# Is there a “nice” discontinuous, bijective homomorphism $f: (\mathbb{R},+) \to (\mathbb{R},+)$?

Consider $(\mathbb{R},+)$ as a topological group. Using the axiom of choice, we can construct a $\mathbb{Q}$-basis for $\mathbb{R}$ and using this basis, we can define a discontinuous, bijective homomorphism from $(\mathbb{R},+)$ to itself.

Is it possible to find such a homomorphism without using the axiom of choice?

• Such a map is necessarily not measurable, so it's gonna be nasty and not nice. – ncmathsadist Dec 28 '15 at 16:08
• – Asaf Karagila Dec 28 '15 at 16:11

The answer is no, we need some choice to construct such homomorphism, because it is consistent with ZF (without choice) that every function $\phi:\Bbb R\rightarrow\Bbb R$ satifying $\phi(x+y)=\phi(x)+\phi(y)$ is continuous. You can get far more details in this MO answer.