Discontinuous group homomorphism from $(\mathbb{R},+)$ to $S^1$,unit circle

Question

It is well known that if $\gamma:(\mathbb{R},+)\rightarrow S^1$ is continuous homomorphism, then $\exists y\in\mathbb{R}$,such that $\gamma(x)=e^{ixy}$.

Show that there is a discontinuous homomorphism $\gamma:(\mathbb{R},+)\rightarrow S^1$, and if $\gamma:(\mathbb{R},+)\rightarrow S^1$ is a homomorphism that is a Borel function, then $\gamma$ is continuous.

Answer 1

As this question popped up recently, let me sketch the solution.

1. Regard $\mathbb R$ as a vector space over $\mathbb Q$. Fix a (necessarily uncountable) Hamel basis. Then you will find plenty of discontinuous $\mathbb Q$-linear functionals (for example by considering evaluations at basis elements). Fix one and call it $h$.

2. Consider the map $x\mapsto e^{i h(x)}$. Show that it is a discontinuous homomorphism.

As for the second part of your question, invoke Pettis' theorem, originally proved in:

B.J. Pettis, On continuity and openness of homomorphisms in topological groups, Ann. of Math. (2) 52, (1950), 293–308.