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?
Best Answer
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:
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 $x_0$.