Call a metric space $Y$ a completion of a metric space $X$ if $X$ is dense in $Y$ and $Y$ is complete. State and prove a uniqueness theorem for completions of metric spaces.
I've read now in several places that completions of metric spaces are unique up to isometry, and also found a detailed proof here. However, the framework followed in the attached link is not the same as what Rudin calls a completion. The PDF attached says that: A completion of a metric space $(X,d)$ is a pair consisting of a complete metric space $(Y,d')$ and an isometry $\phi: X\to Y$ such that $\phi(X)$ is dense in $Y$. On the other hand, Rudin does not bring in a map $\phi$, and directly says (as above in quoted text) that $Y$ is a completion of $X$ if $\overline{X} = Y$ and $Y$ is complete.
Is it possible to avoid the map $\phi: X\to Y$ as given here, or is it necessary? I'm unable to come up with a uniqueness theorem given Rudin's definition of completion of metric spaces. I'd appreciate any help!
Best Answer
We always need such a map $\phi$, but if $X$ is actually a subset of $Y$, then $\phi$ can be taken to be the inclusion map $\iota: X \hookrightarrow Y$.
Another typical choice: $Y$ is taken to be the set of all Cauchy sequences of elements in $X$ (modulo the obvious equivalence relation), and $X$ is identified with the subset of constant sequences (so the map $\phi(x) := (x, x, x, ...)$)