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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 $I_{y_d}:=(\lim\limits_{y \to y_d^-,\ y\in f(\mathbb{R})\\}f^{-1}(y),\lim\limits_{y \to y_d^+,\ y\in f(\mathbb{R})}f^{-1}(y))=(\sup_{y<y_d,\ y\in f(\mathbb{R})}f^{-1}(y),\inf_{y>y_d,\ y\in f(\mathbb{R})} f^{-1}(y))$ 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 $f(\bar{x})<y_d$, 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 $y_1<y_2$: 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 $x_1=f^{-1}(y_1)<x_2=f^{-1}(y_2)$

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  • $\begingroup$ Is the function f continuous? $\endgroup$ Jun 15 at 9:21
  • $\begingroup$ @EduardoMaza no, it is not continuous $\endgroup$
    – lorenzo
    Jun 15 at 9:22
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    $\begingroup$ See also math.stackexchange.com/q/3719441/42969 and math.stackexchange.com/q/4116908/42969. $\endgroup$
    – Martin R
    Jun 15 at 9:28
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    $\begingroup$ 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. $\endgroup$
    – Martin R
    Jun 15 at 18:42
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    $\begingroup$ @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/…)? $\endgroup$
    – lorenzo
    Jun 15 at 20:14
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Consider the function $f^{-1}:\Bbb R\to\Bbb R$ given by $$f^{-1}(y):=\begin{cases}y-1 & y<0\\y & y\ge 0.\end{cases}$$ Here, I use the label "$f^{-1}$" only formally, not as an indication that it is the inverse of some function $f$ (though of course, it is the inverse of a function).

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

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

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

Since all you've concluded about $f^{-1}$ is that it is strictly increasing and is a bijection from the range of its inverse to $\Bbb R,$ then it is entirely possible that $f^{-1}=g,$ in which case your argument falls down.

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 $x<x_0$ (resp., for $x>x_0$) we have $f(x)<y_0$ (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 $y_m<y_0<y_M.$

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 $y_m<y<y_M.$

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 $f(x_0-\varepsilon)=y_m<y=f(x),$ then $x_0-\varepsilon<x.$ Similarly, $x<x_0+\varepsilon,$ 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.

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  • $\begingroup$ I'll think on it! If I can come up with a slick and straightforward proof, I'll add it to my answer. $\endgroup$ Jul 31 at 0:00
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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.

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  • $\begingroup$ Nice argument ! $\endgroup$
    – Medo
    Jul 31 at 19:44
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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

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  • $\begingroup$ It seems to me that $-2<0$ but $f(-2)=2>1=f(0)$ so $f$ is not strictly increasing $\endgroup$
    – lorenzo
    Jul 31 at 0:59
  • $\begingroup$ The typo $-x$ is corrected to $x$. $\endgroup$
    – Medo
    Jul 31 at 3:44
  • $\begingroup$ 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. $\endgroup$
    – lorenzo
    Jul 31 at 10:58
  • $\begingroup$ @ 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 ? $\endgroup$
    – Medo
    Jul 31 at 19:38

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