Let $(X,\rho)$ and $(X,\sigma)$ be metric spaces. If $X$ is $\rho$-complete is $X$ $\sigma$-complete? Justify your answer.
A little bit of reference showed that this is not necessarily the case. So I have to give a counter example of a space which is $\rho$-complete but not $\sigma$-complete.
Any hints/ideas? Thanks in advance.
The example is $\mathbb{R}$, the real numbers. They are complete with the usual metric ($d(x,y) = |x-y|$), but you can change this to $d'(x,y) = |\arctan x-\arctan y|$, if i'm not mistaken, and the resulting space won't be complete... So the idea is that $\mathbb{R}$ and, for instance, $(-\frac\pi{2},\frac\pi{2})$ are homeomorphic, but the latter is incomplete...
Good answers above. But I think a minimal example (if you care about that) is the natural numbers $\mathbb{N}$ with $\rho(m,n)=\lvert m-n\rvert$ and $\sigma(m,n)=\lvert m^{-1}-n^{-1}\rvert$.
