1.1. prerequisites

A firm command of basic point-set topology is necessary to read this post. My go-to reference is Willard’s General Topology, which takes a lot of inspiration from Kelley’s famous text by the same name. Both references are quite comprehensive; in particular, they both include much more extensive treatments of uniform spaces than I give here. For a gentler introduction to point-set topology, another one of my favorites is Gamelin and Greene’s Introduction to Topology. This was the first text on topology that I ever worked though, back in my undergraduate days.

Beyond these topological prerequisites, the reader should also know the basic definitions of various basic algebraic structures, like groups, vector spaces, and algebras.

• T. W. Gamelin, R. E. Greene. Introduction to topology. Second edition. Dover Publications, Inc., Mineola, NY, 1999.

• J. L. Kelley. General topology. D. Van Nostrand Co., Inc., Toronto-New York-London, 1955.

• S. Willard. General topology. Reprint of the 1970 original [Addison-Wesley, Reading, MA]. Dover Publications, Inc., Mineola, NY, 2004.

1.2. uniform structures

Before we begin, let’s fix some notation and terminology:

Now, we give the definitions of the main objects considered in this series of posts:

The next exercise is very important, as it shows that every topological vector space has an underlying topological group, and so all general results proved for the latter type of structure automatically apply to the former type. Therefore, just as a topological algebra is an extra piece of structure placed on top of a topological vector space, so too may a topological vector space be understood as an extra piece of structure placed on top of a topological group. The whole theory is then conceptualized as a hierarchy of structures.

Topological groups have very special types of topologies:

Let $G$ be a set which is simultaneously a group and a topological space, but not necessarily a topological group. If all left- and right-translation maps \eqref{left-eqn} and \eqref{right-eqn} are homeomorphisms, then we shall say the topology on $G$ is homogeneous. This explains the parenthetical description of the theorem, as the theorem shows that all topological groups have homogeneous topologies.

One of the most convenient features of homogeneous topologies is that they are completely determined by the neighborhood system at the identity. To explain this, for completeness and convenience, we first recall a basic construction from point-set topology:

Now, let’s explain the aforementioned feature of homogeneous topologies:

To explain and prove the theorem, let’s fix some notation. For each $g\in G$ and each subset $U\subseteq G$, we define the left and right translates of $U$ to be the sets

\begin{equation}\notag gU = L_g(U) \quad \text{and} \quad Ug = R_g(U). \end{equation}

The key observation is that if $U$ is a neighborhood of $h\in G$, then $gU$ and $Ug$ are neighborhoods of $gh$ and $hg$, respectively; indeed, this follows immediately from the homogeneity property of the topology. But since any topology whatsoever is determined by the neighborhood systems of its points, this observation immediately implies that the topology is determined by the neighborhoods of $e$. Q.E.D.

In view of the theorem, it is therefore very easy to define a homogeneous topology on a group $G$. Indeed, we simply describe a set of subsets of $G$ (satisfying properties (1.)-(5.) in the definition/theorem above) that will serve as the neighborhood system of $e\in G$. We then define the translates of these sets to be the neighborhood systems of all other points. The topology obtained in this way is then homogeneous by construction. I suggest that you work out the details on your own, just to make sure you understand everything.

So, we have two types of mathematical structures at hand: Topological groups, and groups with homogeneous topologies. We have shown that every topological group has a homogeneous topology. Conversely, we might wonder about what extra conditions (if any) must be placed on a homogeneous topology to guarantee that it yields a topological group. Now, clearly, if $G$ is a topological group, then the map

\begin{equation}\label{mult-eqn} G \times G \to G, \quad (g,h) \mapsto gh^{-1}, \end{equation}

is continuous at $(e,e)$. It turns out that this condition is actually sufficient for a homogeneous topology on a group $G$ to yield a topological group:

Throughout this series of posts, the most common types of topological vector spaces that we shall encounter are normed vector spaces. Let’s recall the definition:

Then:

For the proof, let $B(v;\dev)$ denote the open $\dev$-ball centered at $v\in V$, for $\dev>0$:

\begin{equation}\notag B(v;\dev) = \left\{ w \in V : \norm{w-v} < \dev \right\}. \end{equation}

We first need to prove that the addition map

\begin{equation}\notag V \times V \to V, \quad (v,w) \mapsto v+w, \end{equation}

is continuous at an arbitrary $(v_0,w_0)$. But we have

\begin{equation}\label{first-in-eqn} \norm{(v+w)-(v_0+w_0)} \leq \norm{v-v_0} + \norm{w-w_0} < \delta_1 + \delta_2 \end{equation}

provided that $\norm{v-v_0}<\delta_1$ and $\norm{w-w_0}<\delta_2$. Thus, we may make the left-hand side of \eqref{first-in-eqn} smaller than any given $\dev>0$ by choosing $\delta_1,\delta_2>0$ small enough. Proving that the inversion map

\begin{equation}\notag V \to V, \quad v\mapsto -v, \end{equation}

is continuous is even easier, so I’ll skip the details. To finish the proof, we need to show that the scaling map

\begin{equation}\notag k \times V \to V, \quad (\lambda,v) \mapsto \lambda v, \end{equation}

is continuous at an arbitrary $(\lambda_0,v_0)$. But note that

\begin{equation}\label{second-in-eqn} \norm{\lambda v- \lambda_0v_0} \leq |\lambda| \norm{v-v_0} + |\lambda-\lambda_0| \norm{v_0} < (|\lambda_0| + \delta_1)\delta_2 + \delta_1 \norm{v_0} \end{equation}

provided that $|\lambda - \lambda_0| < \delta_1$ and $\norm{v-v_0}<\delta_2$. Thus, by choosing $\delta_1,\delta_2>0$ sufficiently small, we may guarantee that the left-hand side of \eqref{second-in-eqn} is less than any given $\dev>0$. Q.E.D.

By combining a normed vector space and an algebra with one extra condition, we get the following fundamental structure:

As we just saw, the underlying vector space of a normed algebra $A$ is a topological vector space. In fact, the submultiplicativity of the norm is the exact ingredient needed to upgrade $A$ to a topological algebra:

1.3. convergence and completeness

The defining property of a Banach space, that is, the property that distinguishes it from a general normed vector space, is its completeness with respect to the metric induced by its norm. However, completeness is not a topological property, as Cauchy sequences are not defined in general topological spaces.

But neither is completeness satisfactorily and entirely captured by the metric alone. Indeed, let’s recall that two norms $\norm{\cdot}_1$ and $\norm{\cdot}_2$ on a vector space $V$ are called equivalent if there are constants $a,b>0$ such that

\begin{equation}\notag a \norm{v}_1 \leq \norm{v}_2 \leq b \norm{v}_1, \quad \forall v\in V. \end{equation}

Letting

\begin{equation}\notag d_1(v,w) = \norm{v-w}_1 \quad \text{and} \quad d_2(v,w) = \norm{v-w}_2 \end{equation}

be the two metrics induced by the norms, we then have the following important pair of results, the proofs of which I leave to you:

The combination of these two results suggests that completeness should be a topological property, but again, as we already observed, completeness is not a general topological property!

The answer is that completeness is a uniform property, a type of property belonging to mathematical structures called uniform spaces. Between the two extremes of general topological spaces and metric spaces, you should imagine that the class of uniform spaces is a middle ground somewhere in between. So, the class of uniform spaces can be conceptualized as a generalization of metric spaces, but, fascinatingly, it also serves as a generalization of topological groups:

In order to motivate the definition of a uniform space, we shall examine the particular parts of a metric space structure that support the definitions of Cauchy sequences and uniformly continuous functions. In fact, we shall find it convenient to study these definitions in a more general pseudometric space $(X,d)$, which is a set $X$ equipped with a pseudometric $d$, which means that $d$ satisfies all the axioms of a metric except the requirement that

\begin{equation}\notag d(x,y) = 0 \quad \Leftrightarrow \quad x=y, \end{equation}

for $x,y\in X$.

Now, let’s recall the definitions:

The key observation is that these two definitions do not depend solely on single points of the pseudometric space $X$, but rather on pairs of points of $X$. In fact, suppose that to every $\dev>0$ we define the following subset of the cartesian product $X\times X$:

\begin{equation}\notag U_{X,\dev} = \left\{ (x_1,x_2) \in X\times X \mid d_X(x_1,x_2) < \dev\right\}. \end{equation}

Then a completely trivial re-write of the definitions above replaces the implication \eqref{unif-cont-eqn} with

and replaces \eqref{Cauchy-eqn} with

The definition of a general uniform space is obtained by isolating several properties of the sets $U_{X,\dev}$ implied by the pseudometric axioms, and using these properties as the defining axioms for a so-called uniformity. Then, a set equipped with a uniformity will be called a uniform space. In this section, we shall study the definition and basic properties of uniform spaces, leaving the discussion of convergence (and completeness) to the subsequent section.

To formulate these properties properly, we first need to set up some notation:

Equipped with this new notation, let’s return to our considerations of pseudometric spaces:

I want to step through the proof in detail, to show you how the pseudometric axioms imply all four of these properties.

(1): This is just a consequence of the fact $d(x,x)=0$, for all $x\in X$.

(2): Given $\dev>0$, we have $U^{-1}_\dev = U_\dev$ since the pseudometric is symmetric: $d(x,y) = d(y,x)$, for all $x,y\in X$. Thus, we may take $\delta = \dev$.

(3): This is a consequence of the triangle inequality. Indeed, suppose that we take $\delta=\dev/2$. Then, given $(x,z) \in U_{\dev/2} \circ U_{\dev/2}$, there is a $y\in X$ such that

\begin{equation}\notag d(x,z) \leq d(x,y) + d(y,z) \leq \dev/2 + \dev/2 = \dev. \end{equation}

Hence $(x,z) \in U_\dev$.

(4): Take $\delta = \min\{\dev_1,\dev_2\}$. Q.E.D.

The next result is an exact analog for topological groups of the previous theorem, but now it’s not the pseudometric axioms that are used in its proof, but rather the continuity of the group operations.

(1): This follows from the fact $gg^{-1} = e$ for all $g\in G$, and that each $N\in \calb_e$ contains the identity.

(2): Since $G$ is a topological group, the inversion map

\begin{equation}\notag \iota:G\to G, \quad g\mapsto g^{-1}, \end{equation}

is continuous. In particular, it is continuous at $e$, so if $N\in \calb_e$, then there is a neighborhood $M\in \calb_e$ such that $\iota(M) \subseteq N$. But as you may easily check, this implies $U_M \subseteq U_{N}^{-1}$.

(3): Since $G$ is a topological group, the multiplication map

\begin{equation}\notag \mu: G \times G \to G, \quad (g,h) \mapsto gh, \end{equation}

is continuous. In particular, it is continuous at $(e,e)$, so if $N\in \calb_e$, there is a neighborhood $M\in \calb_e$ such that $\mu(M \times M) \subseteq N$. But this implies $U_{M} \circ U_{M} \subseteq U_N$.

(4): One of the properties of bases of neighborhood systems states that if $N_1,N_2\in \calb_e$, there is an $M \in \calb_e$ such that $M\subseteq N_1 \cap N_2$. But then we have $U_M \subseteq U_{N_1} \cap U_{N_2}$. Q.E.D.

As the parenthetical descriptions of these theorems indicate, they show that pseudometric spaces and topological groups both generate uniformities via the collections of subsets \eqref{base-metric-eqn} and \eqref{base-tg-eqn}. The four common properties of these collections highlighted in these theorems are extracted to form the definition of a base of a uniformity. Before giving this general definition, we first need to define uniformities and uniform spaces:

For practice with the definition, I suggest that you work through the following:

Now that we have defined uniformities and uniform spaces, we extract the properties from the theorems above on pseudometric spaces and topological groups and use them as axioms to define the following fundamental structure:

The link between bases for uniformities and actual uniformities is explained in:

Now, using this new language, we may officially say that the sets \eqref{base-metric-eqn} and \eqref{base-tg-eqn} form bases for uniformities on a pseudometric space and topological group, respectively. In the former case, the uniformity obtained in this way is called the pseudometric uniformity. If $(X,\calu)$ is a uniform space and there exists a pseudometric $d$ on $X$ for which $\calu$ is the pseudometric uniformity induced by $d$, then $(X,\calu)$ is called a pseudometrizable uniform space.

Like the association between bases of topologies and topologies, keep in mind that the association between bases of uniformities and uniformities is many-to-one: Two distinct bases for uniformities may generate the same uniformity.

Turning our attention toward induced topologies, we have:

Though every uniformity induces a unique topology, two distinct uniformities may induce the same uniform topology. This means that a uniformity on a set represents strictly more structure than the uniform topology that it induces.

The following result will prove very convenient.

To clarify, openness of an entourage $U$ is with respect to the product topology on $X\times X$, where $X$ is equipped with its uniform topology.

For the proof, first note that the collection of open entourages is nonempty, since it contains $X\times X$. For the rest of the argument, notice that it suffices to prove that

\begin{equation}\label{implication-eqn} U \in \calu \quad \Rightarrow \quad U^\circ \in \calu, \end{equation}

where $U^\circ$ is the interior of $U$. To do this, we first choose an entourage $V$ such that

\begin{equation}\label{contain1-eqn} V \circ V \circ V \subseteq U. \end{equation}

Since the symmetric entourages form a base, we can shrink $V$ if needed and assume both that it is symmetric, and that the containment \eqref{contain1-eqn} holds. Now, if we can prove that $V\subseteq U^\circ$, then \eqref{implication-eqn} will follow. But notice that if $(x,y)\in V$, we have the containment

\begin{equation}\label{contain2-eqn} V_x \times V_y \subseteq U. \end{equation}

Indeed, if $(z,w)\in V_x \times V_y$, then $(z,w) \in V \circ V \circ V$ as well, since $V$ is symmetric and all three $(x,y),(x,z),(y,w)\in V$. Then $(z,w) \in U$ by \eqref{contain1-eqn}, from which \eqref{contain2-eqn} follows. But since $V_x \times V_y$ is open in the product topology, contains $(x,y)$, and is contained in $U$, we have $(x,y) \in U^\circ$. Thus, $V \subseteq U^\circ$, as desired. Q.E.D.

1.4.

Having discussed the definitions of uniform spaces, uniform continuity, bases, and uniform topologies, we turn our attention toward convergence and completeness. As one learns in their first (thorough) course in point-set topology, sequences in the classical sense do not prove fully adequate for capturing and encoding topological notions in spaces that are not first-countable. (See the section on Further Reading.) Rather, a more general convergence theory is needed, two candidates being the theory of nets and the theory of filters.

Unfortunately, a complete and self-contained discussion of either of these latter two convergence theories would take us too far afield. However, we don’t really need them, since our primary examples of uniform spaces will all be pseudometrizable, spaces for which the classical theory of sequences is sufficient. It is enough to know a few basic relationships between sequences and nets, the first being:

One may define a Cauchy net in a uniform space by mimicking the definition of a Cauchy sequence; then, one proves easily that every convergent net is Cauchy, and that every Cauchy sequence (see below) is a Cauchy net. Spaces for which Cauchy nets converge are given a special title:

The link between the uniform notion of completeness and the classical pseudometric notion is given in:

So, as long as the uniform spaces under consideration are pseudometrizable—and all of our examples will be—this theorem allows us to focus our attention on the more familiar concept of completeness for pseudometric spaces. This theorem is the justification for our omission of a detailed description of nets.

We will need one more result before continuing; its metric version should be familiar.

The proof is rather straightforward, but it does involve nets, so it will be omitted. See the section on Further Reading for references containing proofs.

Now, let’s give the official definition of Cauchy sequences in uniform spaces:

We now turn our attention toward completions of uniform spaces. Here’s the definition:

Though technically a completion is a certain map $\iota:X \to Y$, the target $Y$ is often called the completion. As with any object defined via a universal property, a completion $Y$ is unique up to (uniform) isomorphism, provided that it exists. It is a general result of the theory that every uniform space has a completion, and proofs may be found in the references in the section on Further Reading. We will prove this existence result in the special case of pseudometrizable uniform spaces:

If $(x_n)$ is a Cauchy sequence in $X$, we will write $[x_n]$ for its equivalence class. We recall that a metric completion of $(X,d)$ is given by the metric space $\widehat{X}$, which is the set of all equivalence classes of Cauchy sequences in $X$ with metric given by

\begin{equation}\notag \widehat{d}\left([x_n],[x_n’] \right) = \lim_{n\to \infty} d(x_n,x_n’). \end{equation}

As one proves in their course in point-set topology or real analysis, the metric space $\widehat{X}$ equipped with this metric is complete, and the natural inclusion map

\begin{equation}\notag \iota: X \to \widehat{X} \end{equation}

sending each $x\in X$ to the equivalence class $[x,x,\ldots]$ is an isometric embedding onto a dense subspace of $\widehat{X}$.

If we wanted to finish the proof quickly, we could use the extension theorem to prove that the map $\iota$ have the defining universal property of a completion. However, we choose to give a self-contained proof. And actually, a “net” version of the proof we are about to give is more or less how one proves that previous extension theorem.

So, suppose that

\begin{equation}\notag \alpha: X \to Z \end{equation}

is a uniformly continuous map to a complete Hausdorff uniform space; we need to prove that there exists a unique uniformly continuous extension

\begin{equation}\notag \widehat{\alpha}: \widehat{X} \to Z \end{equation}

such that $\alpha = \widehat{\alpha} \circ \iota$. If such an extension $\widehat{\alpha}$ exists, then the usual argument shows that we must have

\begin{equation}\label{ext-eqn} \widehat{\alpha} \left([x_n] \right) = \lim_{n\to \infty} \alpha(x_n). \end{equation}

Thus, $\widehat{\alpha}$ is unique, provided that it exists.

For existence, we define $\widehat{\alpha}$ via this last formula \eqref{ext-eqn}. We must then check that $\widehat{\alpha}$ is well-defined by showing that the limit on the right-hand side of \eqref{ext-eqn} exists (and is unique!) and that it does not depend on the particular representative $(x_n)$ of the equivalence class.

To see that the limit exists, note that because $\alpha$ is uniformly continuous it carries Cauchy sequences to Cauchy sequences, and since $Z$ is complete, the limit must exist. Furthermore, since $Z$ is Hausdorff, the limit is unique.

To see that the limit does not depend on the choice of representative Cauchy sequence $(x_n)$, suppose that $(x_n’)$ is a second Cauchy sequence, equivalent to $(x_n)$. But again, since $\alpha$ is uniformly continuous, it preserves equivalence classes, and so $(\alpha(x_n))$ and $(\alpha(x_n’))$ are equivalent in $Z$. But $Z$ is Hausdorff, and hence

\begin{equation}\notag \lim_{n\to \infty} \alpha(x_n) = \lim_{n\to \infty} \alpha(x_n’). \end{equation}

All that is left is to prove that $\widehat{\alpha}$, defined by \eqref{ext-eqn}, is uniformly continuous. For this, we let an entourage $U$ in $Z$ be given, and we choose a symmetric entourage $V$ in $Z$ such that

\begin{equation}\label{three-eqn} V \circ V \circ V \subseteq U. \end{equation}

Since $\alpha$ is uniformly continuous, there is an $\dev > 0$ such that

\begin{equation}\label{imp-eqn} d(x,x’) < \dev \quad \Rightarrow \quad (\alpha(x),\alpha(x’)) \in V. \end{equation}

I claim then that

\begin{equation}\label{imp2-eqn} \widehat{d} \left([x_n],[x_n’] \right) < \dev \quad \Rightarrow \quad \left( \widehat{\alpha}\left([x_n] \right),\widehat{\alpha}\left([x_n’]\right) \right) \in U \end{equation}

for all $[x_n],[x_n’] \in \widehat{X}$. To prove this, first set

\begin{equation}\notag z = \widehat{\alpha}\left([x_n] \right) \quad \text{and} \quad z’ = \widehat{\alpha}\left([x_n’] \right). \end{equation}

If $\widehat{d} \left([x_n],[x_n’] \right) < \dev$, there is an integer $N \geq 0$ such that

\begin{equation}\notag n\geq N \quad \Rightarrow \quad d(x_n,x_n’) < \dev. \end{equation}

But in view of \eqref{imp-eqn}, this gives

\begin{equation}\notag n\geq N \quad \Rightarrow \quad (\alpha(x_n),\alpha(x_n’)) \in V. \end{equation}

Since $\alpha(x_n) \to z$ and $\alpha(x_n’) \to z’$, we may choose $N$ large enough so that

\begin{equation}\notag n\geq N \quad \Rightarrow \quad (z,\alpha(x_n)), (z’,\alpha(x_n’)) \in V. \end{equation}

Combining these last two implications with the containment \eqref{three-eqn} and the fact $V$ is symmetric, we get that $(z,z’) \in U$. This proves the desired implication \eqref{imp2-eqn} and finishes the proof. Q.E.D.

The completion $\widehat{X}$ of the pseudometrizable uniform space $X$ was constructed by using a particular choice of pseudometric $d$ on $X$. However, the defining universal property of completions shows that $\widehat{X}$ only depends on the uniform structure of $X$, at least up to uniform isomorphism. Indeed, this follows from the fact that the universal property holds in the category of uniform spaces, not just the category of metric spaces.

To end this section, we give a convenient result which lets us identify completions without having to verify the defining universal property.

This is just an application of the extension theorem.