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Let $~f : E \rightarrow \mathbb{R}, ~ E \subset \mathbb{R}~~$ be a periodic fucntion and $ S = \{ T \in \mathbb{R} ~ : ~ \forall x\in \mathbb{R} ~~ f(x+T) = f(x) \} $ be the set of all periods of fucntion f.

We know that if $ ~T~ $ is a period of fucntion $~f~$ then $ \forall n \in \mathbb{N} ~~ nT ~$ is also a period, so the set S is at least countably infinite. We also know that if any real number is a period of $ ~f~ $ than $~ f$ is constant, so for constant fucntion the set of all periods is uncountable.

I have the following question: if the set S of all periods of function $~f~$ is uncountable does it mean that $~f~$ is a constant? Or there is an example of non-constant periodic function which has this property?

Thanks in advance!

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2 Answers 2

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The set of periods of a function $f\colon \mathbb{R}\to\mathbb{R}$ can be any subgroup of the additive real numbers. Indeed, if $S$ is any subgroup of $\mathbb{R}$, then any one-to-one function $\overline{f}\colon\mathbb{R}/S \to \mathbb{R}$ lifts to a periodic function $f\colon\mathbb{R}\to\mathbb{R}$ whose periods are precisely the elements of $S$.

There are a wide variety of different subgroups of the real numbers. For example, the real numbers are a vector space over $\mathbb{Q}$ with uncountable dimension, and any vector subspace is a subgroup. In particular, $\mathbb{R}$ has plenty of proper uncountable subgroups.

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Unfortunately i have some trouble with notation: could you explain, please, what do you mean by $R/S$ and by $\overline{f}$? –  Igor Apr 1 '13 at 11:45
    
@Igor $\mathbb{R}/S$ means the quotient of $\mathbb{R}$ by the subgroup $S$. Also $\overline{f}$ can be any function, which lifts to a function $f$ by composing with the quotient homomorphism $\mathbb{R}\to\mathbb{R}/S$. –  Jim Belk Apr 1 '13 at 15:19
    
Thank you very much! –  Igor Apr 2 '13 at 21:18

Take the characteristic function of any uncountable set closed under addition and difference.

I hope this helps ;-)

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Perhaps, you meant the set of transcendental numbers because the set of algebraic numbers is countable :) Thanks for idea - i'll try to work with it. –  Igor Mar 31 '13 at 21:14
    
@Igor Transcendental no, because $0$ is not transcendental. But yeah, definetely not algebraic (surely it will have to contain some transcendental numbers) ;-) You can get some ideas here. –  dtldarek Mar 31 '13 at 21:20

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