Uniform Spaces 4 - The Bolzano-Weierstrass Theorem

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

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

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

The Bolzano-Weierstrass Theorem

So far in this series on uniform spaces, we have talked about uniform spaces primarily as generalizations of metric spaces, and used ideas from the study of metric spaces to motivate the theory of uniform spaces. However, I haven't yet explained how the study of uniform spaces can be applied to other areas of math. In this post, I will discuss the connection between uniform spaces and compact topological spaces, and end with an application of uniform spaces to a purely topological result (Tychonoff's Theorem).

The key to this proof is the Bolzano-Weierstrass Theorem for uniform spaces which characterizes compactness in terms of uniform space properties that are easier to check.

Theorem (Bolzano-Weierstrass in a Uniform Space): [Assuming choice] A uniform space is compact if and only if it is complete and totally bounded.

To explain the terminology and intuition behind this theorem, we once again turn to metric spaces.

The Bolzano-Weierstrass Theorem is an extremely important theorem in analysis and topology. It is, in a sense, the canonical link between metric spaces and compact spaces.

Theorem (Bolzano-Weierstrass for Metric Spaces): A metric space is compact if and only if it is complete and totally bounded.

Recall that a metric space $M$ is totally bounded if for all $\epsilon > 0$, there exists a finite list of points $x_1 \dots x_m$ such that every point in $M$ is within distance $\epsilon$ of some $x_i$. I tend to think of compact topological spaces as "the small topological spaces". As every totally bounded metric space can be embedded into its completion, a complete totally bounded metric space (and hence, a compact metric space), I view totally bounded metric spaces as being "the small metric spaces". For example, a subset of $\mathbb{R}^n$ is totally bounded if and only if it is bounded.

In terms of entourages, a metric space $M$ is totally bounded if and only if for all entourages $A_\epsilon$, there exists a finite collection $\{x_1, x_2, \dots x_n\}$ of points in $X$ such that

In this form, we can clearly see how to generalize the notion of total boundedness to uniform spaces.

Definition: A uniform space $(X, \Phi)$ is totally bounded if for all $A \in \Phi$, there exists a finite collection $\{x_1, x_2 \dots x_n\}$ such that

Just like metric spaces, I tend think of totally bounded uniform spaces as "the small uniform spaces". For example: every subspace of a totally bounded space is totally bounded, and a union of two totally bounded spaces is totally bounded.

Importantly for the next section, the completion of a totally bounded space is still totally bounded.

Lemma: Let $(X, \Phi)$ be a totally bounded uniform space. Then, $\overline{X}$ is also totally bounded.

Proof: Let $\overline{\Phi}$ be the uniform structure on $\overline{X}$. Fix an entourage $A \in \overline{\Phi}$. Let $B \in \overline{\Phi}$ such that $B \circ B \subset A$. Let $C$ be the restriction of $B$ to $X \times X$. Since $X$ is totally bounded, there exist $x_1 \dots x_n \in X$ such that

Hence,

Since $X$ is dense in $\overline{X}$: for all $x \in \overline{X}$, there exists $y \in X$ such that $(x, y) \in B$. Thus,

$\square$.

There is a nice characterization of totally bounded spaces in terms of Cauchy nets that is similar to the Bolzano-Weierstrass Theorem for uniform spaces.

Theorem: [Assuming choice] Let $(X, \Phi)$ be a uniform space. Then $X$ is totally bounded if and only if every net in $X$ has a Cauchy subnet.

Proof: This can be proven similarly to the Bolzano-Weierstrass Theorem for uniform spaces. $\square$.

We can now prove the Bolzano-Weierstrass Theorem for uniform spaces:

Theorem (Bolzano-Weierstrass for Uniform Spaces): [Assuming choice] A uniform space is compact if and only if it is complete and totally bounded.

Proof: Let $(X, \Phi)$ be a compact uniform space. In the first blog post of this series, we proved that a topological space is compact if and only if every net has a convergent subnet. If a Cauchy net $a_\bullet$ has a convergent subnet $b_\bullet \rightarrow x$, then $a_\bullet \rightarrow x$. Thus, every Cauchy net in $X$ converges. In other words, $X$ is complete.

To prove that $X$ is totally bounded, fix an entourage $A$. For every $x \in X$, $A[x]$ contains an open neighborhood $U_x$ of $x$.

As $X$ is compact, this open cover has a finite subcover $(U_{x_1} \dots U_{x_n})$. Thus,

Hence, $X$ is totally bounded.

Now suppose that $X$ is complete and totally bounded. We need to prove that $X$ is compact. We know from the first blog post of this series that a space is compact if and only if every net has a convergent subnet. Suppose for contradiction that $a_\bullet: J \rightarrow X$ is a net in $X$ with no convergent subnet.

Define a set $\mathcal{S} \subset \mathcal{P}(X)$ (a set of subsets of $X$) to be a "good collection" if the following hold:

1. $\mathcal{S} \neq \varnothing$
2. Whenever $A_1 \dots A_n \in \mathcal{S}$, then $a_\bullet \in (A_1 \cap \dots \cap A_n)$ infinitely often.

For example, $\{X\}$ is a good collection. Whenever $(\mathcal{S}_\alpha)_\alpha$ is an increasing chain of good collections, then $\bigcup_\alpha \mathcal{S_\alpha}$ is a good collection. By Zorn's Lemma, there exists a maximal good collection $\mathcal{S}^*$. It can be shown that $\mathcal{S}^*$ is a filter on $X$, and that for any $A \in \mathcal{S}^*$, $a_\bullet \in A$ infinitely often. Less obviously, it is also true that whenever $A \in \mathcal{S}^*$ and $A = \bigcup_{i=1}^n A_i$, then $A_i \in \mathcal{S}^*$ for some $i$.

We will use $\mathcal{S}^*$ to construct a Cauchy net without a limit. Define

Order $\mathcal{T}$ by: $(A_1, j_1) < (A_2, j_2)$ when $A_2 \subset A_1$. Let $b_\bullet: \mathcal{T} \rightarrow X$ by $b_{(A, j)} = a_j$. It is not hard to show that $b_\bullet$ is naturally a subnet of $a_\bullet$.

We now show that $b_\bullet$ is Cauchy. Fix an entourage $B \in \Phi$. Let $C \in \Phi$ such that $C \circ C \subset B$ and $C$ is symmetric. Since $X$ is totally bounded, there exist $x_1 \dots x_n \in X$ such that

It follows from the properties of $\mathcal{S}^*$ that $C[x_i] \in \mathcal{S}^*$ for some $i$. Thus, $b_\bullet \in C[x_i]$ eventually. Hence, $(b_\bullet, b_\bullet) \in B$ eventually as well. It follows that $b_\bullet$ is Cauchy.

Since $b_\bullet$ is Cauchy and $X$ is complete, $b_\bullet$ converges. Thus, $a_\bullet$ has a convergent subnet. $\square$.

Compact Spaces as Uniform Spaces

In order to apply uniform space techniques to boring old topological spaces, we need to give those topological spaces uniform structures. Thankfully, every compact Hausdorff topological space has a unique uniform structure. The details are omitted, but the uniform structure can be defined follows.

Let $X$ be a compact Hausdorff topological space. For every finite open cover $(U_1 \dots U_n)$ of $X$, construct the entourage

These entourages generate a uniform structure on $X$.

Tychonoff's Theorem (Again)

After a couple lemmas about uniform spaces, we will be ready to apply uniform space theory to Tychonoff's Theorem.

Lemma: Let $(X_\alpha)_{\alpha \in I}$ be a collection of complete uniform spaces. The product $X = \prod_{\alpha \in I} X_\alpha$ is complete.

Proof: Consider a Cauchy net $a_\bullet$ in $X$. The projection of $a_\bullet$ into $X_\alpha$ is a Cauchy net in $X_\alpha$, and has a limit $x_\alpha$. Let $x$ be the point $(x_\alpha)_{\alpha \in I}$. The limit of $a_\bullet$ is $x$. $\square$.

Lemma: Let $(X_\alpha)_{\alpha \in I}$ be a collection of totally bounded uniform spaces. The product $X = \prod_{\alpha \in I} X_\alpha$ is totally bounded.

Proof: If $X$ is empty, then it is trivially totally bounded. So assume that $X$ is non-empty. In particular, for any $x \in X_\alpha$, there exists $y \in X$ whose projection onto $X_\alpha$ is $x$.

Since entourages of the form $E = A_\alpha \times \prod_{\beta \neq \alpha} (X_\beta\times X_\beta)$ generate the uniform structure on $X$, it suffices to check the total boundedness condition on these entourages. If

for some collection $x_1 \dots x_n$ in $X_\alpha$, then

where the projection of $y_i$ onto $X_\alpha$ is $x_i$. $\square$.

Theorem (Tychonoff's Theorem, Hausdorff Case): [Assuming choice] Let $(X_\alpha)_{\alpha \in I}$ be a collection of compact Hausdorff topological spaces. The product $X = \prod_{\alpha \in I} X_\alpha$ is compact.

Proof: Each $X_\alpha$ has a unique uniform structure, and that structure is complete and totally bounded. It suffices to show that $X$ is complete and totally bounded as a uniform space. This follows from the above two lemmas. $\square$.

The use of the Axiom of Choice in this proof (via the Bolzano-Weierstrass Theorem for uniform spaces) can not be avoided. In fact, it has been shown that the Tychonoff's Theorem is equivalent to the Axiom of Choice (in ZF). While this is arguably not the simplest proof of Tychonoff's Theorem, it still (I would argue) displays the power of uniform spaces.

Extensions: You can also define uniform spaces in terms of open covers, and if so, many of the arguments presented in this post become more natural.