3
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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...

$\endgroup$
2
$\begingroup$

Hint: Think of the discrete metric.

$\endgroup$
2
$\begingroup$

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 $\sigma(m,n)=\lvert m^{-1}-n^{-1}\rvert$.

$\endgroup$

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.