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If $F$ is a nonempty closed subset of $\mathbb{R}$ such that $x-y\in F$ for $x,y\in F$, then show that $F=\mathbb{R}$ or $F=\alpha\mathbb{Z}$ for some $\alpha\in \mathbb{R}$.

Anupam
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    Hint: Maybe (i) $F={0}$; (ii) Maybe $F$ has a smallest positive element. (iii) Maybe it doesn't. Case (ii) will give us $\alpha \mathbb{Z}$. Case (iii) is where we need to use the fact that $F$ is closed. – André Nicolas Sep 26 '13 at 02:15

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Hint: Show that if $F$ is not discrete, then $F$ is dense. Do this by assuming $F$ has a limit point $x_0$ (which is necessarily in $F$), and then for every $\varepsilon>0$ find some $x_\varepsilon\in F$ such that $|x_0-x_\varepsilon|<\varepsilon$. Consider translations to show that every neighborhood of radius $\varepsilon$ has an element of $F$ in it.

Thus, if $F$ isn't $\mathbb{R}$, it must be discrete. Then, use the standard technique of showing that $\inf \{x\in F:x>0\}$ generates $F$.

Alex Youcis
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  • First part is fine, but how to show that $\alpha$ generates $F$ in second case? – Anupam Sep 27 '13 at 17:54
  • @Anupam: Let $\alpha$ be the infimum. By discreteness, we have $\alpha \in F$. Let $\beta \in F \setminus {0}$ be arbitrary. By replacing $\beta$ with its inverse, assume that $\beta > 0$. Let $n$ be the largest positive integer such that $n\alpha \leqslant \beta$. (Note that $\alpha \leqslant \beta$, by construction.) Then, $\beta - n\alpha$ is a nonnegative element of $F$ which is strictly less than $\alpha$. Thus, $\beta = n\alpha$ as desired. – Aryaman Maithani Jul 12 '21 at 19:06