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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?

Martin Sleziak
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goblin GONE
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  • $0+\sqrt{2}\cdot0\not\in\Bbb Z+\sqrt{2}\Bbb Z$? – anon Aug 07 '14 at 11:28
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    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 07 '14 at 11:33
  • @blue: understood. –  Aug 07 '14 at 11:34
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    Isn't $\mathbb{Z} + \sqrt{2}\mathbb{Z}$ a dense subgroup of $\mathbb R$ ? – Gabriel Romon Aug 07 '14 at 11:35
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    Combine these two questions: http://math.stackexchange.com/questions/889296/xy-sqrt2-infimum-x-y-in-mathbbz and http://math.stackexchange.com/questions/90177/subgroup-of-mathbbr-either-dense-or-has-a-least-positive-element. – Najib Idrissi Aug 07 '14 at 11:36
  • @blue, good point. But that just leaves me even more unsure of how to prove it... – goblin GONE Aug 07 '14 at 11:37
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    In this question we show that $\Bbb{Z}[\sqrt2]$ is dense in $\Bbb{R}$. That settles it. – Jyrki Lahtonen Aug 07 '14 at 11:39

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$\mathbb{Z}+\sqrt{2}\mathbb{Z}$ 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 $\neq \mathbb{R}$.

Hamou
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  • How can we prove that any non-monogenerated subgroup of $\mathbb{R}$ is dense? – goblin GONE Aug 07 '14 at 11:39
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    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 07 '14 at 11:43