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The set of periods of a real function $f$ is the set of real numbers $r$ such that $f(x+r)=f(x)$ $\forall x\in\mathbb R$. A periodic set is the set of periods for some function $f$. It is easy to prove that every periodic set is an additive subgroup of real numbers. Is the converse true?

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Let $G$ be an additive subgroup of the reals. Let $f(x)=1$ if $x\in G$, and let $f(x)=0$ otherwise. We show that $G$ is the set of periods of $f$.

Suppose that $r\in G$. If $x\in G$, then $x+r\in G$, and therefore $f(x+r)=f(x)=1$. If $x\not\in G$, then similarly $f(x+r)=f(x)=0$. So $r$ is a period of $f$.

Now suppose that $r\not\in G$. We show that $r$ is not a period of $f(x)$. Since $r\not\in G$, we have $f(0)=1$ but $f(0+r)=0$. So $r$ is not a period of $f$.

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