Exercises

Problems to check your understanding of this chapter. The solutions are folded away — try them yourself first.

Medium

1Medium

An irrational $x$ is a Liouville number if for every $m \ge 1$ there are infinitely many integer pairs $(p, q)$ with $q \ge 2$ satisfying $|x - p/q| < 1/q^m$ . For $k \ge 1$ put $U_k = \bigcup_{q \ge 2}\bigcup_{p \in \mathbb{Z}} (p/q - 1/q^k,\ p/q + 1/q^k)$ . (a) Show each $U_k$ is a dense open set in $\mathbb{R}$ . (b) Show $G = \bigcap_{k \ge 1} U_k \cap \bigcap_{j \ge 1} (\mathbb{R} \setminus \{r_j\})$ , where $\{r_j\}$ enumerates the rationals, is a $G_\delta$ set all of whose elements are Liouville numbers. (c) Use the Baire category theorem to conclude the Liouville numbers are dense in $\mathbb{R}$ and uncountable.

U_k = \bigcup_{q \ge 2}\bigcup_{p \in \mathbb{Z}} \left(\frac{p}{q} - \frac{1}{q^k},\ \frac{p}{q} + \frac{1}{q^k}\right)

G = \bigcap_{k \ge 1} U_k \ \cap\ \bigcap_{j \ge 1} (\mathbb{R} \setminus \{r_j\})

Baire Category Theorem

Show solution

(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.

2Medium

Let $T(x) = \frac{\cos x + 1}{4}$ for $x \in [0, 1]$ . (a) Show $T$ maps $[0, 1]$ into itself and is a contraction (name the contraction constant). (b) Conclude $T$ has a unique fixed point $x^* \in [0, 1]$ . (c) With $x_0 = 0$ and $x_{n+1} = T(x_n)$ , use the error bound $|x_n - x^*| \le \frac{q^n}{1-q}|x_1 - x_0|$ to find the smallest $n$ guaranteeing $|x_n - x^*| < 10^{-3}$ .

|T(x) - T(y)| \le \frac{1}{4}|x - y|

|x_n - x^*| \le \frac{2}{3}\left(\frac{1}{4}\right)^n

Show solution

(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
$|T(x) - T(y)| = \frac{|\cos x - \cos y|}{4} \le \frac{1}{4}|x - y|,$
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 $|x_5 - x^*| \le \frac{2}{3 \cdot 1024} = \frac{1}{1536} < 10^{-3}$ .

3Medium

Let $C$ be the set of points in $[0, 1]$ possessing a ternary expansion using only the digits $0$ and $2$ (the standard middle-thirds Cantor set). For $C + C = \{x + y : x, y \in C\}$ , prove $C + C = [0, 2]$ .

C = \left\{ \sum_{n=1}^{\infty} \frac{2a_n}{3^n} : a_n \in \{0, 1\} \right\}

C + C = [0, 2]

Length Zero, Points Uncountable

Show solution

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
$x + y = \sum_{n=1}^{\infty} \frac{2(a_n + b_n)}{3^n} = 2\sum_{n=1}^{\infty} \frac{e_n}{3^n} = s.$
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.

Hard

4Hard

Let $C(f)$ be the set of points where $f: \mathbb{R} \to \mathbb{R}$ is continuous. (a) Using the oscillation $\omega_f(x) = \inf_{\delta > 0} \sup\{|f(y) - f(z)| : y, z \in (x-\delta, x+\delta)\}$ , show $C(f) = \bigcap_{k=1}^{\infty} \{x : \omega_f(x) < 1/k\}$ with each $\{x : \omega_f(x) < 1/k\}$ open; i.e. $C(f)$ is a $G_\delta$ set. (b) Use the Baire category theorem to show $\mathbb{Q}$ is not a $G_\delta$ subset of $\mathbb{R}$ . (c) Conclude no function $f: \mathbb{R} \to \mathbb{R}$ is continuous exactly at the rationals and discontinuous exactly at the irrationals.

C(f) = \bigcap_{k=1}^{\infty} \{x : \omega_f(x) < 1/k\}

\mathbb{Q} = \bigcap_{n=1}^{\infty} G_n \ \ (G_n \text{ open}) \quad \Longrightarrow \quad \bigcap_{j=1}^{\infty}\left(G_j \setminus \{q_j\}\right) = \varnothing

Baire Category Theorem

Show solution

(a) $f$ is continuous at $x$ iff $\omega_f(x) = 0$ : continuity drives the oscillation on $\delta$ -intervals to $0$ , and conversely vanishing oscillation is exactly the $\varepsilon$ - $\delta$ condition. Hence $C(f) = \bigcap_k \{\omega_f < 1/k\}$ . 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 $G_\delta$ .)