Uniform Spaces 2 - Introduction to Uniform Spaces

Pre-requisites: point-set topology, metric spaces, previous post on uniform spaces

This is my second post in a series on uniform spaces. This post goes over the basic definitions of uniform spaces and how they relate to the more familiar metric spaces. My other posts are:

1. Nets
2. Introduction to Uniform Spaces
3. The Completion of a Uniform Space
4. The Bolzano-Weierstrass Theorem
5. Countably Uniform Spaces

What is a Uniform Space?

A uniform space is a topological space with extra structure that provides a well-defined notion of "closeness". In a general topological space, it is possible to talk about when a sequence of points converges (see my previous post), and hence "get arbitrarily close", but it is impossible to quantify how "close" two individual points are. This is possible in uniform spaces. Before we jump into formal definitions, I want to take a moment to consider the most important examples of uniform spaces: metric spaces.

The metric $d$ of a metric space $(X, d)$ automatically gives us a way to define how "close" two points are. Formally, for $\epsilon > 0$, say that two points $p, q \in X$ are "$\epsilon$-close" if $d(p, q) \leq \epsilon$. Importantly, metric spaces don't have a single definition of "close", but rather infinitely many: one for each $\epsilon > 0$. Depending on how exact you want to be, you may consider $p$ and $q$ to be "close" if they are "1-close", or maybe if they are "0.5-close", or maybe only if they are "0.0001-close". The smaller the value of $\epsilon$, the stronger the notion of "$\epsilon$-closeness", but the metric space structure itself doesn't tell you which is the right notion of "closeness" to use. When we talk about "closeness" in a metric space, we have to keep all of these in mind.

To more easily talk about these various notions of "close", define the following sets.

Now, $p$ and $q$ are "$\epsilon$-close" if and only if $(p, q) \in A_\epsilon$. These sets are called entourages. They are also the language of uniform spaces.

Definition: a uniform space is a pair $(X, \Phi)$ where $X$ is a set and $\Phi$ is a collection of subsets of $X \times X$ called entourages satisfying:

1. $X \times X \in \Phi$.
2. If $A \in \Phi$ and $A \subset B$, then $B \in \Phi$.
3. If $A, B \in \Phi$, then $A \cap B \in \Phi$.
4. If $A \in \Phi$, then $A$ contains the diagonal.
5. If $A \in \Phi$, then so is its opposite, $A^{opp}$.
6. If $A \in \Phi$, then there exists $B \subset A$ such that the composition $B \circ B$ is a subset of $A$.

Drawing from our intuition of metric spaces, each entourage captures a "notion of closeness" for points in $X$. For any entourage $A$, say that two points $p$ and $q$ are "$A$-close" if $(p, q) \in A$.

The first axiom states that $X \times X \in \Phi$. This implies that $X$ has some notion of closeness. Since two points $p, q$ will be "$(X \times X)$-close", this represents the weakest possible notion of closeness.

If $A$ and $B$ are subsets of $X \times X$, and $A \subset B$, then whenever two points are "$A$-close", they will also be "$B$-close". In other words, $B$ represents a weaker notion of "closeness" than $A$ does. The second axiom states that if $A$ is an entourage, and $B$ is weaker than $A$, then $B$ is also an entourage. A common theme in uniform spaces is that weaker entourages don't really matter, so there's no harm in always including them. This axiom isn't really needed (see the below discussion on bases), but it does simplify things.

The third axiom states that if $A$ and $B$ are two notions of closeness, then so is $A \cap B$. Note that if two points $p, q$ are "$(A \cap B)$-close", then they are both "$A$-close" and "$B$-close". The first three axioms together define a structure called a filter.

The fourth axiom states that every entourage $A$ contains the diagonal $\Delta = \{(x, x) : x \in X\}$. In other words, every point $x \in X$ is "$A$-close" to itself. It would be pretty weird if any point were not "close" to itself!

The fifth axiom states that for any entourage $A$, its opposite $A^{opp} := \{(x, y) : (y, x) \in A\}$ is also an entourage. This axiom captures the idea that "closeness" should be a symmetrical relation: $p$ is "close" to $q$ if and only if $q$ is "close" to $p$. For technical reasons, we don't force every entourage to be symmetrical. Instead, the fifth axiom in conjunction with the third axiom guarantees the next best thing: every entourage $A$ contains a (stronger) symmetric entourage, $A \cap A^{opp}$.

The sixth axiom is the most complicated, but essentially captures the idea of being "half as close". Given any two entourages $B, C$, define their composition to be

The sixth axiom states that for all entourages $A$, there exists an entourage $B$ such that $B \circ B \subset A$. Informally, we can think of $B$ as being half the size of $A$. This idea is somewhat justified in the context of metric spaces, where $A_{\epsilon/2} \circ A_{\epsilon/2} \subset A_{\epsilon}$.

Every metric space (and more generally every pseudometric space), has a canonical uniform space structure which we now describe. Let $(X, d)$ be a metric space. Let $A \subset X \times X$ be an entourage if there exists some $\epsilon > 0$ such that $A_\epsilon \subset A$. Let $\Phi$ be the set of all entourages of $X$. The reader is encouraged to verify that $(X, \Phi)$ satisfies the axioms of a uniform space.

For an example of a uniform space that can not be obtained from any metric space, consider $X = [0,1]^J$, where $J$ is any uncountable index set. For each $\epsilon > 0$, let $A_\epsilon$ be the entourage of $[0,1]$ described in the previous paragraph. For each $j \in J$, let $A_\epsilon^j \subset [0,1] \times [0,1]$ be $A_\epsilon$, considered as an entourage of the $j^\text{th}$ copy of $[0,1]$. $X$ can be made into a uniform space by defining $A \subset X \times X$ to be an entourage of $X$ if

for some $\epsilon > 0$, some $n$, and some sequence $j_1 \dots j_n \in J$. The reader is again encouraged to check that this is a valid uniform space. We will later prove that this space is not the uniform space of any metric.

Separated Spaces

We say that a uniform space is separated if the uniform structure can tell apart any two distinct points. Formally, a uniform space $(X, \Phi)$ is separated if whenever $a \neq b$, there exists $A \in \Phi$ such that $(a, b) \not\in A$. For example, the uniform structure on a metric space is always separated, but the uniform structure on a pseudometric space may not be.

A non-separated uniform space can usually be replaced by a separated uniform space without any problems. Let $(X, \Phi)$ be a uniform space. Define an equivalence relation on $X$ by: $a \sim b$ if for all entourages $A$, $(a, b) \in A$. Informally, $a \sim b$ if $a$ and $b$ can not be separated. By identifying points that can't be separated, we can turn any uniform space on the set $X$ into a separated uniform space on $X / \sim$. As such, we don't lose a whole lot by only considering separated uniform spaces. In these notes, I only really care about separated uniform spaces.

Bases

All of the uniform spaces that we have encountered so far (metric spaces, $[0,1]^J$, and topological groups) are defined in the same way with reference to some special collection of entourages. This is a very common theme that shows up again and again when dealing with uniform spaces. To deal with this systematically, we introduce the concept of a basis of a uniform space.

Definition: Given a uniform space $(X, \Phi)$, say that some subset $\Gamma \subset \Phi$ is a basis for $\Phi$ if for all $A \in \Phi$, there exists $B \in \Gamma$ such that $B \subset A$.

Note: $\Gamma$ is sometimes called a fundamental system of $\Phi$.

For example: if $(X, d)$ is a metric space, and $(X, \Phi)$ is the corresponding uniform space, then $\Gamma$ is a basis for $\Phi$, where

As we shall see, it is often times easier to deal with a basis for a uniform space than it is to deal with the entire uniform space, in the same way that it is often easier to deal with the basis for a topology than it is to deal with the whole topology. This is possible because a basis uniquely defines its uniform structure.

Note that not every set $\Gamma$ of subsets of $X \times X$ is the basis of uniform structure. For the sake of completeness, the following lemma characterizes when a collection of entourages forms a basis. The reader is encouraged to verify that the basis entourages of a metric space satisfies these requirements.

Lemma: A set $\Gamma$ of subsets of $X \times X$ is the basis for a uniform space if and only if:

1. $\Gamma$ is non-empty
2. If $A, B \in \Gamma$, then there exists $C \in \Gamma$ such that $C \subset A \cap B$.
3. If $A \in \Gamma$, then $A$ contains the diagonal.
4. If $A \in \Gamma$, then there exists $B \in \Gamma$ such that $B \subset A^{opp}$.
5. If $A \in \Gamma$, then there exists $B \in \Gamma$ such that $B \circ B \subset A$.

Now that we have a bit of experience with the definitions, it's time for some examples!

Example: Let $X$ be any set and let $\Gamma = \{\Delta\}$, where $\Delta \subset X \times X$ is the diagonal. $\Gamma$ is the basis for a uniform structure $\Phi$. It can be seen that $A \subset X \times X$ is an entourage if and only if it contains the diagonal. We call $(X, \Phi)$ the discrete uniform space on $X$.

Example: On the opposite end, let $X$ be any set and let $\Gamma = \{X \times X\}$. $\Gamma$ is again the basis for a uniform structure $\Phi$, but now $\Phi = \{X \times X\}$. We call $(X, \Phi)$ the antidiscrete uniform space on $X$.

Example: (For readers familiar with topological groups) Let $G$ be a topological group. For any neighborhood $U$ of the identity, define

Let $\Gamma$ be the set of all $A_U$ sets, and let $\Gamma'$ be the set of all $A_U'$ sets. Then, $\Gamma$ and $\Gamma'$ are the bases for two (generally different) uniform structures on $G$. If $G$ is an abelian topological group, then these two uniform structures agree. We will investigate this uniform space in depth in later blog posts.

Exercise: Let $X = \mathbb{N}$, and let $\Gamma$ be all subsets $A \subset X \times X$ that contain the diagonal and such that $X - A$ is finite. Is $\Gamma$ the basis of a uniform space?

Exercise: Let $X = \mathbb{N}$ and make $X$ into a metric space with $d(x, y) = |x - y|$. Describe the uniform structure on $X$.

Exercise (Harder): Let $X = (0, \infty)$. Let $d_1(x, y) = |x - y|$ and $d_2(x, y) = |\frac{1}{x} - \frac{1}{y}|$ be two metrics on $X$. Let $\Phi_1$ be the uniform structure of $(X, d_1)$. Let $\Phi_2$ be the uniform structure on $(X, d_2)$. Prove that $\Phi_1 \neq \Phi_2$.

The Topology of a Uniform Space

Every uniform space is also a topological space, allowing us to apply techniques and intuition from topology to the study of uniform spaces.

To motivate the general case, we again refer back to the uniform space of a metric space and try to describe its topology in terms of its uniform structure. Let $(X, d)$ be a metric space. A subset $U \subset X$ is open if and only if for all $x \in U$, there exists $\epsilon > 0$ such that if $d(x, y) \leq \epsilon$, then $y \in x$. In other words, $U$ is open if and only if it contains all points that are "$A_\epsilon$-close" to $x$. Since every entourage of $X$ contains $A_\epsilon$ for some $\epsilon$, we actually have the following.

Lemma: Let $(X, d)$ be a metric space. Let $U \subset X$. $U$ is open if and only if for all $x \in U$, there exists some entourage $A$ such that $U$ contains all points "$A$-close" to $x$. $\square$.

This is a definition that works for all uniform spaces! For convenience, whenever $A \subset X \times X$, define $A[x] = \{y : (x, y) \in A\}$. This is the set of all points "$A$-close" to $x$.

Definition: Let $(X, \Phi)$ be a uniform space. The corresponding topological space $(X, \tau)$ is defined by: $U \in \tau$ if and only if for all $x \in U$, there exists $A \in \Phi$ such that $A[x] \subset U$.

The reader is encouraged to check that $(X, \tau)$ is in fact a topological space. In the future, we will often conflate a uniform space and its corresponding topological space.

It is not clear at first what the topological space $(X, \tau)$ looks like, and even whether it contains any nontrivial open sets. The following theorem nicely describes $(X, \tau)$ in terms of its neighborhoods.

Theorem: Let $(X, \Phi)$ be a uniform space. For all $x \in X$, the collection $\{A[x] : A \in \Phi\}$ is a neighborhood basis for $x$.

Proof: If $U$ is an open neighborhood of $x$, then there exists $A \in \Phi$ such that $A[x] \subset U$. So, it suffices to show that for all $A \in \Phi$, there exists an open neighborhood $U$ of $x$ such that $U \subset A[x]$.

Let $A \in \Phi$ and $x \in X$ be arbitrary.

Let $A_0 = A$. Inductively, find $A_{n+1} \in \Phi$ such that $A_{n+1} \circ A_{n+1} \subset A_n$. Let $Z_n = (A_n \circ A_{n-1} \dots A_1)[x]$.

Let $U = \bigcup_n Z_n$. Note that $Z_n \subset (A_n \circ A_n \circ A_{n-1} \circ A_{n-2} \dots A_1)[x]$. Since $A_n \circ A_n \subset A_{n-1}$, $Z_n \subset (A_{n-1} \circ A_{n-1} \circ A_{n-2} \dots A_1)[x]$. Repeating this argument, $Z_n \subset A_0[x]$. If $z \in Z_n$, there exists $A_{n+1}[z] \subset Z_{n+1} \subset U$, so $Z_n$ is open.

$\square$.

Part of the power of uniform spaces is that they can add additional information to topological spaces, allowing you to apply uniform space techniques to a purely topological problem. We will see this in a later blog post with a proof of Tychonoff's Theorem using uniform spaces. In order for this to work, whenever a relevant space $X$ has is both a topological space $(X, \tau)$ and a uniform space $(X, \Phi)$, the canonical topological space derived from $(X, \Phi)$ must be $(X, \tau)$. In this case, we say that the topological and uniform structures are compatible.

For example, every metric space is both a topological space and a uniform space. It is straightforward to check that these topological and uniform structures are compatible.

For another example, it's possible to define a subspace of a uniform space. If $(X, \Phi)$ is a uniform space and $Y \subset X$ is any subset, then $(Y, \Psi)$ is a uniform subspace of $X$, where $\Psi = \{A |_{Y \times Y} : A \in \Phi \}$.

Lemma: Let $(X, \Phi)$ be a uniform space, and let $Y \subset X$. The subspace topology on $Y$ is the same as the topology induced by the uniform structure on $Y$, as a uniform subspace of $X$.

Proof: Let $y \in Y$. In either topology, a neighborhood basis for $y$ is given by $\{A[y] \cap Y : A \in \Phi\}$. $\square$.

It's natural to wonder which topological spaces have a compatible uniform structure. The full answer is not easy. One easy step in that direction is the following.

Lemma: Every separated uniform space is Hausdorff.

Proof: Let $(X, \Phi)$ be a separated uniform space. There exists $A \in \Phi$ such that $(a, b) \not\in A$. Let $B \in \Phi$ such that $B \circ B \subset A$. Without loss of generality, we may take $B$ to be symmetric. If $c \in B[a] \cap B[b]$, then $(a, c) \in B$ and $(c, b) \in B$. Thus, $(a, b) \in A$. So, we must have $B[a]$ and $B[b]$ disjoint. Thus, $a$ and $b$ are separated by disjoint neighborhoods. So, $X$ is Hausdorff as a topological space. $\square$.

Most generally, a topological space has a compatible uniform structure if and only if it is completely regular, but a proof of this fact would take us too far afield.

Uniform Functions

Uniformly continuous functions are very important in analysis. These are functions that are not just continuous, but are equally continuous throughout their entire domain. As you might guess from the name, uniformly continuous functions are also very important to uniform spaces. Recall that a function $f: X \rightarrow Y$ between metric spaces is continuous if and only if for all $x \in X$ and all $\epsilon > 0$, there exists $\delta > 0$ such that if $d(x, x') < \delta$, then $d(f(x'), f(x)) < \epsilon$. Say that $f$ is uniformly continuous if for all $\epsilon > 0$, there exists a single $\delta > 0$ dependent only on $\epsilon$ such that whenever $d(x, x') < \delta$, $d(f(x), f(x')) < \epsilon$.

The function $f: \mathbb{R} \rightarrow \mathbb{R}$ given by $f(x) = x^2$ is not uniformly continuous because $f(x + 1) - f(x)$ can be arbitrarily large. On the other hand, $f(x) = x$ and $f(x) = \sqrt{|x|}$ are both uniformly continuous. It's a well-known fact that every continuous function $f: [0,1] \rightarrow \mathbb{R}$ is uniformly continuous.

You could rephrase uniform continuity with metric spaces as: a function $f: X \rightarrow Y$ is uniformly continuous if and only if for all $\epsilon > 0$, there exists $\delta > 0$ such that $(f \times f)^{-1}(A_\epsilon)$ contains $A_\delta$. This can be generalized to all uniform spaces.

Definition: A function $f: X \rightarrow Y$ between uniform space is uniform if for all entourages $A$ of $Y$, $(f \times f)^{-1}(A)$ is an entourage of $X$.

It is easy to prove that the identity map is uniform, and that the composition of two uniform maps is uniform. Thus, uniform spaces and uniform maps form a category. We will talk more about this category in the next section.

Two uniform spaces are called isomorphic if there exist uniform function $f: X \rightarrow Y$ and $g: Y \rightarrow X$ such that $g \circ f$ is the identity on $Y$ and $f \circ g$ is the identity on $X$. If this is the case, then $f$ and $g$ are called isomorphisms. Note that a uniform map $f: X \rightarrow Y$ is an isomorphism if and only if it is bijective and the inverse $f^{-1}: Y\rightarrow X$ is uniform. For example, $f: \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = x$ is an isomorphism (where $\mathbb{R}$ is here considered as a uniform space). However, $f(x) = x^{1/3}$ is not an isomorphism because its inverse is not uniform. Isomorphic uniform spaces are for all intents and purposes identical.

A uniform map $f: X \rightarrow Y$ is called a uniform embedding if $f$ is an isomorphism onto the image $f(X)$ as a subspace of $Y$. Uniform embeddings capture the idea of "being a subspace": if $X \subset Y$, then the inclusion map $X \rightarrow Y$ is a uniform embedding; if $X \rightarrow Y$ is a uniform embedding, then $X$ can be identified with a subspace of $Y$.

Building on our intuition from metric spaces, we might expect that every uniform map is continuous. This is indeed the case. Because of this, uniform maps (between uniform spaces) are sometimes still called "uniformly continuous".

Theorem (Continuity of Uniform Maps): Let $(X, \Phi)$ and $(Y, \Psi)$ be uniform spaces. Let $f: X \rightarrow Y$ be uniform. Then, $f$ is continuous with respect to the induced topologies on $X$ and $Y$.

Proof: Pick $x \in X$ and $A \in \Psi$. It suffices to prove that there exists $B \in \Phi$ such that $f(B[x]) \subset A[f(x)]$. This holds for $B = (f \times f)^{-1}(A)$.

$\square$.

Exercise: Show that a uniform embedding $X \rightarrow Y$ of metric spaces is also a topological embedding.

Exercise: Show that isomorphism is an equivalence relation between uniform spaces.

Exercise: Give an example of a continuous, bounded, but not uniformly continuous function $f: \mathbb{R} \rightarrow \mathbb{R}$.

Exercise: Let $A = S^2 - \{p\}$ and $B = \mathbb{R}^2$ be metric spaces. Show that $A$ and $B$ are not isomorphic as uniform spaces.

Exercise: Give an example of two metric spaces $A, B$ where $A$ is bounded and $B$ is unbounded, but $A$ and $B$ are isomorphic as uniform spaces.

The Category of Uniform Spaces

Uniform spaces and uniform maps form a category. We will call this the uniform category or $\textbf{Unif}$, although I'm not aware of any official name. This section is written for readers familiar with basic category theory and uses categorical language freely. For the purpose of brevity, this section contains no proofs, and is more a reference for some facts about $\textbf{Unif}$.

$\textbf{Unif}$ contains all products and coproducts. All products and coproducts commute with the functor $T$.

The categorical product can be described as follows. Let $(X_\alpha)_{\alpha \in I}$ be a collection of uniform spaces. Let $X = \prod_\alpha X_\alpha$ be the Cartesian product (as sets). For any finite subset $J \subset I$ and any collection $(A_\alpha)_{\alpha \in J}$ where $A_\alpha$ is an entourage of $X_\alpha$, let $\prod_{\alpha \in J} A_\alpha \times \prod_{\alpha \not\in J} (X_\alpha \times X_\alpha)$ be an entourage of $\prod_{\alpha \in I} X_\alpha$. These entourages form a basis for a uniform structure on $X$. Along with the projection maps $\pi_\alpha: X\rightarrow X_\alpha$, this makes $X$ into the categorical product of $(X_\alpha)_{\alpha \in J}$. In particular, the final object (empty product) is a one-point space with its unique uniform structure.

The categorical coproduct can be described as follows. Let $(X_\alpha)_{\alpha \in I}$ be a collection of uniform spaces. Let $X = \coprod_\alpha X_\alpha$ be the disjoint union (as sets). For any (possibly infinite) collection $(A_\alpha)_{\alpha \in I}$ where $A_\alpha$ is an entourage of $X_\alpha$, let $\bigcup_{\alpha \in I} A_\alpha$ be an entourage of $X$. These entourages form the basis for a uniform structure on $X$. Along with the inclusion maps $\iota_\alpha: X_\alpha \rightarrow X$, this makes $X$ into the categorical coproduct of $(X_\alpha)_{\alpha \in I}$. In particular, the initial object in the uniform category is $\varnothing$, the empty set, with its unique uniform structure.

The product of uniform spaces actually gives some great examples of uniform spaces that are not simply metric spaces. For example, consider $X = [0,1]^J$ where $J$ is some uncountable set. This uniform space does not have a countable basis. However, the uniform space associated to every metric space $(Y, d)$ does have a countable basis, namely $\Gamma = \{A_{1/n}:n \in \mathbb{N}\}$. So, $X$ is not the uniform space of any metric space.

More generally, $\textbf{Unif}$ actually contains all limits and colimits.

The association of a topological space to a uniform space is actually a functor $T: \textbf{Unif} \rightarrow \textbf{Top}$. $T$ commutes with all limits and colimits.

The association of a uniform space to a pseudometric space is also a functor $S: \textbf{PseudoMetric} \rightarrow \textbf{Unif}$. This functor also commutes with all limits and colimits. The composition $T \circ S$ is (naturally isomorphic to) the canonical functor $\textbf{PseudoMetric} \rightarrow \textbf{Top}$.

For more categorical facts, see this mathoverflow.net answer.

Extensions: This series of posts talks about uniform spaces in the context of entourages. There are two other major approaches to uniform spaces. One of these is based on certain open covers of the space, which I find harder to motivate. The other approach uses pseudometrics to define uniform spaces, but for many uniform spaces (such as topological groups), it takes some work to prove that the uniform structure can be defined by pseudometrics.