings);
• compact spaces (and compact Hausdorff spaces, i.e. compact $\ T_{2}$-spaces);
• metrics and pseudo-metrics, metric and pseudo-metric spaces, topology induced by a metric or pseudo-metric.

Definition

Auxiliary set-theoretical notation, notions and properties

Given a set $\ X$, and $V,W\subseteq X\times X$, let's use the notation:

$\Delta _{X}\ :=\ \{(x,x):x\in X\}$

and

$V^{-1}\ :=\ \{(y,x):(x,y)\in V\}$

and

$W\circ \,V:=\{(x,z):\exists _{y\in X}\ \left((x,y)\in V,\ \ (y,z)\in W\right)\}$

Theorem

• $\left(\left(V\subseteq V'\right)\land \left(W\subseteq W'\right)\right)\ \Rightarrow \ \left(W\circ V\subseteq W'\circ V'\right)$
• $\Delta _{X}\circ V\ =\ V\circ \Delta _{X}\ =\ V$
• $\Delta _{X}\subseteq V\ \Rightarrow W\circ V\supseteq W$
• $\Delta _{X}\subseteq W\ \Rightarrow W\circ V\supseteq V$
• $(\Delta _{X}\subseteq V\ \land \ \Delta _{X}\subseteq W)\ \ \Rightarrow \ \ W\circ V\ \supseteq \ V\cup W$
• if $\ A$  and $\ B$  are $\ W$-sets,  where $\Delta _{X}\subseteq W$,  and if $\ A\cap B\neq \emptyset$,  then $\ A\cup B$  is a $\ (W\circ W)$-set; or in the Sikorski's notation:

$A\cup B\neq \emptyset \ \ \Rightarrow \ \ {\mathit {diam}}(A\cup B)\ <\ W\circ W$

for every  $\ V,V',W,W'\subseteq X\times X$,  and $\ A,B\subseteq X$.

Definition  A subset $\ A\$ of $\ X\$ is called a $\ V$-set if  $A\times A\subseteq V$, in which case we may also use Sikorski's notation:

${\mathit {diam}}(A)\ <\ V$

• Let $\ {\mathcal {K}}$  be a family of sets such that the union of any two of them is a $\ V$-set  (where $\ V\subseteq X\times X$).  The union $\ \bigcup {\mathcal {K}}$  is a $\ V$-set.

Uniform space (definition)

An ordered pair $(X,{\mathcal {U}})$, consisting of a set $\ X$ and a family ${\mathcal {U}}$ of subsets of $\ X\times X$, is called a uniform space, and ${\mathcal {U}}$ is called a uniform structure in $\ X$, if the following five properties (axioms) hold:

1. ${\mathcal {U}}\neq \emptyset$
2. $\forall _{W\in {\mathcal {U}}}\ \Delta _{X}\subseteq W$
3. $\forall _{V\in {\mathcal {U}}}\ \forall _{W\subseteq X\times X}\ (V\subseteq W\ \Rightarrow \ W\in {\mathcal {U}})$
4. $\forall _{V,W\in {\mathcal {U}}}\ V\cap W^{-1}\in {\mathcal {U}}$
5. $\forall _{W\in {\mathcal {U}}}\exists _{V\in {\mathcal {U}}}\ V\circ V\subseteq W$

Members of ${\mathcal {U}}$ are called entourages.

Instead of the somewhat long term uniform structure we may also use short term uniformity—it means exactly the same.

Example:   $X\times X\$  is an entourage of every uniform structure in $\ X$.

Two extreme examples

The single element family ${\mathcal {U}}:=\{X\times X\}$ is a uniform structure in $\ X$; it is called the weakest uniform structure (in $\ X$).

Family

${\mathcal {U}}\ :=\ \{W\subseteq X\times X:\Delta _{X}\subseteq W\}$

is a uniform structure in $\ X$  too; it is called the strongest uniform structure or the discrete uniform structure in $\ X$; it contains every other uniform structure in $\ X$.

• $\ {\mathcal {U}}$  is the strongest uniform structure in $\ X$  if and only if  $\ \Delta _{X}\in {\mathcal {U}}$.

Uniform base

A family ${\mathcal {B}}$ is called to be a base of a uniform structure ${\mathcal {U}}$ in $\ X$ if  ${\mathcal {U}}={\mathcal {U}}_{\mathcal {B}}$, where:

${\mathcal {U}}_{\mathcal {B}}\ :=\ \{W\subseteq X\times X:\exists _{B\in {\mathcal {B}}}\ B\subseteq W\}$

Remark  Uniform bases are also called fundamental systems of neighborhoods of the uniform structure (by Bourbaki).

Instead of starting with a uniform structure, we may begin with a family ${\mathcal {B}}$.  If family ${\mathcal {U}}_{\mathcal {B}}\$ is a uniform structure in $\ X$, then we simply say that ${\mathcal {B}}$ is a uniform base (without mentioning explicitly any uniform structure).

Theorem A family ${\mathcal {B}}\$ of subsets of $X\$ is a uniform base if and only if the following properties hold:

1. ${\mathcal {B}}\neq \emptyset$
2. $\forall _{W\in {\mathcal {B}}}\ \Delta _{X}\subseteq W$
3. $\forall _{V,W\in {\mathcal {B}}}\ V\cap W^{-1}\in {\mathcal {U}}_{\mathcal {B}}$
4. $\forall _{W\in {\mathcal {B}}}\exists _{V\in {\mathcal {B}}}\ V\circ V\subseteq W$

Remark  Property 3 above features ${\mathcal {U}}_{\mathcal {B}}\$ (it's not a typo!)--it's simpler this way.

The symmetric base

Let $\ V\subseteq X\times X$. We say that $\ V\$  is symmetric if $\ V^{-1}=V$.

Let $\ V\$ be as above, and let $\ W:=V\cap V^{-1}$. Then $\ W\$ is symmetric, i.e.

$(V\cap V^{-1})^{-1}=V\cap V^{-1}$

Now let ${\mathcal {U}}\$ be a uniform structure in $\ X$. Then

${\mathcal {U}}_{S}\ :=\ \{W\in {\mathcal {U}}:W^{-1}=W\}$

is a base of the uniform structure $\ {\mathcal {U}}$; it is called the symmetric base of $\ {\mathcal {U}}$.  Thus every uniform structure admits a symmetric base.

Example

Notation:  $Fin(X)\$ is the family of all finite subsets of $\ X$.

Let $\ X\$  be an infinite set. Let

$W_{A}\ :=\ \Delta _{X}\cup (A\times A)$

for every $\ A\subseteq X$, and

${\mathcal {A}}\ :=\ \{W_{A}:X\backslash A\in Fin(X)\}$

Each member of ${\mathcal {A}}\$ is symmetric. Let's show that ${\mathcal {A}}\$ is a uniform base:

Indeed, axioms 1-3 of uniform base obviously hold. Also:

$W_{A}\circ W_{A}\ =\ W_{A}\$

hence axiom 4 holds too. Thus $\ {\mathcal {A}}\$ is a uniform base.

The generated uniform structure $\ {\mathcal {U}}_{\mathcal {A}}\$ is different both from the weakest and from the strongest uniform structure in $\ X$,  (because $\ X\$ is infinite).

Metric spaces

Let $\ (X,d)$ be a metric space. Let

$B_{t}\ :=\ \{(x,y):d(x,y)<t\}$

for every real $\ t>0$.  Define now

${\mathcal {B}}_{d}\ :=\ \{B_{t}:t>0\}$

and finally:

${\mathcal {U}}_{d}\ :=\ \{W:\exists _{t}\ B_{t}\subseteq W\subseteq X\times X\}$

Then ${\mathcal {U}}_{d}$ is a uniform structure in $\ X$; it is called the uniform structure induced by metric $\ d$  (in $\ X$).

Family ${\mathcal {B}}_{d}\$ is a base of the structure ${\mathcal {U}}_{d}\$ (see above). Observe that:

• $\Delta _{X}\ \subseteq \ B_{t}$
• $B_{t}^{-1}\ =\ B_{t}$
• $B_{s}\cap B_{t}\ =\ B_{\,\min(s,t)}$
• $B_{t}\circ B_{t}\ \subseteq \ B_{2\cdot t}$

for arbitrary real numbers  $\ s,t>0$.  This is why ${\mathcal {B}}_{d}\$ is a uniform base, and ${\mathcal {U}}_{d}$ is a uniform structure (see the axioms of the uniform structure above).

Remark (!)   Everything said in this text fragment is true more generally for arbitrary pseudo-metric space $\ (X,d)$; instead of the standard metric axiom:

$d(x,y)=0\ \Leftrightarrow x=y\$

a pseudo-metric space is assumed to satisfy only a weaker axiom:

$d(x,x)=0\$

(for arbitrary  $\ x,y\in X$).

The induced topology

First another piece of auxiliary notation--given a set $\ X$, and $W\subseteq X\times X$, let

$W(x)\ :=\ \{y:(x,y)\in W\}=W\cap (\{x\}\times X)$

Let $(X,{\mathcal {U}})$ be a uniform space. Then families

${\mathcal {U}}_{x}\ :=\ \{W(x):W\in {\mathcal {U}}\}$

where $\ x$ runs over $\ X$, form a topology defining system of neighborhoods in $\ X$. The topology itself is defined as:

${\mathcal {T}}_{\mathcal {U}}\ :=\ \{G\subseteq X:\forall _{x\in G}\ G\in {\mathcal {U}}_{x}\}$

• The topology induced by the weakest uniform structure is the weakest topology. Furthermore, the weakest uniform structure is the only one which induces the weakest topology (in a given set).
• The topology induced by the strongest (discrete) uniform structure is the strongest (discrete) topology. Furthermore, the strongest uniform structure is the only one which induces the discrete topology in the given set if and only if that set is finite. Indeed, for any infinite set also the uniform structure $\ {\mathcal {U}}_{\mathcal {A}}\$ (see Example above) induces the discrete topology. Thus different uniform structures (defined in the same set) can induce the same topology.
• The topology ${\mathcal {T}}_{d}\$ induced by a metrics $\ d\$ is the same as the topology induced by the uniform structure induced by that metrics:

${\mathcal {T}}_{{\mathcal {U}}_{d}}\ =\ {\mathcal {T}}_{d}$

Convention  From now on, unless stated explicitly to the contrary, the topology considered in a uniform space is always the topology induced by the uniform structure of the given space. In particular, in the case of the uniform spaces the general topological operations on sets, like interior $\ {\mathit {Int}}(A)$  and closer  $\ {\mathit {Cl}}(A)$,  are taken with respect to the topology induced by the uniform structure of the respective uniform space.

Example  Consider three metric functions in the real line $\ {\mathcal {R}}$:

• $d(x,y):=|x-y|\$
• $\delta (x,y)\ :=\ 2\cdot d(x,y)\$
• $d_{c}(x,y):=|x^{3}-y^{3}|\$

All these three metric functions induce the same, standard topology in $\ {\mathcal {R}}$.  Furthermore, functions $\ d\$ and $\ \delta \$ induce the same uniform structure in $\ {\mathcal {R}}$. Thus different metric functions can induce the same uniform structure. On the other hand, the uniform structures induced by $\ d\$ and $\ d_{c}\$ are different, which shows that different uniform structures, even when they are induced by metric functions, can induce the same topology.

Theorem  Let $(X,{\mathcal {U}})$ be a uniform space. The family of all entourages $\ U\in {\mathcal {U}}$  which are open in $\ X\times X$ is a base of structure ${\mathcal {U}}$

Remark  An equivalent formulation of the above theorem is:

• the interior of every entourage is an entourage.

Proof (of the theorem).  Let $\ W\in {\mathcal {U}}$  be an arbitrary entourage. Let $\ V\in {\mathcal {U}}$  be a symmetric entourage such that $\ V\circ V\circ V\subseteq W$. It is enough to prove that entourage $\ V$  is contained in the topological interior of $\ U$. Let's do it. Let $\ (a,b)\in V$ . Let $\ (x,y)\in V(a)\times V(b)$.  Then, since $\ V$  is symmetric, we have:

$(x,a),\ (a,b),\ (b,y)\ \in \ V$

hence $\ (x,y)\in V\circ V\circ V\subseteq W$.  This proves that

$V(a)\times V(b)\ \subseteq W$

Thus every point $\ (a,b)\in V$  belongs to the topological interior of $\ W$,  i.e. the entire $\ V$  is contained in the interior of $\ W$.

End of proof.

Separation properties

Notation:

$W(A)\ :=\ \bigcup _{x\in A}\ W(x)$

for every entourage $\ W$  and $\ A\subseteq X$  (see above the definition of $\ W(x)$). Thus $\ W(A)$  is a neighborhood of $\ A$.

Warning  $\{W(A):W\in {\mathcal {U}}\}$   does not have to be a base of neighborhoods of $\ A$, as shown by the following example (consult the section about metric spaces, above):

Example  Let $\ {\mathcal {R}}$  be the space of real numbers with its customary Euclidean distance (metric)

$d(x,y)\ :=\ |x-y|$

and the uniformity induced by this metric (see above)—this uniformity is called Euclidean. Let $\ {\mathcal {N}}:=\{1,2,\dots \}$  be the set of natural numbers. Then the union of open intervals:

$U\ :=\ \bigcup _{n\in {\mathcal {N}}}(n-{\frac {1}{n}};n+{\frac {1}{n}})$

is an open neighborhood of $\ {\mathcal {N}}$&nbsd in $\ {\mathcal {R}}$,&nbsd but there does not exist any $\ t>0$  such that $\ B_{t}({\mathcal {N}})\subseteq U$  (see above). It follows that $\ U$  does not contain any set $\ W({\mathcal {N}})\in {\mathcal {U}}$,  where $\ {\mathcal {U}}$  is the Euclidean uniformity in $\ {\mathcal {R}}$.

Definition  Let $\ A,B\subseteq X$,  and $\ W$  be an entourage. We say that $\ A$  and $\ B$  are $\ W$-apart, if

$(A\times B)\ \cap W\ =\ \emptyset$

in which case we write

$\delta (A,B)\ >\ W$

in the spirit of Sikorski's notation (it is an idiom, don't try to parse it).

• Let $\ A,B\subseteq X$  be $\ W$-apart. Let $\ V$  be another entourage, and let it be symmetric (meaning $\ V^{-1}=V)$  and such that  $\ V\circ V\circ V\subseteq W$.  Then $\ V(A)$  and $\ V(B)$  are $\ V$-apart:

$\delta (V(A),V(B))\ >\ V$

We see that two sets which are apart (for an entourage) admit neighborhoods which are apart too. Now we may mimic Paul Urysohn by stating a uniform variant of his topological lemma:

Uniform Urysohn Lemma Let $\ A,B\subseteq X$  be apart. Then there exists a uniformly continuous function $\ f:X\rightarrow [0;1]$  such that $\ f(x)=0$ for every $\ x\in A$,  and $\ f(x)=1$ for every $\ x\in B$.

It is possible to adopt the main idea of the Urysohn's original proof of his lemma to this new uniform situation by iterating the statement just above the Uniform Urysohn Lemma.

Proof (of the Uniform Urysohn Lemma)

Let $\ W$ be an entourage. Let $\ A,B\subseteq X$  be $\ W$-apart.  Let $\ (W_{n}:n=0,1,\dots )$  be a sequence of entourages such that

• $W_{0}\ :=\ W$
• $W_{n}^{-1}\ =\ W_{n}$
• $W_{n}\circ W_{n}\subseteq W_{n-1}\$

for every $\ n=1,2,\dots$.  Next, let $\ A_{r},B_{r}\subseteq X$  for every $\ r:={\frac {k}{2^{n}}}$,  where $\ n=1,2,\dots$  and $\ k=0,1,\dots ,2^{n}$,  be defined, inductively on $\ n$,  as follows:

• $A_{0}:=A\quad {\mathit {and}}\quad A_{1}:=X$
• $B_{0}:=X\quad {\mathit {and}}\quad B_{1}:=B$
• $A_{\frac {2\cdot k-1}{2^{n}}}\ :=\ X\ \backslash \ W_{n}\left(B_{\frac {k}{2^{n-1}}}\right)$
• $B_{\frac {2\cdot k-1}{2^{n}}}\ :=\ X\ \backslash \ W_{n}\left(A_{\frac {k-1}{2^{n-1}}}\right)$

for every $\ n=1,2,\dots$  and $\ k=1,\dots ,2^{n-1}$.  We see that

• $\ A_{\frac {m-1}{2^{n}}}$  and $B_{\frac {m}{2^{n}}}$  are $\ W_{n}$-apart for every $\ n=0,1,\dots$  and $\ m=1,\dots ,2^{m}$;
• $A_{r}\cup B_{r}\ =\ X$  for every $\ r$;
• the assignment $\ r\mapsto A_{r}$  is increasing, while $\ r\mapsto B_{r}$  is decreasing.

The required uniform function can be defined as follows:

$f(x)\ :=\ \inf \ \{r:x\in A_{r}\}$

for every $\ x\in X$.  Obviously, $\ f(x)=0$  for every $\ x\in A$,  and $\ f(x)=1$  for every $\ x\in B$.  Furthermore, let  $\ \epsilon >0$.  Then

$\epsilon \ >\ {\frac {1}{2^{n-1}}}$

for certain positive integer $\ n$. Let $\ x,y\in X$  be such that

$f(x)+\epsilon \ \leq \ f(y)$

Then there exists $\ m\in \{1,\dots ,2^{n}\}$  such that

$f(x)\ <\ {\frac {m-1}{2^{n}}}\ <\ {\frac {m}{2^{n}}}\ <\ f(y)$

Thus $\ x\in A_{\frac {m-1}{2^{n}}}$,  while $\ y\notin A_{\frac {m}{2^{n}}}$,  hence $\ y\in B_{\frac {m}{2^{n}}}$.  Thus points $\ x$  and $\ y$  are $\ W_{n}$-apart.

We have proved that for every $\ (x,y)\in W_{n}$  the images are less then $\ \epsilon$-apart:

$\forall _{(x,y)\in W_{n}}\ |f(x)-f(y)|\ <\ \epsilon$

End of proof.

Now let's consider a special case of one of the two sets being a 1-point set.

• Let $\ p\in X$,  and let $\ G$ be a neighborhood of $\ p$  (with respect to the uniform topology, i.e. with respect to the topology induced by the uniform structure). Then $\ \{p\}$  and $\ X\backslash G$  are apart.

Indeed, there exists an entourage $\ W\in {\mathcal {U}}$  such that $\ W(p)\subseteq G$,  which means that

$\ (\{p\}\times (X\backslash G))\ \cap \ W\ \ =\ \ \emptyset$

i.e.  $\ \{p\}$  and $\ X\backslash G$  are $\ W$-apart.

Thus we may apply the Uniform Urysohn Lemma:

Theorem  Every uniform space is completely regular (as a topological space with the topology induced by the uniformity).

Remark  This only means that there is a continuous function $\ f:X\rightarrow [0;1]$  such that $\ f(p)=0$  and $\ f(x)=1$ for every $\ x\in X\backslash G$,  whenever $\ G$  is a neighborhood of $\ p$.  However, it does not mean that uniform spaces have to be Hausdorff spaces. In fact, uniform space with the weakest uniformity has the weakest topology, hence it's never Hausdorff, not even T₀, unless it has no more than one point.

On the other hand, when one of any two points has a neighborhood to which the other one does not belong then the two 1-point sets, consisting of these two points, are apart, hence they admit disjoint neighborhoods. Thus it is easy to prove the following:

Theorem  The following three topological properties of a uniform space $\ (X,{\mathcal {U}})$  are equivalent

• $\ (X,{\mathcal {T}}_{\mathcal {U}})$  is a T₀-space;
• $\ (X,{\mathcal {T}}_{\mathcal {U}})$  is a T₂-space (i.e. Hausdorff);
• $\bigcap \ {\mathcal {U}}\ =\ \Delta _{X}$.

When a uniform structure induces a Hausdorff topology then it's called separating.

Uniform continuity and uniform homeomorphisms

Let $(X,{\mathcal {U}})$ and $(Y,{\mathcal {V}})$ be uniform spaces. Function $\ f:X\rightarrow Y$  is called uniformly continuous if

$\forall _{V\in {\mathcal {V}}}\ (f\times f)^{-1}(V)\ \in \ {\mathcal {U}}$

A more elementary calculus δε-like equivalent definition would sound like this (UV play the role of δε respectively):

$f\$  is uniformly continuous if (and only if) for every  $V\in {\mathcal {V}}\$  there exists  $U\in {\mathcal {U}}\$  such that for every  $x',x''\in X\$  if  $(x',x'')\in {\mathcal {U}}\$  then  $(f(x'),f(x''))\in {\mathcal {V}}$.

Every uniformly continuous map is continuous with respect to the topologies induced by the ivolved uniform structures.

Example Every constant map from one uniform space to another is uniformly continuous.

A uniform map $\ f:X\rightarrow Y$  of a uniform space $(X,{\mathcal {U}})$  into a uniform space $(Y,{\mathcal {V}})$  is called a uniform homeomorphism of these two spaces) if it is bijective, and the inverse function $\ f^{-1}:Y\rightarrow X$  is a uniform map of$(Y,{\mathcal {V}})$  into $(X,{\mathcal {U}})$.

Constructions and operations

Constructions of new uniform spaces based on already existing uniform spaces are called operations. Otherwise they are called simply constructions. Thus the uniformity induced by a metric (see above) is an example of a construction (of a uniformity).

A full conceptual appreciation of operations and constructions requires the theory of categories (see below).

Partial order of uniformities

The set of uniform structures in a set $\ X$  is (partially) ordered by the inclusion relation; given two uniformities $\ {\mathcal {U}}$  and $\ {\mathcal {V}}$  in $\ X$  such that $\ {\mathcal {U}}\subseteq {\mathcal {V}}$  we say that $\ {\mathcal {U}}$  is weaker than $\ {\mathcal {V}}$  and $\ {\mathcal {V}}$  is stronger than $\ {\mathcal {U}}$.  The set of all uniform structures in $\ X$  has the weakest (smallest) and the strongest (largest) element (uniformity). We will see in the next section, that each set of uniform structures in $\ X$  admits the least upper bound. Thus it follows that each set admits also the greatest lower bound—indeed, the weakest uniformity is one of the lower bounds of a set, and there exists the least upper bound of the set of all lower bounds, which is the required greatest lower bound. In short, the uniformities in arbitrary set $\ X$ form a complete Birkhoff lattice.

The least upper bound

Let  $\ U,U',W,W'\subseteq X\times X$  be such that:

$U\circ U\subseteq U'$     and     $W\circ W\subseteq W'$

Then

$(U\cap W)\circ (U\cap W)\ \subseteq U'\cap W'$

The same holds not just for two but for any finite (or just arbitrary) family of pairs $\ (U,U')$ as above. In particular, let $\ A$  be an arbitrary family of uniformities in $\ X$.  We will construct the least upper bound of such a family:

For each $\ U\in {\mathcal {U}}\in {\mathcal {A}}$  let entourage $\ {\sqrt {U/{\mathcal {U}}}}\in {\mathcal {U}}$  be such that:

${\sqrt {U/{\mathcal {U}}}}\circ {\sqrt {U/{\mathcal {U}}}}\ \subseteq {\mathcal {U}}$

Then, whenever for a finite (or any) family $\ {\mathcal {C}}\subseteq {\mathcal {A}}$  an entourage $\ U_{\mathcal {U}}$  is selected for each $\ {\mathcal {U}}\in {\mathcal {C}}$, we obtain:

$(\bigcap _{{\mathcal {U}}\in {\mathcal {C}}}{\sqrt {U_{\mathcal {U}}/{\mathcal {U}}}})\circ (\bigcap _{{\mathcal {U}}\in {\mathcal {C}}}{\sqrt {U_{\mathcal {U}}/{\mathcal {U}}}})\ \subseteq \ \bigcap _{{\mathcal {U}}\in {\mathcal {C}}}U_{\mathcal {U}}$

Now it is easy to see that the family

${\mathcal {B}}\ :=\ \{\bigcap _{{\mathcal {U}}\in {\mathcal {C}}}U_{\mathcal {U}}:{\mathcal {C}}\in {\mathit {Fin}}({\mathcal {A}})\ \land \ \forall _{{\mathcal {U}}\in {\mathcal {C}}}\ U_{\mathcal {U}}\in {\mathcal {U}}\}$

is a uniform base. It is obvious that the uniformity $\ {\mathcal {U}}_{\mathcal {B}}$,  generated by $\ {\mathcal {B}}$,  is the least upper bound of ${\mathcal {A}}$:

$\ {\mathcal {U}}_{\mathcal {B}}\ =\ {\mathit {lub}}({\mathcal {A}})$

Preimage

Let $\ X$  be a set; let $\ (Y,{\mathcal {V}})$  be a uniform space; let $\ f:X\rightarrow Y$  be an arbitrary function. Then

${\mathcal {B}}_{f}\ :=\ \{(f\times f)^{-1}(V):V\in {\mathcal {V}}\}$

is a base of a uniform structure $\ {\mathcal {U}}_{f}$  in $\ X$.  Uniformity $\ {\mathcal {U}}_{f}$  is called the preimage of uniformity $\ {\mathcal {V}}$  under function $\ f$.  Now $\ f$  became a uniform map of the uniform space $\ (X,{\mathcal {U}}_{f})$  into $\ (Y,{\mathcal {Y}})$.  Moreover, and that's the whole point of the preimage operation, uniformity $\ {\mathcal {U}}_{f}$  is the weakest in $\ X$ , with respect to which function $\ f$  is uniform.

• Let $\ X$  be a set; let $\ (Y,{\mathcal {V}})$  be a uniform space; let $\ f:X\rightarrow Y$  be an arbitrary surjection. Then for every uniform space $\ (Z,{\mathcal {W}})$,  and every function $\ g:Y\rightarrow Z$  such that $\ g\circ f$  is a uniform map of $\ (X,{\mathcal {U}}_{f})$  into $\ (Z,{\mathcal {W}})$,  the function $\ g$  is a uniform map of $\ (Y,{\mathcal {V}})$  into $\ (Z,{\mathcal {W}})$.

The preimage uniformity can be characterized purely in terms of function; thus the following theorem could be a (non-constructive) definition of the preimage uniformity:

Theorem  Let $\ X$  be a set; let $\ (Y,{\mathcal {V}})$  be a uniform space; let $\ f:X\rightarrow Y$  be an arbitrary function. The preimage uniformity is the only uniform structure $\ {\mathcal {U}}={\mathcal {U}}_{f}$  which satisfies the following two conditions:

• $\ f$  is a uniform map of $\ (X,{\mathcal {U}})$  into $\ (Y,{\mathcal {V}})$;
• for every uniform space $(E,{\mathcal {S}})$,  and for every function $\ c:E\rightarrow X$,  if $\ f\circ c$  is a uniform map of $(E,{\mathcal {S}})$,  into $(Y,{\mathcal {V}})$,  then $\ c$  is a uniform map of $\ (E,{\mathcal {S}})$  into $\ (X,{\mathcal {U}})$.

Proof  The first condition means that $\ {\mathcal {U}}$  is stronger than the preimage $\ {\mathcal {U}}_{f}$;  and the second condition, once we substitute $(E,{\mathcal {S}}):=(X,{\mathcal {U}}_{f})$,  and $\ c:={\mathit {Id}}_{X}$, tells us that $\ {\mathcal {U}}$  is weaker than $\ {\mathcal {U}}_{f}$.  Thus $\ {\mathcal {U}}={\mathcal {U}}_{f}$.  Of course $\ {\mathcal {U}}$  satisfies both conditions of the theorem.

End of proof.

Uniform subspace

Let $\ (Y,{\mathcal {V}})$  be a uniform space; let $\ X$  be a subset of $\ Y$.  Let uniformity $\ {\mathcal {U}}$  be the primage of uniformity $\ {\mathcal {V}}$  under the identity embedding $\ i:X\rightarrow Y$  (where $\ \forall _{x\in X}\ i(x):=x$).  Then $\ (X,{\mathcal {U}})$  is called the uniform subspace of the uniform space $\ (Y,{\mathcal {V}})$,  and $\ {\mathcal {U}}$  – the subspace uniformity. It is directly described by the equality:

${\mathcal {U}}\ =\ \{V\cap (X\times X):\ V\in {\mathcal {V}}\}$

The subspace uniformity is the weakest in $\ X$  under which the embedding $\ i:X\rightarrow Y$  is uniform.

The following theorem is a characterization of the subspace uniformity in terms of functions (it is a special case of the theorem about the preimage structure; see above):

Theorem  Let $\ X\subseteq Y$,  where $\ (Y,{\mathcal {V}})$  is a uniform space. The subspace uniformity is the only uniform structure $\ {\mathcal {U}}$  in $\ X$  which satisfies the following two conditions:

• the identity embedding $\ i:X\rightarrow Y$  is a uniform map of $\ (X,{\mathcal {U}})$  into $\ (Y,{\mathcal {V}})$;
• for every uniform space $(E,{\mathcal {S}})$,  and for every function $\ c:E\rightarrow X$,  if $\ i\circ c$  is a uniform map of $(E,{\mathcal {S}})$  into $(Y,{\mathcal {V}})$,  then $\ c$  is a uniform map of $\ (E,{\mathcal {S}})$  into $\ (X,{\mathcal {U}})$.

Uniform (Cartesian) product

Let $\ {\mathcal {X}}:=\left(\left(X_{a},{\mathcal {U}}_{a}\right):a\in A\right)$  be an indexed family of uniform spaces. Let $\ \pi _{a}:X\rightarrow X_{a}$  be the standard projection of the cartesian product

$\ X:=\prod _{a\in A}\ X_{a}$

onto $\ X_{a}$,  for every $\ a\in A$. Then the least upper bound of the preimage uniformities:

${\mathcal {U}}\ :=\ {\mathit {lub}}\,\{{\mathcal {U}}_{\pi _{a}}:a\in A\}$

is called the product uniformity in $\ X$,  and $\ (X,{\mathcal {U}})$  is called the product of the uniform family $\ {\mathcal {X}}$.  Thus the product uniformity is the weakiest under which the standard projections are uniform. It is characterized in terms of functions as follows:

Theorem  The product uniformity ${\mathcal {U}}$  (see above) is the only one in the Cartesian product $\ X$,  which satisfies the following two conditions:

• each projection $\ pi_{a}\ (a\in A)$  is a uniform map of $\ (X_{a},{\mathcal {U}}_{a})$  into $\ (X,{\mathcal {U}})$;
• for every uniform space $(E,{\mathcal {S}})$,  and for every (indexed) family of uniform maps $\ c_{a}:E\rightarrow X_{a}$,  of $(E,{\mathcal {S}})$  into $\ (X_{a},{\mathcal {U}}_{a})$  (for $\ a\in A$)  there exists exactly one uniform map $\ c:E\rightarrow X$  such that:

$\forall _{a\in A}\ c_{a}\ =\ \pi _{a}\circ c$

Remark  The theory of sets tells us that that unique uniform map $\ c$  is, as a function, the diagonal product:

$c=\triangle _{a\in A}\ c_{a}$

Thus the above theorem really says that the diagonal product of uniform maps is uniform.

Remark  In many texts the diagonal product,  $\ \triangle _{a\in A}\ c_{a}$,  is called incorrectly the Cartesian product of functions,  $\prod _{a\in A}c_{a}$;  the correct terminology is used for instance in  "Outline of General Topology"  by Ryszard Engelking.

The category of the uniform spaces

The identity function $\ {\mathcal {I}}_{X}:X\rightarrow X$, which maps every point onto itself, is a uniformly continuous map of $(X,{\mathcal {U}})$ onto itself, for every uniform structure ${\mathcal {U}}$ in $\ X$.

Also, if $\ f:X\rightarrow Y\$ and $\ g:Y\rightarrow Z\$ are uniformly continuous maps of $\ (X,{\mathcal {U}})\$ into $\ (Y,{\mathcal {V}})\$, and of $\ (Y,{\mathcal {V}})\$ into $\ (Z,{\mathcal {W}})\$ respectively, then $\ g\circ f:X\rightarrow Z\$ is a uniformly continuous map of $\ (X,{\mathcal {U}})$ into $\ (Z,{\mathcal {W}})$.

These two properties of the uniformly continuous maps mean that the uniform spaces (as objects) together with the uniform maps (as morphisms) form a category  ${\mathit {US}}\$  (for Uniform Spaces).

Remark A morphism in category  ${\mathit {US}}\$  is more than a set function; it is an ordered triple consisting of two objects (domain and range) and one set function (but it must be uniformly continuous). This means that one and the same function may serve more than one morphism in  $\ {\mathit {US}}$.

Pointers

Pointers play a role in the theory of uniform spaces which is similar to the role of Cauchy sequences of points, and of the Cantor decreasing sequences of closed sets (whose diameters converge to 0) in mathematical analysis. First let's introduce auxiliary notions of neighbors and clusters.

Neighbors

Let $\ (X,{\mathcal {U}})\$ be a uniform space. Two subsets $\ A,B\$ of $\ X\$ are called neighbors – and then we write $\ A\delta B$ – if:

$(A\times B)\ \cap \ U\ \neq \ \emptyset \$

for arbitrary $\ U\in {\mathcal {U}}$.

• Either $\ A\delta B$  or there exists an entourage $\ W$ such that $\ A$  and $\ B$  are $\ W$-apart.

If more than one uniform structure is present then we write $\ A\delta _{\mathcal {U}}B\$ in order to specify the structure in question.

The neighbor relation enjoys the following properties:

• no set is a neighbor of the empty set;
• $A\delta B\ \Rightarrow B\delta A$
• $(A\subseteq A'\ \land \ A\delta B)\ \Rightarrow \ A'\delta B$
• $A\,\delta \,(B\cup C)\ \Rightarrow \ (A\delta B\ \lor \ A\delta C)$
• $\{x\}\,\delta \,A\ \Leftrightarrow \ x\in {\mathit {Cl}}(A)$
• ${\mathit {Cl}}(A)\cap {\mathit {Cl}}(B)\ \neq \ \emptyset \ \ \Rightarrow \ \ A\delta B$

for arbitrary $\ A,A',B\subseteq X$  and  $\ x\in X$.

Remark  Relation $\ A\delta B$,  and a set of axioms similar to the above selection of properties of $\ \delta$,  was the start point of the Efremovich-Smirnov approach to the topic of uniformity.

Also:

• if  $\ W$  is an entourage,  $\ A$  and $\ B$  are both $\ W$-sets, and $\ A$  and $\ B$  are neighbors, then the union $\ A\cup B$  is a $\ (W\circ V\circ W)$-set for every entourage $\ V$; in particular, it is a $\ (W\circ W\circ W)$-set.

Furthermore, if $\ f:X\rightarrow Y\$ is a uniformly continuous map of $\ (X,{\mathcal {U}})\$ into $\ (Y,{\mathcal {V}})$,  then

• $A\delta _{\mathcal {U}}B\ \Rightarrow \ f(A)\,\delta _{\mathcal {V}}f(B)$

for arbitrary $\ A,B\subseteq X$.

Clusters

Let $\ (X,{\mathcal {U}})\$ be a uniform space. A family $\ {\mathcal {K}}\$ of subsets of $\ X\$ is called a cluster if each two members of $\ {\mathcal {K}}\$ are neighbors.

• Every subfamily of a cluster is a cluster.
• If every member of a cluster is a $\ W$-set, then its union is a $\ W\circ W\circ W$-set.
• If $\ f:X\rightarrow Y\$ is a uniformly continuous map of  $(X,{\mathcal {U}})$  into  $\ \ (Y,{\mathcal {V}})$,  and $\ {\mathcal {K}}\$ is a cluster in $\ (X,{\mathcal {U}})$,  then

$\{f(W):W\in {\mathcal {K}}\}$

is a cluster in $\ (Y,{\mathcal {V}})$.

Pointers

A cluster $\ {\mathcal {K}}\$ in a uniform space $\ (X,{\mathcal {U}})\$ is called a pointer if for every entourage $\ U\in {\mathcal {U}}\$ there exists a $\ U$-set $\ A$  (meaning  $A\times A\subseteq U$)  such that

$\forall _{K\in {\mathcal {K}}}\ A\cap K\ \neq \ \emptyset$

If $\ f:X\rightarrow Y\$ is a uniformly continuous map of  $(X,{\mathcal {U}})$  into  $\ \ (Y,{\mathcal {V}})$,  and $\ {\mathcal {K}}\$ is a pointer in $\ (X,{\mathcal {U}})$,  then

$\{f(W):W\in {\mathcal {K}}\}$

is a pointer in $\ (Y,{\mathcal {V}})$.

• Every base of neighborhoods of a point is a pointer. Thus the filter of all neighborhoods of a point is called the pointer of neighborhoods (of the given point).

Equivalence of pointers, maximal and minimal pointers

Let the elunia of two families $\ {\mathcal {K}},{\mathcal {L}}$,  be the family  $\ {\mathcal {K}}\Cup {\mathcal {L}}$  of the unions of pairs of elements of these two families, i.e.

$\ {\mathcal {K}}\Cup {\mathcal {L}}\ :=\ \{K\cup L:K\in {\mathcal {K}},\ L\in {\mathcal {L}}\}$

Definition  Two pointers $\ {\mathcal {K}},{\mathcal {L}}$   are called equivalent if their  $\ {\mathcal {K}}\Cup {\mathcal {L}}$  elunia is a pointer, in which case we write $\ {\mathcal {K}}\sim {\mathcal {L}}$.

This is indeed an equivalence relation: reflexive, symmetric and transitive.

• Two pointers are equivalent if and only if their union is a pointer.

• The union of all pointers equivalent with a given one is a pointer from the same equivalence class. Thus each equivalent class of pointers has a pointer which contains every pointer of the given class. The following three properties of a pointer $\ {\mathcal {P}}$  in a uniform space $\ (X,{\mathcal {U}})$  are equivalent:
  • if $\ A\subseteq X$  is a neighbor of every member of $\ {\mathcal {P}}$  then $\ A\in {\mathcal {P}}$;
  • ${\mathcal {P}}$  is not contained in any pointer different from itself;
  • ${\mathcal {P}}$  contains every pointer equivalent to itself.

• Let $\ {\mathcal {P}}$  be a pointer in $\ (X,{\mathcal {U}})$.  Let

$P_{U}\ :=\ \bigcup \ \{A\in {\mathcal {P}}:A\times A\subseteq U\}$

for every entourage $\ U\in {\mathcal {U}}$.  Then $\ P_{U}$  is a $U\circ U\circ U$-set. It follows that

${\mathcal {Q}}\ :=\ \{P_{U}:U\in {\mathcal {U}}\}$

is a pointer equivalent to $\ {\mathcal {P}}$.

• Let's call a pointer $\ {\mathcal {P}}$  upward full if it has every superset $\ B\subseteq X$  of each of its members $\ A\in {\mathcal {U}}$.  If $\ {\mathcal {P}}$  is an arbitrary pointer, then its upward fulfillment

${\mathcal {P}}'\ :=\ \{B:\exists _{A\in {\mathcal {P}}}\ A\subseteq B\subseteq X\}$

is an upward full pointer equivalent to $\ {\mathcal {P}}$.

• Let $\ {\mathcal {P}}$  be a pointer which is maximal in its equivalence class. Let \mathcal Q</math>  be the pointer defined above. Let \mathcal Q'</math>  be its upward fulfillment. Pointer \mathcal Q'</math>  is the unique upward full pointer of its class, which is contained in any other upward full pointer of this class.

We see that each equivalent class of pointers has two unique pointers: one maximal in the whole class, and one minimal among all upward full pointers.

Convergent pointers

A pointer $\ {\mathcal {P}}$  in a uniform space is said to point to point $\ x$  if it is equivalent to the pointer of the neighborhoods of $\ x$.  When a pointer points to a point then we say that such a pointer id convergent.

• A uniform space is Hausdorff (as a topological space) of and only if no pointer converges to more than one point.

Complete uniform spaces and completions

A uniform space is called complete if each pointer of this space is convergent.

Remark  In mathematical practice (so far) only Hausdorff complete uniform spaces play an important role; it must be due to the fact that in Hausdorff spaces each pointer points to at the most one point, and to exactly one in the case of a Hausdorff complete space.

For every uniform space $\ (X,{\mathcal {U}})$  its completion is defined as a uniform map $\ c:X\rightarrow X'$  of $\ (X,{\mathcal {U}})$  into a Hausdorff complete space $\ (X',{\mathcal {U}}')$,  which has the following universality property:

for every uniform map $\ f:X\rightarrow Y$  of $\ (X,{\mathcal {U}})$  into a Hausdorff complete space $\ (Y,{\mathcal {V}})$ there exists exactly one uniform map $\ f':X'\rightarrow Y$  of $\ (X',{\mathcal {U}}')$  into $\ (Y,{\mathcal {V}})$ such that $\ f=f'\circ c$.

Theorem  For every uniform space$\ (X,{\mathcal {U}})$  there exists a completion $\ c':X\rightarrow X'$  of $\ (X,{\mathcal {U}})$  into a Hausdorff complete space $\ (X',{\mathcal {U}}')$.  Such a completion is unique up to a uniform homeomorphism, meaning that if $\ c'':X\rightarrow X''$  is another completion of $\ (X,{\mathcal {U}})$  into a Hausdorff complete space $\ (X'',{\mathcal {U}}'')$.  then there is exactly one uniform homeomorphism $\ h:X'\rightarrow X''$  such that $\ c''=h\circ c'$.

Remark  The second part of the theorem, about the uniqueness of the completion (up to a uniform homeomorphism) is an immediate consequence of the definition of the completion (it has a uniqueness statement as its part).