# Continuous function and quasi-isometry [closed]

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.

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## closed as too localized by Andreas Blass, Qiaochu Yuan, Chris Godsil, Bill Johnson, MishaAug 1 '12 at 6:23

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Continuous functions from where to where? And what is your definition of quasi-isometry? –  Igor Rivin Jul 31 '12 at 21:27

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.