In my introductory Topology course, a completion of a metric space $(X,d)$ is a complete metric space $(\tilde{X}, \tilde{d})$ together with an isometry $i : (X, d) \rightarrow (\tilde{X}, \tilde{d})$ with the following property:
For every complete metric space $(Y, d_Y)$ and every isometry $f : (X, d) \rightarrow (Y, d_Y)$ there exists a unique isometry $g : (\tilde{X}, \tilde{d}) \rightarrow (Y, d_Y)$ satisfying $f = g ∘ i$.
I want to show that $((\tilde{X}, \tilde{d}), i)$ is a completion of $(X, d)$ if and only if $(\tilde{X}, \tilde{d})$ is complete, $i$ is an isometry and $i[X]$ is dense in $\tilde{X}$. For "$\Rightarrow$", suppose $i[X]$ is not dense in $\tilde{X}$. As $\overline{i[X]}$ is closed in $\tilde{X}$, and $\tilde{X}$ is complete, $\overline{i[X]}$ is complete. Thus it can take the role of the complete space $Y$ in the above definition. Thus, letting $Y := \overline{i[X]}$ and $f : X \rightarrow Y$ any isometry, I am looking for the contradiction that I can now find more than one isometry $g : \tilde{X} \rightarrow Y$ satisfying $f = g ∘ i$. Can I construct such $g$, or is this not a feasible strategy?
Best Answer
If $i[X]$ where not dense in $(\tilde{X}, \tilde{d})$, pick $p$ in $\tilde{X}$ not in the closure of $i[X]$. We know $r>0$ exists so that $B_{\tilde{d}}(p,r) \cap i[X] = \emptyset$.
Define $Y=\overline{i[X]}$ in the restricted metric $d_Y$ from $\tilde{d}$, and note it is a complete metric space (being closed in one) and $f=i$ is well-defined isometry from $(X,d)$ to $(Y,d_Y)$. Now suppose an isometry $g: (\tilde{X}, \tilde{d}) \to (Y,d_Y)$ completing the diagram (so $g \circ i = f$) existed, then where can $p$ go?
$g(p) \in Y$ so lies (in $\tilde{X}$) in the closure of $i[X]$, so for some $x_0 \in X$, $\tilde{d}(i(x_0), g(p)) < r$.
But $i(x_0)= f(x_0) = g(i(x_0))$ and then $d_Y(g(p), g(i(x_0)) < r$ while we knew $\tilde{d}(p, i(x_0)) \ge r$, so $g$ is not an isometry.
So $i[X]$ has to be dense in $\tilde{X}$.