Let $f:[a,b] \rightarrow \Bbb R$ be a strictly monotone increasing. Then $f$ has an inverse function $g:[c,d]\rightarrow \Bbb R,$ where $[c,d]$ is the range of $f$. I'm trying to prove that $g$ is continuous at d.
My intial thoughts for an attempt of a proof:
Strictly monotone functions are injective. So if $\alpha, \beta \in [a,b]$ and $\alpha \not= \beta $ then $\alpha < \beta$. Since $f$ is strictly monotone increasing $f(\alpha) < f(\beta)$ and $f(\alpha) \not= f(\beta)$.
Since $f$ is strictly increasing, so is $f^{-1}$. So if $\alpha < \beta$ then $f^{-1}(f(\alpha)) < f^{-1}(f(\beta))$.
This is because if there exists $\alpha$ and $\beta $ $\in (a,b)$ with $\alpha < \beta$ such that $f^{-1}(\alpha)$ = $\alpha '$ and $f^{-1}(\beta)$ = $\beta '$ and $\alpha ' < \beta '$ then
$\beta = f^{-1}(\beta ') \le f^{-1}(\alpha ') = \alpha$
which is a contradiction if $f$ is strictly increasing.
The remainder of the proof is some form of an epsilon delta proof to show that the inverse function is continuous from the left at the right end point. My attempt:
Let $b$ be the upper limit $ \in [a,b]$ and define $d = f(b)$.
Next, I want to show that $\lim_{x\rightarrow d^{-}}f^{-1}(x) = b$ for any $\epsilon >0$ such that $(b-\epsilon) \subset [a,b]$.
So, $f(b-\epsilon) < f(b)$.
Let $\delta = 1/2 (f(b)-f(b-\epsilon))$
Then $f(x_0-\epsilon) < f(x_0)-\delta$
So if $|x-d| < \delta$, then $|f^{-1}(x)-f^{-1}(d)|<\epsilon$
then continuity holds at $f^{-1}(d)$, which is possible by the Archimedean principle. Currently, I'm having trouble with the epsilon-delta proof. I don't think the argument is strong enough.