I'm having some problems proving the inverse is continuous. The hint in the book is to use the standard epsilon-delta definition of continuity. I believe the easiest route is a proof by contradiction, but with all of the quantifiers in the statement, I may be incorrectly negating the statement I am trying to prove. Also, I have at my disposal the intermediate value theorem, which most of my proof relies on. Below is the proposition:
Let $a < b$ be real numbers, and let $ f:[a, b] \to \mathbb{R} $ be a function which is both continuous and strictly monotone increasing. Then $f$ is a bijection from $[a, b]$ to $[f(a), f(b)]$, and the inverse $f^{-1}: [f(a), f(b)] \to [a, b]$ is also continuous and strictly monotone increasing.
Below is my attempt at a proof:
Let $x_1, x_2 \in [a, b]$ be real numbers such that $f(x_1) = f(x_2)$. From the trichotomy of the real numbers, we have that exactly one of the following is true: $x_1 = x_2$, $x_1 < x_2$, or $x_1 > x_2$. Suppose $x_1 \not = x_2$. Then, by definition of strictly increasing monotone functions, we have that $f(x_1) \not = f(x_2)$. Thus, $x_1 = x_2$, and $f$ is injective. Now let $y \in [f(a), f(b)]$ be a real number. Then, by the intermediate value theorem, there exists a real number $c \in [a, b]$ such that $f(c) = y$. Thus, $f$ is a surjection from $[a, b]$ to $[f(a), f(b)]$. Since $f$ is both injective and surjective, we can conclude that $f$ is a bijection from $[a, b]$ to $[f(a), f(b)]$. To show that $f^{-1}$ is strictly monotone increasing, let $y_1, y_2 \in [f(a), f(b)]$ be real numbers such that $y_1 < y_2$. Then,by the intermediate value theorem, there exist $x_1, x_2 \in [a, b]$ such that $f(x_1) = y_1$ and $f(x_2) = y_2$. Since $f$ is strictly monotone increasing, we have $x_1 < x_2$. Using the definition of an inverse, we have\begin{align*}f^{-1}(y_1) &= f^{-1}(f(x_1)) \\&= x_1 \\&< x_2 \\&= f^{-1}(f(x_2)) \\&=f^{-1}(y_2) \text{,}\end{align*}showing that $f^{-1}$ is strictly monotone increasing. Finally, we will show that $f^{-1}$ is continuous. Let $y_0 \in [f(a), f(b)]$ be a real number, and let $\epsilon > 0 $ be a real number. As before, there exists a real number $x_0 \in [a, b]$ such that $f(x_0) = y_0$. Likewise, for any real number $y \in [f(a), f(b)]$, the intermediate value theorem tells us that there exists a real number $x \in [a, b]$ such that $f(x) = y$. We want to show that there exists a $\delta > 0 $ such that $ | f^{-1}(y) -f^{-1}( y_0) | < \epsilon$ for all $y \in [f(a), f(b)]$ such that $|y - y_0| < \delta$. This is equivalent to showing that there exists a $\delta > 0 $ such that $ | x - x_0 | < \epsilon$ for all $f(x) \in [f(a), f(b)]$ such that $|f(x) - f(x_0)| < \delta$. Written in the order we are more accustomed to, this is equivalent to showing that there exists a $\delta > 0 $ such that $|f(x) - f(x_0)| < \delta$ for all $x \in [a, b]$ such that $|x - x_0| < \epsilon$. Suppose, for the sake of contradiction, that $f^{-1}$ is not continuous. That is, suppose for all $\delta > 0$, there exists an $\epsilon > 0$ such that $|f(x) - f(x_0)| \ge \delta$ for all $x \in [a, b]$ such that $|x - x_0| < \epsilon$.
I am not really sure where to go from here, and I'm not certain I correctly negated the statement that the inverse of $f$ is continuous. Any help is greatly appreciated.
P.S. This is not for any homework, just self study. I've never taken a class in analysis, so please feel free to point out anything I am doing wrong (or that is less than rigorous).
