No continuous function switches $\mathbb{Q}$ and the irrationals Is there a way to prove the following result using connectedness?
Result:
Let $J=\mathbb{R} \setminus \mathbb{Q}$ denote the set of irrational numbers. There is no continuous map $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $f(\mathbb{Q}) \subseteq J$ and $f(J) \subseteq \mathbb{Q}$.
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Here's a way to use connectedness, really amounting to using the intermediate value theorem.
If $f(\mathbb{Q})\subseteq \mathbb R\setminus\mathbb Q$ and $f(\mathbb R\setminus \mathbb Q)\subseteq\mathbb  Q$, then $f(0)\neq f(\sqrt 2)$.  Because intervals are connected in $\mathbb R$ and $f$ is continuous, $f[0,\sqrt 2]$ is connected.  Because connected subsets of $\mathbb R$ are intervals, $f[0,\sqrt 2]$ contains the interval $\left[\min\{f(0),f(\sqrt 2)\},\max\{f(0),f(\sqrt 2)\}\right]$.  The set of irrational numbers in this interval is uncountable, yet contained in the countable set $f(\mathbb Q)$, a contradiction.
A slightly briefer outline: The hypothesis implies that $f$ is nonconstant with range contained in the countable set $\mathbb Q\cup f(\mathbb Q)$, whereas the intermediate value theorem and uncountability of $\mathbb R$ imply that a nonconstant continuous function $f:\mathbb R\to\mathbb R$ has uncountable range.
