Theorem (Uniqueness of the completion). Let \(h: X \to Y\) and \(h': X \to Y'\) be isometric imbeddings of the metric space \((X, d)\) in the complete metric spaces \((Y, D)\) and \((Y', D')\), respectively. Then there is an isometry of \((\overline{h(X)}, D)\) with \((\overline{h'(X)}, D')\) that equals \(h' h^{-1}\) on the subspace \(h(X)\).
Solution
Let \[ T_0=h'\circ h^{-1}:h(X)\longrightarrow h'(X). \] Because both \(h\) and \(h'\) are isometric embeddings, \(T_0\) is an isometry.
For \(y\in\overline{h(X)}\), choose a sequence \(h(x_n)\to y\). Then \((h(x_n))\) is Cauchy, so \[ D'(h'(x_m),h'(x_n))=D(h(x_m),h(x_n))\to0. \] Completeness of \(Y'\) gives a limit of \(h'(x_n)\), and since all terms lie in \(h'(X)\), that limit lies in \(\overline{h'(X)}\). Define \[ T(y)=\lim_n h'(x_n). \]
This is well defined. If also \(h(z_n)\to y\), then \[ D'(h'(x_n),h'(z_n))=D(h(x_n),h(z_n))\to0, \] so the two image sequences have the same limit.
For \(y,z\in\overline{h(X)}\), choose \(h(x_n)\to y\) and \(h(z_n)\to z\). Continuity of the metrics gives \[ D'(T(y),T(z)) =\lim_n D'(h'(x_n),h'(z_n)) =\lim_n D(h(x_n),h(z_n)) =D(y,z). \] Thus \(T\) is an isometry, hence injective.
It is surjective as well. Given \(y'\in\overline{h'(X)}\), choose \(h'(x_n)\to y'\). Then \((h(x_n))\) is Cauchy and therefore converges in the complete subspace \(\overline{h(X)}\) to some \(y\); by definition \(T(y)=y'\).
Finally, if \(y=h(x)\in h(X)\), the constant sequence \(x_n=x\) gives \[ T(h(x))=h'(x)=h'h^{-1}(h(x)). \] Hence \(T\) is the required isometry extending \(h'h^{-1}\).