Analysis Problem of the Day 62

Today’s problem appeared as Problem 9 on the Texas A&M Real Analysis Qual:

Problem 9. Prove or disprove that there exists a continuous function $f: \mathbb{R} \to \mathbb{R}$ such that $f(x)$ is rational for $x$ irrational and $f(x)$ is irrational for $x$ rational.

Solution: Suppose such an $f$ did exist. For $a_1,a_2,... \in \mathbb{Q}$ an enumeration of the rationals, consider the sets $A_i:=f^{-1}(\{a_i\}),$ which are closed in $\mathbb{R}$ since $f$ is continuous. Then, by assumption, $A_i \subseteq \mathbb{R} \setminus \mathbb{Q}$ and $\bigcup_{i=1}^\infty A_i = \mathbb{R} \setminus \mathbb{Q}.$ Thus, the irrationals would be a countable union of closed sets, which we call an $F_\sigma$ set. But the irrationals are not an $F_\sigma$ set (and the rationals are not a $G_\delta$ set) by a classic Baire category theorem argument. Thus, such an $f$ cannot exist.

Proof of the claim: By DeMorgan’s laws, it suffices to show the rationals are not a countable intersection of open sets. Suppose they were, i.e. $\mathbb{Q}=\bigcap_{i=1}^\infty U_i.$ Then, each $U_i$ contains $\mathbb{Q}$ and therefore must be dense in $\mathbb{R}.$ Clearly then, $U_i \setminus \{a_i\}$ is also open and dense. But then, $\bigcap_{i=1}^\infty U_i \setminus \{a_i\} = \varnothing,$ which is impossible since by the Baire category theorem, an intersection of open dense sets has to be dense (and thus cannot be empty).

Remark: In general, there are lots of powerful nonconstructive existence/non-existence arguments that rely on the Baire category theorem. For example, it can be proven that there exists a continuous nowhere differentiable function (without providing a specific example of one).

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