Thomae s Function Variant Points of Continuity

Thomae's function

From HandWiki

Short description

: Function that is discontinuous at rationals and continuous at irrationals

Point plot on the interval (0,1). The topmost point in the middle shows f(1/2) = 1/2

Thomae's function is a real-valued function of a real variable that can be defined as:^[1]

[math]\displaystyle{ f(x) = \begin{cases} \frac{1}{q} &\text{if }x = \tfrac{p}{q}\quad (x \text{ is rational), with } p \in \mathbb Z \text{ and } q \in \mathbb N \text{ coprime}\\ 0 &\text{if }x \text{ is irrational.} \end{cases} }[/math]

It is named after Carl Johannes Thomae, but has many other names: the popcorn function, the raindrop function, the countable cloud function, the modified Dirichlet function, the ruler function,^[2] the Riemann function, or the Stars over Babylon (John Horton Conway's name).^[3] Thomae mentioned it as an example for an integrable function with infinitely many discontinuities in an early textbook on Riemann's notion of integration.^[4]

Since every rational number has a unique representation with coprime (also termed relatively prime) [math]\displaystyle{ p \in \mathbb Z }[/math] and [math]\displaystyle{ q \in \mathbb N }[/math], the function is well-defined. Note that [math]\displaystyle{ q=+1 }[/math] is the only number in [math]\displaystyle{ \mathbb N }[/math] that is coprime to [math]\displaystyle{ p=0. }[/math]

It is a modification of the Dirichlet function, which is 1 at rational numbers and 0 elsewhere.

Properties

Thomae's function [math]\displaystyle{ f }[/math] is bounded and maps all real numbers to the unit interval:[math]\displaystyle{ \;f: \mathbb R\; \rightarrow \;[0,\; 1]. }[/math]
[math]\displaystyle{ f }[/math] is periodic with period [math]\displaystyle{ 1:\; f(x + n) = f(x) }[/math] for all integers $n$ and all real $x$ .

Proof of periodicity
For all [math]\displaystyle{ x \in \mathbb R \smallsetminus \mathbb Q, }[/math] we also have [math]\displaystyle{ x+n \in \mathbb R \smallsetminus \mathbb Q }[/math] and hence [math]\displaystyle{ f(x+n) = f(x)= 0, }[/math] For all [math]\displaystyle{ x \in \mathbb Q,\; }[/math] there exist [math]\displaystyle{ p \in \mathbb Z }[/math] and [math]\displaystyle{ q \in \mathbb N }[/math] such that [math]\displaystyle{ \;x = p/q,\; }[/math] and [math]\displaystyle{ \gcd(p,\;q) = 1. }[/math] Consider [math]\displaystyle{ x + n = (p + nq) / q }[/math]. If [math]\displaystyle{ d }[/math] divides [math]\displaystyle{ p }[/math] and [math]\displaystyle{ q }[/math], it divides [math]\displaystyle{ p + nq }[/math] and [math]\displaystyle{ p }[/math]. Conversely, if [math]\displaystyle{ d }[/math] divides [math]\displaystyle{ p + nq }[/math] and [math]\displaystyle{ q }[/math], it divides [math]\displaystyle{ (p + nq) - nq = p }[/math] and [math]\displaystyle{ q }[/math]. So [math]\displaystyle{ \gcd(p + nq, q) = \gcd(p, q) = 1 }[/math], and [math]\displaystyle{ f(x + n) = 1/q = f(x) }[/math].

Proof of periodicity

For all [math]\displaystyle{ x \in \mathbb R \smallsetminus \mathbb Q, }[/math] we also have [math]\displaystyle{ x+n \in \mathbb R \smallsetminus \mathbb Q }[/math] and hence [math]\displaystyle{ f(x+n) = f(x)= 0, }[/math]

For all [math]\displaystyle{ x \in \mathbb Q,\; }[/math] there exist [math]\displaystyle{ p \in \mathbb Z }[/math] and [math]\displaystyle{ q \in \mathbb N }[/math] such that [math]\displaystyle{ \;x = p/q,\; }[/math] and [math]\displaystyle{ \gcd(p,\;q) = 1. }[/math] Consider [math]\displaystyle{ x + n = (p + nq) / q }[/math]. If [math]\displaystyle{ d }[/math] divides [math]\displaystyle{ p }[/math] and [math]\displaystyle{ q }[/math], it divides [math]\displaystyle{ p + nq }[/math] and [math]\displaystyle{ p }[/math]. Conversely, if [math]\displaystyle{ d }[/math] divides [math]\displaystyle{ p + nq }[/math] and [math]\displaystyle{ q }[/math], it divides [math]\displaystyle{ (p + nq) - nq = p }[/math] and [math]\displaystyle{ q }[/math]. So [math]\displaystyle{ \gcd(p + nq, q) = \gcd(p, q) = 1 }[/math], and [math]\displaystyle{ f(x + n) = 1/q = f(x) }[/math].

[math]\displaystyle{ f }[/math] is discontinuous at all rational numbers, dense within the real numbers.

Proof of discontinuity at rational numbers
Let [math]\displaystyle{ x_0 = p/q }[/math] be an arbitrary rational number, with [math]\displaystyle{ \;p \in \mathbb Z,\; q \in \mathbb N, }[/math] and [math]\displaystyle{ p }[/math] and [math]\displaystyle{ q }[/math] coprime. This establishes [math]\displaystyle{ f(x_0) = 1/q. }[/math] Let [math]\displaystyle{ \;\alpha \in \mathbb R \smallsetminus \mathbb Q\; }[/math] be any irrational number and define [math]\displaystyle{ x_n = x_0 + \frac{\alpha}{n} }[/math] for all [math]\displaystyle{ n \in \mathbb N. }[/math] These [math]\displaystyle{ x_n }[/math] are all irrational, and so [math]\displaystyle{ f(x_n) = 0 }[/math] for all [math]\displaystyle{ n \in \mathbb N. }[/math] This implies [math]\displaystyle{ \|x_0 - x_n\| = \frac{\alpha}{n},\quad }[/math] and [math]\displaystyle{ \quad \|f(x_0) - f(x_n)\| = \frac{1}{q}. }[/math] Let [math]\displaystyle{ \;\varepsilon = 1/q\; }[/math], and given [math]\displaystyle{ \delta \gt 0 }[/math] let [math]\displaystyle{ n = 1 + \left\lceil\frac{\alpha}{\delta }\right\rceil. }[/math] For the corresponding [math]\displaystyle{ \;x_n }[/math] we have [math]\displaystyle{ \|f(x_0) - f(x_n)\|= 1/q \ge \varepsilon\quad }[/math] and [math]\displaystyle{ \|x_0 - x_n\| = \frac{\alpha}{n} = \frac{\alpha}{1 + \left\lceil\frac{\alpha}{\delta}\right\rceil} \lt \frac{\alpha}{\left\lceil\frac{\alpha}{\delta}\right\rceil} \le \delta, }[/math] which is exactly the definition of discontinuity of [math]\displaystyle{ f }[/math] at [math]\displaystyle{ x_0 }[/math].

Proof of discontinuity at rational numbers

Let [math]\displaystyle{ x_0 = p/q }[/math] be an arbitrary rational number, with [math]\displaystyle{ \;p \in \mathbb Z,\; q \in \mathbb N, }[/math] and [math]\displaystyle{ p }[/math] and [math]\displaystyle{ q }[/math] coprime.

This establishes [math]\displaystyle{ f(x_0) = 1/q. }[/math]

Let [math]\displaystyle{ \;\alpha \in \mathbb R \smallsetminus \mathbb Q\; }[/math] be any irrational number and define [math]\displaystyle{ x_n = x_0 + \frac{\alpha}{n} }[/math] for all [math]\displaystyle{ n \in \mathbb N. }[/math]

These [math]\displaystyle{ x_n }[/math] are all irrational, and so [math]\displaystyle{ f(x_n) = 0 }[/math] for all [math]\displaystyle{ n \in \mathbb N. }[/math]

This implies [math]\displaystyle{ |x_0 - x_n| = \frac{\alpha}{n},\quad }[/math] and [math]\displaystyle{ \quad |f(x_0) - f(x_n)| = \frac{1}{q}. }[/math]

Let [math]\displaystyle{ \;\varepsilon = 1/q\; }[/math], and given [math]\displaystyle{ \delta \gt 0 }[/math] let [math]\displaystyle{ n = 1 + \left\lceil\frac{\alpha}{\delta }\right\rceil. }[/math] For the corresponding [math]\displaystyle{ \;x_n }[/math] we have

[math]\displaystyle{ |f(x_0) - f(x_n)|= 1/q \ge \varepsilon\quad }[/math] and

[math]\displaystyle{ |x_0 - x_n| = \frac{\alpha}{n} = \frac{\alpha}{1 + \left\lceil\frac{\alpha}{\delta}\right\rceil} \lt \frac{\alpha}{\left\lceil\frac{\alpha}{\delta}\right\rceil} \le \delta, }[/math]

which is exactly the definition of discontinuity of [math]\displaystyle{ f }[/math] at [math]\displaystyle{ x_0 }[/math].

[math]\displaystyle{ f }[/math] is continuous at all irrational numbers, also dense within the real numbers.

Proof of continuity at irrational arguments
Since [math]\displaystyle{ f }[/math] is periodic with period [math]\displaystyle{ 1 }[/math] and [math]\displaystyle{ 0 \in \Q, }[/math] it suffices to check all irrational points in [math]\displaystyle{ I=(0,\;1).\; }[/math] Assume now [math]\displaystyle{ \varepsilon \gt 0,\; i \in \N }[/math] and [math]\displaystyle{ x_0 \in I \smallsetminus \Q. }[/math] According to the Archimedean property of the reals, there exists [math]\displaystyle{ r \in \N }[/math] with [math]\displaystyle{ 1/r \lt \varepsilon , }[/math] and there exist [math]\displaystyle{ \; k_i \in \N, }[/math] such that for [math]\displaystyle{ i = 1, \ldots, r }[/math] we have [math]\displaystyle{ 0 \lt \frac{k_i}{i} \lt x_0 \lt \frac{k_i +1}{i}. }[/math] The minimal distance of [math]\displaystyle{ x_0 }[/math] to its i-th lower and upper bounds equals [math]\displaystyle{ d_i := \min\left\{\left\|x_0 - \frac{k_i}{i}\right\|,\; \left\|x_0 - \frac{k_i + 1}{i}\right\| \right\}. }[/math] We define [math]\displaystyle{ \delta }[/math] as the minimum of all the finitely many [math]\displaystyle{ d_i. }[/math] [math]\displaystyle{ \delta := \min_{1\le i\le r}\{d_i\},\; }[/math] so that for all [math]\displaystyle{ i = 1, ..., r, }[/math] [math]\displaystyle{ \quad \|x_0 - k_i/i\| \ge \delta\quad }[/math] and [math]\displaystyle{ \quad\|x_0 - (k_i+1)/i\| \ge \delta. }[/math] This is to say, all these rational numbers [math]\displaystyle{ k_i/i,\;(k_i + 1)/i,\; }[/math] are outside the [math]\displaystyle{ \delta }[/math]-neighborhood of [math]\displaystyle{ x_0. }[/math] Now let [math]\displaystyle{ x \in \mathbb{Q} \cap (x_0 - \delta, x_0 + \delta) }[/math] with the unique representation [math]\displaystyle{ x = p/q }[/math] where [math]\displaystyle{ p, q \in \mathbb N }[/math] are coprime. Then, necessarily, [math]\displaystyle{ q \gt r,\; }[/math] and therefore, [math]\displaystyle{ f(x)=1/q \lt 1/r \lt \varepsilon. }[/math] Likewise, for all irrational [math]\displaystyle{ x \in I, \; f(x) = 0 = f(x_0),\; }[/math] and thus, if [math]\displaystyle{ \varepsilon \gt 0 }[/math] then any choice of (sufficiently small) [math]\displaystyle{ \delta \gt 0 }[/math] gives [math]\displaystyle{ \|x - x_0\| \lt \delta \implies \|f(x_0) - f(x)\| = f(x) \lt \varepsilon. }[/math] Therefore, [math]\displaystyle{ f }[/math] is continuous on [math]\displaystyle{ \mathbb R \smallsetminus \mathbb Q.\quad }[/math]

Proof of continuity at irrational arguments

Since [math]\displaystyle{ f }[/math] is periodic with period [math]\displaystyle{ 1 }[/math] and [math]\displaystyle{ 0 \in \Q, }[/math] it suffices to check all irrational points in [math]\displaystyle{ I=(0,\;1).\; }[/math] Assume now [math]\displaystyle{ \varepsilon \gt 0,\; i \in \N }[/math] and [math]\displaystyle{ x_0 \in I \smallsetminus \Q. }[/math] According to the Archimedean property of the reals, there exists [math]\displaystyle{ r \in \N }[/math] with [math]\displaystyle{ 1/r \lt \varepsilon , }[/math] and there exist [math]\displaystyle{ \; k_i \in \N, }[/math] such that

for [math]\displaystyle{ i = 1, \ldots, r }[/math] we have [math]\displaystyle{ 0 \lt \frac{k_i}{i} \lt x_0 \lt \frac{k_i +1}{i}. }[/math]

The minimal distance of [math]\displaystyle{ x_0 }[/math] to its i-th lower and upper bounds equals

[math]\displaystyle{ d_i := \min\left\{\left|x_0 - \frac{k_i}{i}\right|,\; \left|x_0 - \frac{k_i + 1}{i}\right| \right\}. }[/math]

We define [math]\displaystyle{ \delta }[/math] as the minimum of all the finitely many [math]\displaystyle{ d_i. }[/math]

[math]\displaystyle{ \delta := \min_{1\le i\le r}\{d_i\},\; }[/math] so that

for all [math]\displaystyle{ i = 1, ..., r, }[/math] [math]\displaystyle{ \quad |x_0 - k_i/i| \ge \delta\quad }[/math] and [math]\displaystyle{ \quad|x_0 - (k_i+1)/i| \ge \delta. }[/math]

This is to say, all these rational numbers [math]\displaystyle{ k_i/i,\;(k_i + 1)/i,\; }[/math] are outside the [math]\displaystyle{ \delta }[/math]-neighborhood of [math]\displaystyle{ x_0. }[/math]

Now let [math]\displaystyle{ x \in \mathbb{Q} \cap (x_0 - \delta, x_0 + \delta) }[/math] with the unique representation [math]\displaystyle{ x = p/q }[/math] where [math]\displaystyle{ p, q \in \mathbb N }[/math] are coprime. Then, necessarily, [math]\displaystyle{ q \gt r,\; }[/math] and therefore,

[math]\displaystyle{ f(x)=1/q \lt 1/r \lt \varepsilon. }[/math]

Likewise, for all irrational [math]\displaystyle{ x \in I, \; f(x) = 0 = f(x_0),\; }[/math] and thus, if [math]\displaystyle{ \varepsilon \gt 0 }[/math] then any choice of (sufficiently small) [math]\displaystyle{ \delta \gt 0 }[/math] gives

[math]\displaystyle{ |x - x_0| \lt \delta \implies |f(x_0) - f(x)| = f(x) \lt \varepsilon. }[/math]

Therefore, [math]\displaystyle{ f }[/math] is continuous on [math]\displaystyle{ \mathbb R \smallsetminus \mathbb Q.\quad }[/math]

[math]\displaystyle{ f }[/math] is nowhere differentiable.

Proof of being nowhere differentiable
For rational numbers, this follows from non-continuity. For irrational numbers: For any sequence of irrational numbers [math]\displaystyle{ (a_n)_{n=1}^\infty }[/math] with [math]\displaystyle{ a_n \ne x_0 }[/math] for all [math]\displaystyle{ n \in \mathbb{N}_{+} }[/math] that converges to the irrational point [math]\displaystyle{ x_0,\; }[/math] the sequence [math]\displaystyle{ (f(a_n))_{n=1}^\infty }[/math] is identically [math]\displaystyle{ 0,\; }[/math] and so [math]\displaystyle{ \lim_{n \to \infty}\left\|\frac{f(a_n)-f(x_0)}{a_n - x_0}\right\| = 0. }[/math] According to Hurwitz's theorem, there also exists a sequence of rational numbers [math]\displaystyle{ (b_n)_{n=1}^{\infty} = (k_n/n)_{n=1}^\infty,\; }[/math] converging to [math]\displaystyle{ x_0,\; }[/math] with [math]\displaystyle{ k_n \in \mathbb Z }[/math] and [math]\displaystyle{ n \in \mathbb N }[/math] coprime and [math]\displaystyle{ \|k_n/n - x_0\| \lt \frac{1}{\sqrt{5}\cdot n^2}.\; }[/math] Thus for all [math]\displaystyle{ n, }[/math] [math]\displaystyle{ \left\|\frac{f(b_n)-f(x_0)}{b_n - x_0} \right\| \gt \frac{1/n - 0}{1/(\sqrt{5}\cdot n^2)} =\sqrt{5}\cdot n \ne 0\; }[/math] and so [math]\displaystyle{ f }[/math] is not differentiable at all irrational [math]\displaystyle{ x_0. }[/math]

Proof of being nowhere differentiable

For rational numbers, this follows from non-continuity.

For irrational numbers:

For any sequence of irrational numbers [math]\displaystyle{ (a_n)_{n=1}^\infty }[/math] with [math]\displaystyle{ a_n \ne x_0 }[/math] for all [math]\displaystyle{ n \in \mathbb{N}_{+} }[/math] that converges to the irrational point [math]\displaystyle{ x_0,\; }[/math] the sequence [math]\displaystyle{ (f(a_n))_{n=1}^\infty }[/math] is identically [math]\displaystyle{ 0,\; }[/math] and so [math]\displaystyle{ \lim_{n \to \infty}\left|\frac{f(a_n)-f(x_0)}{a_n - x_0}\right| = 0. }[/math]

According to Hurwitz's theorem, there also exists a sequence of rational numbers [math]\displaystyle{ (b_n)_{n=1}^{\infty} = (k_n/n)_{n=1}^\infty,\; }[/math] converging to [math]\displaystyle{ x_0,\; }[/math] with [math]\displaystyle{ k_n \in \mathbb Z }[/math] and [math]\displaystyle{ n \in \mathbb N }[/math] coprime and [math]\displaystyle{ |k_n/n - x_0| \lt \frac{1}{\sqrt{5}\cdot n^2}.\; }[/math]

Thus for all [math]\displaystyle{ n, }[/math] [math]\displaystyle{ \left|\frac{f(b_n)-f(x_0)}{b_n - x_0} \right| \gt \frac{1/n - 0}{1/(\sqrt{5}\cdot n^2)} =\sqrt{5}\cdot n \ne 0\; }[/math] and so [math]\displaystyle{ f }[/math] is not differentiable at all irrational [math]\displaystyle{ x_0. }[/math]

[math]\displaystyle{ f }[/math] has a strict local maximum at each rational number.

See the proofs for continuity and discontinuity above for the construction of appropriate neighbourhoods, where [math]\displaystyle{ f }[/math] has maxima.

[math]\displaystyle{ f }[/math] is Riemann integrable on any interval and the integral evaluates to [math]\displaystyle{ 0 }[/math] over any set.

The Lebesgue criterion for integrability states that a bounded function is Riemann integrable if and only if the set of all discontinuities has measure zero.^[5] Every countable subset of the real numbers - such as the rational numbers - has measure zero, so the above discussion shows that Thomae's function is Riemann integrable on any interval. The function's integral is equal to [math]\displaystyle{ 0 }[/math] over any set because the function is equal to zero almost everywhere.

Empirical probability distributions related to Thomae's function appear in DNA sequencing.^[6] The human genome is diploid, having two strands per chromosome. When sequenced, small pieces ("reads") are generated: for each spot on the genome, an integer number of reads overlap with it. Their ratio is a rational number, and typically distributed similarly to Thomae's function.

If pairs of positive integers [math]\displaystyle{ m,n }[/math] are sampled from a distribution [math]\displaystyle{ f(n,m) }[/math] and used to generate ratios [math]\displaystyle{ q=n/(n+m) }[/math], this gives rise to a distribution [math]\displaystyle{ g(q) }[/math] on the rational numbers. If the integers are independent the distribution can be viewed as a convolution over the rational numbers, [math]\displaystyle{ g(a/(a+b))=\sum_{t=1}^\infty f(ta)f(tb) }[/math]. Closed form solutions exist for power-law distributions with a cut-off. If [math]\displaystyle{ f(k)=k^{-\alpha} e^{-\beta k}/\mathrm{Li}_\alpha(e^{-\beta}) }[/math] (where [math]\displaystyle{ \mathrm{Li}_\alpha }[/math] is the polylogarithm function) then [math]\displaystyle{ g(a/(a+b)) = (ab)^{-\alpha} \mathrm{Li}_{2\alpha}(e^{-(a+b)\beta})/\mathrm{Li}^2_{\alpha}(e^{-\beta}) }[/math]. In the case of uniform distributions on the set [math]\displaystyle{ \{1,2,\ldots , L\} }[/math] [math]\displaystyle{ g(a/(a+b)) = (1/L^2) \lfloor L/\max(a,b) \rfloor }[/math], which is very similar to Thomae's function. Both their graphs have fractal dimension 3/2.^[6]

The ruler function

Main page: Ruler function

For integers, the exponent of the highest power of 2 dividing [math]\displaystyle{ n }[/math] gives 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, ... (sequence A007814 in the OEIS). If 1 is added, or if the 0s are removed, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 2, 1, ... (sequence A001511 in the OEIS). The values resemble tick-marks on a 1/16th graduated ruler, hence the name. These values correspond to the restriction of the Thomae function to the dyadic rationals: those rational numbers whose denominators are powers of 2.

A natural follow-up question one might ask is if there is a function which is continuous on the rational numbers and discontinuous on the irrational numbers. This turns out to be impossible. the set of discontinuities of any function must be an $F σ$ set. If such a function existed, then the irrationals would be an $F σ$ set. The irrationals would then be the countable union of closed sets [math]\displaystyle{ \textstyle\bigcup_{i = 0}^\infty C_i }[/math], but since the irrationals do not contain an interval, nor can any of the [math]\displaystyle{ C_i }[/math]. Therefore, each of the [math]\displaystyle{ C_i }[/math] would be nowhere dense, and the irrationals would be a meager set. It would follow that the real numbers, being the union of the irrationals and the rationals (which, as a countable set, is evidently meager), would also be a meager set. This would contradict the Baire category theorem: because the reals form a complete metric space, they form a Baire space, which cannot be meager in itself.

A variant of Thomae's function can be used to show that any $F σ$ subset of the real numbers can be the set of discontinuities of a function. If [math]\displaystyle{ A =\textstyle \bigcup_{n=1}^{\infty}F_n }[/math] is a countable union of closed sets [math]\displaystyle{ F_n }[/math], define

[math]\displaystyle{ f_A(x) = \begin{cases} \frac{1}{n} & \text{if } x \text{ is rational and } n \text{ is minimal so that } x \in F_n\\ -\frac{1}{n} & \text{if } x \text{ is irrational and } n \text{ is minimal so that } x \in F_n\\ 0 & \text{if } x \notin A \end{cases} }[/math]

Then a similar argument as for Thomae's function shows that [math]\displaystyle{ f_A }[/math] has A as its set of discontinuities.

For a general construction on arbitrary metric space, see this article Kim, Sung Soo. "A Characterization of the Set of Points of Continuity of a Real Function." American Mathematical Monthly 106.3 (1999): 258-259.

Notes

↑ Beanland, Roberts & Stevenson 2009, p. 531
↑ "…the so-called ruler function, a simple but provocative example that appeared in a work of Johannes Karl Thomae … The graph suggests the vertical markings on a ruler—hence the name." (Dunham 2008)
↑ John Conway. "Topic: Provenance of a function". The Math Forum. Archived from the original on 13 June 2018. https://web.archive.org/web/20180613235037/mathforum.org/kb/message.jspa?messageID=1375516.
↑ Thomae 1875, p. 14, §20
↑ Spivak 1965, p. 53, Theorem 3-8
↑ ^6.0 ^6.1 Trifonov, Vladimir; Pasqualucci, Laura; Dalla-Favera, Riccardo; Rabadan, Raul (2011). "Fractal-like Distributions over the Rational Numbers in High-throughput Biological and Clinical Data". Scientific Reports 1 (191): 191. doi:10.1038/srep00191. PMID 22355706. Bibcode: 2011NatSR...1E.191T.

References

Thomae, J. (1875) (in german), Einleitung in die Theorie der bestimmten Integrale, Halle a/S: Verlag von Louis Nebert
Abbott, Stephen (2016), Understanding Analysis (Softcover reprint of the original 2nd ed.), New York: Springer, ISBN 978-1-4939-5026-3
Bartle, Robert G.; Sherbert, Donald R. (1999), Introduction to Real Analysis (3rd ed.), Wiley, ISBN 978-0-471-32148-4, https://archive.org/details/introductiontore00bart_1 (Example 5.1.6 (h))
Beanland, Kevin; Roberts, James W.; Stevenson, Craig (2009), "Modifications of Thomae's Function and Differentiability", The American Mathematical Monthly 116 (6): 531–535, doi:10.4169/193009709x470425
Dunham, William (2008), The Calculus Gallery: Masterpieces from Newton to Lebesgue (Paperback ed.), Princeton: Princeton University Press, ISBN 978-0-691-13626-4, https://books.google.com/books?id=aYTYBQAAQBAJ&pg=PA149
Spivak, M. (1965), Calculus on manifolds, Perseus Books, ISBN 978-0-8053-9021-6

External links

Hazewinkel, Michiel, ed. (2001), "Dirichlet-function", Encyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN 978-1-55608-010-4, https://www.encyclopediaofmath.org/index.php?title=p/d032860
Weisstein, Eric W.. "Dirichlet Function". http://mathworld.wolfram.com/DirichletFunction.html.

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Source: https://handwiki.org/wiki/Thomae%27s_function

Thomae s Function Variant Points of Continuity