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Let $f:\mathbb{R}\mapsto\mathbb{R}$ be a one-to-one function with $f(\mathbb{R})=\mathbb{R}$. If $f^{-1}(x)$ is continuous $\forall x\in\mathbb{R}$, prove or disprove that $f(x)$ is continuous $\forall x\in\mathbb{R}$.

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    Can you prove that such a function $f$ must be strictly monotone? – Milo Brandt Aug 25 '15 at 16:47
  • If $f$ is a bijection, then so is $f^{-1}$. Are bijections (on $\mathbb{R}$) continuous? – copper.hat Aug 25 '15 at 16:54
  • @MiloBrandt I don't think I can do so. Any hint? –  Aug 25 '15 at 16:55
  • @Jason What is the definition of continuity? –  Aug 25 '15 at 17:12
  • @EugenCovaci A function f is continuous $\forall x\in\mathbb{R}$ if and only if $$\lim_\limits{x\to x_0}{f(x)}=f(x_0)$$ for all $x_0\in\mathbb{R}$ –  Aug 25 '15 at 17:21
  • I think it a duplicate, see http://math.stackexchange.com/questions/368824/is-the-inverse-of-a-continuous-bijective-function-also-continuous?rq=1 –  Aug 25 '15 at 17:31
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    When you figured out what Milo Brandt suggested and don't know where to go from there, I suggest you have a look at this. – Ben Aug 25 '15 at 17:53
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    @MiloBrandt I found out that by using the IVT many times with a<c<b by reductio ad absurdum we end up with $f^{-1}$ being strictly monotone, so f is also strictly monotone. –  Aug 26 '15 at 09:33

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