0
$\begingroup$

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?

$\endgroup$
7
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Very simple answer. $\endgroup$ – Grešnik Jun 10 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 at 8:40
  • $\begingroup$ Well, what about $x+1$ for rationals and $x+2$ for irrationals? $\endgroup$ – Kavi Rama Murthy Jun 10 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 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 at 8:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.