# Proof of continuity of Thomae Function at irrationals.

\begin{align} t(x) = \begin{cases} 0 & \text{if x is irrational}\\ \frac{1}{n} & \text{if x = \frac{m}{n} where \gcd(m,n) = 1} \end{cases} \end{align}

I can prove the discontinuity at rational $b$ by taking a sequence of irrationals $x_n$ which converge to $b$.

But while going through an argument for continuity at irrationals. I found this in a book.

On the other hand if $b$ is an irrational number and $\epsilon > 0$ then there is a natural number $n_0$ such that $1/n_0 < \epsilon$. There are only finite number of rationals with denominator less than $n_0$ in the interval $(b-1,b+1)$. Hence we can find a $\delta > 0$ such that $\delta$ neighbourhood of $b$ contains no rational with denominator less than $n_0$.

I understand the rest of the proof. But I am unable to prove the emphasized text. Although I find it intuitive.

Let $m=n_0-1$, so we want to consider rationals with denominators $1,\cdots,m$ in the interval $(b-1,b+1)$. Since consecutive rationals with denominator k differ by $1/k$ and the interval $(b-1,b+1)$ has length 2, there are at most 2k rationals with denominator k in $(b-1,b+1)$.

Therefore there are at most $2\cdot1+2\cdot2+\cdots+2m$ rationals in $(b-1,b+1)$ with denominator less than $n_0$, so we can choose a $\delta$ with $0<\delta<|b-r|$, where r is the rational with denominator less than $n_0$ in $(b-1,b+1)$ which is closest to b.

• what if we take $b = 1$ and $k = 4$? You see the "consecutive" rationals with denominator $k$ are $1/4$, $3/4$, $5/4$, and $7/4$, which are at a distance of $2/4 = 1/2$. So don't you think your argument breaks down? – Saaqib Mahmood Mar 30 '17 at 17:36
• @SaaqibMahmuud In your example, I am arguing that there are at most 8 rationals of the form $\frac{m}{4}$ in the interval $(0,2)$, since the distance between $\frac{m}{4}$ and $\frac{m+1}{4}$ is $\frac{1}{4}$ for each $m$. – user84413 Apr 1 '17 at 19:13

Let $b \in \mathbb{R} \setminus \mathbb{Q}$. Given $\epsilon > 0$ let $K = \left \lceil \frac{1}{\epsilon} \right \rceil$. Thus, $\frac{1}{K} < \epsilon$.

Note that $K$ is a finite number and the number of integers less than $K$ is also finite. This means the number of rationals of the form $\frac{1}{q} > \frac{1}{K}$ is also finite.

Shrink the interval $(b-1, b+1)$ down to $(b-q, b+q)$ such that all these $\frac{1}{q}$ are tossed out, leaving only rationals $\frac{1}{q} < \frac{1}{K} < \epsilon$.

It follows that if $|x -b| < \delta$ then $|f(x) - f(b) | = |f(x)| \leq \frac{1}{K} < \epsilon$.

• Will this be a good approach to shrink the interval ? $Let E_{n}:=\{x \in Q | f(x) \geq 1/n\}$ now $n\{E_{n}\} \leq \frac{n(n+1)}{2} = m \; (say)$ clearly m $\in$ $N$ $E_{n}=\{b_{1},b_{2} \ldots b_{m}\}$ let $q=inf\{|b-b_{i}|i=1,2 \ldots m\}$ then if $|x-b|<q$ $\Rightarrow$ $|f(x)-f(b)|=|f(x)|=f(x) <1/n< \epsilon$ – Vishweshwar Tyagi Nov 2 '17 at 14:36

The natural extension of Thomae's function to the hyperreals is defined by the same formula: \begin{align} t(x) = \begin{cases} \frac{1}{n} & \text{if x = \frac{m}{n} where m,n\in{}^\ast\mathbb N and \gcd(m,n) = 1} \\ 0 & \text{otherwise.}\\ \end{cases} \end{align} To show that $t(x)$ is continuous at $c\in\mathbb R\setminus \mathbb Q$, suppose $q$ is a hyperrational infinitely close to $c$. Clearly $q\not\in\mathbb R$. Since $q$ is appreciable (i.e., finite but not infinitesimal), its denominator $n$ is necessarily an infinite hyperinteger. Hence $t(q)=\frac{1}{n}\approx 0$ where $\approx$ is the relation of infinite proximity. Thus $t(x)$ is infinitesimal at all points infinitely close to $c$, proving the continuity of $t(x)$ at $c$.

Let $\frac{m}{n}$ be a rational number such that $$b-1 \leq \frac{m}{n} \leq b+1,$$ where $m$ and $n$ are integers such that $n> 0$ and $\gcd (m, n) = 1$.

Then we see that $$n ( b-1) \leq m < \leq n(b+1).$$ Thus, $$m \in \mathbb{Z} \cap \left[ \ n(b-1), \ n(b+1) \ \right].$$ Therefore, $$m \in \left\{ \ \lceil n(b-1) \rceil, \ldots, \lfloor n(b+1) \rfloor \ \right\},$$ where $\lfloor 2.5 \rfloor = 2$ and $\lceil 2.5 \rceil = 3$, and $\lfloor 2 \rfloor = 2 = \lceil 2 \rceil$.

Hence, for each natural number $n$, there are at most $$N \colon= \lfloor n(b+1) \rfloor - \lceil n(b-1) \rceil$$ rational numbers with denominator $n$ in the closed interval $[b-1, b+1]$ and hence in the open interval $(b-1, b+1)$.