# Inverse function $f^{-1}:f(\mathbb{R})\to\mathbb{R}$ of a strictly increasing function $f:\mathbb{R}\to\mathbb{R}$ is continuous

I have proved the following statement and I would like to know if my proof is correct and/or/if/how it can be improved.

"Suppose $$f:\mathbb{R}\to\mathbb{R}$$ is a strictly increasing function.

Prove that the inverse function $$f^{-1}:f(\mathbb{R})\to\mathbb{R}$$ is a continuous function."

My proof:

Let $$f:\mathbb{R}\to\mathbb{R}$$ be a strictly increasing function: then it is injective and as a function $$f:\mathbb{R}\to f(\mathbb{R})$$ it must be surjective so it has an inverse $$f^{-1}:f(\mathbb{R})\to\mathbb{R}$$ which must be strictly increasing too$$^{(1)}$$.

Suppose now that $$f^{-1}$$ were discontinuous at a point $$y_d\in f(\mathbb{R})$$: then, being an increasing function, $$y_d$$ must be a jump discontinuity so the interval $$I_{y_d}:=(\lim\limits_{y \to y_d^-,\ y\in f(\mathbb{R})\\}f^{-1}(y),\lim\limits_{y \to y_d^+,\ y\in f(\mathbb{R})}f^{-1}(y))=(\sup_{yy_d,\ y\in f(\mathbb{R})} f^{-1}(y))$$ must be nonempty and we can pick an element $$\bar{x}\neq x_d=f^{-1}(y_d)$$ in it so $$f(\bar{x})=y_d$$, but being $$f$$ strictly increasing by hypothesis it is also either $$f(\bar{x})>y_d$$ or $$f(\bar{x}), a contradiction.

So, $$f^{-1}(\mathbb{R})\to\mathbb{R}$$ cannot be discontinuous at any point i.e. it must be continuous on $$f(\mathbb{R})$$. $$\square$$

$$^{(1)}$$ let $$y_1,y_2\in f(\mathbb{R})$$ and suppose wlog $$y_1: then $$f^{-1}(y_1)=f^{-1}(f(x_1))=x_1$$ and $$f^{-1}(y_2)=f^{-1}(f(x_2))=x_2$$ and if $$x_1\geq x_2$$ then $$f(x_1)=y_1\geq y_2=f(x_2)$$ contradiction, so it must be $$x_1=f^{-1}(y_1)

• Is the function f continuous? Jun 15, 2021 at 9:21
• @EduardoMaza no, it is not continuous Jun 15, 2021 at 9:22
• Jun 15, 2021 at 9:28
• Note that the left-sided and/or the right-sided limit may not exist, e.g. if $y_d$ is an isolated point of $f(\Bbb R)$. Similarly, the sets $\{ y< y_d \mid y \in f(\Bbb R) \}$ and/or $\{ y> y_d \mid y \in f(\Bbb R) \}$ may be empty, so that their supremum resp. infimum is not defined. Jun 15, 2021 at 18:42
• @Martin R thank you for your comment; regarding your first remark, being $f^{-1}$ increasing, shouldn't the left and right-sides limit always exist ( math.stackexchange.com/q/4171854 ) (even if $y_d$ is an isolated point, see math.stackexchange.com/questions/27429/…)? Jun 15, 2021 at 20:14

Consider the function $$g:\Bbb R\to\Bbb R$$ given by $$g(y):=\begin{cases}y-1 & y<0\\y & y\ge 0.\end{cases}$$

Clearly, $$g$$ is strictly increasing, is a bijection from the range of its inverse to $$\Bbb R,$$ and has a jump disontinuity at $$y=0.$$ Letting $$y_d=0,$$ we see that $$\left(\lim_{y\to y_d^-,\ y\in\operatorname{dom}g}g(y),\lim_{y\to y_d^+,\ y\in\operatorname{dom}g}g(y)\right)=(-1,0)$$ is certainly nonempty, but contains no elements of the range of $$g$$--that is, no element of the domain of the inverse of $$g.$$

That's the flaw in your argument. Just because an interval is non-empty doesn't mean that it contains an element in the domain of an arbitrary, strictly increasing function. Since all you've concluded about $$f^{-1}$$ is that it is strictly increasing, is a bijection from the range of its inverse to $$\Bbb R,$$ and it has a jump discontinuity, then it is entirely possible that $$f^{-1}=g,$$ in which case your argument falls down.

In other words, you haven't actually justified the statement

we can pick an element $$\bar{x}\neq x_d=f^{-1}(y_d)$$ in [the nonempty interval] so $$f(\bar{x})=y_d,$$

Added: A better approach would be to proceed directly. Take an arbitrary $$y_0\in f[\Bbb R],$$ and let $$x_0:=f^{-1}(y_0).$$ Since $$f$$ is strictly increasing, then for $$x (resp., for $$x>x_0$$) we have $$f(x) (resp. $$f(x)>y_0$$).

Take an arbitrary $$\varepsilon>0,$$ let $$y_m:=f(x_0-\varepsilon),$$ and let $$y_M:=f(x_0+\varepsilon),$$ so that $$y_m,y_M\in f[\Bbb R]$$ and $$y_m

Letting $$\delta=\min\{y_0-y_m,y_M-y_0\},$$ we have $$\delta>0,$$ and for all $$y\in\Bbb R,$$ if $$|y-y_0|<\delta,$$ then $$y_m

In particular, take any $$y\in f[\Bbb R]$$ such that $$|y-y_0|<\delta,$$ and let $$x=f^{-1}(y).$$ Since $$f$$ is strictly increasing and $$f(x_0-\varepsilon)=y_m then $$x_0-\varepsilon Similarly, $$x and so $$|x-x_0|<\varepsilon,$$ or equivalently, $$\bigl|f^{-1}(y)-f^{-1}(y_0)\bigr|<\varepsilon,$$ whence we have showed that $$f^{-1}$$ is continuous at $$y_0,$$ as desired.

Call $$B=f(\mathbb{R})$$ and let $$g : B \to \mathbb{R}$$ be $$f$$'s inverse. Note that $$g$$ is also stricly increasing. Suppose $$g$$ is discontinuous at a point $$y \in B$$. Let $$\alpha = \lim_{w \in B, w \to y^{-}} g(w)$$ and $$\beta = \lim_{z \in B, z \to y^{+}} g(z)$$. Since $$g$$ is increasing and discontinuous at $$y$$, we have $$\alpha < \beta$$.

Take $$x_0 \in \left [\alpha + \frac{\beta - \alpha}{3}, \beta - \frac{\beta - \alpha}{3} \right ]$$.

Then, for small enough $$\delta > 0$$, it happens that $$g(y - \delta) < x_0 < g(y + \delta)$$. Note that $$\delta$$ doesn't depend on $$x_0$$.

Applying $$f$$ to the above inequality yields $$y - \delta < f(x_0) < y + \delta$$. Letting $$\delta \to 0$$, we find that $$f(x_0)=y$$. So $$f$$ is actually constant at the nondegenerate interval $$\left [\alpha + \frac{\beta - \alpha}{3}, \beta - \frac{\beta - \alpha}{3} \right ]$$!!

Here I assumed $$y$$ can be approximated with points of $$B$$ from both sides. But the same argument can be used if $$y$$ is only a left limit point or right limit point of $$B$$, to do this you just need to use $$g(y)$$ in place of $$\beta$$ and $$\alpha$$, respectively.

• Nice argument !
– Medo
Jul 31, 2021 at 19:44

First I would like to thank @Cameron Buie for his effort to point out the flaw in the OP proof.

Of course the statement is false. Monotonicity implies continuity almost everywhere. Even a strictly monotone function is not necessarily continuous. Neither is its inverse which is strictly monotone too.

Here is a simple counterexample:

$$f(x)=\left\{ \begin{array}{ll} x+2, & \hbox{x>0;} \\ 1, & \hbox{x=0;} \\ x, & \hbox{x<0.} \end{array} \right.$$ The function $$f:\mathbb{R}\rightarrow ]-\infty,0[\,\cup \,\{1\}\cup\,]2,\infty[\$$ is a strictly increasing bijection.

The strictly increasing inverse $$f^{-1}: ]-\infty,0[\,\cup \,\{1\}\,\cup\,]2,\infty[ \rightarrow \mathbb{R}$$

$$f^{-1}(x)=\left\{ \begin{array}{ll} x-2, & \hbox{x>2;} \\ 0, & \hbox{x=1;} \\ x, & \hbox{x<0.} \end{array} \right.$$ is discontinuous

• It seems to me that $-2<0$ but $f(-2)=2>1=f(0)$ so $f$ is not strictly increasing Jul 31, 2021 at 0:59
• The typo $-x$ is corrected to $x$.
– Medo
Jul 31, 2021 at 3:44
• Let $g=f^{-1}$: then $g$ is clearly true for $x<0$ and also $\lim_{x\to 0^-}=0=\lim_{x\to 0^+}g(x)$ (it is vacuously true that the right-hand limit exists and is equal to $0$) so it is also continuous at $0$; it is also vacuously true that $\lim_{x\to 1^-}g(x)=\lim_{x\to 1^+}g(x)=g(1)=0$ (take any $\varepsilon >0$ and $\delta =\frac{1}{2}$); it is clearly continuous for $x>2$ and $\lim_{x\to 2^-}g(x)=\lim_{x\to 2^+}g(x)=g(2)=0$ so $g=f^{-1}$ is indeed continuous. Jul 31, 2021 at 10:58
• @ lorenzo. No, because the limits $\lim_{x\rightarrow 0^{+}}g$ and $\lim_{x\rightarrow 2^{-}}g$ do not exist. Let me prove discontinuity in another way for you. What is the inverse image of the open interval $]1/2,3/2[$ ? It is the set $\{1\}\cup\,]5/2,7/2[$ which is not an open set. Convinced ?
– Medo
Jul 31, 2021 at 19:38
• The reasoning @lorenzo used is not correct, as you say, but their conclusion is correct. There are some subtle points that you missed, Medo. First, the preimage of $(1/2,3/2)$ under $g$ is the open set $(5/2,7/2).$ On the other hand, the preimage of $(-1,1)$ under $g$ is $(-1,0)\cup\{1\}\cup(2,3),$ which is not open in $\Bbb R,$ but is open in $dom(g)$ taken as a subspace of $\Bbb R,$ since (for example) $$dom(g)\cap(-1,3)=(-1,0)\cup\{1\}\cup(2,3).$$ The kicker is that $g$ is (vacuously) continuous at $1$ even though the limits don't exist, since $1$ is isolated in $dom(g).$ Aug 21, 2023 at 18:16

I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
This exercise is Exercise 23 on p.40 in Exercises 2B in this book.

Exercise 23
Suppose $$f:\mathbb{R}\to\mathbb{R}$$ is a strictly increasing function. Prove that the inverse function $$f^{-1}:f(\mathbb{R})\to\mathbb{R}$$ is a continuous function.
[Note that this exercise does not have as a hypothesis that $$f$$ is continuous.]

My proof:

I use this Ramiro's idea.
Obviously, $$f^{-1}:f(\mathbb{R})\to\mathbb{R}$$ is also a strictly increasing function.
Assume that $$f^{-1}$$ is not continuous at $$y_0\in f(\mathbb{R})$$.
Since $$f^{-1}$$ is not continuos at $$y_0$$, $$y_0$$ is a left limit point of $$f(\mathbb{R})$$ or a right limit point of $$f(\mathbb{R})$$.
If $$y_0$$ is a left limit point of $$f(\mathbb{R})$$, then $$\lim_{y\to y_0-} f^{-1}(y)$$ exists and $$\lim_{y\to y_0-} f^{-1}(y)=\sup\{f^{-1}(y):y
If $$y_0$$ is a right limit point of $$f(\mathbb{R})$$, then $$\lim_{y\to y_0+} f^{-1}(y)$$ exists and $$\lim_{y\to y_0+} f^{-1}(y)=\inf\{f^{-1}(y):y_0

(1) We consider the case in which $$y_0$$ is a left limit point of $$f(\mathbb{R})$$ and a right limit point of $$f(\mathbb{R}).$$
$$\lim_{y\to y_0-} f^{-1}(y)\leq f^{-1}(y_0)\leq\lim_{y\to y_0+} f^{-1}(y)$$ holds.
Since $$f^{-1}$$ is not continuous at $$y_0$$, $$\lim_{y\to y_0-} f^{-1}(y)=\lim_{y\to y_0+} f^{-1}(y)=f^{-1}(y_0)$$ does not hold.
So, $$\lim_{y\to y_0-} f^{-1}(y)< f^{-1}(y_0)$$ or $$f^{-1}(y_0)<\lim_{y\to y_0+} f^{-1}(y)$$ holds.

(2) We consider the case in which $$y_0$$ is a left limit point of $$f(\mathbb{R})$$, but is not a right limit point of $$f(\mathbb{R}).$$
$$\lim_{y\to y_0-} f^{-1}(y)\leq f^{-1}(y_0)$$ holds.
Since $$f^{-1}$$ is not continous at $$y_0$$, $$\lim_{y\to y_0-} f^{-1}(y)=f^{-1}(y_0)$$ does not hold.
So, $$\lim_{y\to y_0-} f^{-1}(y)< f^{-1}(y_0)$$ holds.

(3) We consider the case in which $$y_0$$ is a right limit point of $$f(\mathbb{R})$$, but is not a left limit point of $$f(\mathbb{R}).$$
$$f^{-1}(y_0)\leq\lim_{y\to y_0+} f^{-1}(y)$$ holds.
Since $$f^{-1}$$ is not continous at $$y_0$$, $$f^{-1}(y_0)=\lim_{y\to y_0+} f^{-1}(y)$$ does not hold.
So, $$f^{-1}(y_0)<\lim_{y\to y_0+} f^{-1}(y)$$ holds.

Therefore, at least, (a) or (b) holds:

(a) $$y_0$$ is a left limit point and $$\lim_{y\to y_0-} f^{-1}(y)
(b) $$y_0$$ is a right limit point and $$f^{-1}(y_0)<\lim_{y\to y_0+} f^{-1}(y).$$

First we consider the case (a):
Let $$x_0\in (\lim_{y\to y_0-} f^{-1}(y), f^{-1}(y_0))$$.
Then, since $$x_0, $$f(x_0)
And $$f(x_0)\in f(\mathbb{R})$$.
So, $$x_0=f^{-1}(f(x_0))\in\{f^{-1}(y):y
Note that $$\lim_{y\to y_0-} f^{-1}(y)=\sup\{f^{-1}(y):y holds.
So, $$x_0\leq\lim_{y\to y_0-} f^{-1}(y).$$
Let $$x_0\in (f^{-1}(y_0), \lim_{y\to y_0+} f^{-1}(y))$$.
Then, since $$f^{-1}(y_0), $$y_0=f(f^{-1}(y_0))
And $$f(x_0)\in f(\mathbb{R})$$.
So, $$x_0=f^{-1}(f(x_0))\in\{f^{-1}(y):y_0
Note that $$\lim_{y\to y_0+} f^{-1}(y)=\inf\{f^{-1}(y):y_0 holds.
So, $$\lim_{y\to y_0+} f^{-1}(y)\leq x_0.$$
So, $$f^{-1}$$ is continuous at $$y$$ for any $$y\in f(\mathbb{R}).$$