# Any epsilon-delta proof of the continuity of the inverse of a real-valued strictly monotonic continuous function on an open interval?

Let $a$ and $b$ be real numbers such that $a < b$, and let $f \colon (a, b) \to \mathbb{R}$ be a strictly monotonic, continuous function. Then I can show that the range of $f$ is also an open interval $(c, d)$, where $-\infty \leq c < d \leq +\infty$. Thus the inverse function $g \colon (c, d) \to (a, b)$ exists and the inverse function $g$ is also strictly monotonic: $g$ is strictly increasing if $f$ is strictly increasing, whereas $g$ is strictly decreasing if $f$ is strictly decreasing. Moreover, $g$ is also continuous on all of (c, d). I can show this through the inverse images of open intervals argument.

Now my question is, how to prove the continuity of $g$ at an arbitrary point $q \in (c, d)$, using the $\varepsilon$-$\delta$ argument?

My Attempt:

Suppose $q \in (c, d)$, and let $\varepsilon > 0$ be given. Then we can find a unique point $p \in (a, b)$ for which $q = f(p)$. As $f$ is continuous at $p$, so we can find a real number $\delta > 0$ such that $$\lvert f(x) - f(p) \rvert < \varepsilon$$ for every point $x \in (a, b)$ for which $$\lvert x-p \rvert < \delta.$$ That is, $$f(p) - \varepsilon < f(x) < f(p) + \varepsilon$$ whenever $$a < x < b \ \mbox{ and } \ p-\delta < x < p + \delta.$$ The last statement is equivalent to saying that $$f(x) \in \big( f(p) - \varepsilon, f(p) + \varepsilon \big)$$ whenever $$x \in (a, b) \cap \big( p - \delta, p + \delta \big).$$

What next? How to proceed from here and show that the function $g$ is continuous at the point $q \in (c, d)$?

• @JoséCarlosSantos what have you edited in my post? Commented May 8, 2018 at 11:22
• I removed the analysis tag (it was redundant, since you are already using the real-analysis tag) and I have added the inverse-function tag. Commented May 8, 2018 at 11:26

You can assume without loss of generality that $f$ is increasing. Take $\varepsilon>0$. Again, you can assume without loss of generality that $\varepsilon$ is so small that $(p-\varepsilon,p+\varepsilon)\subset(a,b)$. Let $\delta>0$ be so small that $(q-\delta,q+\delta)\subset\bigl(f(p-\varepsilon),f(p+\varepsilon)\bigr)$. Then\begin{align}|x-q|<\delta&\implies x\in\bigl(f(p-\varepsilon),f(p+\varepsilon)\bigr)\\&\implies g(x)\in(p-\varepsilon,p+\varepsilon)\\&\iff\bigl|g(x)-g(q)\bigr|<\varepsilon.\end{align}
• there is one small point, however, that I'd like to raise. From your proof, it seems that in showing that $g$ is continuous at $q$, we needn't use the continuity of the function $f$ at the point $p$. Is it really so? Commented May 8, 2018 at 11:21
• @SaaqibMahmood Indeed, the continuity of $f$ is not needed. Commented May 8, 2018 at 11:25
• I hope that this example will persuade you: let $f\colon[-1,1]\longrightarrow\mathbb R$ be defined by$$f(x)=\begin{cases}x&\text{ if }x<0\\&x+1\text{ otherwise.}\end{cases}$$It is discontinuous at $0$. Its inverse is$$\begin{array}{rccc}g\colon&[-1,0)\cup[1,2]&\longrightarrow&[-1,1]\\&x&\mapsto&\begin{cases}x&\text{ if }x<0\\x-1&\text{ otherwise.}\end{cases}\end{array}$$It is continuous at $1(=f(0))$, right?! Commented May 8, 2018 at 11:40