Is there a periodic function without minimum period such that all the possible periods are irrationals? Let $f:\mathbb R\to\mathbb R$ be a periodic function. We say $f$ is without minimum period if, $\forall t$ such that $f(x+t)=f(x)\forall x$, there is a $t'$ such that $0<t'<t$ and $f(x+t')=f(x)\forall x$.
The easiest examples of such functions are constant functions.
Dirichlet's function ($1$ if $x\in\mathbb Q$ and $0$ if $x\not\in\mathbb Q$) too is a periodic function without minimum period, cause $\forall q\in\mathbb Q$ it's true that $f(x+q)=f(x)$.
Let's say that $P_f$ is the set of all possible periods of $f$. (example: $P_{constant}=\mathbb R$, $P_{dirichlet's}=\mathbb Q$)
Is there a periodic function without minimum period such that $P_f\cap\mathbb Q=\emptyset$?
 A: I guess what you can get as a set of periods is exactly any additive subgroup of the reals. Certainly the periods are closed under addition. On the other hand, for any subgroup $G$ or $\mathbb R$, mimic the Dirichlet function by defining $f$ to be the indicator function of $G$. Here the set of periods is exactly $G$ itself.
Another example would be the additive subgroup generated by $\sqrt 2$ and $\sqrt 3$.
A: this is the complete question, sorry.
Let $f:\mathbb R\to\mathbb R$ be a periodic function. We say $f$ is without minimum period if, $\forall t$ such that$f(x+t)=f(x)\forall x$, there is a $t'$ such that $0 < t' < t$ and $f(x+t')=f(x)\forall x$.
The easiest examples of such functions are constant functions.
Dirichlet's function ($1$ if $x\in\mathbb Q$ and $0$ if $x\not\in\mathbb Q$) too is a periodic function without minimum period, cause $\forall q\in\mathbb Q$ it's true that $f(x+q)=f(x)$.
Let's say that $P_f$ is the set of all possible periods of $f$. (example: $P_{constant}=\mathbb R$, $P_{dirichlet's}=\mathbb Q$)
Is there a periodic function without minimum period such that $P_f\cap\mathbb Q=\emptyset$?
