Thomae Function is Continuous at Irrational Numbers
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 $\delta \in \R_{>0}$.
Then from Between two Real Numbers exists Irrational Number:
- $\exists z \in \R \setminus \Q: x < z < x + \delta$
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 for all $\delta \in \R_{>0}$ it is possible to find $z$ 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 $\dfrac 1 q < \epsilon$.
Lemma
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 $q$.
$\Box$
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.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $1$: Review of some real analysis: Exercise $1.5: 21$