TITLE: topologically equivalent spaces
QUESTION [5 upvotes]: I was wondering if there exist a metric $d:X\times X\longrightarrow\mathbb{R}$ (with $X:=(-1,1)$) such that $(-1,1)$ is complete and that $(d,X)$, $(d_{usual},X)$ are topologically equivalent?
I know for example that $(-1,1)\longrightarrow\mathbb{R}$ is a homeomorphism, because we can find a one-to-one function $f$ such that $f$ and $f^{-1}$ is continuous. So we can conclude that $(-1,1)$ and $\mathbb{R}$ are homeomorphic. But what about $(d,X)$ and $(d_{usual},X)$?

REPLY [5 votes]: Any homeomorphism $f : (-1, 1) \to \mathbb{R}$ yields a complete metric for $(-1, 1)$, because $\mathbb{R}$ is complete. Just put $d(x_0, x_1) = d(f(x_0), f(x_1))$. Note that $d$ is obviously symmetric and satisfies the triangle inequality. The only point where you need to use the fact that $f$ is a bijection is to show that $d(x_0, x_1) = 0$ implies $x_0 = x_1$; and to show that $d$ is topologically equivalent to the standard metric on $(-1, 1)$, you need to use the fact that $f$ is a homeomorphism.