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 $\mathcal L_{f > 0}$ 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 $<$ forms a right-total relation on $\mathcal L_{f > 0}$, for if not then $f$ would have a period.

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

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 $\sequence {L_n}$ is Cauchy, the sequence $\sequence {d_n}_{n \mathop \ge 1}$ formed by taking $d_n = L_n - L_{n - 1}$ is null.

From Combination of Periodic Elements it follows that $\sequence {d_n}_{n \mathop \ge 1}$ is contained in $\mathcal L_{f > 0}$, for every $d_n$ is a periodic element of $f$.

Consider the sequence $\sequence {\paren {x - y} \bmod d_n}_{n \mathop \ge 1}$.

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

Let $\epsilon \in \R_{> 0}$ such that $\epsilon < \size {\map f x - \map f y}$.

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

But:

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

But our choice of $x$ was completely arbitrary.

Hence the result.