No function $\mathbb{R} \to \mathbb{R}$ is continuous at every rational and discontinuous at every irrational, because $\mathbb{Q}$ is not $G_\delta$

There is no function $f : \mathbb{R} \to \mathbb{R}$ that is continuous at every rational and discontinuous at every irrational (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$: the $\varepsilon$-$\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point, The rationals embed densely in the reals).

Equivalently: $\mathbb{Q}$ is not the continuity set of any function $\mathbb{R} \to \mathbb{R}$.

The contrast with Thomae's function is the point. There is a function continuous exactly at the irrationals, namely $t$ (The Dirichlet function is continuous at no point of $\mathbb{R}$, and Thomae's function is continuous at every irrational and at no rational, so its set of continuity points is exactly the set of irrationals and its oscillation at $c$ equals $t(c)$), and one might expect the two arrangements to be symmetric. They are not, because the classes $F_\sigma$ and $G_\delta$ are exchanged by complementation while $\mathbb{Q}$ and the irrationals are, and only one of the two sets is $G_\delta$ ($\mathbb{Q}$ is $F_\sigma$, meager and not $G_\delta$, while the irrationals are $G_\delta$, residual and not $F_\sigma$).

• Where the work actually is. Nothing in this corollary is hard; all of it was done earlier. The $G_\delta$ theorem is For $f : A \to \mathbb{R}$ the set of points of $A$ at which $f$ is discontinuous is the intersection with $A$ of an $F_\sigma$ subset of $\mathbb{R}$, and the set of points at which $f$ is continuous is the intersection with $A$ of a $G_\delta$ subset; for $A = \mathbb{R}$ the two sets are $F_\sigma$ and $G_\delta$ outright, which rests on the oscillation, and the failure of $\mathbb{Q}$ to be $G_\delta$ is $\mathbb{Q}$ is $F_\sigma$, meager and not $G_\delta$, while the irrationals are $G_\delta$, residual and not $F_\sigma$, which is where the Baire category theorem is spent. The corollary is the place where those two meet.

• A weaker statement is true and much cheaper, and is not what is proved here. That no monotone function is continuous exactly at the rationals follows from Froda's theorem: the set of discontinuities of a monotone function on an interval is at most countable, the injection into $\mathbb{N}$ being built from one fixed enumeration of the rationals by least index, so no choice principle is used alone, since the irrationals are uncountable. The statement above is about all functions and needs category.