# $f(\mathbb{R}\setminus \mathbb{Q}) \subseteq \mathbb{Q}$ and $f(\mathbb{Q}) \subseteq \mathbb{R}\setminus \mathbb{Q}$ imply that $f$ is not continuous [duplicate]

Possible Duplicate:
No continuous function that switches $\mathbb{Q}$ and the irrationals

Let $f: \mathbb{R} \to \mathbb{R}$ be function satisfying the two conditions: $f(\mathbb{R}\setminus \mathbb{Q}) \subseteq \mathbb{Q}$ and $f(\mathbb{Q}) \subseteq \mathbb{R}\setminus \mathbb{Q}$. Then,

Show that $f$ cannot be continuous.

I'm trying this problem for some time but can't make any useful progress. I will appreciate any help. Even some good hints will do. Regards.

Hint: The conditions imply the range of $f$ is countable and that $f$ is non-constant.
If $f$ is continuous we have that $f(\mathbb{R})$ is an interval. Thus $f(\mathbb{R})$ is uncountable. On the other hand, we have $$f(\mathbb{R}) \subset f(\mathbb{R}\setminus \mathbb{Q}) \cup f(\mathbb{Q})$$ Thus $f(\mathbb{R})$ is countable as union finite of countables. Contradiction.
• $f(\mathbb Q)$ is not contained in $\mathbb Q$, since $f(\mathbb Q) \subseteq \mathbb R \setminus \mathbb Q$, but of course $f(\mathbb R)$ is countable. – Hans Giebenrath Dec 5 '12 at 19:20