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It is well known that any metric space $(X,d)$ can be isometrically embedded onto a dense subspace of a complete metric space $X^*$ which comprises all Cauchy sequences in $X$ modulo the equivalence relationship: $\{x_n\}\sim\{y_n\}$ if $d(x_n,y_n)\to 0$, and it is also well known that if there are two such embeddings $\iota: X\to X^*$ and $\iota':X\to X'^*$ then $X^*$ and $X'^*$ must be isometrically isomorphic, and hence need not be distinguished.

However, if all we know is $X$ is a dense subspace of a complete metric space $Y$, can we conclude that $Y$ is the completion of $X$? After all, $Y$ may manifest no apparent relation to Cauchy sequences in $X$.

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  • $\begingroup$ I'd say yes since $X$ being dense means that $cl(X) = Y$ $\endgroup$ – cronos2 Sep 24 '16 at 9:18
  • $\begingroup$ @cronos2 closure is taken in $Y$. But completion relies solely on $X$. $\endgroup$ – Vim Sep 24 '16 at 9:27
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Yes, we can. Let $d$ be the complete metric on $Y$. For each $y\in Y$ let $S(y)$ be the set of sequences in $X$ converging to $y$. Each $d$-Cauchy sequence in $X$ belongs to $S(y)$ for exactly one $y\in Y$, and it’s easy to check that if $\langle x_n:n\in\Bbb N\rangle$ and $\langle y_n:n\in\Bbb N\rangle$ are $d$-Cauchy sequences in $X$, they belong to the same $S(y)$ if and only if $\langle d(x_n,y_n):n\in\Bbb N\rangle$ converges to $0$ in $\Bbb R$. Thus, the sets $S(y)$ are precisely the equivalence classes of Cauchy sequences used to construct $X^*$, and the map $X^*\to Y:S(y)\mapsto y$ is is an isometry.

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