# Welcome to $\mathsf{Pr} \infty \mathsf{fWiki}$

 ProofWiki is an online compendium of mathematical proofs! Our goal is the collection, collaboration and classification of mathematical proofs. If you are interested in helping create an online resource for math proofs feel free to register for an account. Thanks and enjoy! If you have any questions, comments, or suggestions please post on the discussion page, or contact one of the administrators. Also, feel free to take a look at the frequently asked questions because you may not be the first with your idea. To see what's currently happening in the community, visit the community portal.
21,520 Proofs 17,147 Definitions Help

# Featured Proof

## Theorem

Let $D_M: \R \to \R$ denote the Thomae function:

$\forall x \in \R: \map {D_M} x = \begin {cases} 0 & : x = 0 \text { or } x \notin \Q \\ \dfrac 1 q & : x = \dfrac p q : p, q \in \Z, p \perp q, q > 0 \end {cases}$

where:

$\Q$ denotes the set of rational numbers
$\Z$ denotes the integers
$p \perp q$ denotes that $p$ and $q$ are coprime (that is, $x$ is a rational number expressed in canonical form)

Then $\map {D_M} x$ is:

continuous at all irrational $x$ and at $x = 0$
discontinuous at all rational $x$ such that $x \ne 0$.

## Proof

### Rational $x$

Let $x = \dfrac p q \in \Q \setminus \set 0$ such that $\dfrac p q$ is the canonical form of $x$.

Then we have:

$\map {D_M} x = \dfrac 1 q$

Let $\epsilon = \dfrac 1 {2 q}$.

Let $\delta \in \R$.

$\exists y \in \Q: x < y < \delta$
$\exists z \in \R \setminus \Q: x < z < y$

Hence there exists $z \in \R$ such that:

$z: \size {x - z} < \delta: \map {D_M} z = 0$

that is, such that:

$\size {\map {D_M} x - \map {D_M} z} = \dfrac 1 q > \epsilon$

That is, there exists an $\epsilon \in \R_{>0}$ such that it is always possible to find $\delta \in \R_{>0}$ such that there exists $z$ such that:

$\size {x - z} < \delta$

but such that:

$\size {\map {D_M} x - \map {D_M} z} > \epsilon$

Hence when $x$ is rational $\map {D_M} x$ is discontinuous.

$\Box$

### Irrational $x$

In the following it is to be understood that all rational numbers expressed in the form $\dfrac p q$ are in canonical form.

Let $x \in \R \setminus \Q$ or $x = 0$.

Let $\Q$ be ordered in the following way:

$\dfrac {p_1} {q_1} \prec \dfrac {p_2} {q_2} \iff \begin {cases} q_1 < q_2 & : q_1 \ne q_2 \\ p_1 < p_2 & : q_1 = q_2 \end {cases}$

and so we can denote $\Q$ with this ordering as $\struct {\Q, \prec}$

Let $\epsilon$ be arbitrary.

Let $q$ be the smallest positive integer such that $\dfrac 1 q < \epsilon$.

Let $S \subseteq \struct {\Q, \prec}$ defined as:

$S = \set {z \in \Q: z \prec \dfrac 1 q}$

That is, $S$ is the set of all rational numbers whose denominators are all less than or equal to $q$

Let $a$ be the supremum of the set:

$\set {z \in S: z < x}$

Let $b$ be the infimum of the set:

$\set {z \in S: x < z}$

Then we have that the open interval:

$C = \openint a b$

contains $x$ and no rational numbers whose denominators are less than $q$.

Thus:

$\forall y \in C: \map {D_M} y \le \dfrac 1 q$

and so:

$\forall y \in C: \size {\map {D_M} y - \map {D_M} x} \le \epsilon$

because $\map {D_M} x = 0$ by definition.

Letting $\delta = \min \set {\size {x - a}, \size {b - x} }$ gives us our $\delta$.

Thus we have shown that:

$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall y \in \R: \size {y - x} < \delta \implies \size {\map {D_M} y - \map {D_M} x} < \epsilon$

That is, $D_M$ is continuous at $x$.

Hence, when $x$ is irrational or $0$, $\map {D_M} x$ is continuous.

$\blacksquare$