## How to Complete Your Metric Space

Suppose you have a metric space $(X,d)$, and you would like to solve exercises 23, 24, and 25 of Chapter 3 in Baby Rudin. You would like your metric space to be complete (i.e. every Cauchy sequence is convergent), as it makes verifying convergence easier: you don’t actually need to find the limiting value first, and then verify that quantity is indeed the limit. Rather, in a complete metric space, it suffices to show that terms of sequence $(x_{n})_{n=1}^{\infty}$ with sufficiently high indices become arbitrarily close. Mathematically speaking, for any $\epsilon > 0$, there exists an integer $N$ such that

$\displaystyle m,n \geq N \Longrightarrow d(x_{n},x_{m}) < \epsilon$

But what if $(X,d)$ is not complete? Not to worry, we can always construct its completion $(\tilde{X},\tilde{d})$, which is a complete metric space with the property that there is an injective isometry (i.e. metric-preserving map) $\varphi: X \rightarrow \tilde{X}$ such that $\varphi(X)$ is dense in $\tilde{X}$. Moreover, $\varphi: X \rightarrow \tilde{X}$ satisfies a universal property. If $(Y,\rho)$ is a metric space and $\psi: X \rightarrow Y$ is a continuous map, then there exists a unique continuous map $\Phi: \tilde{X} \rightarrow Y$ such that the following diagram commutes:

Now let’s prove these claims. We begin with a lemma on the real sequence $(d(x_{n},x'_{n}))_{n=1}^{\infty}$ for Cauchy sequences $(x_{n}), (x'_{n})$ in $X$.

Lemma 1. If $(x_{n})_{n=1}^{\infty}, (x'_{n})_{n=1}^{\infty}$ are Cauchy sequences in $X$, then $(d(x_{n},x'_{n}))_{n=1}^{\infty}$ converges in $\mathbb{R}$.

Proof. If suffices to show the sequence is Cauchy. Let $\epsilon > 0$ be given. We can choose a positive integer $N$ such that for $n,m \geq N$, we have $\max\left\{d(x_{n},x_{m}), d(x_{n}',x_{m}')\right\} < \frac{\epsilon}{2}$. By the triangle inequality,

$\displaystyle d(x_{n},x_{n}') \leq d(x_{n},x_{m}) + d(x_{m},x_{m}') + d(x_{m}',x_{n}'),$

so that if $d(x_{n},x_{n}') \geq d(x_{m},x_{m}')$, then

$\displaystyle \left|d(x_{n},x_{n}') - d(x_{m},x_{m}')\right| \leq d(x_{n},x_{m}) + d(x_{m}',x_{n}') < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

$\Box$

We use the preceding lemma to define an equivalence relation on Cauchy sequences in $X$. Call two Cauchy sequences $(x_{n})_{n=1}^{\infty}, (x'_{n})_{n=1}^{\infty}$ in $X$ equivalent, which we denote by $(x_{n}) \sim (x'_{n})$, if and only if

$\displaystyle \lim_{n \rightarrow \infty} d(x_{n},x_{n}') = 0$

That $\sim$ is, in fact, an equivalence relation is immediate from the axioms of a metric. Denote the set of equivalence relations with respect to $\sim$ by $\tilde{X}$. For Cauchy sequences $(x_{n})_{n=1}^{\infty} \sim (x_{n}')_{n=1}^{\infty}$ and $(z_{n})_{n=1}^{\infty}$ in $X$, I claim that

$\displaystyle \lim_{n \rightarrow \infty} d(x_{n},z_{n}) = \lim_{n \rightarrow \infty} d(x_{n}',z_{n}),$

where we know the LHS and RHS exist by Lemma 1. Indeed, by the triangle inequality,

$\displaystyle d(x_{n},z_{n}) \leq d(x_{n},x_{n}') + d(x_{n}',z_{n}), \indent d(x_{n}',z_{n}) \leq d(x_{n}',x_{n}) + d(x_{n},z_{n})$

so that $\lim_{n \rightarrow \infty} \left|d(x_{n},z_{n}) - d(x_{n}',z_{n})\right| \leq \lim_{n \rightarrow \infty} d(x_{n},x_{n}') = 0$. If $\tilde{x}, \tilde{x'}$ denote the equivalence classes containing $(x_{n})$ and $(x_{n}')$, respectively, then we can define a function $\Delta: \tilde{X} \times \tilde{X} \rightarrow [0,\infty)$ by

$\displaystyle \Delta(\tilde{x},\tilde{x'}) = \lim_{n \rightarrow \infty} d(x_{n},x_{n}')$

It follows from the metric axioms for $d$ and the definition of our equivalence relation that $\Delta$ is a metric on $\tilde{X}$.

We now show that $(\tilde{X},\Delta)$. First, a general lemma.

Lemma 2. If $A$ is a dense subset of a complete metric space $(Y,\rho)$ such that every Cauchy sequence in $A$ converges to an element in $Y$, then $(Y,\rho)$ is a complete metric space.

Proof. Let $(y_{n})_{n=1}^{\infty}$ be a Cauchy sequence in $Y$. Let $\epsilon > 0$ be given, and let $N$ be a positive integer such that $m,n \geq N$ implies that $d(y_{m},y_{n}) < \frac{\epsilon}{4}$. For each $n$, we can choose $x_{n} \in A$ such that $d(x_{n},y_{n}) < \frac{\epsilon}{2^{n}}$. I claim that $(x_{n})_{n=1}^{\infty}$ is Cauchy and therefore converges to a limit $x \in Y$. Indeed, by the triangle inequality

$\displaystyle n,m \geq N \Longrightarrow d(x_{n},x_{m}) \leq d(x_{n},y_{n}) + d(y_{n},y_{m}) + d(y_{m},x_{m}) < \frac{\epsilon}{2^{n}} + \frac{\epsilon}{4}+\frac{\epsilon}{2^{m}} < \epsilon$

We complete the proof of the lemma by showing that $y_{n} \rightarrow x$. Taking $N$ to be larger, if necessary, we may assume that $n \geq N$ implies that $d(x_{n},x) < \frac{\epsilon}{2}$. By the triangle inequality,

$\displaystyle n \geq N \Longrightarrow d(y_{n},x) \leq d(y_{n},x_{n}) + d(x_{n},x) < \frac{\epsilon}{2^{n}} + \frac{\epsilon}{2} \leq \epsilon$

$\Box$

Define $\varphi: X \rightarrow \tilde{X}$ by $x \mapsto (x_{n})_{n=1}^{\infty}$, where $x_{n} := x$ for all $n$.

Lastly, we verify the universal property. Let $(Y,\rho)$ and $\psi: X \rightarrow Y$ be as above. It is clear that $\varphi$ is injective. To see that $\varphi$ is an isometry, observe that

$\Delta(\varphi(x),\varphi(x')) = \lim_{n \rightarrow \infty} d(x_{n},x_{n}') = d(x,x'), \indent \forall x,x' \in X$

I claim that $\varphi(X)$ is dense in $\tilde{X}$. Suppose $\tilde{x} \in \tilde{X}$ contains the sequence $(x_{n})_{n=1}^{\infty}$. There exists a positive integer $N$ such that $m,n \geq N$ implies that $d(x_{n},x_{m}) < \frac{\epsilon}{2}$. Then

$\Delta(\tilde{x},\varphi(x_{N})) = \lim_{n \rightarrow \infty} d(x_{n},x_{N}) \leq \epsilon$

Let $(\varphi(x_{n}))_{n=1}^{\infty}$ be a Cauchy sequence in $\varphi(X)$. Let $\tilde{x} \in \tilde{X}$ be the equivalence class containing the sequence $(x_{n})_{n=1}^{\infty}$. Then for $n \geq N$,

$\displaystyle \Delta(\varphi(x_{n}),\tilde{x}) = \lim_{m \rightarrow \infty} d(x_{n},x_{m}) \leq \frac{\epsilon}{2} < \epsilon$

So $(\tilde{X},\Delta)$ is a complete metric space by Lemma 2.

Lastly, we prove the universal property. It is tautological that $\varphi: X \rightarrow \tilde{X}$ is continuous. For a continuous map $\psi: X \rightarrow Y$, we can define $\Phi: \tilde{X} \rightarrow Y$ as follows. For $\tilde{x} \in \tilde{X}$, let $(\varphi(x_{n}))_{n=1}^{\infty}$ be a sequence in $\varphi(X)$ that converge to $\tilde{x}$. Define

$\displaystyle \Phi(\tilde{x}) = \lim_{n \rightarrow \infty} \psi(x_{n})$

We need to verify the limit in the RHS actually exists and that $\Phi$ is independent of the choice of sequence $(\varphi(x_{n}))_{n=1}^{\infty}$. The second assertion follows from the continuity of $\psi$. Observe that $(x_{n})_{n=1}^{\infty}$ is a Cauchy sequence in $X$ by the triangle inequality, and by the continuity of $\psi$, it follows that $(\psi(x_{n}))_{n=1}^{\infty}$ is Cauchy in $Y$. Since $Y$ is complete, $\psi(x_{n})$ converges to some $y \in Y$.

The universal property allows us to deduce that any complete metric spaces, in which $X$ isometrically embeds densely, are isometrically isomorphic. Indeed, suppose $\psi: X \rightarrow Y$ is an isometric embedding into a complete metric space $(Y,\rho)$. Our work above shows that there exists a unique continuous map $\Psi: Y \rightarrow \tilde{X}$ such that the following diagram commutes:

By the universal property, $\Phi \circ \Psi: Y \rightarrow Y$ and $\Psi \circ \Phi: \tilde{X} \rightarrow \tilde{X}$ must both be the identity maps, which shows that $\Phi = \Psi^{-1}$. We see that the completion of a metric space is only defined up to isometric isomorphism.

Completing a metric space gives us another way to construct the real numbers. Let $(X,d)$ be the rational numbers with the Euclidean metric $d(x,y) := \left|x-y\right| \ \forall x,y \in X$. It is not hard to see that this metric space is not complete; the reader, as an exercise, should try to come up with some examples of Cauchy sequences in $\mathbb{Q}$ which do not have rational limits. It is a consequence of the Archimedean property of the reals that $\mathbb{Q}$ is dense in $\mathbb{R}$. Any Cauchy sequence $(x_{n})_{n=1}^{\infty}$ in $\mathbb{R}$ is bounded, and the Bolzano-Weierstrass theorem tells us that every bouded sequence has a convergent subsequence, from which we can conclude that $(x_{n})$ converges. Hence, $\mathbb{R}$ is complete and is therefore isometrically isomorphic to the completion of the rationals.

In a follow-up post, we will sketch the connection between completing the $\sigma$-algebra of a probability space and completing a metric space.