I have started a proof for the following problem. I am not sure that I have completely shown everything I need to show. Can anyone validate or invalidate my proof?
Let $(Y,d)$ be a complete metric space and $X$ a dense subset of $Y$. The set $X$ is a also a metric space with respect to the same metric. Let $X^*$ be the completion of $X$. $X^*$ is the set of equivalence classes of Cauchy sequences $\{x_n\}$ in $X$. Since $Y$ is complete, $ \lim_n x_n$ exists in $Y$ . Equivalent Cauchy sequences have the same limit, hence $f(\{x_n\}) = \lim_n x_n$ is a well defined map $f : X^* \rightarrow Y$ . Show that $f$ is an isomorphism of metric spaces, i.e. a surjective isometry.
Proof:
Let $\{x_n\}$ be a Cauchy sequence such that $x_n \in X$ for all $n$. Let $\{ \tilde{x} \}$ denote an equivalence class of Cauchy sequences in X.
$X$ is dense in $Y$ means for all $y \in Y$ and $\epsilon > 0$ there exists a sequence $\{x_n\}$ with elements in $X$ and $N \in \mathbb{N}$ such that $d(x_n,y) < \epsilon/2$ if $n > N$, i.e. $\lim_nx_n = y \in Y$. By the triangle inequality we have $d(x_n,x_m) \leq d(x_n, y) + d(y, x_m) = \epsilon$ if $n,m > N$. Therefore $\{x_n\}$ is a Cauchy sequence and hence belongs to an equivalence class $\{\tilde{x}\} \in X^*$. This shows that $f$ is surjective.
For $f$ to be an isometry $d^*(\{\tilde{x}\}, \{\tilde{y}\}) = d(x,y)$ for all $\{\tilde{x}\}, \{\tilde{y}\} \in X^*$. The distance $d^*$ is defined as the distance between any two representative elements from each equivalence class. Pick a Cauchy sequence out of each equivalence class, say $\{x_n\}$ from $\{\tilde{x}\}$ and $\{y_n\}$ from $\{\tilde{y}\}$. Since $Y$ is complete the limits of each of these sequences exist in $Y$. Therefore $d(\{x_n\}, \{y_n\}) = d(\lim_nx_n, \lim_ny_n) = d(x,y)$. Thus $f$ is isometric since distance is preserved. Q.E.D.