Section 2.3 Uniqueness of the Completion
Note on the exam.
The statement of the theorem below is very much examinable, but, this year, its proof is not. Nevertheless, the proof is an excellent exercise in applying Lemma 2.4, and so may be worth looking at for that reason alone.
Theorem 2.11.
Any two completions of the same metric space are isometric.
Proof.
Suppose that \((\hat{X}, \hat{d})\) and \((\tilde{X}, \tilde{d})\) are completions of the metric space \((X,d)\text{.}\) Then there exist sets \(\hat{X}_0 \subseteq
\hat{X}\) and \(\tilde{X}_0 \subseteq\tilde{X}\) that are dense in \(\hat{X}\) and \(\tilde{X}\text{,}\) respectively, and isometric to \(X\text{.}\) Let \(\hat{f} \maps X \to \hat{X}_0\) and \(\tilde{f} \maps X \to \tilde{X}_0\) be bijective isometries.
Let \(f = \hat{f} \circ \tilde{f}^{-1} \maps \tilde{X}_0 \to \hat{X}_0\text{.}\) This is a bijective isometry; in particular it is uniformly continuous. Lemma 2.4 asserts that there exists a unique continuous extension \(F \maps \tilde{X} \to \hat{X}\text{.}\) We want to show that \(F\) is a bijective isometry as well (and therefore \(\tilde{X}\) and \(\hat{X}\) are isometric).
Let \(\tilde{x}, \tilde{y} \in \tilde{X}\text{.}\) Choose two sequences \((\tilde{x}_n)_{n \in \N}\) and \((\tilde{y}_n)_{n \in \N}\) in \(\tilde{X}_0\) converging to \(\tilde{x}\) and \(\tilde{y}\text{,}\) respectively. Then by the continuity of \(F\text{,}\)
\begin{equation*}
\begin{split}
\hat{d}(F(\tilde{x}), F(\tilde{y})) \amp = \lim_{n \to \infty} \hat{d}(F(\tilde{x}_n), F(\tilde{y}_n)) \\
\amp = \lim_{n \to \infty} \hat{d}(f(\tilde{x}_n), f(\tilde{y}_n)) \\
\amp = \lim_{n \to \infty} \tilde{d}(\tilde{x}_n, \tilde{y}_n) \\
\amp = \tilde{d}(\tilde{x}, \tilde{y}).
\end{split}
\end{equation*}
Hence \(F\) is an isometry. It follows automatically that \(F\) is injective. In order to show that it is surjective, consider \(\hat{x} \in \hat{X}\text{.}\) Choose a sequence \((\hat{x}_n)_{n \in \N}\) in \(\hat{X}_0\) that converges to \(\hat{x}\text{.}\) Let \(\tilde{x}_n = f^{-1}(\hat{x}_n)\text{.}\) Then
\begin{equation*}
\tilde{d}(\tilde{x}_m, \tilde{x}_n) = \hat{d}(\hat{x}_m, \hat{x}_n) \to 0
\end{equation*}
as \(m, n \to \infty\text{.}\) Thus \((\tilde{x}_n)_{n \in \N}\) is a Cauchy sequence in \(\tilde{X}\text{.}\) Since this space is complete, there exists a limit \(\tilde{x} = \lim_{n \to \infty} \tilde{x}_n \in \tilde{X}\text{.}\) Now by the continuity of \(F\text{,}\)
\begin{equation*}
F(\tilde{x}) = \lim_{n \to \infty} F(\tilde{x}_n) = \lim_{n \to \infty} \hat{x}_n = \hat{x}.
\end{equation*}
Therefore, the map \(F\) is surjective. It follows that the two completions are isometric.
This means that any two completions of a given metric space are essentially the same. For this reason, we normally identify all of them and speak of the completion of a metric space.
