# Prove that if $f$ is strictly increasing on $I$, then $f$ has a continuous inverse.

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.

• Maybe ask where you found the proof, or at least provide us with a link to the proof you are not understanding? If we just go and guess which website/forum you are talking about, this will end in chaos.
– Dirk
Commented May 29, 2017 at 12:05
• You don't need $f$ to be continuous: if $f$ is strictly monoton on an interval, then $f^{-1}$ will be continuous. Commented May 29, 2017 at 12:08
• @Bemte For example, the one i just added to the OP. I don't understand why they've chosen that $\delta$ and then the rest of the proof and how it follows :( Commented May 29, 2017 at 12:10
• That proof is not quite correct: suppose $I=[a,b]$ and let $x_0=a$. Then there is no $f(x_0-\epsilon)$ ... Commented May 29, 2017 at 12:17
• @MichaelHoppe I think it is just for $x_0 \neq a, b$, as that is the difficult part... Commented May 29, 2017 at 12:20

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.

• Excellent, helps a lot! Commented May 29, 2017 at 15:00
• You're definitely welcome. Commented May 29, 2017 at 17:16
• The prove is due to my teacher Peter Dombrowski. Commented May 29, 2017 at 17:32

To show that $f^{-1}$ is continuous in $y_0$, they use the definition of continuity: for any $\epsilon$, if $y$ is 'close enough' from $y_0$ (i.e. $|y-y_0|<\delta$ for some $\delta$), then $f^{-1}(y)$ is close enough to $f^{-1}(y_0)$ (i.e. $|f^{-1}(y)-f^{-1}(y_0)|<\epsilon$).

Their choice for $\delta$ is just 'the one that works' (but a smaller one would work too, obviously). They prove it works thanks to the strict monotony of $f$. In practice, you often find the $\delta$ by thinking in reverse order: this is the $\epsilon$ inegality I want to reach, what $\delta$ do I need to reach it?

Here, what we need is that the image of $[x_0-\epsilon;x_0+\epsilon]$ by f contains all of $[y_0-\delta;y_0+\delta]$. Once you write it, you got the $\delta$ they chose...