3

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?

Huy
  • 6,867
  • 1
    Such a map is necessarily not measurable, so it's gonna be nasty and not nice. – ncmathsadist Dec 28 '15 at 16:08
  • 1
    http://math.stackexchange.com/questions/369736/cauchy-functional-equation-with-non-choice and http://math.stackexchange.com/questions/166176/a-question-concerning-on-the-axiom-of-choice-and-cauchy-functional-equation/ and http://math.stackexchange.com/questions/110125/is-there-a-non-trivial-example-of-a-mathbb-q-endomorphism-of-mathbb-r and http://math.stackexchange.com/questions/115486/what-is-operatornameaut-mathbbr and probably a few others – Asaf Karagila Dec 28 '15 at 16:11

1 Answers1

4

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.

Here you can see a handful of collected facts about the nontrivial solutions, provided any exist.

Community
  • 1
Wojowu
  • 27,526