# Existence of a metric which is complete and equivalent to standard metric

Does there exist a metric $d: (-1,1)\times (-1,1) \to \mathbb R$ which is complete and is equivalent to the standard metric (Euclidean metric) on $(-1,1)$?

I think there does not exist such a metric and my current approach has been to assume both conditions on the metric and try to find a contradiction in terms of a Cauchy sequence not converging to a point in $(-1,1)$ but this hasn't been very succesful so far. I think I am supposed to use the fact that $(-1,1)$ and are $\mathbb R$ are topologically equivalent but I dont know exactly how.

Yes, there is; this is immediate from the fact that $(-1,1)$ is homeomorphic to $\Bbb R$, and the standard metric on $\Bbb R$ is complete. To get a concrete example of such a metric, let

$$f:(-1,1)\to\Bbb R:x\mapsto\tan\frac{\pi x}2\;,$$

and for $x,y\in(-1,1)$ define

$$d(x,y)=|f(x)-f(y)|\;.$$

• How can I infer from this that the metric $d$ and the standard metric are equivalent in the sense that they determine the same family of open sets on $(-1,1)$? – Joogs Sep 22 '16 at 18:57
• @Joogs: Use the fact that $f$ and $f^{-1}$ take open intervals to open intervals. – Brian M. Scott Sep 22 '16 at 19:07
• Okay, $(-1,1)$ is homeomorphic to $\mathbb R$ so there exist a bijection $f$ between them, both $f$ and $f^{-1}$ are continuous so they map open intervals to open intervals. But then I cant seem to connect the dots any further. – Joogs Sep 22 '16 at 19:22
• @Joogs: Let $\tau_d$ be the topology generated by $d$, and let $\tau$ be the usual topology on $(-1,1)$. Each open $d$-ball is an open interval, so $\tau_d\subseteq\tau$. Now show that each open interval in $(-1,1)$ is a union of open $d$-balls, so that $\tau\subseteq\tau_d$. – Brian M. Scott Sep 22 '16 at 19:33
• How does it follow from the completeness of $\mathbb R$ that $(-1,1)$ is also complete with respect to $d$ ? – Joogs Sep 22 '16 at 20:08