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Feb 9 at 19:36 comment added Emil Jeřábek If you fix $0<u\in H$, then for any $x$, $x\in H\implies(x-u,x+u)\subseteq H$ and $x\notin H\implies(x-u,x+u)\cap H=\varnothing$.
Feb 9 at 19:08 comment added Jeyrome Sapin "$H = \{b:\exists n\in\mathbb N\,(-na\le b\le na)\}$ is a nontrivial convex subgroup, hence clopen". Here, does convex mean that $\forall x,y \in G : (0 \leq x \leq y) \wedge (y \in H) \Rightarrow x \in H$? How does this imply that $H$ is clopen for, I suppose, the order topology?
Aug 27, 2014 at 14:05 history answered Emil Jeřábek CC BY-SA 3.0