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.

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One Response to How to Complete Your Metric Space

  1. Pingback: How to Complete Your Metric Space II | Math by Matt

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