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.
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.
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.
