Dirichlet Function is Discontinuous

Theorem
Let $D: \R \to \R$ denote the Dirichlet function:
 * $\forall x \in \R: \map D x = \begin {cases} c & : x \in \Q \\ d & : x \notin \Q \end {cases}$

where $\Q$ denotes the set of rational numbers.

Then $D$ is discontinuous at every $x \in \R$.

Proof
Let $\epsilon = \dfrac {\size {c - d} } 2$.

Let $x \in \Q$.

Let $\delta \in \R_{>0}$ be arbitrary.

Let $y \in \Q$ such that $\size {x - y} < \delta$.

, let $y > x$.

From Between two Rational Numbers exists Irrational Number:
 * $\exists z \in \R \setminus \Q: x < z < y$

and so:
 * $\size {\map D x - \map D z} = \size {c - d} > \epsilon$

Similarly if $y < x$:
 * $\exists z \in \R \setminus \Q: y < z < x: \size {\map D x - \map D z} > \epsilon$

and by definition of continuity $D$ is discontinuous at $x$.

Let $x \in \R \setminus \Q$.

Let $\delta \in \R_{>0}$ be arbitrary.

Let $y \in \R \setminus \Q$ such that $\size {x - y} < \delta$.

, let $y > x$.

From Between two Real Numbers exists Rational Number:
 * $\exists z \in \Q: x < z < y$

and so:
 * $\size {\map D x - \map D z} = \size {c - d} > \epsilon$

Similarly if $y < x$:
 * $\exists z \in \Q: y < z < x: \size {\map D x - \map D z} > \epsilon$

and by definition of continuity $D$ is discontinuous at $x$.

$D$ has been shown to be discontinuous at all $x \in \R$ whether $x$ is rational or irrational.

Hence the result.