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The set $\{m+n \sqrt{2}: m,n \in \mathbb{Z} \} $ is dense in $\mathbb{R}$, and the set $\{m-n \sqrt{2}: m,n \in \mathbb{N} \} $ is dense in the negative reals.

Let $S=\{m+n\sqrt{2}:m,n \in \mathbb{N} \}$. I have proved that $S$ is not dense in $\mathbb{R}$ as neither $0$ is in $S$ nor it is a limit point of $S$ (since, $1+\sqrt{2}$ is the smallest element in $S$. But I am not sure that $S$ doesn't have any limit point.

Question: $S$ has limit points or not?

Learning
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1 Answers1

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$S$ does not have limit points.

Let $\ell\in\mathbb R$. You can show that there are only finitely many elements $s\in S$ such that $s\le \ell+1$. Thus $\ell$ is not a limit point.

Astyx
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