Continuous only on the Rationals?

We first show a necessary condition for a function to be discontinuous on a certain set and then show that $\mathbb{Q}^c = \mathbb{R} \setminus \mathbb{Q}$ doesn't satisfy this condition.

Lemma: The set of points where a real function $f$ is discontinuous is a countable union of closed sets.

Proof of lemma:

Define $\omega_f(x)$ as follows:

$\omega_f(a) = \inf_{\substack{a \in I \\ I \text{ is an} \\ \text{open} \\ \text{interval}}} \sup_{x, y \in I} |f(x) - f(y)|$

Note that $\omega_f(x) = 0 \iff f \text{ is continuous at } x$

Hence, set of points where $f$ is discontinuous is exactly the following : $\{x : \omega_f(x) > 0\} = \bigcup_{n = 1}^{\infty}\left \{x : \omega_f(x) \geq \frac{1}{n} \right \}$

So, we'll be done if we show the following are open : $\{x : \omega_f(x) < r\}$

If $\omega_f(a) < r$ then there is an open interval $I$ such that $a$ is in $I$ and $\sup_{x, y \in I } |f(x) - f(y)| < r$ . But then, $\omega_f(x) < r$ for every $x \in I$ . This concludes the proof.

Bonus: On a metric space $X$ with no isolated points, deduce a necessary and sufficient condition for the existance of a function $f: X \rightarrow \mathbb{R}$ such that the function is continuous on $A \subset X$ and discontinuous everywhere else.

---

By the lemma, we'll be done if we show that $\mathbb{Q}^c$ can't be written as a countable union of closed sets.

We proceed by contradiction.

So, suppose $\mathbb{Q}^c = \cup_{n = 1}^{\infty} F_n$ where $F_n$ are closed.

As $\mathbb{Q}^c$ has empty interior, so do $F_n$ .

As $\mathbb{Q} = \cup_{n =1}^{\infty} \{r_n\}$ where $r_n$ is an enumeration of rationals, we can write $\mathbb{R}$ as a countable union of closed sets, each of which has empty interior. This contradicts Baire Category Theorem .

A related result is Froda's theorem : Let $f$ be a monotone function defined on an interval $I$ . Then the set of discontinuities is at most countable.