I know that the key is to use $\tan$ and $\arctan$ to do it. Take any $(a,b) \subset \mathbb{R}$, $(a,b)$ is open. Now I want to show $\tan^{-1}(a,b)$ is open. I need a hint for the next step (just a hint suffices) Thanks.
Topological Equivalence between $(- \pi/2, \pi/2)$ to $R$
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general-topology
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0@countinghaus yes I was thinking about the continuity of $\tan$ and $arctan$ too. I think I will take it as granted they are continuous since I have never actually proved the continuity of trigonometric functions before – 2012-10-10
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Look at the definition of continuous function in terms of open sets.
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1@jsk, well, you have to note that they are bijective on that interval. Then it is an homeomorphism since tan is continuous and arctan is also continuous (both are differentiable, hence continuous) – 2012-10-10