I think I have a correct proof of the above statement. If it is wrong, I'd like to know why. If it is correct, any suggestions on how to make the proof more concise would be greatly appreciated. Here is how I went about proving the above statement:

Lemma: Any discrete subgroup $G$ of $\mathbb{R}$ is cyclic:

Proof: Assuming that $G$ is a discrete subgroup of $\mathbb{R}$, let $H = G \cap (0, \infty)$. Let us show that $\theta = \inf (H)$ generates $G$. First, we need to show that $\theta$ is in fact an element of $H$. To do this we recognize that any open ball $B(\theta, \epsilon)$ must contain an element $x \in H$. Since $G$ is discrete, so is $H$, and hence there is an $\epsilon_0 > 0$ such that $B(\theta, \epsilon_0) \cap H = \{ \theta \}$. Thus $\theta \in H$.

Now we are to show that any $h \in H$ is a multiple of $\theta$. Suppose for contradiction that there is an $h \in H$ such that $h \neq n \theta$ for any $n \in \mathbb{Z}$. Let $S = \{ n \in \mathbb{N} : n \theta > h \}$. By the well-ordering of the naturals, there is an element $k \in S$ such that $(k-1) \theta < h < k \theta$. Then, by the closure of $G$, $h-(k-1) \theta \in G$. But $0 < h - (k-1) \theta < \theta$ so this contradicts the minimality of $ \theta$ in $H$. Similarly, we can show that $- \theta $ generates $-H$ and since $- \theta$ is itself an integer multiple of $ \theta$, we have that $\theta$ generates all of $G$. So $G$ is indeed cyclic.

Now, by the above lemma, if we show that $X = \{ m + n \sqrt{2} : m,n \in \mathbb{Z} \}$ is not cyclic then it is not discrete. Suppose for contradiction that $X$ is cyclic. Then there is a $t = i + j \sqrt{2}$ that generates $X$. But $kt \neq i + (j+1) \sqrt{2}$ for any $k \in \mathbb{Z}$. So we have a contradiction.

Now, since we know that $G$ is not discrete, let us show that it is in fact dense in $\mathbb{R}$. Since $G$ is not discrete, we know that there is a point $y \in G$ such that, for each $\epsilon > 0$, $B(y, \epsilon)$ contains an element $x \in G$ with $x \neq y$. Therefore, by the closure of $G$, each ball $B(0,\epsilon)$ contains an $x \in G$ with $x \neq 0$.

We will use this fact about the open balls centered at $0$ to show that $G$ is dense in $\mathbb{R}$. First, assume that $r >0$ is not an element of $G$. We want to show that, for any $\delta > 0$, the ball $B(r, \delta)$ contains some element $g \in G$. Well, we know that there is a positive $h \in G \cap B(0, \delta / 2)$. Using the well-ordering of the naturals, we can find a $m \in \mathbb{N}$ such that $(m-1)h < r < mh$. But since $|mh - (m-1)h| = h < \delta$, we get that $mh \in B(r, \delta)$. Similarly, we can show that for any $r < 0$ $(r \notin G)$, $\delta >0$, $B(r, \delta)$ contains a $g \in G$. So $G$ is dense in $R$

This also shows that, in general, any noncyclic subgroup $G \leqslant \mathbb{R}$ is dense in $\mathbb{R}$.

  • $\begingroup$ Using pigeonhole and the fractional part you can get a smaller proof. $\endgroup$ – Souza Sep 29 '20 at 5:10

Very short proof:

  • every term of the sequence $a_r = (-1 + \sqrt{2})^r$ is of the form $m + n \sqrt{2}$ for $r \in \Bbb{N}$

  • any real number $x$ is at most $|a_r|/2$ away from an element of $a_r \Bbb{Z}$

  • $a_r \to 0$ as $r \to \infty$, because $0 < -1 + \sqrt{2} < 1$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.