# Uniform Spaces 5 - Countably Uniform Spaces

This is the fifth post in my series on uniform spaces. See my other posts here:

- Nets
- Introduction to Uniform Spaces
- The Completion of a Uniform Space
- The Bolzano-Weierstrass Theorem
- Countably Uniform Spaces

## Introduction

So far in this series of posts, I have introduced uniform spaces as a generalization of metric spaces, and I have drawn all of my inspiration for uniform spaces from metric spaces. Perhaps, I have even been too focused on metric spaces in these blog posts. In this post, I hope to rectify this situation by determining exactly when a uniform space arises from a metric space, and when it does not.

## Countably Uniform Spaces

I don't know an official term for what I will call "countably uniform spaces", despite how important they are to the question of when a uniform space is metrizable. If anyone knows of a more accepted term, feel free to let me know.

**Definition:** A *countably uniform space* is a uniform space with a countable basis.

Recall that a basis for a uniform space $(X, \Phi)$ is a subset $\Gamma \subset \Phi$ such that for any entourage $A \in \Phi$, there is some $B \in \Gamma$ such that $B \subset A$. (Note: this is completely separate from the notion of a topological space basis.) A uniform space can be entirely described in terms of its basis, and because of this, a smaller basis often corresponds to a less complicated space. For example, if $(X, d)$ is a metric space, then the entourages

form a countable basis for $X$, as a uniform space. Thus, every metric space is countably uniform.

For an example of a space that is not countably uniform, let $J$ be an uncountable set and consider $X = [0,1]^J$, the product of uncountably many copies of $[0,1]$. If $[0,1]^J$ had a countable basis, then $[0,1]^J$ would be first-countable as a topological space, but it is not. I leave it as an exercise to prove moreover that any basis for $[0,1]^J$ must have at least $|J|$ elements.

In general, an uncountable product or coproduct of uniform spaces is usually not countably uniform.

However, countably uniform spaces are well-behaved with respect to to all the uniform space operations that we've discussed so far as long as those operations only involve countable collections of spaces. The following proofs have been omitted since they are a bit long to write up, but they do not require any new machinery,

**Lemma:** Let $(X_n)_{n \in \mathbb{N}}$ be a countable collection of countably uniform spaces. Then, $\prod_{n \in \mathbb{N}} X_n$ is also countably uniform.

**Lemma:** Let $(X_n)_{n \in \mathbb{N}}$ be a countable collection of countably uniform spaces. Then, $\coprod_{n \in \mathbb{N}} X_n$ is also countably uniform.

**Lemma:** If $X$ is a countably uniform space, then so is $\overline{X}$.

## Metrizable Uniform Spaces.

Define a uniform space $(X, \Phi)$ to be *metrizable* if there exists a metric $d$ on $X$ such that the uniform space $(X, \Phi)$ and the metric space $(X, d)$ are compatible. In other words, the uniform space associated to $(X, d)$ is $(X, \Phi)$. The metric $d$ need not be unique, and usually isn't. It is not hard to show that every metrizable space is separable and countably uniform.

**Theorem** (Metrizability of Uniform Spaces): A uniform space is metrizable if and only if it is separable and countably uniform.

It suffices to prove that a separable, countably uniform space $X$ is metrizable. Suppose that $(A_n)_{n \in \mathbb{N}}$ is a countable uniform basis for $X$. By replacing $A_n$ with $A_n \cap A_{n-1} \cap \dots \cap A_0$, we may assume that $A_{k+1} \subset A_k$ for all $k$. By taking a subsequence of this, we may further assume that $A_{k+1} \circ A_{k+1} \subset A_k$ for all $k$.

The key to constructing a metric on $X$ is treat $A_{n}$ as a notion of "$2^{-n}$-closeness". In particular, we would like to have $d(x,y) \leq 2^{-n}$ whenever $x, y \in A_{n}$.

For any sequence $(a_1, a_2 \dots a_m)$, define the *length* of the sequence to be:

Define $d(x,y)$ to be the minimum length of any sequence whose first term is $x$ and whose second term is $y$. Although it may not be obvious, this function *is* a metric. This metric also satisfies: $d(x, y) \leq 2^{-n}$ if and only if $(x, y) \in A_{n}$. From here, it can be shown that $(X, d)$ is indeed compatible with $(X, \Phi)$. $\square$.

For me, this theorem challenges the utility of uniform spaces in general because it implies that almost every uniform space encountered in practice is just a metric space, since the vast majority of practical uniform spaces are countably generated. After all, if almost every uniform space is a metric space, why not just deal with metric spaces? I have two answers to this.

First, metric spaces carry extra information that may be irrelevant. For example, you can make $\mathbb{R}^2$ into a metric space in a number of ways, such as the Euclidean norm and the taxicab norm. Sometimes this is relevant, but sometimes all you need is the uniform structure on a metric space, and it makes more sense to not bother with assigning a metric at all.

Second, although most uniform spaces are countably uniform, not every uniform space is. For example, there are plenty of wild topological vector spaces (which I will describe in the next two posts) which require arbitrarily large bases.

## Consequences

The most important consequence to the classification of metrizable uniform spaces, in my opinion, is that it allows you to quote well-known theorems about metric spaces without re-proving them in the consequence of countably uniform spaces.

For example, consider the Arzelà–Ascoli Theorem. In its classical form, the Arzelà–Ascoli Theorem states that if $(f_n)_{n \in \mathbb{N}}$ is a sequence of continuous functions $f_n: [0,1] \rightarrow \mathbb{R}$ which is both uniformly bounded and uniformly equicontinuous, then a subsequence $(f_{n_k})_{k \in \mathbb{N}}$ converges uniformly.

While this is its most classical form, the Arzelà–Ascoli Theorem has been generalized to functions between complete metric spaces. Let $X$ and $Y$ be metric spaces, and let $S$ be a collection of functions $X \rightarrow Y$. Define $S$ to be *totally bounded valued* if $\{f(x) : x \in X, f \in S\}$ a totally bounded $Y$. Define $S$ to be *uniformly equicontinuous* if for all $\epsilon > 0$, there exists a $\delta > 0$ such that whenever $d(x_1,x_2) < \delta$ and $f \in S$, then $d(f(x_1), f(x_2)) < \epsilon$.

**Theorem** (Arzelà–Ascoli Theorem for Metric Spaces): Let $(f_n)_{n \in \mathbb{N}}$ be a sequence of functions $f_n: X \rightarrow Y$ where $X$ is a compact metric space and $Y$ is a complete metric space. If this sequence is uniformly totally bounded valued and equicontinuous, then a subsequence $(f_{n_k})_{k \in \mathbb{N}}$ converges uniformly.

Now we can generalize from metric spaces to countably uniform spaces. Recall from the last post that a separable uniform space is a compact if and only if it is totally bounded and complete. So, we may replace $X$ by a complete, totally bounded, countably uniform space and replace $Y$ by a complete, countably uniform space.

Recall also that every continuous function on a compact uniform space is automatically uniformly continuous. So, the function $f_n: X \rightarrow Y$ should be uniformly continuous.

There are also clear ways to write the notions of "uniformly equicontinuous" and "uniformly convergent" without reference to a metric.

Let $X, Y$ be uniform space, and let $S$ be a collection of uniform functions $X \rightarrow Y$. Define $S$ to be *uniformly equicontinuous* if for all entourage $A$ of $X$, there exists an entourage $B$ of $Y$ such that $(f \times f)(B) \subset A$ for all $f \in S$.

Let $X, Y$ be uniform space, and let $(f_n)_{n \in \mathbb{N}}$ be a sequence of uniform functions $X \rightarrow Y$. Say that $(f_n)_{n \in \mathbb{N}}$ *converges uniformly* to $f: X \rightarrow Y$ if for any entourage $B$ of $Y$, there exists $N \in \mathbb{N}$ such that for all $n, m > N$, we have $(f_n(x), f_m(x)) \in B$.

**Theorem** (Arzelà–Ascoli Theorem for Uniform Spaces): Let $X$ be a complete, totally bounded countably uniform space, and let $Y$ be a complete, countably uniform space. Let $(f_n)_{n \in \mathbb{N}}$ be a sequence of uniform functions $f_n: X \rightarrow Y$. If this sequence is totally bounded valued uniformly equicontinuous, then it is converges uniformly.

On the one hand, this sort of proof is fairly easy to construct and can tell you a lot about uniform spaces and uniform continuity. On the other hand, the construction of a metric space on a countably uniform space is not trivial, and I find proofs that bootstrap metric space results directly to uniform space results (as I have just done) to be less instructive and less aesthetically pleasing. This is why I haven't discussed the metrizability of uniform spaces before now.

## An Application to Topology

Lastly, because I can't resist, I once again conclude with an application to compact topological spaces.

**Theorem:** Let $X$ be a compact Hausdorff topological space. $X$ is metrizable if and only if it is second-countable.

*Proof:* Suppose that $X$ is a compact metric space. Then, $X$ is second-countable by a simple topological argument. For each $r > 0$, $X$ can be covered by a finite number of open balls of radius $r$, say $B_r(x_1) \dots B_r(x_m)$. Let $S_r = \{B_{r}(x_i) : 1 \leq i \leq m\}$. If $x \in X$ and $U$ is a neighborhood of $X$, then for some $r > 0$, $B_{2r}(x) \subset U$. Thus, for some $i$, $x \in B_r(x_i)$ and $B_r(x_i) \subset U$. We can see that $\bigcup_{n \in \mathbb{N}} S_{1/n}$ is a topological basis for $X$.

Now suppose that $X$ is compact, second-countable space. We know from the previous post that $X$ has a unique uniform structure, and we even know a basis for this uniform structure. Given a finite open cover $C =(U_1 \dots U_m)$ of $X$, define:

The collection of entourages $\{A_C : C \text{ is a finite open cover of } X\}$ is a basis for the uniform structure on $X$. We just need to show that this uniform structure is countably uniform.

Let $S$ be a countable topological basis for $X$. Since $X$ is compact, every finite open cover can be refined to a finite open cover using only sets in $S$. As a result, the uniform structure on $X$ has a basis of entourages of the form $A_C$, where $C$ is a finite open cover of $X$ using only sets in $S$. There are a countable number of finite subsets of $S$, so a countable number of finite open covers using $S$. Thus, $X$ is countably uniform. $\square$.

*Extensions:* The metrizability of topological spaces is a very important area of topology; the two most important results here are (perhaps) Urysohn's Theorem, which applies only to second-countable spaces, and the general Nagata–Smirnov Theorem. I do not know if the version of the Arzelà–Ascoli Theorem for Uniform Spaces I gave can be generalized to non-countably uniform spaces, but my intuition says that it can't.