Proving that an open subgroup of $\mathbb{R}$ must be all of $\mathbb{R}$ Let $G$ be an open subset of $\mathbb{R}$ which is also a subgroup of the group $(\mathbb{R}, +)$. Show that $G = \mathbb{R}$.
(Hint: $0$ belongs to $G$ and hence $(-\epsilon,\epsilon)$ is a subset of $G$ for some $\epsilon > 0$. Use the fact that $G$ is closed under addition.)
 A: The hint pretty much tells you what to do, but I’ll expand it a bit. Start with the fact that $0\in G$, since it’s the identity element. Since $G$ is open, there must be an open interval $(-E,E)$ around $0$ that is contained in $G$. For any $g\in G$ we have $g+g\in G$, and $g+g+g\in G$, and in general $ng\in G$ for each positive integer $n$.
Let $I=(-E,E)$, and for $n\ge 1$ let $I_n=\{ng:g\in I\}$. Each $I_n\subseteq G$. What is $\bigcup_\limits{n\ge 1}I_n$?
A: Suppose for contradiction that $G$ is a proper subset of $\mathbb{R}$. Following the hint, let
$$
s=\sup\{E\in \mathbb{R}_+\,|\, (E,-E)\in G\}<\infty
$$
Now, clearly $\frac{s}{2}\in G$ and therefore $\frac{s}{2}+\frac{s}{2}=s\in G$ (the same argument holds for $-s$). How does this allow us to complete the argument?
Full solution:

 But since $G$ is open, it must contain an open neighborhood of $s$ and of $-s$, contradicting the supremum property of $s$, hence yielding a contradiction. Conclude $G=\mathbb{R}$.

A: Proof: Let if possible $\mathbb R-G\neq \phi$,then take $r>0$ from $\mathbb R-G$.Then $\frac{r}{n}\in \mathbb R-G$ for all $n\in \mathbb N$.
Since $G$ is open,$\mathbb R-G$ is closed and $\{\frac{r}{n}\}_n$ is a sequence in $\mathbb R-G$ which converges to $0\in \mathbb R$.But,$0\notin \mathbb R-G$.Thus we have a contradiction.So,$\mathbb R-G$ is empty and hence $G=\mathbb R$.
A: There is a more general fact:

Lemma. Let $G$ be a topological group and $H\subseteq G$ a subgroup with nonempty interior. Then $H$ is both closed and open in $G$.

Proof. First we will show that $H$ is open. Let $x\in H$. By assumption there is $U\subseteq H$ nonempty and open. Let $u\in U$. Then $x\in xu^{-1}U$. Of course $xu^{-1}U$ is open as well (because $g\mapsto hg$ is a homeomorphism for any $h$) and $xu^{-1}U\subseteq H$ since $H$ is a subgroup. This shows that $H$ is open.
Now we will show that $H$ is closed by showing that its complement is open. Let $x\in G$ be such that $x\not\in H$. Since $H$ is open then so is $xH$ but obviously $H\cap xH=\emptyset$ and thus $x$ has an open neighbourhood fully contained in the complement of $H$. Which completes the proof. $\Box$

Corollary. Let $G$ be a connected topological group and $H\subseteq G$ a subgroup with nonempty interior. Then $H=G$.

Proof. By our lemma $H$ is both closed and open. But since $G$ is connected then the only possibility is that $H=G$. $\Box$
And so all you need to know is that $\mathbb{R}$ is a connected topological group.
