I have a problem. I don't know if every continuos function is necessarily a quasiisometry. 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.

closed as too localized by Andreas Blass, Qiaochu Yuan, Chris Godsil, Bill Johnson, Misha Aug 1 '12 at 6:23
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The map $\mathbb{R}\rightarrow \mathbb{S}^1$ given by $f(x) = \exp(2\pi i x)$ is continuous, yet not a quasiisometry. 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 quasiisometry. Any map (not necessarily continuous) between compact sets is a quasiisometry. 


And how can you prove the last statement about compact spaces. I already thought about that, but was able to establish onl one side of the inequality in the definition of quasiisometry. 

