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