Thomae's Function | Real Analysis

The popcorn function

f(p/q) = 1/q

(reduced),

f(x) = 0

(irrational): continuous at every irrational, discontinuous at every rational. Given

\varepsilon

, only finitely many dots reach height

\varepsilon

— so near an irrational a

\delta

can always be found.

Examples & Exercises

Prove completely that Thomae's function $f$ ( $f(p/q) = 1/q$ in lowest terms, $f = 0$ at irrationals) is discontinuous at every rational $x_0$ .

x_0 = \frac{p}{q}\ \text{(lowest terms)} \Rightarrow f(x_0) = \frac{1}{q} > 0, \quad \varepsilon = \frac{1}{2q}

View centered at the rational

x_0 = 1/2

f(1/2) = 1/2

, yet irrationals with

f = 0

sit arbitrarily close, so no

\delta

works for

\varepsilon = 1/4

Show proof

Write $x_0 = p/q$ in lowest terms, so $f(x_0) = 1/q > 0$ . Set $\varepsilon = 1/(2q) > 0$ . By density of the irrationals, for every $\delta > 0$ the interval $(x_0 - \delta, x_0 + \delta)$ contains an irrational $x$ .
At an irrational $x$ , $f(x) = 0$ , so $|f(x) - f(x_0)| = 1/q > 1/(2q) = \varepsilon$ . That is, for every $\delta > 0$ there is an $x$ with $|x - x_0| < \delta$ but $|f(x) - f(x_0)| \ge \varepsilon$ ; by the $\varepsilon$ - $\delta$ definition, $f$ is not continuous at $x_0$ .
Since $x_0$ was an arbitrary rational, Thomae's function is discontinuous at every rational. In the scene, centering on a rational (e.g. $1/2$ ) gives $\delta = 0$ , exhibiting this argument directly.

Prove completely, by an $\varepsilon$ - $\delta$ argument, that Thomae's function is continuous at every irrational $x_0$ .

S_\varepsilon = \left\{x \in [0,1] : f(x) \ge \varepsilon\right\} = \left\{\frac{p}{q} : q \le \frac{1}{\varepsilon}\right\}\ \text{finite}

View centered at the irrational

x_0 \approx 0.618

: only finitely many dots (those with

q \le 10

) reach height

\varepsilon = 0.1

, and

\delta

is the distance to the nearest one.

Show proof

Let $x_0$ be irrational and $\varepsilon > 0$ arbitrary. Consider $S_\varepsilon = \{x \in [0,1] : f(x) \ge \varepsilon\}$ . By the definition of $f$ , members of $S_\varepsilon$ are only lowest-terms fractions $p/q$ with $1/q \ge \varepsilon$ , i.e. $q \le 1/\varepsilon$ ; hence $S_\varepsilon$ is a finite set.
Since $x_0$ is irrational, $x_0 \notin S_\varepsilon$ . Thus $\delta = \min_{x \in S_\varepsilon} |x - x_0|$ , the minimum of finitely many positive numbers, is strictly positive. If $|x - x_0| < \delta$ then $x$ is at distance at least $\delta$ from every member of $S_\varepsilon$ , so $x \notin S_\varepsilon$ — i.e. $f(x) < \varepsilon$ (for $x = x_0$ this is trivial since $f(x_0) = 0 < \varepsilon$ ).
Since $|f(x) - f(x_0)| = |f(x) - 0| = f(x) < \varepsilon$ , $f$ is continuous at $x_0$ . As $x_0$ was an arbitrary irrational, Thomae's function is continuous at every irrational — the $\delta$ shown in the scene is exactly this minimum.

Show that Thomae's function is a pointwise limit of continuous functions — i.e. prove completely that it is of Baire class 1.

f_n(x) = \max_{\substack{p/q\ \text{lowest terms},\ q \le n}} \max\left(0, \frac{1}{q} - n\left|x - \frac{p}{q}\right|\right) \to f(x)\ \forall x

Q = 30

\varepsilon = 0.05

: a pointwise limit of continuous tent functions draws these popcorn dots — a Baire-1 function must have a dense set of continuity points, and Thomae's is the irrationals (a dense

G_\delta

Show proof

For each $n \ge 1$ let $S_n = \{p/q : 1 \le q \le n,\ \gcd(p,q) = 1\}$ (a finite set) and define $f_n(x) = \max_{r \in S_n} \max(0, 1/q(r) - n|x - r|)$ , where $q(r)$ is the denominator of $r = p/q$ . Each tent $x \mapsto \max(0, 1/q - n|x - r|)$ is continuous, and a finite maximum of continuous functions is continuous, so each $f_n$ is continuous.
Let $x = a/b$ in lowest terms. For $n \ge b$ , the tent at $r = x$ itself gives the value $1/b$ , while for any other $r' = p/q \ne x$ with $q \le n$ we have $|x - r'| \ge 1/(bq)$ (distinct rationals with denominators $b, q$ differ by at least $1/(bq)$ ), so $n|x - r'| \ge n/(bq) \ge 1/q$ — the tent value is 0. Hence $f_n(x) = 1/b = f(x)$ for $n \ge b$ .
If $x$ is irrational, $f(x) = 0$ . Fix $M \ge 1$ : tents with denominator $q \le M$ all vanish for large $n$ (there are only finitely many rationals with denominator $\le M$ , each at positive distance from the irrational $x$ ), and tents with $q > M$ are worth at most $1/q < 1/M$ . Hence $\limsup_{n \to \infty} f_n(x) \le 1/M$ ; since $M$ is arbitrary, $f_n(x) \to 0 = f(x)$ . So $f_n \to f$ at every $x$ : Thomae's function is Baire class 1.