# Subgroup of Real Numbers is Discrete or Dense

## Theorem

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.

## Proof

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.

We have that $G \cap \R_{>0}$ is bounded below by $0$.

Hence, by the Continuum Property, $G \cap \R_{>0}$ admits an infimum.

So, 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 $\epsilon > 0$.

As $a = 0$ is the infimum, by Characterizing Property of Infimum of Subset of Real Numbers:

$\exists g \in G \cap \R_{>0}: 0 \le g < \epsilon$

Because $g \in \R_{>0}$:

$0 < g < \epsilon$

Then:

$y = g \cdot \floor {\dfrac x g} \in G$

where $\floor {\, \cdot \,}$ denotes the floor function.

We have:

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

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

### Case 2

Suppose $a > 0$.

First we need to show that $a \in G$.

Aiming for a contradiction, suppose $a \notin G$.

$\forall \epsilon > 0: \exists g \in G: a \le g < a + \epsilon$

In particular, let $\epsilon = a$.

Then:

$\exists g_2 \in G: a \le g_2 < 2 a$

Then let $\epsilon = g_2$.

Thus:

$\exists g_1 \in G: a \le g_1 < g_2$

Thus we have that there exists $g_1, g_2 \in \R$ such that:

$a \le g_1 < g_2 < 2 a$

Then:

 $\ds 0$ $<$ $\ds g_2 - g_1$ $\ds$ $<$ $\ds 2 a - g_1$ $\ds$ $<$ $\ds 2 a - a$ $\ds$ $=$ $\ds a$

but:

$g_2 - g_1 \in G$

This contradicts $a$ being the infimum.

$a \in G$

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

Let $g \in G$.

Then:

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

By Real Number minus Floor, we have:

 $\ds 0 \le g - a \cdot \floor {\frac g a}$ $=$ $\ds a \cdot \paren {\dfrac g a - \floor {\dfrac g a} }$ $\ds$ $<$ $\ds a$

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

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

Thus:

$g \in a \Z$

$\blacksquare$