Does there exist a bijective function $f:{\mathbb R}\rightarrow{\mathbb R}$ that is nowhere-continuous, assuming that both domain and range have the "standard topology"? 1

1 By this I mean the one generated by the open intervals $(a, b) \subset {\mathrm R}$. BTW, if this topology has a name more readily recognized than the standard topology (on ${\mathbb R}$), please toss me a comment!

EDIT: the original version of this question allowed for the possibility that $f$ be only injective, but shortly after I posted the following injective function came to mind: let $n:{\mathbb Q}\rightarrow {\mathbb N}$ be an ordering of the rationals, and define

$$f(x)=\begin{cases} n(x) & x\in\mathbb Q\\ x& x\notin\mathbb Q\end{cases}$$

It is clear that this $f$ is injective, and it seems to me that the proof of the nowhere-continuity of the Dirichlet function applies to this case as well.

EDIT2: OK, I was next going to try modifying the candidate above to make the function bijective, but Asaf Karagila got there first, with a much neater solution than what I was heading for...

  • $\begingroup$ You can call the topology the euclidean topology, because it's induced by the euclidean metric $d(x,y) = |x-y|$. $\endgroup$ – Dario Feb 3 '13 at 12:25

How about: $$f(x)=\begin{cases} x+1 & x\in\mathbb Q\\ x& x\notin\mathbb Q\end{cases}$$

  • $\begingroup$ Nice, thanks!!! $\endgroup$ – kjo Feb 2 '13 at 15:35

This function should work.

$ f(x) = \begin{cases} x & \text{if x $\in \mathbf{Q}$ and $x \neq 0$} \\ -x & \text{if x $\in \mathbf{R-Q}$}\\ \sqrt{2} & \text{if $ x = 0 $} \end{cases}$

You can find more counterexaples of this type in "Counterexamples in analysis" by Gelbaum and Holmsted.

By the way, as far as I know, the standard topology on $\mathbf{R}^{n}$ is usually called "euclidean topology"

  • 2
    $\begingroup$ Not a bijection, $f(0)=f(-\sqrt{2})=\sqrt{2}$. $\endgroup$ – Julien Feb 2 '13 at 15:39
  • 2
    $\begingroup$ does not the case $x\in\mathbb Q$ include $x=0$? $\endgroup$ – Stefan Hamcke Feb 2 '13 at 15:48
  • $\begingroup$ You are both right: I edited the answer including $x \in \mathbf{Q}$ and $x \neq 0$. By the way @Asaf example is really nice, instead my function works for $x \geq 0 $ $\endgroup$ – Alessandro Feb 3 '13 at 11:43

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