Suppose you have a metric space , 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 with sufficiently high indices become arbitrarily close. Mathematically speaking, for any , there exists an integer such that
But what if is not complete? Not to worry, we can always construct its completion , which is a complete metric space with the property that there is an injective isometry (i.e. metric-preserving map) such that is dense in . Moreover, satisfies a universal property. If is a metric space and is a continuous map, then there exists a unique continuous map such that the following diagram commutes:
Now let’s prove these claims. We begin with a lemma on the real sequence for Cauchy sequences in .
Lemma 1. If are Cauchy sequences in , then converges in .
Proof. If suffices to show the sequence is Cauchy. Let be given. We can choose a positive integer such that for , we have . By the triangle inequality,
so that if , then
We use the preceding lemma to define an equivalence relation on Cauchy sequences in . Call two Cauchy sequences in equivalent, which we denote by , if and only if
That is, in fact, an equivalence relation is immediate from the axioms of a metric. Denote the set of equivalence relations with respect to by . For Cauchy sequences and in , I claim that
where we know the LHS and RHS exist by Lemma 1. Indeed, by the triangle inequality,
so that . If denote the equivalence classes containing and , respectively, then we can define a function by
It follows from the metric axioms for and the definition of our equivalence relation that is a metric on .
We now show that . First, a general lemma.
Lemma 2. If is a dense subset of a complete metric space such that every Cauchy sequence in converges to an element in , then is a complete metric space.
Proof. Let be a Cauchy sequence in . Let be given, and let be a positive integer such that implies that . For each , we can choose such that . I claim that is Cauchy and therefore converges to a limit . Indeed, by the triangle inequality
We complete the proof of the lemma by showing that . Taking to be larger, if necessary, we may assume that implies that . By the triangle inequality,
Define by , where for all $n$.
Lastly, we verify the universal property. Let and be as above. It is clear that is injective. To see that is an isometry, observe that
I claim that is dense in . Suppose contains the sequence . There exists a positive integer such that implies that . Then
Let be a Cauchy sequence in . Let be the equivalence class containing the sequence . Then for ,
So is a complete metric space by Lemma 2.
Lastly, we prove the universal property. It is tautological that is continuous. For a continuous map , we can define as follows. For , let be a sequence in that converge to . Define
We need to verify the limit in the RHS actually exists and that is independent of the choice of sequence . The second assertion follows from the continuity of . Observe that is a Cauchy sequence in by the triangle inequality, and by the continuity of , it follows that is Cauchy in . Since is complete, converges to some .
The universal property allows us to deduce that any complete metric spaces, in which isometrically embeds densely, are isometrically isomorphic. Indeed, suppose is an isometric embedding into a complete metric space . Our work above shows that there exists a unique continuous map such that the following diagram commutes:
By the universal property, and must both be the identity maps, which shows that . 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 be the rational numbers with the Euclidean metric . 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 which do not have rational limits. It is a consequence of the Archimedean property of the reals that is dense in . Any Cauchy sequence in is bounded, and the Bolzano-Weierstrass theorem tells us that every bouded sequence has a convergent subsequence, from which we can conclude that converges. Hence, 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 -algebra of a probability space and completing a metric space.