Why homeomorphism between $R$ and $(0,1)$ makes $(0,1)$ completely metrizable space?

Question

I know that we can have few homeomorphisms between $R$ and $(0,1)$ such as $f(t)=\tan(-\frac{\pi}{2}+t\pi)$ but I can not understand why this metric $d(x,y)=|f(x)-f(y)|$ makes $(0,1)$ a complete metric space? I can not understand why now the Cauchy sequence such this one $\{\frac{1}{n}\}_n$ has limit in $(0,1)$. How the measure affects limit points?

Answer 1

Because now the distance from $\frac1m$ to $\frac1n$ is $\left|f\left(\frac1m\right)-f\left(\frac1n\right)\right|$, which is not the original distance. And now $\left(\frac1n\right)_{n\in\Bbb N}$ is not a Cauchy sequence anymore, and therefore the fact that it doesn't converge is not an obstacle to the fact that $(0,1)$ is complete.

Answer 2

With this new metric, $\left\{1\over n\right\}$ is not a Cauchy sequence anymore.