Exercises

(a) Each $U_k$ is a union of open intervals, hence open. Every rational $p/q$ lies in the interval centered at itself, so $\mathbb{Q} \subset U_k$; since $\mathbb{Q}$ is dense, each $U_k$ is dense.

(b) $G$ is a countable intersection of open sets, hence $G_\delta$ (each $\mathbb{R} \setminus \{r_j\}$ is open too). Let $x \in G$. Then $x$ is irrational, and for each $k$ there is a pair $(p_k, q_k)$ with $|x - p_k/q_k| < 1/q_k^k$.

If only finitely many distinct pairs occurred, some pair $(p, q)$ would repeat for infinitely many $k$, forcing $|x - p/q| = 0$, i.e. $x = p/q$, contradicting irrationality. So infinitely many distinct pairs occur. For fixed $m$, the pairs with $k \ge m$ satisfy $|x - p_k/q_k| < 1/q_k^k \le 1/q_k^m$, so $x$ meets the definition of a Liouville number.

(c) $G$ is a countable intersection of dense open sets, so by the Baire category theorem it is dense in $\mathbb{R}$; hence so are the Liouville numbers. Moreover $G$ is uncountable: if $G = \{x_n\}$ were countable, then $G \cap \bigcap_n (\mathbb{R} \setminus \{x_n\}) = \varnothing$ would be a countable intersection of dense open sets, hence dense by Baire — a contradiction.

(a) $\cos$ decreases on $[0, \pi]$ and $1 < \pi/3$, so $\cos 1 > \cos(\pi/3) = 1/2$. Hence for $x \in [0,1]$, $\cos x \in (\cos 1, 1] \subset (1/2, 1]$ and $T(x) \in (3/8, 1/2] \subset [0, 1]$.

For the contraction property, the mean value theorem gives $|\cos x - \cos y| \le |x - y|$ (since $|\sin| \le 1$), so

So $T$ is a contraction with constant $q = 1/4$.

Since $[0, 1]$ is a complete metric space, the Banach fixed point theorem gives a unique fixed point $x^* \in [0, 1]$. This is (b).

(c) With $x_0 = 0$, $x_1 = T(0) = 1/2$, so $|x_1 - x_0| = 1/2$. The error bound with $q = 1/4$ gives $|x_n - x^*| \le \frac{(1/4)^n}{3/4} \cdot \frac{1}{2} = \frac{2}{3}4^{-n}$.

Now $\frac{2}{3}4^{-n} < 10^{-3}$ iff $4^n > 2000/3 \approx 666.7$, and $4^4 = 256 < 666.7 < 1024 = 4^5$, so the smallest such $n$ is $5$. Indeed $|x_5 - x^*| \le \frac{2}{3 \cdot 1024} = \frac{1}{1536} < 10^{-3}$.

Dividing the digits by $2$ rewrites the set as $C = \{\sum_{n \ge 1} 2a_n 3^{-n} : a_n \in \{0,1\}\}$. If $x, y \in C$ then $x + y \le 2$, so $C + C \subset [0, 2]$ is clear.

We prove the reverse inclusion.

Let $s \in [0, 2]$. Then $s/2 \in [0, 1]$ has a ternary expansion $s/2 = \sum_{n \ge 1} e_n 3^{-n}$ with $e_n \in \{0, 1, 2\}$.

Split each digit: put $(a_n, b_n) = (0,0)$ if $e_n = 0$, $(1,0)$ if $e_n = 1$, $(1,1)$ if $e_n = 2$; then $a_n, b_n \in \{0, 1\}$ and $e_n = a_n + b_n$. With $x = \sum 2a_n 3^{-n}$ and $y = \sum 2b_n 3^{-n}$, we have $x, y \in C$ and

Hence $s \in C + C$. Since $s$ was arbitrary, $[0, 2] \subset C + C$, and the two inclusions give $C + C = [0, 2]$.

Remarkably, the sum of two "hole-ridden" sets of measure zero fills an entire interval.

(a) Put $\Omega(\delta) := \sup\{|f(y) - f(z)| : y, z \in (x-\delta, x+\delta)\}$, so $\omega_f(x) = \inf_{\delta > 0} \Omega(\delta)$. $(\Rightarrow)$ Suppose $f$ is continuous at $x$. For every $\varepsilon > 0$ there is $\delta > 0$ with $|y - x| < \delta \Rightarrow |f(y) - f(x)| < \varepsilon/2$.

Then for $y, z \in (x-\delta, x+\delta)$, $|f(y) - f(z)| \le |f(y) - f(x)| + |f(x) - f(z)| < \varepsilon$, so $\Omega(\delta) \le \varepsilon$ and $\omega_f(x) \le \varepsilon$. Since $\varepsilon$ was arbitrary, $\omega_f(x) = 0$.

$(\Leftarrow)$ Suppose $\omega_f(x) = 0$. For every $\varepsilon > 0$, $\inf_{\delta > 0} \Omega(\delta) = 0 < \varepsilon$, so some $\delta > 0$ has $\Omega(\delta) < \varepsilon$. Then $|y - x| < \delta \Rightarrow |f(y) - f(x)| \le \Omega(\delta) < \varepsilon$ (the case $z = x$) — $f$ is continuous at $x$.

Hence $x \in C(f) \iff \omega_f(x) = 0 \iff \omega_f(x) < 1/k \ (\forall k) \iff x \in \bigcap_k \{\omega_f < 1/k\}$; that is, $C(f) = \bigcap_k \{\omega_f < 1/k\}$.

Each $\{\omega_f < 1/k\}$ is open. Let $\omega_f(x_0) < 1/k$. By definition of the infimum, some $\delta > 0$ gives $M := \sup\{|f(y) - f(z)| : y, z \in (x_0 - \delta, x_0 + \delta)\} < 1/k$.

If $x_1 \in (x_0 - \delta/2, x_0 + \delta/2)$ then $(x_1 - \delta/2, x_1 + \delta/2) \subset (x_0 - \delta, x_0 + \delta)$, so $\omega_f(x_1) \le M < 1/k$. Thus $\{\omega_f < 1/k\}$ contains a neighborhood of $x_0$: it is open.

(b) Suppose $\mathbb{Q} = \bigcap_n G_n$ with $G_n$ open. Each $G_n \supset \mathbb{Q}$ is dense, hence open dense. Enumerate $\mathbb{Q} = \{q_1, q_2, \dots\}$ and put $H_j = G_j \cap (\mathbb{R} \setminus \{q_j\})$. Each $H_j$ is open dense (removing one point preserves density).

By Baire, $\bigcap_j H_j$ is dense in $\mathbb{R}$, but $\bigcap_j H_j \subset \mathbb{Q} \setminus \mathbb{Q} = \varnothing$ — a contradiction. So $\mathbb{Q}$ is not $G_\delta$.

(c) If some $f$ had $C(f) = \mathbb{Q}$, then (a) would make $\mathbb{Q}$ a $G_\delta$, contradicting (b). Hence no function is continuous exactly on the rationals.

(For contrast, Thomae's function realizes the reverse — continuous exactly at the irrationals — and the irrationals are $G_\delta$.)