Subgroups of $(\mathbb R, +)$ are either dense or cyclic.

Solution 1:

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 $\mathbb R$.

Solution 2:

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 $G=\langle a\rangle$)