TITLE: Can a non complete metric space be homeomorphic to its completion?
QUESTION [5 upvotes]: Suppose we have a non complete metric space. Can it be homeomorphic to its metric completion?

REPLY [7 votes]: Consider the metric space $\left(\mathbb R\setminus\{0\}\right)\times\mathbb Z\subset\mathbb R^2$

This metric space is not complete and its completion is $\mathbb R\times\mathbb Z$.  The two spaces are homeomorphic.  I hope that's easy to see, but here's an explicit pair of continuous inverses:
$f\;\colon\;\left(\mathbb R\setminus \{0\}\right)\times\mathbb Z\to\mathbb R\times \mathbb Z$ given by:
$$
f(x, n)=\begin{cases}
(\log(x), 2n) & x>0 \\
(\log(-x), 2n+1) & x<0
\end{cases}
$$
$g\;\colon\;\mathbb R\times\mathbb Z\to\left(\mathbb R\setminus\{0\}\right)\times\mathbb Z$ given by
\begin{align}
g(y, &2k) = (e^y, k)\\
g(y, 2k&+1) = (-e^y, k)
\end{align}
You can check for yourself that these functions are continuous and that their compositions are the identity maps.