A question about Thomae's function: trouble in understanding continuity at irrationals

Question

I am trying to understand the well-known example of Thomae's function $f:\mathbb{R}\rightarrow\mathbb{R}$ which is continuous only at irrationals.

I am unable to understand the continuity at irrationals. The main trouble is in understanding some fact about irrationals, their open neighborhoods, and rationals in that neighborhood with some condition on their denominators.

Q. What is the crucial property of irrationals, their (bounded) open neighborhoods and the rationals in that neighborhood used in the this example?

Answer 1

Really, the only germane property that the rationals/irrationals have, in the construction of Thomae's functions, is that the rationals are countable and the two sets are complementary. You can generalise the construction like so:

1. Let $(x_n)$ be an injective sequence in $\Bbb{R}$.
2. Let $(y_n)$ be any sequence in $\Bbb{R}$ such that $y_n \neq 0$ for all $n$, and $y_n \to 0$.
3. Define $$f(x) = \begin{cases}y_n & \text{if there exists } n \in \Bbb{N} \text{ such that } x = x_n \\ 0 & \text{otherwise.}\end{cases}$$

Thomae's function can be recovered by taking $(x_n)$ to be the sequence $$0, 1, \color{blue}{\frac{1}{2}}, \color{green}{\frac{1}{3}}, \color{green}{\frac{2}{3}}, \color{red}{\frac{1}{4}}, \color{red}{\frac{3}{4}}, \color{purple}{\frac{1}{5}}, \color{purple}{\frac{2}{5}}, \color{purple}{\frac{3}{5}}, \color{purple}{\frac{4}{5}}, \color{orange}{\frac{1}{6}}, \color{orange}{\frac{5}{6}}, \ldots$$ i.e. all rational numbers in $[0, 1]$ ordered lexicographically, first by the denominator of their lowest form, then by the numerator of their lowest form. Match it to a sequence $(y_n)$, $$0, 1, \color{blue}{\frac{1}{2}}, \color{green}{\frac{1}{3}}, \color{green}{\frac{1}{3}}, \color{red}{\frac{1}{4}}, \color{red}{\frac{1}{4}}, \color{purple}{\frac{1}{5}}, \color{purple}{\frac{1}{5}}, \color{purple}{\frac{1}{5}}, \color{purple}{\frac{1}{5}}, \color{orange}{\frac{1}{6}}, \color{orange}{\frac{1}{6}}, \ldots$$ and you get Thomae's function. It's not too difficult to see that $y_n \to 0$; note that only finitely many terms will be larger than any given $\frac{1}{m}$ where $m \in \Bbb{N}$.

The generalised Thomae's function $f$ will be continuous at every point except at the $(x_n)$ points.

At each point $x_m$, any $\delta$-neighbourhood is an uncountable set, meaning a sequence like $(x_n)$ cannot cover the whole neighbourhood. Thus, in any $\delta$-neighbourhood, we will find values of $x$ such that $x \neq x_n$ for all $n$, and so $f(x) = 0$.

If we had continuity at $x_m$, then there would have to exist a $\delta > 0$, corresponding to $\varepsilon = |y_n| > 0$, i.e. such that $$|x - x_m| < \delta \implies |f(x) - f(x_m)| < \varepsilon \implies |f(x) - y_n| < |y_n|.$$ But, if we let $x$ be an element of the $\delta$-neighbourhood around $x_m$ that doesn't belong to the sequence $(x_n)$, then $f(x) = 0$, so $$|x - x_m| < \delta \implies |0 - y_n| < |y_n|,$$ a contradiction. Thus, $f$ is discontinuous at all points in the sequence $(x_n)$.

On the other hand, pick a point $x_0$ not in the sequence $(x_n)$ and $\varepsilon > 0$. Since $y_n \to 0$, there exists an $N \in \Bbb{N}$ such that $$n \ge N \implies |y_n| < \varepsilon.$$ Consider the set of "bad" points $\{x_1, \ldots, x_{N - 1}\}$. This set contains all points $x$ where $|f(x)| \ge \varepsilon$, and it's finite! Take $\delta$ to be the smallest distance from $x_0$ to any of the $N - 1$ points above, and then $$|x - x_0| < \delta \implies x \notin \{x_1, \ldots, x_{N - 1}\} \implies |f(x) - f(x_0)| = |f(x)| < |y_n| < \varepsilon,$$ proving continuity at $x_0$.

Answer 2

The key property here is the following :

Given any real number $\alpha$ and any positive integer $n$, there is a neighborhood $I$ of $\alpha$ such that if $x\in\mathbb {Q} \cap I, x\neq \alpha$ then $x$ has denominator greater than $n$.

The property above can be proved very easily. Consider two integers $a, b$ such that $a<\alpha <b$ and then consider all the rationals in interval $(a, b) $ with denominator less than equal to $n$. Clearly such rationals are finite in number and $k$ is their total count (exclude $\alpha$ if it is one of these rationals). Now the interval $[a, b] $ is divided into finitely many ($k+1$) subintervals and exactly one of these subintervals say $[c, d] $ contains $\alpha$ in its interior. The neighborhood $I=(c, d) $ satisfies the property mentioned above.

The continuity of Thomae function at irrationals is an immediate consequence of the above property. If $\alpha $ is an irrational point under consideration and $\epsilon>0$ be given then we choose integer $n>1/\epsilon$ and apply above property to get a neighborhood $I$ of $\alpha$. If $x\in I$ and $x=p/q$ is rational then $q>n$ and $f(x) = 1/q<1/n<\epsilon$ and if $x\in I$ is irrational then $f(x) =0$. In any case we have $x\in I\implies |f(x) - f(\alpha) |<\epsilon $ and we are done.