Discontinuous additive functions via a Hamel basis

Example

View $\R$ as a vector space over the field $\Q$ of rational numbers. The numbers $1$ and $\sqrt{2}$ are linearly independent over $\Q$, because $\sqrt{2}$ is irrational. By the axiom of choice, this independent set extends to a basis $B$ of $\R$ over $\Q$, called a Hamel basis. Every real $x$ is then a unique finite sum $x = \sum_{b \in B} q_b(x)\, b$ with rational coefficients $q_b(x)$, almost all zero. Define $f \colon \R \to \R$ by $f(x) = q_1(x)$, the coefficient of the basis vector $1$.

Check of additivity: the coefficients of $x + y$ are the sums of the coefficients of $x$ and $y$, so $f(x + y) = f(x) + f(y)$. So $f$ is an additive function. It is not of the form $f(x) = cx$: $f(1) = 1$ would force $c = 1$, but $f(\sqrt{2}) = 0 \ne \sqrt{2}$. Hence $f$ is not continuous, since every continuous additive function is linear. One can also see this directly: $f$ takes only rational values, while a continuous function on $\R$ with only rational values is constant, and $f(0) = 0 \ne 1 = f(1)$.