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 $X$.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.