Section 2.2 Existence of the Completion
The metric subspace \(\Q\) of \(\R\) (with the usual metric) is an example of an incomplete metric space. We can use it as an inspiration for the following definition of the completion of a metric space. In this case, we can get around the incompleteness of \(\Q\) by working in \(\R\) instead. In general, given an incomplete metric space, we want to do something similar.
Definition 2.7.
A completion of a metric space \((X,d)\) is a complete metric space \((\hat{X}, \hat{d})\) with a dense subset \(\hat{X}_0\) isometric to \(X\text{.}\)
Remark 2.8.
It is usual to identify \(\hat{X}_0\) with \(X\) and view \(X\) as a subset of \(\hat{X}\text{.}\)
Example 2.9.
\(\R\) is a completion of \(\Q\text{.}\)
This concept would be of no use if we didn’t know whether a given metric space has a completion. Fortunately, it turns out that every metric space has one.
Theorem 2.10.
Any metric space has a completion.
Proof.
Let \((X,d)\) be a metric space, and define a relation \(\sim\) between Cauchy sequences \(\seq xn, \seq yn\) in \((X,d)\) by
\begin{equation*}
\seq xn \sim \seq yn \iff \lim_{n \to \infty} d(x_n,y_n) = 0\text{.}
\end{equation*}
You are asked to show in Exercise 2.2.1 that \(\sim\) is an equivalence relation. This allows us to define
\begin{equation}
\hat X = \{ \text{Cauchy sequences in }(X,d) \} / \sim\text{.}\tag{2.1}
\end{equation}
Thus elements of \(\hat X\) are equivalence classes \([\seq xn]\) where \(\seq xn\) is a Cauchy sequence in \((X,d)\text{.}\)
Intuitively, we are thinking of \([\seq xn]\) as standing in for the ‘limit’ \(\lim_{n \to \infty} x_n\) which may have been ‘missing’ from the set \(X\text{.}\) The quotient in (2.1) then allows us to think of any two Cauchy sequences ‘with the same limit’ as the same point in \(\hat X\text{.}\) With this intuition in mind, we want to associate an element \(x \in
X\) with the equivalence class of the constant (and hence Cauchy) sequence \([(x)_{n\in\N}] \in \hat X\text{.}\) To make this precise, we set
\begin{equation}
\hat X_0 = \{ [(x)_{n\in\N}] : x \in X \}\tag{2.2}
\end{equation}
and let \(f \maps X \to \hat X_0\) be the bijection defined by
\begin{equation}
f(x) = [(x)_{n \in \N}]\text{.}\tag{2.3}
\end{equation}
Finally, we define \(\hat d \maps \hat X \times \hat X \to \R\) by
\begin{equation}
\hat d\big([\seq xn],[\seq yn]\big) = \lim_{n\to \infty} d(x_n,y_n) \text{.}\tag{2.4}
\end{equation}
\(\hat d\) is well-defined and a metric.
Let \(\seq xn\) and \(\seq yn\) be Cauchy sequences in \((X,d)\text{.}\) By Exercise 1.3.4, the limit on the right hand side of (2.4) exists. To show that \(\hat d\) is well-defined, we must further show that the right hand side of (2.4) does not depend on which representatives we choose of the equivalence classes on the left hand side. To this end, let \(\seq{x'}n\) and \(\seq {y'}n\) be Cauchy sequences in \(X\) with \(\seq xn \sim \seq{x'}n\) and \(\seq
yn \sim \seq{y'}n\text{.}\) Then using the triangle and reverse triangle inequalities (as in Exercise 1.3.4), we estimate
\begin{equation*}
\abs{d(x_n,y_n)-d(x_n',y_n')} \le d(y_n,y_n')+d(x_n,x_n') \to 0
\end{equation*}
as \(n \to \infty\text{,}\) which implies that
\begin{equation*}
\lim_{n \to \infty} d(x_n,y_n) = \lim_{n\to \infty} d(x_n',y_n')
\end{equation*}
as desired. The axioms in Definition 1.1 for \(\hat d\) now follow easily from the corresponding axioms for \(d\) and basic properties of limits of real numbers. We request the details for the triangle inequality in Exercise 2.2.2.
\(\hat X_0\) is dense in \(\hat X\) and isometric to \(X\).
The mapping \(f \maps X \to \hat X_0\) defined in (2.3) is clearly bijective, and it is also straightforward to check that it is an isometry, so it remains to show that \(\hat X_0\) is dense in \(\hat X\text{.}\) Let \(\seq xn\) be a Cauchy sequence in \((X,d)\) and let \(\varepsilon \gt 0\text{.}\) Since \(\seq xn\) is Cauchy, we can find \(N \in \N\) such that \(d(x_N,x_n) \lt
\varepsilon/2\) for all \(n \ge N\text{.}\) Thus the distance between \([\seq xn]\) and the equivalence class \(f(x_N)=[(x_N)_{n\in \N}] \in \hat X_0\) satisfies
\begin{equation*}
\hat d\big([\seq xn],f(x_N)\big)
= \lim_{n \to \infty} d(x_n,x_N) \le \frac \varepsilon2 \lt \varepsilon\text{.}
\end{equation*}
\((\hat X,\hat d)\) is complete.
Let \(\seq{\hat x}n\) be a Cauchy sequence in \((\hat X, \hat d)\text{.}\) Our goal is to show that this sequence converges to some limit \(\hat y \in
\hat X\text{.}\) Since \(\hat X_0\) is dense in \(\hat X\text{,}\) for each \(n \in
\N\) we can find \(y_n \in X\) such that
\begin{equation}
\hat d(\hat x_n, f(y_n))
\lt \frac 1n\text{.}\tag{2.5}
\end{equation}
We claim that \(\seq yn\) is a Cauchy sequence in \((X,d)\text{.}\) For any \(n,m \in \N\text{,}\) the triangle inequality gives
\begin{align*}
d(y_n,y_m)
\amp = \hat d(f(y_n),f(y_m))\\
\amp \le
\hat d(f(y_n),\hat x_n)
+ \hat d(\hat x_n,\hat x_m)
+ \hat d(\hat x_m,f(y_m))\\
\amp \le \frac 1n + \hat d(\hat x_n, \hat x_m) + \frac 1m\text{.}
\end{align*}
The claim now follows from the fact that \(\seq{\hat x}n\) is a Cauchy sequence in \((\hat X, \hat d)\text{.}\)
Since \(\seq yn\) is a Cauchy sequence, it gives rise to an element \(\hat y = [\seq yn]\) of \(\hat X\text{.}\) Moreover, by the triangle inequality and (2.5),
\begin{align*}
\hat d(\hat x_n, \hat y)
\amp =
\hat d(\hat x_n, [\seq yk])\\
\amp \le
\hat d(\hat x_n, f(y_n))
+ \hat d(f(y_n),[\seq yk])\\
\amp \le
\frac 1n
+ \lim_{k\to \infty} d(y_n,y_k)\text{.}
\end{align*}
As \(n \to \infty\text{,}\) the first term on the right hand side clearly vanishes, and the second term also vanishes since \(\seq yk\) is Cauchy (see Exercise 1.3.5).
Exercises Exercises
1. (PS6) Equivalence relation in the proof of Theorem 2.10.
Let \((X,d)\) be a metric space. Define a relation \(\sim\) between Cauchy sequences \(\seq xn, \seq yn\) in \((X,d)\) by
\begin{equation*}
\seq xn \sim \seq yn \iff \lim_{n \to \infty} d(x_n,y_n) = 0\text{.}
\end{equation*}
Show that \(\sim\) is transitive, i.e. that \(\seq xn \sim \seq yn\) and \(\seq yn \sim \seq zn\) implies \(\seq xn
\sim \seq zn\text{.}\) Conclude that \(\sim\) is an equivalence relation.
Hint.
Don’t forget that many of your tricks from Analysis 1 for manipulating limits require you to know ahead of time that some of the limits involved actually exist. Exercise 1.3.4 and Exercise 2.2.2 are be helpful here.
Solution 1. Argument using Exercise 2.2.2
Since we have already done Exercise 1.3.4 and Exercise 2.2.2, we can give a very short proof. Let \(\seq xn,
\seq yn, \seq zn\) be Cauchy sequences in \((X,d)\text{.}\) Then by Exercise 2.2.2 we have
\begin{align*}
\lim_{n\to\infty} d(x_n,z_n)
\amp \le \lim_{n\to \infty} d(x_n,y_n) + \lim_{n\to\infty} d(y_n,z_n),
\end{align*}
where the limits all exist thanks to Exercise 1.3.4. If \(\seq xn \sim \seq yn\) and \(\seq yn \sim \seq zn\text{,}\) then the two limits on the right hand side are both \(0\text{,}\) and so (since it must be non-negative) the limit on the left hand side is also \(0\text{,}\) i.e. \(\seq xn \sim \seq zn\) as desired. Thus \(\sim\) is transitive.
The symmetry of \(\sim\) follows from the symmetry of \(d\text{,}\) while reflexivity follows from the fact that \(d(x,x)=0\) for any \(x \in X\text{.}\) Therefore \(\sim\) is an equivalence relation.
Solution 2. Direct argument
Let \(\seq xn, \seq yn, \seq zn\) be Cauchy sequences in \((X,d)\text{,}\) and suppose that \(\seq xn \sim \seq yn\) and \(\seq yn \sim \seq zn\text{.}\) Using the triangle inequality we have, for any \(n \in \N\text{,}\)
\begin{equation*}
d(x_n,z_n) \le d(x_n,y_n) + d(y_n,z_n),
\end{equation*}
and hence
\begin{align*}
d(x_n,z_n)
\amp \le d(x_n,y_n) + d(y_n,z_n)\\
\amp \to 0 + 0 = 0
\end{align*}
as \(n \to \infty\text{,}\) which implies that \(\seq xn \sim \seq zn\text{.}\) Thus \(\sim\) is transitive.
Note that here we can get actually away without using Exercise 1.3.4. The limits on the right hand side are zero by assumption. Since the left hand side is nonnegative, this then proves that its limit also exists and is also zero.
The symmetry of \(\sim\) follows from the symmetry of \(d\text{,}\) while reflexivity follows from the fact that \(d(x,x)=0\) for any \(x \in X\text{.}\) Therefore \(\sim\) is an equivalence relation.
Comment 1.
Both of the solutions above explain why the limit \(\lim_{n\to\infty}
d(x_n,z_n)\) exists. Simply assuming that it exists, without providing any justification whatsoever, would probably have cost marks on an exam.
Comment 2.
This question ends with ‘Conclude that…’. This means that you should write something about how this conclusion can in fact be reached. Failing to do so on an exam would likely cost marks.
2. (PS5) Triangle inequality in Theorem 2.10.
Let \(\seq xn\text{,}\) \(\seq yn\) and \(\seq zn\) be Cauchy sequences in a metric space \((X,d)\text{.}\) Show that
\begin{equation*}
\lim_{n \to \infty} d(x_n,z_n)
\le
\lim_{n \to \infty} d(x_n,y_n)
+ \lim_{n \to \infty} d(y_n,z_n)\text{.}
\end{equation*}
Hint.
Don’t forget that many of your tricks from Analysis 1 for manipulating limits require you to know ahead of time that some of the limits involved actually exist. Exercise 1.3.4 may be helpful here.
Solution.
Using the triangle inequality we have, for any \(n \in \N\text{,}\)
\begin{equation*}
d(x_n,z_n) \le d(x_n,y_n) + d(y_n,z_n).
\end{equation*}
Taking \(n \to \infty\text{,}\) we deduce that
\begin{align*}
\lim_{n\to\infty} d(x_n,z_n)
\amp \le \lim_{n\to \infty} \Big(d(x_n,y_n) + d(y_n,z_n)\Big),\\
\amp = \lim_{n\to \infty} d(x_n,y_n) + \lim_{n\to\infty} d(y_n,z_n),
\end{align*}
where the limits all exist thanks to Exercise 1.3.4.
Comment 1.
Recall from Analysis 1 that, for sequences \(\seq an\) and \(\seq bn\) are sequences of real numbers, it is not always true that
\begin{equation*}
\lim_{n\to\infty} (a_n+b_n) = \lim_{n\to\infty} a_n + \lim_{n\to\infty} b_n\text{.}
\end{equation*}
For instance we could have \(a_n = n\) and \(b_n = -n\text{.}\) Then neither \(\seq an\) nor \(\seq bn\) converge (in the sense of Definition 1.29), whereas the limit on the left hand side is \(0\text{.}\)
If this had been an exam question, explaining why the limits exist in the first place would have counted for non-negligible portion of the total marks.
Comment 2.
Note that there is no reason whatsoever for the Cauchy sequences \(\seq xn, \seq
yn, \seq zn\) to be convergent, as we have not assumed that \((X,d)\) is complete. Indeed, this exercise is a step in the proof of Theorem 2.10, which is only interesting in the case where \((X,d)\) is not complete.
3. (PS5) Completion vs. closure.
Suppose \((X,d)\) is a complete metric space and let \(S \subset X\) be a subset of \(X\) with closure \(\overline S\text{.}\) Let \(d_S\) and \(d_{\overline S}\) denote restrictions of \(d\) to pairs of points in \(S\) and \(\overline S\text{,}\) respectively. Show that \((\overline
S,d_{\overline S})\) is a completion of \((S,d_S)\text{.}\)
Hint.
Use Theorem 1.23 and Theorem 1.41.
Solution.
Thanks to Theorem 1.23, \(\overline S\) is closed in \((X,d)\text{.}\) Since \((X,d)\) is complete, Theorem 1.41 then implies that \((\overline
S,d_{\overline S})\) is also complete. It remains to show that there is a subset \(S_0 \subset \overline S\) which is both isometric to \(S\) and dense in the metric space \((\overline S,d_{\overline S})\text{.}\) We simply take \(S_0=S\text{,}\) which is clearly isometric to \(S\text{.}\) To see that \(S\) is dense in \((\overline S,d_{\overline S})\text{,}\) let \(x \in
\overline S\) and \(\varepsilon\gt 0\text{.}\) Since \(x \in \overline S\) we know that \(B_{\varepsilon}(x)\cap S \ne \varnothing\text{.}\) It follows that there exists an \(s \in S\) with \(d_{\overline S}(s,x) = d(s,x) \lt \varepsilon\text{.}\)
Comment.
We can slightly shorten the last step above argument by appealing to Lemma 2.3, but there is a tiny wrinkle. When we originally took the closure of \(S\text{,}\) we were thinking of it as a subset of \((X,d)\text{.}\) To use Lemma 2.3, we need this to be the same thing as taking the closure of \(S\) as a subset of \((\overline S, d_{\overline S})\text{.}\) Thankfully this turns out to be the case, but in the official solutions we prefer to just give a direct argument and sidestep this whole issue.
If you did want to talk about talk about the closure of \(S\) as a subset of \((X,d)\) and as a subset of \((\overline S, d_{\overline S})\) in the same sentence, probably the least confusing thing to do is to give them different names: “Let \(C \subseteq X\) be the closure of \(S\) as a subset of \((X,d)\text{,}\) and let \(C' \subseteq \overline S\) be the closure of \(S\) as a subset of \((\overline S, d_{\overline S})\)”.
