Proof. We construct \(\overline {X}\) out of Cauchy sequences in \(X\), in exactly the way \(\mathbb {R}\) is built from \(\mathbb {Q}\). The proof has four steps.

Step 1: constructing \(\overline {X}\)

Call two Cauchy sequences \((x_n)\) and \((x_n')\) in \(X\) equivalent, written \((x_n)\sim (x_n')\), if \[\lim _{n\rightarrow \infty }d(x_n,x_n')=0 .\] This is an equivalence relation: reflexivity and symmetry are immediate, and transitivity follows from \(d(x_n,x_n'')\leq d(x_n,x_n')+d(x_n',x_n'')\). Let \(\overline {X}\) be the set of equivalence classes, and for classes \(\overline {x},\overline {y}\) with representatives \((x_n),(y_n)\) define \[\overline {d}(\overline {x},\overline {y})=\lim _{n\rightarrow \infty }d(x_n,y_n).\]

The limit exists. By the quadrilateral inequality, \[\big |d(x_n,y_n)-d(x_m,y_m)\big |\leq d(x_n,x_m)+d(y_n,y_m),\] so \(\big (d(x_n,y_n)\big )\) is a Cauchy sequence of real numbers and converges.

It does not depend on the representatives. If \((x_n)\sim (x_n')\) and \((y_n)\sim (y_n')\) then the same inequality gives \[\big |d(x_n,y_n)-d(x_n',y_n')\big |\leq d(x_n,x_n')+d(y_n,y_n')\rightarrow 0 ,\] so the two limits agree.

That \(\overline {d}\) is a metric now follows by taking limits in the corresponding statements for \(d\); only definiteness needs a word, and \(\overline {d}(\overline {x},\overline {y})=0\) means precisely that the representatives are equivalent, i.e. \(\overline {x}=\overline {y}\).

Step 2: \(X\) sits isometrically inside \(\overline {X}\)

For \(a\in X\) let \(T(a)\) be the class of the constant sequence \((a,a,a,\dots )\), which is certainly Cauchy. Then \[\overline {d}\big (T(a),T(b)\big )=\lim _{n\rightarrow \infty }d(a,b)=d(a,b),\] so \(T\) is an isometry, and in particular injective. Write \(W=T(X)\).

\(W\) is dense. Let \(\overline {x}\in \overline {X}\) have representative \((x_n)\) and let \(\varepsilon >0\). Since \((x_n)\) is Cauchy there is \(N\) with \(d(x_n,x_N)<\varepsilon \) for all \(n\geq N\), and then \[\overline {d}\big (\overline {x},T(x_N)\big )=\lim _{n\rightarrow \infty }d(x_n,x_N) \leq \varepsilon .\] So every point of \(\overline {X}\) has points of \(W\) arbitrarily close to it, that is \(\overline {W}=\overline {X}\).

Step 3: \(\overline {X}\) is complete

Let \((\overline {x}^{(k)})\) be a Cauchy sequence in \(\overline {X}\). By Step 2 choose \(z_k\in X\) with \[\overline {d}\big (\overline {x}^{(k)},T(z_k)\big )<\frac {1}{k} .\] Then \((z_k)\) is Cauchy in \(X\), because \[d(z_k,z_l)=\overline {d}\big (T(z_k),T(z_l)\big ) \leq \frac 1k+\overline {d}\big (\overline {x}^{(k)},\overline {x}^{(l)}\big )+\frac 1l ,\] and all three terms are small for \(k,l\) large. Let \(\overline {z}\in \overline {X}\) be the class of \((z_k)\). Then \[\overline {d}\big (\overline {x}^{(k)},\overline {z}\big ) \leq \frac 1k+\overline {d}\big (T(z_k),\overline {z}\big ) =\frac 1k+\lim _{l\rightarrow \infty }d(z_k,z_l)\longrightarrow 0 ,\] so \(\overline {x}^{(k)}\rightarrow \overline {z}\). Hence \(\overline {X}\) is complete.

Step 4: uniqueness up to isometry

Suppose \(\widetilde {X}\) is another complete metric space containing a dense isometric copy \(\widetilde {W}\) of \(X\), say by the isometry \(S:X\rightarrow \widetilde {W}\). Then \(S\circ T^{-1}:W\rightarrow \widetilde {W}\) is an isometry between dense subsets. An isometry is uniformly continuous, so it extends to \(\overline {W}=\overline {X}\): given \(\overline {x}\in \overline {X}\) take \(w_n\in W\) with \(w_n\rightarrow \overline {x}\); the images form a Cauchy sequence in \(\widetilde {X}\), which converges by completeness, and the limit does not depend on the chosen sequence because two such sequences interleave into one. The extension is again an isometry, since \(\overline {d}\) and the metric of \(\widetilde {X}\) are continuous, and it is onto because its image is complete, hence closed, and contains the dense set \(\widetilde {W}\).

So the completion is unique up to isometry, which is why one speaks of the completion of \(X\). □