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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.

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marked as duplicate by Jonas Meyer, Henry T. Horton, amWhy, Thomas, Micah Dec 5 '12 at 22:31

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2 Answers 2

up vote 12 down vote accepted

Hint: The conditions imply the range of $f$ is countable and that $f$ is non-constant.

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Supercool hint. Got it. Thanks a thousand tons! :-) –  Sayantan Dec 5 '12 at 17:29

If $ f $ is continuous we have that $ f(\mathbb{R}) $ is an interval. Thus $ f(\mathbb{R}) $ is uncontable. 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 contable as union finite of contables. Contradiction.

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$f(\mathbb Q)$ 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
Sorry, I writed fast and did this mistake. But I will correct. –  user29999 Dec 5 '12 at 21:20

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