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Could you help me and explain to me how to prove that if a certain function $f: (a,b) \rightarrow R$ is continuous and injective, then $f^{-1}$ is also continuous ?

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how can a constant function be injective? – Ram Jan 2 '13 at 11:07
I'm sorry, it was a typo. – Bilbo Jan 2 '13 at 13:01
up vote 2 down vote accepted

A constant function can't be injective so I suppose this is a typo.

Here is a brief proof that $f^{-1}$ is continuous. Many details are missing so you should try and fill them.

First without loss of generality $f$ is strictly increasing in $(a,b)$ (why?).

In addition, $f((a,b))$ is an interval with endpoints $m<M$ (again why?).

Let $y_0\in (m,M)$. Since $f$ is injective, there exists a unique $x_0\in (a,b)$ so that $f(x_0)=y_0$.

Choose $\epsilon>0$ such as that $a<x_0-\epsilon<x_0<x_0+\epsilon<b$. Then, since $f$ is strictly increasing we have that \begin{equation}f(x_0-\epsilon)<f(x_0)=y_0<f(x_0+\epsilon)\end{equation} Let $\delta=\min {\left\{y_0-f(x_0-\epsilon),f(x_0+\epsilon)-y_0\right\}}>0$ and $y\in f((a,b))$. Then, \begin{equation}\left|y-y_0\right|<\delta\implies f(x_0-\epsilon)<y<f(x_0+\epsilon)\end{equation} (why?). I think you can finish the rest.

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I have one question about the last point. – Bilbo Jan 2 '13 at 13:50
@Anna The last point being...? – Nameless Jan 2 '13 at 13:51
I have one question about the last one. Suppose $\delta= y_0-f(x_0-\epsilon)$. I have problenms showing that $y>y_0-\delta=2f(x_0)-f(x_0+\epsilon)$ – Bilbo Jan 2 '13 at 13:58
@Anna $\left|y-y_0\right|<\delta\implies y_0-\delta<y$ – Nameless Jan 2 '13 at 14:03
I'm sorry, I still don't get that. We know that $y_0 - \delta <y$ and $y_0 < f(x_0+ \epsilon)$. How can we deduce from this that $y<f(x_0+ \epsilon)$? – Bilbo Jan 2 '13 at 14:36

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