Let $f$ be a continuous function on $\mathbb{R}\to \mathbb{R}$ which
is strictly increasing in the sense that if $x'<x''$ then
$f(x')<f(x'')$. Prove that $f$ is injective and that its inverse
function $f^{-1}$ is continuous and strictly increasing.
I know that since $f$ is strictly increasing and $a<b$, then $f$ maps the open interval $(a,b)$ injectively onto the open interval $(f(a),f(b))$, from which it follows that $f^{-1}$ is continuous, but how can I prove this in a more formal manner?
Best Answer
Suppose $f(x_1) = f(x_2)$. If $x_1<x_2$, then $f(x_1) < f(x_2)$, and similarly, if $x_1>x_2$, then $f(x_1) > f(x_2)$. It follows that $x_1 = x_2$ and hence $f$ is injective.
Since $f$ is injective, it is invertible on its range $B=f(\mathbb{R})$. Let $f^{-1}$ denote this inverse. Suppose $x_1 < x_2$, and suppose $y_1 = f^{-1}(x_1)$, $y_2 = f^{-1}(x_2)$. We must have $y_1 \neq y_2$, otherwise we would have $f(y_1) = f(y_2)$, a contradiction. We must have $y_1 < y_2$, because otherwise, if $y_1>y_2$, then $f(y_1) > f(y_2)$, again a contradiction. Hence $f^{-1}$ is strictly increasing.
Let $x \in B$, and $\epsilon>0$. Let $y = f^{-1}(x)$. Let $x_+ = f(y+\epsilon)$, $x_- = f(y-\epsilon)$. Note that if $x_- < z < x_+$, then since $f^{-1}$ is strictly increasing, we have $y-\epsilon < f^{-1}(z) < y+\epsilon$. To finish, let $\delta = \min(x_+-x, x-x_-)$. Then if $|z-x| < \delta$, we have $|f^{-1}(z)-f^{-1}(x)| < \epsilon$, hence $f^{-1}$ is continuous.
Aside: I left out a minor detail above. Since $f$ is strictly increasing, it follows that $B$ is open (since $\mathbb{R}$ is open). When showing continuity of $f^{-1}$, the $\epsilon$ must be taken small enough so that $B(x,\epsilon) \subset B$.