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I started to do some research on an unanswered highly upvoted question from this site and on a piece of paper I wrote:

Suppose $f$ is discontinuous at every point and $f(\mathbb R)=\mathbb R$ and $f$ is bijection

...and I stopped writing. Although I did not settle that question, I realized that I supposed that there (and here) exist some functions with properties mentioned above, and started to do a research without really knowing are there such functions?

So, in order to not to discuss and think about functions that I even do not know do they exist, I decided to ask you:

Let $f$ be an everywhere discontinuous (meaning, discontinuous at every point) bijection that maps $\mathbb R$ onto $\mathbb R$. Do such functions exist?

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$f(x)=x$ if $x$ is rational and $x+1$ if $x$ is irrational gives such a function.

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  • $\begingroup$ Very simple answer. $\endgroup$
    – Grešnik
    Jun 10, 2019 at 8:25
  • $\begingroup$ Do you have an example if we suppose that we also have $f(x) \neq x$ everywhere? $\endgroup$
    – Grešnik
    Jun 10, 2019 at 8:40
  • $\begingroup$ Well, what about $x+1$ for rationals and $x+2$ for irrationals? $\endgroup$
    – Kavi Rama Murthy
    Jun 10, 2019 at 8:44
  • $\begingroup$ Yes. I realized that we have when I added a comment, but did not want to delete a comment. What if we also have that for every interval $I$ that $I$ and $f(I)$ are disjoint? $\endgroup$
    – Grešnik
    Jun 10, 2019 at 8:45
  • $\begingroup$ What type of intervals are you talking about? If $I=(-\infty,\infty)$ then $I$ and $f(I)$ cannot be disjoint. $\endgroup$
    – Kavi Rama Murthy
    Jun 10, 2019 at 8:47

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