8.6 A Table of Strange Functions

The later volumes will define continuity, limits of functions and the integral, and each definition will be tested on functions that resist being drawn. This section collects those functions in one place. For each we prove what can be proved with the tools of this chapter: the order, the floor, suprema and finite counting. What needs a limit we only ask, and the questions go into a table at the end.

8.6.1. Irrationals between any two reals

Most of the functions below treat rational and irrational points differently, so we need both kinds in every interval. The rationals are dense (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 4.2.5). The irrationals are dense too. Number Systems left this as an exercise (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Exercise (C2)), and we prove it here. A real number is irrational if it is not in $\mathbb{Q}$ .

Lemma 8.6.1 (Irrationals Are Dense)

For all $a, b \in \mathbb{R}$ with $a < b$ there is an irrational $x$ with $a < x < b$ .

Proof. Let $a < b$ . The real numbers form a complete ordered field (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 5.5.2). So there is $\sqrt 2 \ge 0$ with $(\sqrt 2)^2 = 2$ , and $\mathbb{R}$ is Archimedean. Since $a - \sqrt 2 < b - \sqrt 2$ , the density of the rationals gives $q \in \mathbb{Q}$ with $a - \sqrt 2 < q < b - \sqrt 2$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 4.2.5). Put $x \mathrel{:=} q + \sqrt 2$ ; then $a < x < b$ .

Suppose $x$ were rational. Then $\sqrt 2 = x - q$ would be rational, as a difference of rationals. But $\sqrt 2$ is irrational, as Section 8.3 showed from Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 3.3.1 and the copy of $\mathbb{Q}$ inside $\mathbb{R}$ , a contradiction. Hence $x$ is irrational.

Uses: Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 6.1.4 (square roots exist); Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 6.1.2 ( $\mathbb{R}$ is Archimedean); Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 4.2.2 ( $\mathbb{Q}$ sits inside $\mathbb{R}$ ).

The argument for irrationality works for every rational $q$ : $q + \sqrt 2$ is irrational.

Quick check. Why does the proof shift by $\sqrt 2$ , and not by $1/2$ ?

Solution. For rational $q$ , the number $q + 1/2$ is rational, so the shift would land on a rational. The proof needs a shift that turns every rational into an irrational, and an irrational shift does that.

8.6.2. The Dirichlet function

The simplest function that no pencil can draw is the indicator of the rationals.

Definition 8.6.1 (The Dirichlet Function)

The Dirichlet function is $D \mathrel{:=} \mathbf{1}_{\mathbb{Q}} \colon \mathbb{R} \to \mathbb{R}$ (Sets and Functions, Definition 2.3.2): $D(x) = 1$ for $x \in \mathbb{Q}$ , and $D(x) = 0$ for $x \notin \mathbb{Q}$ .

Proposition 8.6.2 (Properties of the Dirichlet Function)

(a) A real $p \ne 0$ is a period of $D$ (Definition 8.4.3) if and only if $p \in \mathbb{Q}$ .
(b) For all $a < b$ , $\sup_{(a, b)} D = 1$ and $\inf_{(a, b)} D = 0$ , both attained, and $\sup \{ \left| D(x) - D(y) \right| : x, y \in (a, b) \} = 1$ .
(c) For all $a < b$ , $D$ is neither increasing nor decreasing on $(a, b)$ .

Proof.

(a) Let $p \in \mathbb{Q}$ , $p \ne 0$ . Since $\mathbb{Q}$ is closed under sums and negatives, $x \in \mathbb{Q}$ if and only if $x + p \in \mathbb{Q}$ , so $D(x + p) = D(x)$ for every $x$ . Let $p \notin \mathbb{Q}$ . Then $D(0) = 1$ and $D(0 + p) = 0$ , so $p$ is not a period.
(b) Every value of $D$ is $0$ or $1$ . The interval $(a, b)$ contains a rational, where $D$ is $1$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 4.2.5). It also contains an irrational, where $D$ is $0$ (Lemma 8.6.1). An upper bound of $D((a, b))$ that is itself a value is its supremum, so $\sup_{(a, b)} D = 1$ is attained; likewise $\inf_{(a, b)} D = 0$ . Since $D$ is bounded on the non-empty set $(a, b)$ , Proposition 8.3.4 gives $\sup \{ \left| D(x) - D(y) \right| : x, y \in (a, b) \} = 1 - 0 = 1$ .
(c) By Lemma 8.6.1 there is an irrational $t \in (a, b)$ . The rationals are dense, so there are rationals $q_1 \in (a, t)$ and $q_2 \in (t, b)$ . Then $q_1 < t$ and $D(q_1) = 1 > 0 = D(t)$ , so $D$ is not increasing on $(a, b)$ . Also $t < q_2$ with $D(t) = 0 < 1 = D(q_2)$ , so it is not decreasing.

Uses: Sets and Functions, Definition 5.3.5 (a bound that is a value); Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 4.2.5 (the rationals are dense).

The Dirichlet function on $[0, 1]$ : the rational points sit on the upper dotted line, the irrational points on the lower one, and every stretch of either line contains points of both kinds.

Multiplying by $x$ . Let $g(x) \mathrel{:=} x D(x)$ on $\mathbb{R}$ : $g(x) = x$ on $\mathbb{Q}$ and $0$ off $\mathbb{Q}$ . Then $\left| g(x) \right| \le \left| x \right|$ for every $x$ , so near $0$ the values of $g$ are small. Away from $0$ they are not. Let $a \ne 0$ and $\delta > 0$ . If $a \in \mathbb{Q}$ , an irrational $x$ with $\left| x - a \right| < \delta$ (Lemma 8.6.1) gives $\left| g(x) - g(a) \right| = \left| a \right|$ . If $a \notin \mathbb{Q}$ , a rational $x$ with $\left| x - a \right| < \min\{\delta, \left| a \right|/2\}$ gives $\left| g(x) - g(a) \right| = \left| x \right| \ge \left| a \right| - \left| x - a \right| > \left| a \right|/2 ,$ by the triangle inequality (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.3 (e)). So every interval around $a \ne 0$ , however short, holds points where $g$ differs from $g(a)$ by at least $\left| a \right|/2$ .

8.6.3. Thomae’s function

The Dirichlet function gives every rational the same value. Thomae’s function gives a rational a value that shrinks with its denominator. For example, it gives $1/2$ the value $1/2$ , both $1/3$ and $2/3$ the value $1/3$ , and $3/4$ the value $1/4$ . At an irrational point it is $0$ .

Definition 8.6.2 (Thomae’s Function)

For $x \in \mathbb{Q}$ , let $x = p/q$ with $p \in \mathbb{Z}$ , $q \in \mathbb{Z}_{>0}$ and $\gcd(p, q) = 1$ ; the pair $(p, q)$ is unique (Proposition 2.2.6 (a)). Thomae’s function $T \colon \mathbb{R} \to \mathbb{R}$ is $T(x) \mathrel{:=} \begin{cases} 1/q & \text{if } x = p/q \in \mathbb{Q} \text{ in lowest terms,} \\ 0 & \text{if } x \notin \mathbb{Q} . \end{cases}$

Since $0 = 0/1$ and $\gcd(0, 1) = 1$ , $T(0) = 1$ ; likewise $T(n) = 1$ for every $n \in \mathbb{Z}$ . On $\mathbb{Q}$ , $0 < T \le 1$ , since $q \ge 1$ .

Warning. The denominator is read after cancelling: $T(2/4) = T(1/2) = 1/2$ , not $1/4$ . Without lowest terms, “ $1/q$ at $p/q$ ” is not a function at all, since $1/2 = 2/4$ would get the two values $1/2$ and $1/4$ .

Quick check. Compute $T(3/4)$ , $T(-6/8)$ , $T(0.6)$ and $T(\sqrt 2/2)$ .

Solution.

$T(3/4) = 1/4$ .
$T(-6/8) = 1/4$ , since $-6/8 = -3/4$ and $\gcd(-3, 4) = 1$ .
$T(0.6) = 1/5$ , since $0.6 = 3/5$ .
$T(\sqrt 2/2) = 0$ , since $\sqrt 2/2$ is irrational. Indeed, if $\sqrt 2/2 = r \in \mathbb{Q}$ , then $\sqrt 2 = 2r \in \mathbb{Q}$ . But $\sqrt 2$ is irrational, as shown in Section 8.3.

Large values of $T$ are rare. On $[0, 1]$ , the points with $T(x) \ge 1/3$ have denominator $1$ , $2$ or $3$ . They are $0$ , $1$ , $1/2$ , $1/3$ and $2/3$ : only five.

Proposition 8.6.3 (Thomae’s Function Has Few Large Values)

(a) $1$ is a period of $T$ .
(b) For all real $a \le b$ and every $\varepsilon > 0$ , the set $E \mathrel{:=} \{ x \in [a, b] : T(x) \ge \varepsilon \}$ is finite.
(c) For all $a < b$ , the interval $(a, b)$ contains points with $T > 0$ and points with $T = 0$ .

Idea. A large value needs a small denominator: $T(x) \ge \varepsilon$ forces $q \le 1/\varepsilon$ . There are finitely many such $q$ , and for each $q$ only finitely many multiples of $1/q$ lie in $[a, b]$ . The same point appears under several $q$ ( $1/2 = 2/4$ ), so we take a union of finite sets rather than add up counts.

Proof.

(a) For $x \notin \mathbb{Q}$ , also $x + 1 \notin \mathbb{Q}$ , and both values are $0$ . Let $x = p/q$ in lowest terms. Then $x + 1 = (p + q)/q$ . A positive common divisor of $p + q$ and $q$ divides $(p + q) - q = p$ . It divides $q$ as well, so it is a common divisor of $p$ and $q$ and is at most their greatest common divisor $\gcd(p, q) = 1$ ; hence it equals $1$ . So $\gcd(p + q, q) = 1$ and $T(x + 1) = 1/q = T(x)$ .
(b) Let $\varepsilon > 0$ and $Q \mathrel{:=} \left\lfloor 1/\varepsilon \right\rfloor$ (Definition 8.4.1). For $q \in \mathbb{Z}_{>0}$ let $S_q \mathrel{:=} \{ p/q : p \in \mathbb{Z},\ \left\lfloor aq \right\rfloor \le p \le \left\lfloor bq \right\rfloor \} .$ $E \subseteq S_1 \cup \dots \cup S_Q$ . Let $x \in E$ . Then $T(x) \ge \varepsilon > 0$ , so $x \in \mathbb{Q}$ ; write $x = p/q$ in lowest terms. From $1/q \ge \varepsilon$ we get $q \le 1/\varepsilon$ , so $q \le Q$ by Proposition 8.4.1 (g). From $a \le p/q \le b$ and $q > 0$ we get $aq \le p \le bq$ . Hence $\left\lfloor aq \right\rfloor \le aq \le p$ , and $p \le \left\lfloor bq \right\rfloor$ by Proposition 8.4.1 (g). So $x \in S_q$ with $1 \le q \le Q$ . (If $Q = 0$ , no $q$ qualifies and $E = \emptyset$ .)

Each $S_q$ is finite. If $\left\lfloor aq \right\rfloor > \left\lfloor bq \right\rfloor$ , then $S_q = \emptyset$ . Otherwise let $m \mathrel{:=} \left\lfloor aq \right\rfloor$ and $k \mathrel{:=} \left\lfloor bq \right\rfloor - m + 1$ . The map $\{1, \dots, k\} \to S_q$ , $j \mapsto (m + j - 1)/q$ , is onto by the definition of $S_q$ , and injective because $q \ne 0$ . So $S_q$ has $k$ elements (Sets and Functions, Definition 4.1.2).

A finite union of finite sets is finite. Let $A$ and $B$ be finite sets. Then $A \cup B$ is the union of the disjoint sets $A$ and $B \setminus A$ . The second is finite, as a subset of $B$ . So $A \cup B$ is finite by Sets and Functions, Proposition 4.1.5. Induction on the number of sets gives $S_1 \cup \dots \cup S_Q$ finite.

Hence $E$ , a subset of a finite set, is finite by Sets and Functions, Proposition 4.1.6.
(c) The interval $(a, b)$ contains a rational, where $T > 0$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 4.2.5). It also contains an irrational, where $T = 0$ (Lemma 8.6.1).

Uses: Proposition 8.4.1 (a) (the floor is at most the number); Sets and Functions, Proposition 4.1.6 (subsets of finite sets); Sets and Functions, Proposition 4.1.5 (a disjoint union of finite sets is finite); Logic and Proof, Axiom 4.2.1 (induction).

Thomae’s function on $[0, 1]$ , drawn for the denominators $q \le 6$ : the dot over $p/q$ has height $1/q$ , and the irrational points lie on the dotted axis. Only finitely many dots rise above the dashed level $\varepsilon$ .

8.6.4. More strange functions

An involution. A function $f$ with $f \circ f = \operatorname{id}$ is an involution. Exercise C2 of Section 8.2 (Exercise (C2, used later)) showed that an increasing involution of $\mathbb{R}$ is the identity. Here is how far that fails without “increasing”.

Example (An Involution That Is Monotone Nowhere)

Let $J \colon \mathbb{R} \to \mathbb{R}$ , $J(x) \mathrel{:=} x$ for $x \in \mathbb{Q}$ and $J(x) \mathrel{:=} -x$ for $x \notin \mathbb{Q}$ . Show that $J \circ J = \operatorname{id}_{\mathbb{R}}$ , that $J$ is a bijection, and that $J$ is monotone on no interval $(a, b)$ with $a < b$ .

Solution. For $x \in \mathbb{Q}$ , $J(J(x)) = J(x) = x$ . For $x \notin \mathbb{Q}$ , also $-x \notin \mathbb{Q}$ , so $J(J(x)) = J(-x) = -(-x) = x$ . So $J$ is its own two-sided inverse, and it is a bijection by Sets and Functions, Theorem 2.4.4.

Let $a < b$ . By Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 4.2.5, used twice, there are rationals $q_1 < q_2$ in $(a, b)$ , and $J(q_1) = q_1 < q_2 = J(q_2)$ , so $J$ is not decreasing there. By Lemma 8.6.1, used twice, there are irrationals $t_1 < t_2$ in $(a, b)$ , and $J(t_1) = -t_1 > -t_2 = J(t_2)$ , so $J$ is not increasing there.

The map $x \mapsto -x$ was already a non-identity involution; $J$ is one that agrees with the identity on a dense set.

A jump. The function $H$ of Example (A strictly increasing function with a jump), $H(x) = x$ for $x < 0$ and $H(x) = x + 1$ for $x \ge 0$ , is strictly increasing on $\mathbb{R}$ , yet its image $(-\infty, 0) \cup [1, \infty)$ skips the whole interval $[0, 1)$ .

A staircase. The floor function makes a step function out of a straight line.

Example (A Staircase Along the Diagonal)

Let $s \colon [0, 1] \to \mathbb{R}$ , $s(0) \mathrel{:=} 0$ and $s(x) \mathrel{:=} 1/\left\lfloor 1/x \right\rfloor$ for $0 < x \le 1$ . Describe $s$ on $(1/(n + 1), 1/n]$ for $n \in \mathbb{Z}_{>0}$ .

Solution. For $0 < x \le 1$ we have $1/x \ge 1$ , so $\left\lfloor 1/x \right\rfloor \ge 1$ by Proposition 8.4.1 (g), and $s(x)$ is defined. For $n \in \mathbb{Z}_{>0}$ , $\left\lfloor 1/x \right\rfloor = n \iff n \le 1/x < n + 1 \iff \frac{1}{n + 1} < x \le \frac{1}{n} .$ The first equivalence is the definition of the floor. The second takes reciprocals of positive numbers, which reverses inequalities (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 3.2.2). So $s = 1/n$ on $(1/(n + 1), 1/n]$ : a step at height $1/n$ whose right end lies on the line $y = x$ .

Uses: Definition 8.4.1 (the floor).

The staircase $s$ and the line $y = x$ (dashed), drawn for $n \le 7$ : each step starts just above the line and ends on it.

Proposition 8.6.4 (The Staircase Hugs the Diagonal)

(a) $s$ is increasing on $[0, 1]$ .
(b) $s$ takes infinitely many values.
(c) For $0 < x \le 1$ , $0 \le s(x) - x < 2x^2$ . (At $x = 0$ the left side is $0$ and so is the right, so the strict inequality fails there.)

Proof.

(a) Let $0 < x < y \le 1$ . Taking reciprocals reverses the inequality, so $1/y < 1/x$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 3.2.2). Then $\left\lfloor 1/y \right\rfloor \le \left\lfloor 1/x \right\rfloor$ by Proposition 8.4.1 (c). Both are at least $1$ , so taking reciprocals again gives $s(x) = 1/\left\lfloor 1/x \right\rfloor \le 1/\left\lfloor 1/y \right\rfloor = s(y)$ . For $x = 0 < y$ , $s(0) = 0 < s(y)$ .
(b) For $n \in \mathbb{Z}_{>0}$ , $s(1/n) = 1/\left\lfloor n \right\rfloor = 1/n$ (Proposition 8.4.1 (a)). Suppose $s$ took finitely many values. The set $V \mathrel{:=} \{ 1/n : n \in \mathbb{Z}_{>0} \}$ consists of some of them, so it would be finite (Sets and Functions, Proposition 4.1.6). It is non-empty, so it would have a minimum $1/N$ (Sets and Functions, Proposition 3.3.8). But $1/(N + 1) \in V$ and $1/(N + 1) < 1/N$ .
(c) Let $0 < x \le 1$ and $n \mathrel{:=} \left\lfloor 1/x \right\rfloor \ge 1$ . By the solution of Example (A Staircase Along the Diagonal), $1/(n + 1) < x \le 1/n$ , so $s(x) - x = 1/n - x \ge 0$ , and $\frac{1}{n} - x < \frac{1}{n} - \frac{1}{n + 1} = \frac{1}{n(n + 1)} \le \frac{2}{(n + 1)^2} < 2x^2 .$ The middle inequality is $n + 1 \le 2n$ , true for $n \ge 1$ . The last holds because $x > 1/(n + 1) > 0$ gives $x^2 > 1/(n + 1)^2$ (Lemma 1.5.1 (a) with exponent $2$ ).

So near $0$ , $s(x)$ agrees with $x$ up to an error below $2x^2$ , although $s$ is constant on each step.

Quick check. Is $s$ a step function on $[0, 1]$ (Definition 8.4.4)?

Solution. No. A step function with a partition into $n$ open pieces takes at most $2n + 1$ values: one on each piece and one at each of the $n + 1$ partition points. By Proposition 8.6.4 (b), $s$ takes infinitely many values.

A zigzag. Replace the floor by the distance to the nearest integer, and the steps become tents.

Example (A Zigzag That Never Settles)

Let $Z \colon \mathbb{R} \to \mathbb{R}$ , $Z(0) \mathrel{:=} 0$ and $Z(x) \mathrel{:=} \operatorname{dist}(1/x, \mathbb{Z})$ for $x \ne 0$ (Definition 8.4.2). Describe $Z$ on $[1/(n + 1), 1/n]$ for $n \in \mathbb{Z}_{>0}$ .

Solution. For $x$ in that interval, $1/x \in [n, n + 1]$ , and $Z(x) = \min\{ 1/x - n,\ n + 1 - 1/x \} :$ for $1/x < n + 1$ this is Proposition 8.4.2 (a) with $\left\lfloor 1/x \right\rfloor = n$ , and at $x = 1/(n + 1)$ both sides are $0$ . So $Z = 0$ at both ends, $Z = 1/2$ at $x = 2/(2n + 1)$ , where $1/x = n + 1/2$ , and in between $Z$ climbs from $0$ to $1/2$ and falls back. Each interval $[1/(n + 1), 1/n]$ carries one full tent, and the tents crowd toward $0$ .

$Diagram. Labels: x, tfrac 12, 1, tfrac 16$

The zigzag $Z$ on $[1/6, 1]$ : five tents, each rising from $0$ to $1/2$ and back, narrower as they approach $0$ .

Proposition 8.6.5 (The Zigzag Takes Every Value Near 0)

(a) For every $\delta > 0$ and every $t \in [0, 1/2]$ there is $x \in (0, \delta)$ with $Z(x) = t$ .
(b) For every $x \in \mathbb{R}$ , $\left| x Z(x) \right| \le \left| x \right|/2$ .

Proof.

(a) Let $\delta > 0$ and $t \in [0, 1/2]$ . By Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.2.3 there is $n \in \mathbb{Z}_{>0}$ with $1/n < \delta$ . Put $x \mathrel{:=} 1/(n + t)$ . Since $n + t \ge n > 0$ , we have $0 < x \le 1/n < \delta$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 3.2.2). Since $\operatorname{dist}(\cdot, \mathbb{Z})$ has period $1$ (Proposition 8.4.2 (c)), induction on $n$ gives $Z(x) = \operatorname{dist}(n + t, \mathbb{Z}) = \operatorname{dist}(t, \mathbb{Z})$ . Finally $\left\lfloor t \right\rfloor = 0$ , as $0 \le t < 1$ , so by Proposition 8.4.2 (a), $\operatorname{dist}(t, \mathbb{Z}) = \min\{ t, 1 - t \} = t$ , because $t \le 1/2 \le 1 - t$ .
(b) For $x = 0$ both sides are $0$ . For $x \ne 0$ , $\left| x Z(x) \right| = \left| x \right| \operatorname{dist}(1/x, \mathbb{Z}) \le \left| x \right|/2$ by Proposition 8.4.2 (b).

So on every interval $(0, \delta)$ the zigzag sweeps through all of $[0, 1/2]$ , while $x Z(x)$ is pressed between $-\left| x \right|/2$ and $\left| x \right|/2$ .

Two functions on a closed interval. The last two are defined at every point of $[0, 1]$ and still misbehave.

Example (Unbounded on a Closed Interval)

Let $R \colon [0, 1] \to \mathbb{R}$ , $R(0) \mathrel{:=} 0$ and $R(x) \mathrel{:=} 1/x$ for $0 < x \le 1$ . Show that $R$ is not bounded above on $[0, 1]$ (Definition 8.3.1).

Solution. Let $C \in \mathbb{R}$ , and put $C' \mathrel{:=} \max\{C, 0\} + 1 \ge 1$ and $x \mathrel{:=} 1/C' \in (0, 1]$ . Then $R(x) = C' > C$ . So no $C$ is an upper bound of $R([0, 1])$ .

Example (A Supremum Never Reached)

Let $M \colon [0, 1] \to \mathbb{R}$ , $M(x) \mathrel{:=} x$ for $0 \le x < 1$ and $M(1) \mathrel{:=} 0$ . Show that $\sup_{[0, 1]} M = 1$ and that it is not attained.

Solution. Every value is at most $1$ : it is $x < 1$ or it is $0$ . For $\varepsilon > 0$ , put $x \mathrel{:=} \max\{0, 1 - \varepsilon/2\} \in [0, 1)$ ; then $M(x) = x \ge 1 - \varepsilon/2 > 1 - \varepsilon$ . By Proposition 8.3.2, $\sup_{[0, 1]} M = 1$ . Every value is in fact strictly below $1$ , so the supremum is not attained.

Both $R$ and $M$ are defined everywhere on the closed bounded interval $[0, 1]$ . The table below asks which further hypothesis would rule out both behaviors.

8.6.5. The table

Example (A Table of Strange Functions)

Each row names a function, recalls its definition, lists what has been proved about it, and asks what the later volumes will answer.

function	definition	proved here	question for later
$D$ (Definition 8.6.1)	$1$ on $\mathbb{Q}$ , $0$ off $\mathbb{Q}$	periods exactly the non-zero rationals; on every interval sup $1$ , inf $0$ , monotone nowhere	continuous at any point? integrable on $[0, 1]$ ?
$x D(x)$	$x$ on $\mathbb{Q}$ , $0$ off $\mathbb{Q}$	at most $\left\| x \right\|$ in size; near each $a \ne 0$ , points where it differs from its value at $a$ by at least $\left\| a \right\|/2$	continuous exactly where?
$T$ (Definition 8.6.2)	$1/q$ at $p/q$ in lowest terms, $0$ off $\mathbb{Q}$	period $1$ ; finitely many points of $[a, b]$ where $T \ge \varepsilon$ ; positive and zero points in every interval	continuous exactly where? integrable on $[0, 1]$ ?
$J$ (Example (An Involution That Is Monotone Nowhere))	$x$ on $\mathbb{Q}$ , $-x$ off $\mathbb{Q}$	$J \circ J = \operatorname{id}$ ; a bijection; monotone on no interval	continuous at any point?
$H$ (Example (A strictly increasing function with a jump))	$x$ for $x < 0$ , $x + 1$ for $x \ge 0$	strictly increasing; image skips $[0, 1)$	which property forces the image of an interval to be an interval?
$s$ (Example (A Staircase Along the Diagonal))	$1/\left\lfloor 1/x \right\rfloor$ on $(0, 1]$ , $0$ at $0$	increasing; infinitely many values; $0 \le s(x) - x < 2x^2$ for $0 < x \le 1$	continuous at $0$ ? differentiable at $0$ ? integrable?
$Z$ , $x Z(x)$ (Example (A Zigzag That Never Settles))	$\operatorname{dist}(1/x, \mathbb{Z})$ for $x \ne 0$ , $0$ at $0$	$Z$ takes every value of $[0, 1/2]$ on every $(0, \delta)$ ; $\left\| x Z(x) \right\| \le \left\| x \right\|/2$	$Z$ continuous at $0$ ? $x Z(x)$ continuous at $0$ ? differentiable there?
$R$ (Example (Unbounded on a Closed Interval))	$1/x$ on $(0, 1]$ , $0$ at $0$	not bounded above on $[0, 1]$	which hypothesis on a function on $[a, b]$ forces it to be bounded?
$M$ (Example (A Supremum Never Reached))	$x$ on $[0, 1)$ , $0$ at $1$	sup $1$ , not attained	which hypothesis forces the sup to be attained?

Remark (Looking ahead). The analysis volumes answer these questions. Analysis on the Real Line settles the continuity and extreme-value questions once it has continuous functions. Integration, Series, and Function Spaces settles the integrability questions once it has the integral, and once it has the sine, it replaces $\operatorname{dist}(1/x, \mathbb{Z})$ in the zigzag by the sine. No proof in the main text of this section uses a limit, continuity or a derivative.

8.6.6. Exercises

A. Check your understanding

Exercise (A1)

Find the supremum and infimum of $D$ on $(0, 1)$ and of $T$ on $[0, 1]$ . Which of them are attained, and where?

Solution. For $D$ on $(0, 1)$ : supremum $1$ , infimum $0$ , both attained (Proposition 8.6.2 (b)), for example at $1/2$ and at the irrational $\sqrt 2/2$ (irrational by the Quick check after Definition 8.6.2, and $(\sqrt 2/2)^2 = 1/2 < 1$ gives $\sqrt 2/2 < 1$ , and it is positive).

For $T$ on $[0, 1]$ : every value is at most $1$ , since $q \ge 1$ , and $T(0) = T(0/1) = 1$ , $T(1) = T(1/1) = 1$ ; so the supremum is $1$ , attained at $0$ and $1$ . Every value is at least $0$ , and $T(\sqrt 2/2) = 0$ ; so the infimum is $0$ , attained at every irrational point of $[0, 1]$ .

Exercise (A2)

Let $g(x) \mathrel{:=} x D(x)$ .

(a) Is $g$ bounded on $[-5, 5]$ ?
(b) Is $g$ monotone on $(0, 1)$ ?

Solution.

(a) Yes: $\left| g(x) \right| \le \left| x \right| \le 5$ for $x \in [-5, 5]$ .
(b) No. The points $1/2 < \sqrt 2/2 < 3/4$ lie in $(0, 1)$ : squaring, $1/4 < 1/2 < 9/16$ , and squaring is strictly increasing on $[0, \infty)$ (Lemma 1.5.1 (a)). The values are $g(1/2) = 1/2$ , $g(\sqrt 2/2) = 0$ , $g(3/4) = 3/4$ . The first pair shows $g$ is not increasing on $(0, 1)$ , the second that it is not decreasing.

B. Practice

Exercise (B1)

Find all $x \in [0, 1]$ with $T(x) \ge 1/4$ .

Solution. $T(x) \ge 1/4$ forces $x$ rational, $x = p/q$ in lowest terms with $q \le 4$ and $0 \le p \le q$ . Listing $q = 1, 2, 3, 4$ and keeping the pairs with $\gcd(p, q) = 1$ : $0,\ 1,\ \tfrac12,\ \tfrac13,\ \tfrac23,\ \tfrac14,\ \tfrac34 .$ Seven points; $2/4$ is not new, since it is $1/2$ , with value $1/2$ .

Exercise (B2: Prove or disprove)

(a) $D(x) + D(x + \sqrt 2) \le 1$ for every $x \in \mathbb{R}$ .
(b) The restriction of $J$ to $\mathbb{Q}$ is increasing.
(c) Every bijection $\mathbb{R} \to \mathbb{R}$ is monotone on some interval $(a, b)$ with $a < b$ .
(d) $T(x + 1) = T(x)$ for every $x \in \mathbb{R}$ .

Solution.

(a) True. If $x$ and $x + \sqrt 2$ were both rational, so would be their difference $\sqrt 2$ , which is irrational, as shown in Section 8.3. So at most one of the two values is $1$ .
(b) True. On $\mathbb{Q}$ , $J$ is the identity, and $q_1 < q_2$ gives $J(q_1) = q_1 < q_2 = J(q_2)$ .
(c) False. $J$ is a bijection that is monotone on no such interval (Example (An Involution That Is Monotone Nowhere)).
(d) True. This is Proposition 8.6.3 (a).

Exercise (B3)

Compute $s(0.4)$ , and check the inequality of Proposition 8.6.4 (c) at $x = 0.4$ .

Solution. $1/0.4 = 2.5$ and $\left\lfloor 2.5 \right\rfloor = 2$ , so $s(0.4) = 1/2$ . Then $s(0.4) - 0.4 = 0.1$ , and $2 \cdot 0.4^2 = 0.32$ . Indeed $0 \le 0.1 < 0.32$ .

C. Going deeper

Exercise (C1: Additive and increasing)

Let $f \colon \mathbb{R} \to \mathbb{R}$ satisfy $f(x + y) = f(x) + f(y)$ for all $x, y$ , and let $f$ be increasing. Prove that $f(x) = cx$ for every $x \in \mathbb{R}$ , where $c \mathrel{:=} f(1)$ .

Solution (Hint 1). Exercise C1 of Section 8.1 gives $f$ on $\mathbb{Q}$ . Trap $x$ between rationals.

Solution (Hint 2). Suppose $c > 0$ and $f(x) > cx$ for some $x$ . Where can a rational $q$ sit so that monotonicity and $f(q) = cq$ collide?

Solution. By Exercise (C1: Additive functions, used later) (a) and (c), $f(0) = 0$ and $f(q) = cq$ for every $q \in \mathbb{Q}$ . Since $f$ is increasing and $0 < 1$ , $c = f(1) \ge f(0) = 0$ . Fix $x \in \mathbb{R}$ .

Case $c > 0$ . Suppose $f(x) > cx$ . Then $x < f(x)/c$ , and Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 4.2.5 gives $q \in \mathbb{Q}$ with $x < q < f(x)/c$ . Then $f(q) = cq < f(x)$ , although $x < q$ and $f$ is increasing, so $f(x) \le f(q)$ : a contradiction. Suppose $f(x) < cx$ . Then $f(x)/c < x$ , and a rational $q$ with $f(x)/c < q < x$ gives $f(q) = cq > f(x)$ , against $f(q) \le f(x)$ . So $f(x) = cx$ .

Case $c = 0$ . Take rationals $p < x < q$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 4.2.5, in $(x - 1, x)$ and $(x, x + 1)$ ). Then $0 = f(p) \le f(x) \le f(q) = 0$ , so $f(x) = 0 = cx$ .

Without “increasing”, additivity alone does not force $f(x) = cx$ in general. But an additive function that is not of this form cannot be written down without the axiom of choice. This series builds no such function, and it does not prove that none can be written down without choice.

What a complete answer needs. (10 points) - 2: obtains $f(q) = cq$ on $\mathbb{Q}$ and $c \ge 0$ , each with its reason. - 4: for $c > 0$ , refutes $f(x) > cx$ with a rational between $x$ and $f(x)/c$ , using monotonicity in the right direction. - 2: the symmetric case $f(x) < cx$ . - 2: the case $c = 0$ . Another correct route earns the same points: a two-sided squeeze with rationals $p < x < q$ closer than $1/n$ , then Archimedes (2 for $f(q) = cq$ on $\mathbb{Q}$ and $c \ge 0$ , 4 for $cp \le f(x) \le cq$ , 4 for closing the gap). Slips that lose the step: dividing by $c$ without $c > 0$ ; assuming $f$ continuous or taking a limit of rationals.

Exercise (C2)

Let $a < b$ and let $f \colon [a, b] \to \mathbb{R}$ be increasing. Prove that $f$ is bounded and attains its supremum and its infimum on $[a, b]$ . Why does Example (A Supremum Never Reached) not contradict this?

Solution. For $x \in [a, b]$ , $a \le x \le b$ , so $f(a) \le f(x) \le f(b)$ because $f$ is increasing. So $f$ is bounded, $f(b)$ is an upper bound of $f([a, b])$ that is itself a value, hence $\sup_{[a, b]} f = f(b)$ is attained; likewise $\inf_{[a, b]} f = f(a)$ .

The function $M$ is not increasing on $[0, 1]$ : $1/2 < 1$ but $M(1/2) = 1/2 > 0 = M(1)$ . Its supremum fails to be attained exactly because the value at the right end drops.

R. Review

Exercise (R1)

Let $p \in \mathbb{R}[x]$ , and suppose its polynomial function satisfies $p(x + 1) = p(x)$ for every $x \in \mathbb{R}$ . Prove that $p$ is a constant polynomial.

Solution. Let $c \mathrel{:=} p(0)$ and $r \mathrel{:=} p - c \in \mathbb{R}[x]$ . Evaluation respects differences (Proposition 6.4.4 (b)), so $r(x) = p(x) - c$ . By induction on $n \in \mathbb{N}$ , $r(n) = 0$ : $r(0) = p(0) - c = 0$ , and $r(n + 1) = p(n + 1) - c = p(n) - c = r(n)$ .

Suppose $r \ne 0$ , of degree $d$ . Then $0, 1, \dots, d$ are $d + 1$ distinct roots of $r$ , against Theorem 7.4.1, which allows at most $d$ . So $r = 0$ , and $p = c$ is constant.

Exercise (R2)

Prove that the strictly increasing bijections $\mathbb{R} \to \mathbb{R}$ form a subgroup of the group of all bijections $\mathbb{R} \to \mathbb{R}$ under composition. Do the strictly decreasing bijections form one?

Solution. The bijections of $\mathbb{R}$ form a group under $\circ$ (Example (Groups) (c)). Let $G$ be the set of strictly increasing bijections. We use Theorem 5.2.1.

$G \ne \emptyset$ , since $\operatorname{id}_{\mathbb{R}} \in G$ . Let $f, g \in G$ . Since $g$ is strictly increasing and onto, Theorem 8.2.3 (b) gives that $g^{-1} \colon g(\mathbb{R}) = \mathbb{R} \to \mathbb{R}$ is strictly increasing. It is a bijection by Sets and Functions, Theorem 2.4.4. Then $f \circ g^{-1}$ is strictly increasing by Proposition 8.2.1 (e). It is bijective by Sets and Functions, Theorem 2.2.1 (c). So $f \circ g^{-1} \in G$ , and $G$ is a subgroup.

The strictly decreasing bijections do not form a subgroup. A subgroup contains the identity, and $\operatorname{id}_{\mathbb{R}}$ is not strictly decreasing ( $0 < 1$ , $\operatorname{id}(0) < \operatorname{id}(1)$ ). They are not even closed under composition: $x \mapsto -x$ composed with itself is $\operatorname{id}_{\mathbb{R}}$ .

Exercise (R3)

Let $x \in \mathbb{R}$ , and let $a_0 \mathrel{:=} 0$ and $a_n \mathrel{:=} \left\lfloor nx \right\rfloor/n$ for $n \ge 1$ . Prove that $a_n \to x$ .

Solution. For $n \ge 1$ , Proposition 8.4.1 (a) gives $nx - 1 < \left\lfloor nx \right\rfloor \le nx$ ; dividing by $n > 0$ , $0 \le x - a_n < 1/n$ . Let $\varepsilon > 0$ . By Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.2.3 there is $N \in \mathbb{Z}_{>0}$ , depending on $\varepsilon$ , with $1/n < \varepsilon$ for every $n \ge N$ . For such $n$ , $n \ge 1$ and $\left| a_n - x \right| < 1/n < \varepsilon$ . So $a_n \to x$ by Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Definition 4.3.2.

Exercise (R4)

Is there a polynomial $p \in \mathbb{R}[x]$ whose polynomial function is $x \mapsto \left| x \right|$ on $\mathbb{R}$ ?

Solution. No. Suppose there were. Then $p \ne 0$ , since $p(1) = 1$ . Let $d \mathrel{:=} \max\{\deg p, 1\}$ . The polynomials $p$ and $x$ both have degree at most $d$ and agree at the $d + 1$ distinct points $0, 1, \dots, d$ , where $\left| x \right| = x$ . By Theorem 7.4.2 (a), $p = x$ . Then $p(-1) = -1 \ne 1 = \left| -1 \right|$ , a contradiction.

The tool is that a polynomial is fixed by finitely many values. Two look-alikes fail. Comparing growth gives nothing: $\left| x \right|$ grows like a polynomial of degree $1$ . “Since $\left| x \right|$ is even, $p$ has only even powers” needs the same identity principle to compare $p(x)$ with $p(-x)$ , and then still needs an argument.