# How do we know that $\mathbb{Z} + \sqrt{2}\mathbb{Z}$ isn't closed?

The Minkowski sum of closed sets needn't be closed; $\mathbb{Z} + \sqrt{2}\mathbb{Z}$ is the canonical example. However, its not clear to me how to prove this.

Question. How can we prove that $\mathbb{Z} + \sqrt{2}\mathbb{Z}$ isn't closed?

• $0+\sqrt{2}\cdot0\not\in\Bbb Z+\sqrt{2}\Bbb Z$? – blue Aug 7 '14 at 11:28
• There was recently a question about such sequence. I don't think I'll find it though. The idea is to use pigeonhole to show that two elements of $\sqrt 2 \mathbb Z$ can have fractional parts arbitrarily close to each other and thus their difference has a fractional part arbitrarily close to an integer. – Karolis Juodelė Aug 7 '14 at 11:33
• Isn't $\mathbb{Z} + \sqrt{2}\mathbb{Z}$ a dense subgroup of $\mathbb R$ ? – Gabriel Romon Aug 7 '14 at 11:35
• Combine these two questions: math.stackexchange.com/questions/889296/… and math.stackexchange.com/questions/90177/…. – Najib Idrissi Aug 7 '14 at 11:36
• In this question we show that $\Bbb{Z}[\sqrt2]$ is dense in $\Bbb{R}$. That settles it. – Jyrki Lahtonen Aug 7 '14 at 11:39

$\mathbb{Z}+\sqrt{2}\mathbb{Z}$ is a subgroup of $(\mathbb{R},+)$, as any subgroup of $\mathbb{R}$ is dense or mono-gene (generated by one element), and it is easy to show that it is not mono-gene, hence dense, so not closed because it is $\neq \mathbb{R}$.
• How can we prove that any non-monogenerated subgroup of $\mathbb{R}$ is dense? – goblin Aug 7 '14 at 11:39
• If $G$ is a such subgroup, consider $\alpha=\inf G\cap \mathbb{R}_+^*$, and show that $\alpha=0$. And by definition of the $\inf$ you can show the result. – Hamou Aug 7 '14 at 11:43