Dirichlet Function is Periodic

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Theorem

Let $D: \R \to \R$ be a Dirichlet function:

$\forall x \in \R: D \left({x}\right) = \begin{cases} c & : x \in \Q \\ d & : x \notin \Q \end{cases}$


Then $D$ is periodic.

Namely, every non-zero rational number is a periodic element of $D$.


Proof

Let $x \in \R$.

Let $L \in \Q$.


If $x \in \Q$, then:

\(\displaystyle \map D {x + L}\) \(=\) \(\displaystyle c\) rationals are closed under addition
\(\displaystyle \) \(=\) \(\displaystyle \map D x\)

If $x \notin \Q$, then:

\(\displaystyle \map D {x + L}\) \(=\) \(\displaystyle d\) Rational Number plus Irrational Number is Irrational
\(\displaystyle \) \(=\) \(\displaystyle \map D x\)

Combining the above two shows that:

$\forall x \in \R: \map D x = \map D {x + L}$


Hence the result.

$\blacksquare$