Let $f$ be strictly increasing on $I$. Prove that $f$ has an inverse defined on its range and $f^{-1}$ is continuous.
The existence of the inverse is simple (if $y = f(x)$, set $x = f^{-1}(y)$, since $f$ is strictly increasing).
However, I'm difficult understanding the proof for the fact that the inverse is continuous. I have looked it up on the internet and I don't follow their choices of $\delta$ and what they're doing. If someone could explain the proof, that'd be brilliant.
Best Answer
Let $I$ be an interval and $f\colon I\to\mathbb R$ strictly increasing. (Note that $f$ is not supposed to be continuous.) We show that $f^{-1}\colon f(I)\to\mathbb R$ is continuous in $y_0\in f(I)$. For that purpose let $\epsilon\in\mathbb R_+$. Let $x_0:=f^{-1}(y_0)\in I$.
Case (1). $x_0\notin\{\inf(I),\sup(I)\}$. As $I$ is an interval, there are numbers $x_1$, $x_2$ such that $$x_0-\epsilon<x_1<x_2<x_0+\epsilon.$$ As $f$ is strictly increasing we have $$f(x_1)=:y_1<y_0<y_2:=f(x_2).$$
Now let $\delta$ be so small that
$$y_1<y_0-\delta<y_0<y_0+\delta<y_2.$$ Hence if $|y-y_0|<\delta$ we have $y_1<y<y_2$ for all $y\in f(I)$. As $f^{-1}$ is increasing we get $x_1<f^{-1}(y)<x_2$ and that means $|f^{-1}(y)-f^{-1}(y_0)|<\epsilon$.
The other two cases are proven analogously.