Show that if $G$ is a non-trivial subgroup of $\Bbb R$ then either $G$ is dense in $\Bbb R$ or $G=l\Bbb Z$, where $l=\inf\{x\in G:x>0\}$.

My try:

If $G=\Bbb Q$ then $\Bbb Q$ is dense in $\Bbb R$.

If $G=n\Bbb Z$ then $G$ is not dense but $G$ takes the form $G=l\Bbb Z$.

But how should I do the sum for any subgroup of $\Bbb R$?Please help.

  • $\begingroup$ When $l=\inf\{x\in G:x>0\} = 0$, the subgroup will be dense. Take any $t \in \mathbb{R}$ and consider an open interval $(t-1/n, t+1/n)$. Choose $x \in G$ such that $0< x < \frac{1}{2n}$. Then one can find an integer $m$ such that $mx \in (t-1/n, t+1/n)$. $\endgroup$ – user348749 Nov 2 '16 at 7:02

First, note that $n G \subset G$ for every integer $n$.

If $G$ is cyclic, then $G = k \mathbb{Z}$ for some real nonzero $k$. (Because there are no elements of finite order.)

If $G$ is not cyclic, then $$\inf_{\substack{x, y \in G\\ x \neq y}}|x-y| = 0,$$ (otherwise the positive infimum would be your $k$). Thus, if $\epsilon > 0$ and $r$ are real numbers, then $G$ has an element $g$ such that for some integer $n$ the integral multiple $ng$ (which is also in $G$) is within distance $\epsilon$ of $r$.


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