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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 ie 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? $\endgroup$ – Eduardo Maza 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
  • $\begingroup$ @MartinR thank you for you interest in my question and for these links, I have learned a new proof of this statement by reading the second one. It seems to me however that my proof is a bit different from these two: would you mind checking it out and telling me if you think it is correct? $\endgroup$ – lorenzo Jun 15 at 9:51
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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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