Is there a continuous function defined on $\mathbb{R}$ which is a bijection on $\mathbb{R}\backslash\mathbb{Q}$ but not a bijection on $\mathbb{Q}$.

I tried to argue in this way. Let $p,q\in\mathbb{Q}$ with $f(p)=f(q)$ but $p\neq q$. Suppose $a_n\rightarrow p, b_n\rightarrow q$. Then $\lim f(a_n)=\lim f(b_n)$. But the limit case seems of no use.

Another idea is to prove $f$ is strictly monotone in $\mathbb{R}$, but I do not know how to begin.

Appreciate any help or hint!

  • $\begingroup$ I'm not sure I understand. Do you want a continuous map $f : \mathbb{R} \to \mathbb{R}$ so that $f \restriction_{\mathbb{R} \setminus \mathbb{Q}}$ is a bijection onto its image, but $f \restriction_\mathbb{Q}$ is not a bijection onto its image? $\endgroup$ – HallaSurvivor May 17 at 6:58
  • $\begingroup$ Exactly. I want to prove such map does not exist. $\endgroup$ – user823011 May 17 at 7:02
  • $\begingroup$ If $f$ is not constant on $[p,q]$ then there is some $r\in (p,q)$ such that $f(r)\ne f(p)$ which implies that $f([p,r])\supset [f(p),f(r)],f([q,r])\supset [f(q),f(r)]$. $\endgroup$ – reuns May 17 at 10:04

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