another proof of the non-existence of a continuous function that switches the rational and the irrational numbers

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

Proof

Suppose $f$ is such a function. Since $\mathbb{Q}$ is countable, $f(\mathbb{Q})$ and $f(\mathbb{J})$ are also countable. Therefore the image of $f$ is countable. If $f$ is not a constant function, then by the intermediate value theorem the image of $f$ contains a nonempty interval, so the image of $f$ is uncountable. We have just shown that this isn’t the case, so there must be some $c$ such that $f(x)=c$ for all $x\in\mathbb{R}$ . Therefore $f(\mathbb{Q})=\{c\}\subset\mathbb{J}$ and $f(\mathbb{J})=\{c\}\subset\mathbb{Q}$ . Obviously no number is both rational and irrational, so no such $f$ exists.

Title	another proof of the non-existence of a continuous function that switches the rational and the irrational numbers
Canonical name	AnotherProofOfTheNonexistenceOfAContinuousFunctionThatSwitchesTheRationalAndTheIrrationalNumbers
Date of creation	2013-03-22 16:23:54
Last modified on	2013-03-22 16:23:54
Owner	neapol1s (9480)
Last modified by	neapol1s (9480)
Numerical id	7
Author	neapol1s (9480)
Entry type	Proof
Classification	msc 54E52