Let $\J=\R\setminus\Q$ denote the irrationals. There is no continuous function $f\colon \R \to \R$ such that $f(\Q)\subseteq \J$ and $f(\J)\subseteq\Q$ .

Proof

Suppose there is such a function $f$ .

First, $\Q = \bigcup\limits_{i \in \N } {\{q_i\}}$ implies $$ f(\Q) = \bigcup \limits_{i \in \N } {\{f(q_i)\}}. $$ This is because functions preserve unions (see properties of functions).

And then, $f(\Q)$ is first category, because every singleton in $\R$ is nowhere dense (because $\R$ with the Euclidean metric has no isolated points, so the interior of a singleton is empty).

But $f(\J) \subseteq \Q$ , so $f(\J)$ is first category too. Therefore $f(\R)$ is first category, as $f(\R) = f(\Q) \cup f(\J)$ . Consequently, we have $f(\R ) = \bigcup \limits_{i \in \N } {\{z_i\}}$ .

But functions preserve unions in both ways, so

\begin{equation} \R = f^{-1}(\bigcup \limits_{i \in \N } {\{z_i\}}) = \bigcup \limits_{i \in \N } {f^{-1}(\{z_i\})}. \end{equation} Now, $f$ is continuous, and as $\{z_i\}$ is closed for every $i \in \N $ , so is $f^{-1}(\{z_i\})$ . This means that $\overline{f^{-1}(\{z_i\})} = f^{-1}(\{z_i\})$ . If $\operatorname{int}(f^{-1}(\{z_i\})) \neq \varnothing$ , we have that there is an open interval $(a_i, b_i) \subseteq f^{-1}(\{z_i\})$ , and this implies that there is an irrational number $x_i$ and a rational number $y_i$ such that both lie in $f^{-1}(\{z_i\})$ , which is not possible because this would imply that $f(x_i) = f(y_i) = z_i$ , and then $f$ would map an irrational and a rational number to the same element, but by hypothesis $f(\Q)\subseteq \J $ and $f(\J)\subseteq \Q$ .

Then, it must be $\operatorname{int}(f^{-1}(\{z_i\})) = \varnothing$ for every $i \in \N$ , and this implies that $\R$ is first category (by (1)). This is absurd, by the Baire Category Theorem.