# The Unapologetic Mathematician

## Complete Uniform Spaces

Okay, in a uniform space we have these things called “Cauchy nets”, which are ones where the points of the net are getting closer and closer to each other. If our space is sequential — usually a result of assuming it to be first- or second-countable — then we can forget the more complicated nets and just consider Cauchy sequences. In fact, let’s talk as if we’re looking at a sequence to build up an intuition here.

Okay, so a sequence is Cauchy if no matter what entourage we pick to give a scale of closeness, there’s some point along our sequence where all of the remaining points are at least that close to each other. If we pick a smaller entourage we might have to walk further out the sequence, but eventually every point will be at least that close to all the points beyond it. So clearly they’re all getting pressed together towards a limit, right?

Unfortunately, no. And we have an example at hand of where it can go horribly, horribly wrong. The rational numbers $\mathbb{Q}$ are an ordered topological group, and so they have a uniform structure. We can give a base for this topology consisting of all the rays $(a,\infty)=\{x\in\mathbb{Q}|a, the rays $(-\infty,a)=\{x\in\mathbb{Q}|x, and the intervals $(a,b)=\{x\in\mathbb{Q}|a, which is clearly countable and thus makes $\mathbb{Q}$ second-countable, and thus sequential.

Okay, I’ll take part of that back. This is only “clear” if you know a few things about cardinalities which I’d thought I’d mentioned but it turns out I haven’t. It was also pointed out that I never said how to generate an equivalence relation from a simpler relation in a comment earlier. I’ll wrap up those loose ends shortly, probably tomorrow.

Back to the business at hand: we can now just consider Cauchy sequences, instead of more general Cauchy nets. Also we can explicitly give entourages that comprise a base for the uniform structure, which is all we really need to check the Cauchy condition: $E_a=\{(x,y)\in\mathbb{Q}\times\mathbb{Q}|\left|x-y\right|. I did do absolute values, didn’t I? So a sequence $x_i$ is Cauchy if for every rational number $a$ there is an index $N$ so that for all $i\geq N$ and $j\geq N$ we have $\left|x_i-x_j\right|.

We also have a neighborhood base $\mathcal{B}(q)$ for each rational number $q$ given by the basic entourages. For each rational number $r$ we have the neighborhood $\{x\in\mathbb{Q}|\left|x-q\right|. These are all we need to check convergence. That is, a sequence $x_i$ of rational numbers converges to $q$ if for all rational $r$ there is an index $N$ so that for all $i\geq N$ we have $\left|x_i-q\right|.

And finally: for each natural number $n\in\mathbb{N}$ there are only finitely many square numbers less than $2n^2$. We’ll let $a_n^2$ be the largest such number, and consider the rational number $x_n=\frac{a_n}{n}$. We can show that this sequence is Cauchy, but it cannot converge to any rational number. In fact, if we had such a thing this sequence would be trying to converge to the square root of two.

The uniform space $\mathbb{Q}$ is shot through with holes like this, making tons of examples of Cauchy sequences which “should” converge, but don’t. And this is all just in one little uniform space! Clearly Cauchy nets don’t converge in general. But we dearly want them to. If we have a uniform space in which every Cauchy sequence does converge, we call it “complete”.

Categorically, a complete uniform space is sort of alike an abelian group. The additional assumption is an extra property which we may forget when convenient. That is, we have a category $\mathbf{Unif}$ of uniform spaces and a full subcategory $\mathbf{CUnif}$ of complete uniform spaces. The inclusion functor of the subcategory is our forgetful functor, and we’d like an adjoint to this functor which assigns to each uniform space $X$ its “completion” $\overline{X}$. This will contain $X$ as a dense subspace — the closure $\mathrm{Cl}(X)$ in $\overline{X}$ is the whole of $\overline{X}$ — and will satisfy the universal property that if $Y$ is any other complete uniform space and $f:X\rightarrow Y$ is a uniformly continuous map, then there is a unique uniformly continuous $\bar{f}:\overline{X}\rightarrow Y$ extending $f$.

To construct such a completion, we’ll throw in the additional assumption that $X$ is second-countable so that we only have to consider Cauchy sequences. This isn’t strictly necessary, but it’s convenient and gets the major ideas across. I’ll leave you to extend the construction to more general uniform spaces if you’re interested.

What we want to do is identify Cauchy sequences in $X$ — those which should converge to something in the completion — with their limit points in the completion. But more than one sequence might be trying to converge to the same point, so we can’t just take all Cauchy sequences as points. So how do we pick out which Cauchy sequences should correspond to the same point? We’ll get at this by defining what the uniform structure (and thus the topology) should be, and then see which points have the same neighborhoods.

Given an entourage $E$ of $X$ we can define an entourage $\overline{E}$ as the set of those pairs of sequences $(x_i,y_j)$ where there exists some $N$ so that for all $i\geq N$ and $j\geq N$ we have $(x_i,y_j)\in E$. That is, the sequences which get eventually $E$-close to each other are considered $\overline{E}$-close.

Now two sequences will be equivalent if they are $\overline{E}$-close for all entourages $E$ of $X$. We can identify these sequences and define the points of $\overline{X}$ to be these equivalence classes of Cauchy sequences. The entourages $\overline{E}$ descend to define entourages on $\overline{X}$, thus defining it as a uniform space. It contains $X$ as a uniform subspace if we identify $x\in X$ with (the equivalence class of) the constant sequence $x, x, x, ...$. It’s straightforward to show that this inclusion map is uniformly continuous. We can also verify that the second-countability of $X$ lifts up to $\overline{X}$.

Now it also turns out that $\overline{X}$ is complete. Let’s consider a sequence of Cauchy sequences $(x_k)_i$. This will be Cauchy if for all entourages $\overline{E}$ there is an $\bar{N}$ so that if $i\geq\bar{N}$ and $j\geq\bar{N}$ the pair $((x_k)_i,(x_k)_j)$ is in $\overline{E}$. That is, there is an $N_{i,j}$ so that for $k\geq N_{i,j}$ and $l\geq N_{i,j}$ we have $((x_k)_i,(x_l)_j)\in E$. We can’t take the limits in $X$ of the individual Cauchy sequences $(x_k)_i$ — the limits along $k$ — but we can take the limits along $i$! This will give us another Cauchy sequence, which will then give a limit point in $\overline{X}$.

As for the universal property, consider a uniformly continuous map $f:X\rightarrow Y$ to a complete uniform space $Y$. Then every point $\bar{x}$ in $\overline{X}$ comes from a Cauchy sequence $x_i$ in $X$. Being uniformly continuous, $f$ will send this to a Cauchy sequence $f(x_i)$ in $Y$, which must then converge to some limit $\bar{f}(\bar{x})\in Y$ since $Y$ is complete. On the other hand, if $x_i'$ is another representative of $\bar{x}$ then the uniform continuity of $f$ will force $\lim f(x_i)=\lim f(x_i')$, so $\bar{f}$ is well-defined. It is unique because there can be only one continuous function on $\overline{X}$ which agrees with $f$ on the dense subspace $X$.

So what happens when we apply this construction to the rational numbers $\mathbb{Q}$ in an attempt to patch up all those holes and make all the Cauchy sequences converge? At long last we have the real numbers $\mathbb{R}$! Or, at least, we have the underlying complete uniform space. What we don’t have is any of the field properties we’ll want for the real numbers, but we’re getting close to what every freshman in calculus thinks they understand.

November 29, 2007