Nonconstant Periodic Function with no Period is Discontinuous Everywhere

Theorem
Let $f$ be a real periodic function that does not have a period.

Then $f$ is either constant or discontinuous everywhere.

Proof
Let $f$ be a real periodic function that does not have a period.

Let $x \in \R$.

If there is no $y \in \R$ such that $\map f x \ne \map f y$, then the result is verified by Constant Function has no Period.

If $f$ is non-constant, then let $y$ be such a value.

Let $\mathcal L_{f \gt 0}$ be the set of all positive periodic elements of $f$.

This set is non-empty by Absolute Value of Periodic Element is Peroidic.

It is seen that $\lt$ forms a right-total relation on $\mathcal L_{f \gt 0}$, for if not then $f$ would have a period.

By the Axiom of Dependent Choice there must exist a strictly-decreasing sequence $\left\langle{L_n}\right\rangle$ in $\mathcal L_{f \gt 0}$.

Since this sequence is bounded below by zero, it follows via Monotone Convergence Theorem that this sequence converges.

Also, by Convergent Sequence is Cauchy Sequence the sequence is Cauchy.

Since $\left\langle{L_n}\right\rangle$ is Cauchy, the sequence $\left\langle{d_n}\right\rangle_{n \ge 1}$ formed by taking $d_n = L_n - L_{n-1}$ is null.

From Combination of Periodic Elements it follows that $\left\langle{d_n}\right\rangle_{n \ge 1}$ is contained in $\mathcal L_{f \gt 0}$, for every $d_n$ is a periodic element of $f$.

Consider the sequence $\left\langle{\left({x - y}\right) \bmod d_n }\right\rangle_{n \ge 1}$.

It is seen by Limit of Modulo Operation that this sequence is also null.

Let $\epsilon \in \R_{\gt 0}$ such that $\epsilon \lt \left\lvert{\map f x - \map f y}\right\rvert$.

For any $\delta \in \R_{\gt 0}$, there is a $n \in \N_{\gt 0}$ such that:

But

And so $f$ is not continuous at $x$.

But our choice of $x$ was completely arbitrary.

Hence the result.