TITLE: Two metrics induce the same topology, but one is complete and the other isn't
QUESTION [41 upvotes]: I'm looking for an example of two metrics that induce the same topology, but so that one metric is complete and the other is not (Since it is known that completeness isn't a topological invariant).
Thanks in advance for any hints or ideas.

REPLY [2 votes]: With the usual metric, $(0,1]$ is not complete. 
Let us define another metric $\rho$ on this space using the homeomorphism $\phi: [1,\infty) \to (0,1]$. Define $\rho(x,y) = \vert\frac{1}{x}-\frac{1}{y}\vert$. It is easy to check that $\rho$ is indeed a metric. Also, note that the usual metric and $\rho$ are equivalent and hence induce the same topology. 
Now consider a Cauchy sequence in $((0,1],\rho)$. 
$\rho(x_n,x_m) < \epsilon$ for all $n \geq N$ or,
$\vert\frac{1}{x}-\frac{1}{y}\vert < \epsilon$ for all $n \geq N$. 
Hence $(\frac{1}{x_n})$ is a Cauchy sequence in ($[1,\infty),$usual metric), a complete metric space. Let us say, it converges to $y \in [1,\infty)$. It is easy to check that $(x_n)$ converges to $x = \frac{1}{y}$ in $((0,1],\rho)$.