Are there periodic functions without a smallest period?

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The Wikipedia page for periodic functions states that the smallest positive period $P$ of a function is called the fundamental period of the function (if it exists). I was intrigued by the condition that the function actually has a smallest period, so my question is, what properties of a function would cause it to be periodic but not have a smallest period?

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For a nontrivial example, consider the Dirichlet function, which has $$\delta(x) = \begin{cases}0 & \text{ if $x$ is rational}\\1 & \text{ if $x$ is irrational}\end{cases}$$

Then $\delta(x)$ is periodic with period $r$ for every rational number $r$.

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Yes, for example constant function.

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In fact, a continuous function of a real variable having arbitrarily small periods is necessarily a constant. Indeed, the set of periods is then a dense additive subgroup of the real line, and the function is constantly equal to its value at any point.

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You can show that if a periodic function is continuous at at least one point, and doesn't have a minimum period, then it is constant.

In other words: in order for a function to have arbitrarily small periods, it must be either constant or everywhere discontinuous.

The reasoning is simple: if $f$ doesn't have a smallest period, then the image of $f$ must be the same in every open interval. Therefore, no matter how small is $\varepsilon$, the oscillation of $f$ in any $(a, a+\varepsilon) $ (defined as $\sup f - \inf f$ restricted to that interval) is going to be the same as the global oscillation. Since the oscillation is constant as $\varepsilon\to 0$, and we know that the oscillation should converge to $0$ if we want $f$ to be continuous at $a$, we reach the conclusion that $f$ is nowhere continuous (allowing that $f$ is constant in the case that this global constant oscillation is $0$).

It is easy to obtain "examples" of constant functions. As MJD's answer says, an example of a non-constant function with arbitrarily small periods is the indicator function of the rationals ($1$ at the rationals and $0$ at the irrationals). As noted by Hagen von Eitzen in MJD's post, this function doesn't accept a smallest period because a real number $x$ is rational if and only if $x+r$ is also rational for any rational $r$ (hence any rational number is a period of this function).