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Let $C_f$ and $D_f$ mean sets where a function is continuous and discontinuous. I’m trying to prove there is no function $f:[0,1] \to \mathbb{R}$ such that $C_f = [0,1] \cap \mathbb{Q}$ and $D_f = [0,1] \setminus \mathbb{Q}$.

I have seen a proof using the Baire category theorem that there cannot be two functions $f$ and $g$ where $C_f$ and $C_g$ are both dense and $C_f = D_g$. Thomae’s function is continuous at the irrationals and discontinuous at the rationals, so it is impossible to have a function that is discontinuous on irrationals and continuous on rationals. The proof uses a lemma that if $C_f$ is dense then $D_f$ is first category, and some of the steps are not clear to me.

Is there a more elementary proof that there is no function discontinuous on irrationals and continuous on rationals?

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The set of points of continuity of a function $f$ is a $G_\delta$ (i.e. the intersection of a sequence of open sets), because it can be written as

$$ \bigcap_{n \in \mathbb N} \bigcup_{\delta > 0} \{x: \text{diam}(f((x-\delta,x+\delta)))<1/n\}$$ and $\bigcup_{\delta>0} \{x: \text{diam}(f((x-\delta,x+\delta))<1/n\}$ is open.

$[0,1] \cap \mathbb Q$ is not a $G_\delta$ by the Baire Category Theorem. But you can prove this in an elementary way. Suppose $G = \bigcap_{n=1}^\infty U_n$ is a $G_\delta$ in $[0,1]$ that contains all rationals in that interval. Let $r_n$ be an enumeration of those rationals. You can construct inductively a nested sequence of closed intervals $[a_n, b_n]$ such that $[a_n, b_n] \subset U_n$ and does not contain $r_n$. Then the intersection of these intervals is a nonempty subset of $G$ that contains no rationals.

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