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 $a \notin G$.
By Characterizing Property of Infimum of Subset of Real Numbers:
- $\forall \epsilon > 0: \exists g \in G: a \le g < a + \epsilon$
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.
Therefore, by Proof by Contradiction:
- $a \in G$
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:
- $g \in a \Z$
$\blacksquare$