TITLE: $f(\mathbb{R}\setminus \mathbb{Q}) \subseteq \mathbb{Q}$ and $f(\mathbb{Q}) \subseteq \mathbb{R}\setminus \mathbb{Q}$ imply that $f$ is not continuous
QUESTION [5 upvotes]: 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.

REPLY [5 votes]: 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
\begin{equation}
  f(\mathbb{R})  \subset f(\mathbb{R}\setminus \mathbb{Q})  \cup f(\mathbb{Q})   
\end{equation}
Thus $f(\mathbb{R})$ is countable as union finite of countables.
 Contradiction.