harmonic analysis - The dual group of $mathbb Q$

In fact, uniform convergence on compact subsets of $mathbb{Q}subsetmathbb{R}$ induces the usual topology on its group of (continuous) characters $mathbb{R}simeq{tmapstoexp(ixt)}_{xinmathbb{R}}$.

Namely, consider $K={0}cup{1/n,ngeq 1}$. For $xinmathbb{R}$, the corresponding character is uniformly $epsilon$-close on $K$ to the trivial character iff $$|exp(ix/n)−1|<epsilon;;;; (*)$$ for all integers $ngeq1$. Then for $epsilon<1/sqrt{2}$, $x$ must be small : $|x|<2epsilon/pi$. Indeed, consider $kinmathbb{Z}$ such that $|x−kpi|leqpi/2$ , and take $n=|k|$; if $kneq 0$ we reach a contradiction in $(*)$. Hence $k=0$, and the claim follows easily.

This implies that uniform convergence on compact subsets of $mathbb{Q}$ (in fact the one compact subset $K$) induces the usual topology on $mathbb{R}simeqmathrm{Hom}(mathbb{Q},S^1)$.