TITLE: Subgroup of $\mathbb{R}$ either dense or has a least positive element?
QUESTION [36 upvotes]: Let's say $G$ is some additive subgroup of $\mathbb{R}$ that has at least two elements. 
From what I understand, $G$ is then either dense in $\mathbb{R}$, or has some least positive element. What is the reason for this?

REPLY [10 votes]: I once used this fact to solve a problem in the American Mathematical Monthly.  Here's the proof I gave, which is rather similar to Paul's.
We first show that if $G$ has a limit point, then it is dense.  Fix
  $\epsilon > 0$ and let $y$ be a limit point of $G$.  Choose $x_1, x_2 \in
  G$ with $0 < |{x_1-y}| < \epsilon/2$, and $|{x_2 - y}| < |{x_1 -
  y}|$.  Then $x := x_1 - x_2 \in G$ with $0 < |{x}| < \epsilon$.
  $G$ now contains all integer multiples of $x$, so every interval of
  length at least $\epsilon$ contains an element of $G$.  $\epsilon$
  was arbitrary so $G$ is dense.
As the contrapositive of this, if $G$ is not dense then it has no
  limit points.  In this case, let $a := \inf \{ x \in G, x > 0\}$.  (This set is nonempty since $G$ is nontrivial and hence contains at
  least one positive number.)  Since neither $0$ nor $a$ are limit
  points of $G$, $a > 0$ and $a \in G$.  By construction, $a$ is the least positive element of $G$.  (Moreover, $G$ is generated by $a$.)
For those curious about the Monthly problem, it's #11345, volume 115 number 2, page 166, February 2008:

11345. Proposed by Roger Cuculière, France.  Find all nondecreasing functions $f$ from $\mathbb{R}$ to $\mathbb{R}$ such that $f(x+f(y)) = f(f(x)) + f(y)$ for all real $x$ and $y$.