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I am looking at Measure, Integral and Probability book and that is what they say (about the title). Although, I fail to understand this.

If we take a function $f(x) = x^2$ and restrict its domain to $x \geq 0$. Then the inverse is $f^{-1}(x) = \sqrt{x}$. This function's output will be open for union of open intervals.

But why are we concerning ourselves with the inverse of a function to check its continuity? This definition feels like cheating to me. Because if the function $f$ is continuous, then $f^{-1}$ will be continuous, why does this have anything to do with open intervals? And so a proper definition of continuity has to be used, in that a function is differentiable at every point in the domain.

Naz
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    I think your definition says "$;f^{-1}(O);$ is open for each..." – DonAntonio Apr 22 '19 at 10:09
  • @JoséCarlosSantos, $f^{-1}$ has the meaning of a reverse image when we pass the set into the argument of a function? Thank you for defending my question :) – Naz Apr 22 '19 at 10:29
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    Yes, that's it. But if $f$ actually has an inverse, that both meanings of $f^{-1}(O)$ are the same. And you're welcome. ;-) – José Carlos Santos Apr 22 '19 at 10:33

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If $f$ is a function from $A$ to $B$ and $C \subset B$ then $f^{-1}(C)$ is defined as $\{a\in A:f(a)\in C\}$. This does not require the existence of an inverse for $f$.

Kavi Rama Murthy
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  • how does this make a function continuous if that set is open? – Naz Apr 22 '19 at 10:13
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    Do you know the definition of continuity using $\epsilon-\delta$? Then you can take it as a good exercise to try to prove that $f:\mathbb{R}\to\mathbb{R}$ is continuous if and only if $f^{-1}(O)$ is open for each open set $O\subseteq\mathbb{R}$. I'm pretty sure this was also proved on this forum before. – Mark Apr 22 '19 at 10:23
  • Thanks @Mark, I will attempt the proof. – Naz Apr 22 '19 at 10:25
  • can't seem to find the proof on here. I got up to the point where I show that if $f^{-1}(O)$ is open then $\forall x \in f^{-1}(O)$, there is a $\delta > 0$ such that $\left{ y \in \mathbb{R} | |y-x| < \delta \right} \subset f^{-1}(O)$. Now just need to show that this leads to: $|f(y)-f(x)| < \epsilon$. Can't think of a way. – Naz Apr 22 '19 at 11:17
  • I also found this: https://math.stackexchange.com/questions/1468774/f-is-continuous-iff-f-1u-is-open-for-all-open-u, where they give a counterexample – Naz Apr 22 '19 at 11:21
  • Also, I have not proved that $f^{-1}(O)$ is open if $O$ is open. I just assumed that it is. – Naz Apr 22 '19 at 11:22
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That statement has nothing to do with inverse functions. It's about reverse images: if $f\colon A\longrightarrow B$ is a function and $C\subset B$, then$$f^{-1}(C)=\{a\in A\mid f(a)\in C\}.$$

José Carlos Santos
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  • Of course, though this notation is standard, it is unfortunate, since beginners often assume it means that it is the image of some function "$f^{-1}$" as applied to $C$. – Angina Seng Apr 22 '19 at 10:29
  • self-studying measure theory is hard. I have a year worth of formal math education, but it is still hard sometimes. Not just the notation. My personal feeling is that there is a problem with the accessibility of the more abstract math topics from the self-learners perspective, unfortunately. – Naz Apr 22 '19 at 10:31
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    I fully agree. Bourbaki puts the “$-1$” above the $f$ in the case of reverse image and only uses $f^{-1}$ for the inverse function, but the idea was never caught up. – José Carlos Santos Apr 22 '19 at 10:32