Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question

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

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

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

2 Answers 2

up vote 1 down vote accepted

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.

share|improve this answer

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 quasi-isometry.

share|improve this answer
Which side of the inequality? Both sides are proven using the same argument. –  Misha Aug 1 '12 at 5:06

Not the answer you're looking for? Browse other questions tagged or ask your own question.