Proposition 1

Every closed subgroup of the additive group $\mathbf{R}$, other than $\mathbf{R}$ and $\{0\}$, is a discrete group of the form $a.\mathbf{Z}$, where $a > 0$ (in other words, it consists of the integer multiples of $a$).

We begin by showing that every non-discrete subgroup of $\mathbf{R}$ is dense in $\mathbf{R}$. If a subgroup $G$ of $\mathbf{R}$ is not discrete, then for every $\varepsilon > 0$ there is a point $x \neq 0$ in $G$ which belongs to the interval $[-\varepsilon, +\varepsilon]$; since all integer multiples of $x$ belong to $G$, every interval of length $> \varepsilon$ contains an element of $G$, and therefore $G$ is dense in $\mathbf{R}$.

Every closed subgroup of $\mathbf{R}$ other than $\mathbf{R}$ itself is therefore discrete. It remains to show that every discrete subgroup $G$ of $\mathbf{R}$ other than $\{0\}$ is of the form $a.\mathbf{Z}$, where $a > 0$. Now the relation $-G = G$ shows that the set $H$ of elements $> 0$ in $G$ is not empty; if $b \in H$, the intersection of the interval $[0, b]$ and $G$ is compact and discrete, and is therefore finite. Let $a$ be the smallest element of $H$ contained in $[0, b]$, and for every $x \in G$ put $m = [x/a]$, the integer part of $x/a$; then we have $x - ma \in G$ and $0 \leq x - ma < a$. By the definition of $a$ it follows that $x - ma = 0$ and therefore $G = a.\mathbf{Z}$.