no continuous function switches the rational and the irrational numbers

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

Proof

Suppose there is such a function $f$.

First, $\mathbb{Q}=\bigcup _{i\in \mathbb{N}}\{q_i\}$ implies

$$f(\mathbb{Q})=\bigcup _{i\in \mathbb{N}}\{f(q_i)\}.$$

This is because functions preserve unions (see properties of functions).

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

But functions preserve unions in both ways, so

$$ℝ=f^{-1}(\bigcup _{i\in \mathbb{N}}\{z_i\})=\bigcup _{i\in \mathbb{N}}{f}^{-1}(\{z_i\}).$$(1)

Now, $f$ is continuous, and as $\{z_i\}$ is closed for every $i\in \mathbb{N}$, so is $f^{-1}(\{z_i\})$. This means that $\overline{f^{-1}(\{z_i\})}=f^{-1}(\{z_i\})$. If $\mathrm{int}(f^{-1}(\{z_i\}))\ne \mathrm{\varnothing }$, 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 $f(\mathbb{Q})\subseteq 𝕁$ and $f(𝕁)\subseteq \mathbb{Q}$.

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