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I have thought without a solution. Are there actually examples of a function $f:\Bbb{R}\to \Bbb{R}$ such that $f$ is discontinuous at every point but $f\circ f$ is continuous?

Answers will be highly appreciated.

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    $\begingroup$ $f(x)=0$ for $x$ irrational and $f(x)=1$ for $x$ rational. Then $f\circ f(x)=1$. $\endgroup$
    – user551819
    May 4 '18 at 12:49
  • $\begingroup$ @ totoro: Please, does it work? $\endgroup$ May 4 '18 at 12:50
  • $\begingroup$ @ totoro: I've just seen that! Thanks! $\endgroup$ May 4 '18 at 12:51
  • $\begingroup$ You have an example in front of your nose. $\endgroup$
    – user551819
    May 4 '18 at 12:51
  • $\begingroup$ No what you are looking for but a nice, related concept: en.wikipedia.org/wiki/Cantor_function $\endgroup$ May 4 '18 at 13:03
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Consider $$ f(x)=\left\{ \begin{array}{ll} x,&x\in\mathbb{Q},\\ -x,&x\in\mathbb{R}\setminus\mathbb{Q}. \end{array} \right. $$ This function yields $$ f\circ f(x)=x. $$

Edit:

Let me fix the bug. Thanks to @totoro, the above example does not work, because it is continuous at $x=0$.

Considering this, let us make it as follows. $$ f(x)=\left\{ \begin{array}{ll} 1/x,&x\in\mathbb{Q}\setminus\left\{0\right\},\\ 0,&x=0,\\ -1/x,&x\in\mathbb{R}\setminus\mathbb{Q}. \end{array} \right. $$ Now this function is everywhere discontinuous, and yields $$ f\circ f(x)=x. $$

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    $\begingroup$ This function is continuous at $x=0$. $\endgroup$
    – user551819
    May 4 '18 at 12:53
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    $\begingroup$ @totoro: You are right. So what about the template $1/x$? $\endgroup$
    – hypernova
    May 4 '18 at 12:56
  • $\begingroup$ @ hypernova: Thanks for the edit. You have my respect for fixing this! @ totoro and hypernova: Thanks for the eagle eye. You are both good! $\endgroup$ May 4 '18 at 13:32
  • $\begingroup$ Thank you. And I really appreciate @totoro for the crucial comment. I will keep my mistake as it was, in case it could be helpful here :-) $\endgroup$
    – hypernova
    May 4 '18 at 14:13

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