Subgroup of Real Numbers is Discrete or Dense

From ProofWiki
Jump to navigation Jump to search


Let $G$ be a subgroup of the additive group of real numbers.

Then one of the following holds:

$G$ is dense in $\R$.
$G$ is discrete and there exists $a \in \R$ such that $G = a \Z$, that is, $G$ is cyclic.


If $G$ is trivial, then $G$ is discrete and cyclic.

Let $G$ be non-trivial.

Because $x \in G \iff -x \in G$, $G$ has a strictly positive element.

Thus $G \cap \R_{>0}$ is non-empty.

Let $a = \map \inf {G \cap \R_{>0} }$.

Case 1

Suppose $a = 0$.

We will show that $G$ is dense in $\R$.

Let $x \in \R$ and $\epsilon > 0$.

Because $a = 0$, there exists $g \in G$ such that $0 < g \le \epsilon$.


$y = g \cdot \floor {\dfrac x g} \in G$, where $\floor {\, \cdot \,}$ denotes the floor function.

We have:

\(\displaystyle \size {x - y}\) \(=\) \(\displaystyle x - g \cdot \floor {\dfrac x g}\)
\(\displaystyle \) \(=\) \(\displaystyle g \cdot \paren {\frac x g - \floor {\frac x g} }\)
\(\displaystyle \) \(\le\) \(\displaystyle g \cdot 1\) Real Number minus Floor
\(\displaystyle \) \(\le\) \(\displaystyle \epsilon\)

Thus $G$ is dense in $\R$.

Case 2

Suppose $a > 0$.

We show that $G = a \Z$, from which it follows that $G$ is discrete.

Let $g \in G$.


$g - a \cdot \floor {\dfrac g a} \in G$

By Real Number minus Floor, we have:

\(\displaystyle 0 \le g - a \cdot \floor {\frac g a}\) \(=\) \(\displaystyle a \cdot \paren {\dfrac g a - \floor {\dfrac g a} }\)
\(\displaystyle \) \(<\) \(\displaystyle a\)

Because $a = \map \inf {G \cap \R_{>0} }$, we have:

$g = a \cdot \floor {\dfrac g a}$


$g \in a \Z$