Everywhere discontinuous bijections, do they exist?

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?

$$f(x)=x$$ if $$x$$ is rational and $$x+1$$ if $$x$$ is irrational gives such a function.
• Do you have an example if we suppose that we also have $f(x) \neq x$ everywhere? – Grešnik Jun 10 at 8:40
• Well, what about $x+1$ for rationals and $x+2$ for irrationals? – Kavi Rama Murthy Jun 10 at 8:44
• 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? – Grešnik Jun 10 at 8:45
• What type of intervals are you talking about? If $I=(-\infty,\infty)$ then $I$ and $f(I)$ cannot be disjoint. – Kavi Rama Murthy Jun 10 at 8:47