12
$\begingroup$

Possible Duplicate:
Subgroup of $\mathbb{R}$ either dense or has a least positive element?

Let $(\mathbb{R},+)$ be the group of Real Numbers under addition. Let $H$ be a proper subgroup of $\mathbb{R}$. Prove that either $H$ is dense in $\mathbb{R}$ or there is an $a \in \mathbb{R}$ such that $H=\{ na : n=0, \pm{1},\pm{2},\dots\}$.

I am not able to proceed.

$\endgroup$
  • 1
    $\begingroup$ This is not about analysis. $\endgroup$ – Rasmus Aug 9 '10 at 14:06
  • 2
    $\begingroup$ I think the title should be 'Characterizing non-dense subgroups of R' $\endgroup$ – AgCl Aug 10 '10 at 9:29
  • 1
    $\begingroup$ I had to prove this once to solve a Monthly problem; it seemed like something that should be well-known, but I never found it in print. Does anyone know of a reference? $\endgroup$ – Nate Eldredge Aug 11 '10 at 4:35
  • 1
    $\begingroup$ @Nate Eldredge: "Principles of Real Analysis" by Charlambos Aliprantis. This question is somewhere in the first chapter. $\endgroup$ – anonymous Aug 14 '10 at 13:42
16
$\begingroup$

If there is a smallest positive element, then we are done, since any positive element must be an integer multiple of it, or otherwise we could use a euclidean-type-algorithm to get a positive element with smaller value. (I.e., suppose $a$ is the smallest positive element, and $b$ a positive element which is not an integer multiple of $a$---keep subtracting copies of $a$ until you get something that is strictly between $0 $ and $a$.)

So assume there is a sequence $a_n$ contained in the group that consists of positive numbers tending to zero. Then the group contains each ${\mathbb{Z} a_n}$. This means that for each $n$, any number in $\mathbb{R}$ is within $|a_n|$ of an element of the group. Since the $|a_n|$ can be small, we find that the group is dense.

$\endgroup$
  • $\begingroup$ How do you go about your thinking, and what makes you to think about the least element. $\endgroup$ – anonymous Aug 9 '10 at 18:11
  • $\begingroup$ It's a fairly standard trick (cf. the proof that any discrete subgroup of $\mathbb{R}^n$ of full rank is a lattice, for instance). $\endgroup$ – Akhil Mathew Aug 9 '10 at 18:59