I have a problem. I don't know if every continuos function is necessarily a quasi-isometry. I was trying to prove that, but failled for now. I also can't find a counterexample. If you have any hints, i would be very happy if you could help me. Thank you.



Continuous function and quasi-isometry
The map $\mathbb{R}\rightarrow \mathbb{S}^1$ given by $f(x) = \exp(2\pi i x)$ is continuous, yet not a quasi-isometry.


The map from the open semicircle $(\-\pi/2, \pi/2)$ to $\mathbb{R}$ given by $f(\theta) = \tan(\theta)$ is a homeomorhpism, but not a quasi-isometry.


Any map (not necessarily continuous) between compact sets is a quasi-isometry.


