Proof of continuity of Thomae Function at irrationals.

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.

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 $|f(x) - f(b) | = |f(x)| \leq \frac{1}{K} < \epsilon$.

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 $(b-1, b+1)$.