Thomae's function

Example

Define $f \colon \mathbb{R} \to \mathbb{R}$ as follows. For irrational $x$, $f(x) = 0$. For rational $x = p/q$ in lowest terms with $q > 0$, $f(x) = 1/q$. Then $f$ is continuous at each irrational number and discontinuous at each rational number.

Remarks

Let $a$ be irrational and let $\varepsilon > 0$. Only finitely many $p/q$ near $a$ have $1/q \ge \varepsilon$. So $f(x) < \varepsilon$ for all $x$ close enough to $a$. At a rational $a$, irrational points near $a$ have value $0$, but $f(a) > 0$. The function is periodic with period $1$. No function on $\mathbb{R}$ is continuous exactly at the rationals.

Explanations

L1 human

The value $1/q$ is small when the denominator $q$ is large. Near an irrational point, all rationals have large denominators, so the values there are small.