Thomae's function

For all ${\displaystyle x\in \mathbb {R} \setminus \mathbb {Q} ,}$ we also have ${\displaystyle x+n\in \mathbb {R} \setminus \mathbb {Q} }$ and hence ${\displaystyle f(x+n)=f(x)=0,}$

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

Let ${\displaystyle x_{0}=p/q}$ be an arbitrary rational number, with ${\displaystyle \;p\in \mathbb {Z} ,\;q\in \mathbb {N} ,}$ and ${\displaystyle p}$ and ${\displaystyle q}$ coprime.

This establishes ${\displaystyle f(x_{0})=1/q.}$

Let ${\displaystyle \;\alpha \in \mathbb {R} \setminus \mathbb {Q} \;}$ be any irrational number and define ${\displaystyle x_{n}=x_{0}+{\frac {\alpha }{n}}}$ for all ${\displaystyle n\in \mathbb {N} .}$

These ${\displaystyle x_{n}}$ are all irrational, and so ${\displaystyle f(x_{n})=0}$ for all ${\displaystyle n\in \mathbb {N} .}$

This implies ${\displaystyle |x_{0}-x_{n}|={\frac {\alpha }{n}},}$ and ${\displaystyle |f(x_{0})-f(x_{n})|={\frac {1}{q}}.}$

Let ${\displaystyle \;\varepsilon =1/q\;}$, and given ${\displaystyle \delta >0}$ let ${\displaystyle n=1+\left\lceil {\frac {\alpha }{\delta }}\right\rceil .}$ For the corresponding ${\displaystyle \;x_{n}}$ we have $${\displaystyle |f(x_{0})-f(x_{n})|=1/q\geq \varepsilon }$$ and $${\displaystyle |x_{0}-x_{n}|={\frac {\alpha }{n}}={\frac {\alpha }{1+\left\lceil {\frac {\alpha }{\delta }}\right\rceil }}<{\frac {\alpha }{\left\lceil {\frac {\alpha }{\delta }}\right\rceil }}\leq \delta ,}$$

which is exactly the definition of discontinuity of ${\displaystyle f}$ at ${\displaystyle x_{0}}$.

Since ${\displaystyle f}$ is periodic with period ${\displaystyle 1}$ and ${\displaystyle 0\in \mathbb {Q} ,}$ it suffices to check all irrational points in ${\displaystyle I=(0,1).\;}$ Assume now ${\displaystyle \varepsilon >0,\;i\in \mathbb {N} }$ and ${\displaystyle x_{0}\in I\setminus \mathbb {Q} .}$ According to the Archimedean property of the reals, there exists ${\displaystyle r\in \mathbb {N} }$ with ${\displaystyle 1/r<\varepsilon ,}$ and there exist ${\displaystyle \;k_{i}\in \mathbb {N} ,}$ such that

for ${\displaystyle i=1,\ldots ,r}$ we have ${\displaystyle 0<{\frac {k_{i}}{i}}<x_{0}<{\frac {k_{i}+1}{i}}.}$

The minimal distance of ${\displaystyle x_{0}}$ to its i-th lower and upper bounds equals $${\displaystyle d_{i}:=\min \left\{\left|x_{0}-{\frac {k_{i}}{i}}\right|,\;\left|x_{0}-{\frac {k_{i}+1}{i}}\right|\right\}.}$$

We define ${\displaystyle \delta }$ as the minimum of all the finitely many ${\displaystyle d_{i}.}$ $${\displaystyle \delta :=\min _{1\leq i\leq r}\{d_{i}\},\;}$$ so that for all ${\displaystyle i=1,\dots ,r,}$ ${\displaystyle |x_{0}-k_{i}/i|\geq \delta }$ and ${\displaystyle |x_{0}-(k_{i}+1)/i|\geq \delta .}$

This is to say, all these rational numbers ${\displaystyle k_{i}/i,\;(k_{i}+1)/i,\;}$ are outside the ${\displaystyle \delta }$-neighborhood of ${\displaystyle x_{0}.}$

Now let ${\displaystyle x\in \mathbb {Q} \cap (x_{0}-\delta ,x_{0}+\delta )}$ with the unique representation ${\displaystyle x=p/q}$ where ${\displaystyle p,q\in \mathbb {N} }$ are coprime. Then, necessarily, ${\displaystyle q>r,\;}$ and therefore, $${\displaystyle f(x)=1/q<1/r<\varepsilon .}$$

Likewise, for all irrational ${\displaystyle x\in I,\;f(x)=0=f(x_{0}),\;}$ and thus, if ${\displaystyle \varepsilon >0}$ then any choice of (sufficiently small) ${\displaystyle \delta >0}$ gives $${\displaystyle |x-x_{0}|<\delta \implies |f(x_{0})-f(x)|=f(x)<\varepsilon .}$$

Therefore, ${\displaystyle f}$ is continuous on ${\displaystyle \mathbb {R} \setminus \mathbb {Q} .}$

• For rational numbers, this follows from non-continuity.
• For irrational numbers:

  For any sequence of irrational numbers ${\displaystyle (a_{n})_{n=1}^{\infty }}$ with ${\displaystyle a_{n}\neq x_{0}}$ for all ${\displaystyle n\in \mathbb {N} _{+}}$ that converges to the irrational point ${\displaystyle x_{0},\;}$ the sequence ${\displaystyle (f(a_{n}))_{n=1}^{\infty }}$ is identically ${\displaystyle 0,\;}$ and so ${\displaystyle \lim _{n\to \infty }\left|{\frac {f(a_{n})-f(x_{0})}{a_{n}-x_{0}}}\right|=0.}$

  According to Hurwitz's theorem, there also exists a sequence of rational numbers ${\displaystyle (b_{n})_{n=1}^{\infty }=(k_{n}/n)_{n=1}^{\infty },\;}$ converging to ${\displaystyle x_{0},\;}$ with ${\displaystyle k_{n}\in \mathbb {Z} }$ and ${\displaystyle n\in \mathbb {N} }$ coprime and ${\displaystyle |k_{n}/n-x_{0}|<{\frac {1}{{\sqrt {5}}\cdot n^{2}}}.\;}$

  Thus for all ${\displaystyle n,}$ ${\displaystyle \left|{\frac {f(b_{n})-f(x_{0})}{b_{n}-x_{0}}}\right|>{\frac {1/n-0}{1/({\sqrt {5}}\cdot n^{2})}}={\sqrt {5}}\cdot n\neq 0\;}$ and so ${\displaystyle f}$ is not differentiable at all irrational ${\displaystyle x_{0}.}$