# No function continuous at rational points and discontinuous at irrational points in $[0,1]$

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?

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.