Subgroups of $(\mathbb R, +)$ are either dense or cyclic. I was trying to prove that any subgroup of $(\mathbb R, +)$ is either dense in $\mathbb R$ or is a cyclic subgroup of $(\mathbb R, +)$. 
Thanks in advance for any help.
 A: Let $H$ be an additive subgroup of $\mathbb R$. 
We have $H\cap \mathbb R^+\ne \emptyset$.
Let's define 
$$\eta:=\inf \{h\in H\cap \mathbb R^+\}.$$
We can distinguish two cases:


*

*If  $\eta>0$. 
Let $h\in H$, and $k\in\mathbb Z$ such that
$$k\eta\leq \vert h\vert< (k+1)\eta.$$
We have $\vert h\vert-k \eta\in H$, and $0\leq \vert h\vert-k\eta < (k+1)\eta-k\eta=\eta$.
So by the definition of $\eta$, $\vert h\vert-k\eta=0$, so $h=\pm k\eta$.
So $H=[\eta]$, in particular, $H$ is monogene.

*If $\eta=0$.
Let $r\in \mathbb R$, $\epsilon>0$.
Because $\eta=0$, there exists $h\in ]0,\epsilon]\cap H$.
We can consider $r\ge 0$, the case $r\leq 0$ can be treated the same way.
Let $k\in\mathbb N$ so that
$$kh\le r<(k+1).$$
We  do have $kh\in H$, and
\begin{align*}
 0
  &\le r-kh \\
  &\le (k+1)h-kh \\
  &=h \\
  &\le \epsilon
 \end{align*}
So  $\vert{r-kh}\vert\le \epsilon$, which shows that $H$ is dense in $\mathbb R$.
A: Assume $G<\Bbb R$ is not  dense, say no element of $G$ is in the non-empty open interval $(a,a+\epsilon)$ with $\epsilon>0$.
Then show that for every $g\in G$, we have $G\cap (g-\epsilon,g+\epsilon) =\{g\}$.
Then show that $G\cap(0,\infty)$ is either empty (in which case $G=\{0\}$ is cyclic) or has a minimal element $a$ (in which case $G=\langle a\rangle$)
