In fact, uniform convergence on compact subsets of $\mathbb{Q}$ \mathbb{Q}\subset\mathbb{R}$induces the usual topology on its group of (continuous) characters $\mathbb{R}\simeq\{t\mapsto\exp(ixt)\}_{x\in\mathbb{R}}$. Namely, consider $K=\{0\}\cup\{1/n,n\geq 1\}$. If For$x$x\in\mathbb{R}$, the corresponding character is real and uniformly $\epsilon$-close on $K$ to the trivial character iff $$|exp(ix/n)−1|<\epsilon\;\;\;\; (*)$$ for all integers $n\geq1$, then n\geq1$. Then for small enough$\epsilon$($<1/\sqrt{2}$will do), \epsilon<1/\sqrt{2}$, $x$ must be small : $|x|<2\epsilon/\pi$ (hint: <2\epsilon/\pi$. Indeed, consider$k\in\mathbb{Z}$such that$|x−k\pi|\leq\pi/2$, and take$n=|k|$n=|k|$; if $k\neq 0$ to we reach a contradiction in $(*)$. Hence $k=0$, and the claim follows easily)easily.
This implies that uniform convergence on compact subsets of $\mathbb{Q}$ (in fact the one compact subset ) $K$) induces the usual topology on $\mathbb{R}$.\mathbb{R}\simeq\mathrm{Hom}(\mathbb{Q},S^1)$. 1 In fact, uniform convergence on compact subsets of$\mathbb{Q}$induces the usual topology on its group of (continuous) characters $\mathbb{R}\simeq\{t\mapsto\exp(ixt)\}_{x\in\mathbb{R}}$. Namely, consider $K=\{0\}\cup\{1/n,n\geq 1\}$. If$x$is real and $$|exp(ix/n)−1|<\epsilon\;\;\;\; (*)$$ for all integers$n\geq1$, then for small enough$\epsilon$($<1/\sqrt{2}$will do),$x$must be small :$|x|<2\epsilon/\pi$(hint: consider$k\in\mathbb{Z}$such that$|x−k\pi|\leq\pi/2$, and take$n=|k|$if$k\neq 0$to 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 one compact subset) induces the usual topology on$\mathbb{R}\$.