Open subgroups of $\mathbb{R}$ Let $G$ be a nonempty open subset of $\mathbb{R}$ (with usual topology on $\mathbb{R}$) such that $x,y\in G$ implies that $x-y\in G$. Show that $G=\mathbb{R}$.
Clearly $0\in G$. Now how to show that all real numbers are there in $G$? Please help.
 A: Since $G$ is open, we have $(-2\varepsilon,2\varepsilon)\subset G$ for some small positive $\varepsilon$. If $0\leqslant x\leqslant n\varepsilon$, then $x/n\in G$ hence $n\cdot x/n\in G$, that is $x\in G$. Do the same for $x\leqslant 0$. 
A: The complement of an open subgroup is a union of its cosets, all of which are open, so it is itself open. Therefore an open subgroup is also closed. This leaves very few options, as $\mathbb R$ is a connected space.
A: Hint: Every nonempty open subset of $\Bbb R$ contains at least one open interval.
A: More generally, every additive subgroup $G$ of $\mathbb R$ is either discrete or dense (*). Since a discrete set in $\mathbb R$ is not open, $G$ must be dense. Now, the only open and dense subset of $\mathbb R$ is $\mathbb R$ itself.
(*) Consider $\alpha = \inf \{ x \in G : x>0 \}$. Then $G$ is discrete iff $\alpha >0$, in which case $G=\alpha \mathbb Z$. 
A: Use the following facts:


*

*Since $G$ is a subgroup hence $0 \in G$.

*Since $G$ is open there is an $\epsilon >0$ such that $(-\epsilon,\epsilon) \subset G$.
