§4 The Completeness Axiom

This section gives the axiom that guarantees $\mathbb{R}$ has no gaps. Almost every significant result in this book depends on it, and most of them would be false if our world of numbers were only $\mathbb{Q}$.

Maximum and minimum

4.1 Definition.

Let $S$ be a nonempty subset of $\mathbb{R}$.

1. If $S$ contains a largest element $s_0$ (that is, $s_0 \in S$ and $s \le s_0$ for all $s \in S$), then $s_0$ is the maximum of $S$, written $s_0 = \max S$.
2. If $S$ contains a smallest element, it is the minimum of $S$, written $\min S$.

Example 1.

1. Every finite nonempty subset of $\mathbb{R}$ has a maximum and a minimum. For instance, $\max\{0, \pi, -7, e, 3, \frac43\} = \pi$ and $\min\{0, \pi, -7, e, 3, \frac43\} = -7$. Also $\max\{n \in \mathbb{Z} : -4 < n \le 100\} = 100$ and $\min\{n \in \mathbb{Z} : -4 < n \le 100\} = -3$.
2. For real $a < b$ we use the interval notation $[a,b] = \{x : a \le x \le b\},\quad (a,b) = \{x : a < x < b\},\quad [a,b) = \{x : a \le x < b\},\quad (a,b] = \{x : a < x \le b\}.$ $[a,b]$ is a closed interval, $(a,b)$ an open interval, and the others are half-open. Here $\max[a,b] = b$ and $\min[a,b] = a$, while $(a,b)$ has neither a maximum nor a minimum, and $[a,b)$ has minimum $a$ but no maximum.
3. $\mathbb{Z}$ and $\mathbb{Q}$ have no maximum or minimum. $\mathbb{N}$ has no maximum, but $\min \mathbb{N} = 1$.
4. $\{r \in \mathbb{Q} : 0 \le r \le \sqrt2\}$ has minimum $0$ but no maximum: $\sqrt2$ is not in the set, yet the set contains rationals arbitrarily close to $\sqrt2$.
5. $\{n^{(-1)^n} : n \in \mathbb{N}\} = \{1, 2, \frac13, 4, \frac15, 6, \frac17, \dots\}$ has no maximum and no minimum.

Bounds

4.2 Definition.

Let $S$ be a nonempty subset of $\mathbb{R}$.

1. If $M \in \mathbb{R}$ satisfies $s \le M$ for all $s \in S$, then $M$ is an upper bound of $S$, and $S$ is bounded above.
2. If $m \in \mathbb{R}$ satisfies $m \le s$ for all $s \in S$, then $m$ is a lower bound of $S$, and $S$ is bounded below.
3. $S$ is bounded if it is bounded above and below, i.e. $S \subseteq [m, M]$ for some real $m, M$.

Example 2.

1. A maximum is always an upper bound, and a minimum is always a lower bound.
2. For $a < b$, the number $b$ is an upper bound for $[a,b]$, $(a,b)$, $[a,b)$ and $(a,b]$. So is every number larger than $b$, but $b$ is the least upper bound.
3. None of $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{N}$ is bounded above. $\mathbb{N}$ is bounded below, by $1$ or anything smaller, and $1$ is its greatest lower bound.
4. Every nonpositive number is a lower bound for $\{r \in \mathbb{Q} : 0 \le r \le \sqrt2\}$, and $0$ is the greatest. The least upper bound is $\sqrt2$.
5. $\{n^{(-1)^n} : n \in \mathbb{N}\}$ is not bounded above. Its greatest lower bound is $0$.

Supremum and infimum

4.3 Definition.

Let $S$ be a nonempty subset of $\mathbb{R}$.

1. If $S$ is bounded above and has a least upper bound, we call it the supremum of $S$, written $\sup S$.
2. If $S$ is bounded below and has a greatest lower bound, we call it the infimum of $S$, written $\inf S$.

Unlike $\max S$ and $\min S$, the numbers $\sup S$ and $\inf S$ need not belong to $S$. We use the Latin names because of the standard notations $\limsup$ and $\liminf$ that come later; nobody writes “lim lub”.

The following characterization is used constantly. If $S$ is bounded above, then $M = \sup S$ if and only if

$$\text{(i) } s \le M \text{ for all } s \in S, \qquad \text{and} \qquad \text{(ii) whenever } M_1 < M, \text{ there is } s_1 \in S \text{ with } s_1 > M_1.$$

Intuitionsup as a ceiling the set touches

Taking $M_1 = M - \varepsilon$ in (ii) gives the form you will use most often:

$$\begin{gathered} M = \sup S \iff \text{(i) } s \le M \text{ for all } s \in S, \text{ and} \\ \text{(ii}'\text{) for every } \varepsilon > 0 \text{ there is } s \in S \text{ with } s > M - \varepsilon. \end{gathered}$$

Picture $M$ as a ceiling. Condition (i) says no element pokes through it. Condition (ii$'$) says the set comes within every distance $\varepsilon$ of the ceiling, so the ceiling cannot be lowered at all. Proving $\sup S = M$ therefore always takes two separate arguments: one that $M$ is an upper bound, and one that nothing smaller is.

Try it: hunt for the supremum

Drag the blue line $M$ and the orange line $m$. The widget tells you whether $M$ is an upper bound (and if not, which element of $S$ is above it) and whether it is the least one. Then compare $\sup S$ with $\max S$.

Keyboard: Tab to the M or m marker, then use the arrow keys to move it.

elements of Sdrag M (upper bound?)drag m (lower bound?)

M = 1.140 is an upper bound, but not the least one: try sliding it left.

m = -0.1400 is a lower bound, but not the greatest one: try sliding it right.

Snap M to sup SSnap m to inf S

Example 3.

1. If $S$ has a maximum, then $\max S = \sup S$. Similarly for minimum and infimum.
2. If $a < b$, then $\sup[a,b] = \sup(a,b) = \sup[a,b) = \sup(a,b] = b$.
3. $\inf \mathbb{N} = 1$.
4. If $A = \{r \in \mathbb{Q} : 0 \le r \le \sqrt2\}$, then $\sup A = \sqrt2$ and $\inf A = 0$.
5. $\inf\{n^{(-1)^n} : n \in \mathbb{N}\} = 0$.

Example 4.

1. $A = \{\frac1{n^2} : n \in \mathbb{N},\ n \ge 3\}$ is bounded. It has maximum $\frac19$ but no minimum; $\sup A = \frac19$ and $\inf A = 0$.
2. $B = \{r \in \mathbb{Q} : r^3 \le 7\}$ is bounded above (by $2$, say). It has no maximum, because $r^3 \ne 7$ for every rational $r$ by Theorem 2.2. However, $\sup B = \sqrt[3]7$. $B$ is not bounded below; think of the graph of $y = x^3$.
3. $C = \{m + n\sqrt2 : m, n \in \mathbb{Z}\}$ is bounded neither above nor below.
4. $D = \{x \in \mathbb{R} : x^2 < 10\} = (-\sqrt{10}, \sqrt{10})$ is bounded, has no maximum or minimum, and $\inf D = -\sqrt{10}$, $\sup D = \sqrt{10}$.

Common pitfall

• $\sup S$ need not belong to $S$. When it does, it is $\max S$; when it does not, $S$ has no maximum at all. Compare $\sup B$ in Example 4(b), and $\inf A$ in Example 4(a).
• Strict inequalities do not survive taking a supremum. If $s < M$ for every $s \in S$, you may only conclude $\sup S \le M$: every element of $(0, 1)$ is less than $1$, and $\sup(0, 1) = 1$.
• To show $\sup S \le M$, show that $M$ is an upper bound. To show $\sup S \ge M$, you need elements of $S$ near $M$, or an argument that every upper bound is at least $M$. Mixing these up is the most common error in sup proofs.

In every example above, each set that is bounded above has a least upper bound. This is no accident. Otherwise there would be a “gap” between the set and its upper bounds.

4.4 Completeness Axiom.

Every nonempty subset $S$ of $\mathbb{R}$ that is bounded above has a least upper bound. In other words, $\sup S$ exists and is a real number.

The corresponding statement for $\mathbb{Q}$ is false. The set $A = \{r \in \mathbb{Q} : 0 \le r \text{ and } r^2 \le 2\}$ is described entirely in terms of rationals and is bounded above by rationals such as $\frac32$, but no rational number is its least upper bound.

The version for sets bounded below comes for free.

Why this matters

Every axiom in §3 holds in $\mathbb{Q}$ as well as in $\mathbb{R}$. The Completeness Axiom is the first property that fails in $\mathbb{Q}$, and it is the only thing that separates the two. So whenever a later proof uses a supremum of a set that is not finite, look for the step where the existence of that supremum is essential: that is where the theorem would break down over $\mathbb{Q}$.

4.5 Corollary.

Every nonempty subset $S$ of $\mathbb{R}$ that is bounded below has a greatest lower bound $\inf S$.

Used in Exercise 4.9, Exercise 4.14, §5

Proof

Let $-S = \{-s : s \in S\}$. If $m \le s$ for all $s \in S$, then $-m \ge -s$ for all $s \in S$, so $-S$ is bounded above by $-m$. By the Completeness Axiom, $s_0 = \sup(-S)$ exists. Reflecting the picture suggests $\inf S = -s_0$, and to prove it we must show

$$-s_0 \le s \text{ for all } s \in S, \tag{1}$$$$\text{if } t \le s \text{ for all } s \in S, \text{ then } t \le -s_0. \tag{2}$$

(1) says $-s_0$ is a lower bound for $S$, and (2) says it is the greatest one. The proofs of (1) and (2) are Exercise 4.9.

∎

The Archimedean property and denseness of ℚ

It is useful to know that

$$\text{if } a > 0, \text{ then } \tfrac1n < a \text{ for some } n \in \mathbb{N}, \tag{$*$}$$ $$\text{if } b > 0, \text{ then } b < n \text{ for some } n \in \mathbb{N}. \tag{$**$}$$

These are not as obvious as they look: there are ordered fields with elements $a, b > 0$ satisfying $a < \frac1n$ and $n < b$ for every $n$. Such strange elements cannot exist in $\mathbb{R}$, and we must expect to use completeness to prove it.

4.6 Archimedean Property.

If $a > 0$ and $b > 0$, then $na > b$ for some positive integer $n$.

Even if $a$ is tiny and $b$ is huge, some multiple of $a$ exceeds $b$: given enough time, you can empty a large bathtub with a small spoon. Taking $b = 1$ gives $(*)$ and taking $a = 1$ gives $(**)$.

Intuition

The Archimedean property says that $\mathbb{R}$ contains no infinitely large and no infinitely small positive numbers. Its form $(*)$ is the step “choose $n$ with $\frac1n < \varepsilon$”, which appears in almost every limit proof from §7 on.

Proof

Suppose the property fails: there are $a, b > 0$ with $na \le b$ for all $n \in \mathbb{N}$. Then $b$ is an upper bound for $S = \{na : n \in \mathbb{N}\}$, so $s_0 = \sup S$ exists. (This is where completeness is used.) Since $a > 0$, $s_0 - a < s_0$, so $s_0 - a$ is not an upper bound for $S$: there is $n_0 \in \mathbb{N}$ with $s_0 - a < n_0 a$. Then $s_0 < (n_0 + 1)a$, and $(n_0 + 1)a \in S$, contradicting that $s_0$ is an upper bound.

∎

4.7 Denseness of ℚ.

If $a, b \in \mathbb{R}$ and $a < b$, then there is a rational $r \in \mathbb{Q}$ with $a < r < b$.

Used in Exercise 17.12

Proof

We need integers $m$ and $n > 0$ with $a < \frac mn < b$, that is,

$$an < m < bn. \tag{1}$$

Since $b - a > 0$, the Archimedean property gives $n \in \mathbb{N}$ with

$$n(b - a) > 1, \quad\text{so}\quad bn - an > 1. \tag{2}$$

It is plausible that an integer $m$ lies between $an$ and $bn$, but proving it takes some care. By the Archimedean property again there is an integer $k > \max\{|an|, |bn|\}$, so $-k < an < bn < k$. The set $\{j \in \mathbb{Z} : -k \le j \le k \text{ and } an < j\}$ is finite and contains $k$, so it has a minimum $m$. Then $an < m$, and since $m > -k$, the integer $m - 1$ is at least $-k$ but fails $an < m - 1$ by the choice of $m$. So $m - 1 \le an$, and by (2), $m \le an + 1 < bn$. Thus (1) holds.

∎

Intuitionanother face of completeness: nested intervals

Suppose $[a_1, b_1] \supseteq [a_2, b_2] \supseteq [a_3, b_3] \supseteq \cdots$ are closed bounded intervals. Then some real number lies in all of them.

To see why, let $x = \sup\{a_n : n \in \mathbb{N}\}$, which exists by completeness since every $a_n \le b_1$. Certainly $a_n \le x$ for each $n$. For any fixed $m$ and any $n$, nesting gives $a_n \le a_{\max\{m,n\}} \le b_{\max\{m,n\}} \le b_m$, so $b_m$ is an upper bound for the $a_n$’s and therefore $x \le b_m$. Thus $x \in [a_m, b_m]$ for every $m$.

Both hypotheses matter. The open intervals $(0, \frac1n)$ are nested with nothing in common. And over $\mathbb{Q}$, the intervals $[1, 2], [1.4, 1.5], [1.41, 1.42], \dots$ that close in on $\sqrt2$ share no rational point. This “nested interval” form of completeness is what makes repeated halving work, as in the bisection method of §18. Theorem 13.10 is the same idea in $\mathbb{R}^k$.

Reading alongside Abbott

These ideas are in Abbott’s Sections 1.3 and 1.4. Abbott proves the nested interval property first, as the opening consequence of completeness, and uses it to show $\mathbb{R}$ is uncountable. The Archimedean property and the density of $\mathbb{Q}$ follow with essentially the same proofs as here. Abbott also proves that $\sqrt2$ exists as a real number directly from the Axiom of Completeness. Here that fact comes more cheaply in §18, from the Intermediate Value Theorem.

Practice: put the proof in order

Theorem 4.6 is a proof by contradiction that leans on completeness. Rebuild its argument.

Put the proof in order: Archimedean Property (Theorem 4.6)

These are the steps of the proof, shuffled. Use the buttons to move each step into place, then check.

1. Argue by contradiction: suppose there are $a, b > 0$ such that $na \le b$ for every positive integer $n$.
2. Then $S = \{na : n \in \mathbb{N}\}$ is nonempty and bounded above by $b$.
3. By the Completeness Axiom, $s_0 = \sup S$ exists.
4. Because $a > 0$, we have $s_0 - a < s_0$, so $s_0 - a$ is too small to be an upper bound of $S$.
5. So some element of $S$ exceeds it: $s_0 - a < n_0 a$ for some $n_0 \in \mathbb{N}$.
6. Adding $a$ to both sides gives $s_0 < (n_0 + 1)a$.
7. But $(n_0 + 1)a$ belongs to $S$, so $s_0$ is not an upper bound of $S$. This contradiction proves the property.

Practice: fill in the gaps

The skeleton of the proof of Theorem 4.7 is given. The gaps are the result it leans on and the integer it picks.

Fill in the gaps: Denseness of ℚ (Theorem 4.7)

This is the skeleton of the proof. In each gap choose the option that makes the argument work, then check.

1. We want integers $m$ and $n > 0$ with $a < \frac mn < b$. Multiplying by $n$, that means $an < m < bn$.
2. Since $b - a > 0$, the Archimedean propertyTriangle InequalityDenseness of ℚ itself gives $n \in \mathbb{N}$ with $n(b-a) > 1$$n(b-a) < 1$$n(b+a) > 1$.
3. The interval $(an, bn)$ is longer than $1$, so it should contain an integer. Take $m$ to be the least integer greater than $an$the greatest integer less than $an$any integer greater than $an$.
4. Then $m - 1 \le an$, so $m \le an + 1 < bn$ by the choice of $n$. Hence $an < m < bn$, and dividing by $n > 0$ gives $a < \frac mn < b$.

Practice: spot the error

A proof that sounds fine until you ask what it has actually shown.

Spot the error: the supremum of a set of fractions

One line of this argument is wrong. Press the button on the line you think is the mistake.

1. Let $S = \left\{\frac12, \frac34, \frac78, \dots\right\}$, the set of numbers $1 - 2^{-n}$ for $n \in \mathbb{N}$. We claim $\sup S = 2$.
2. Every element of $S$ is less than $1$, hence less than $2$.
3. So $2$ is an upper bound of $S$, and therefore $\sup S = 2$.

Check your understanding

Take the quiz (3 questions)

Let $S = \{\frac1n : n \in \mathbb{N}\}$. Which is right?

$\sup S = \max S = 1$, and $\inf S = 0$, which is not in $S$.$\sup S = 1$ and $\inf S = \min S = 0$.$S$ has no infimum, because it has no smallest element.

Let $S$ be nonempty with upper bound $M$. Which must be true?

$\sup S \le M$$\sup S \in S$$\sup S < M$

Which subset of $\mathbb{Q}$ has no least upper bound in $\mathbb{Q}$?

$\{r \in \mathbb{Q} : r^2 < 2\}$$\{r \in \mathbb{Q} : r < 2\}$$\{\frac1n : n \in \mathbb{N}\}$

Exercises

Exercise 4.1

For each set below that is bounded above, list three upper bounds for the set. Otherwise write “NOT BOUNDED ABOVE” or “NBA.” (An integer $p \ge 2$ is prime if its only positive factors are $1$ and $p$.)

1. $[0, 1]$
2. $(0, 1)$
3. $\{2, 7\}$
4. $\{\pi, e\}$
5. $\{\frac1n : n \in \mathbb{N}\}$
6. $\{0\}$
7. $[0,1] \cup [2,3]$
8. $\bigcup_{n=1}^\infty [2n, 2n+1]$
9. $\bigcap_{n=1}^\infty [-\frac1n, 1 + \frac1n]$
10. $\{1 - \frac1{3^n} : n \in \mathbb{N}\}$
11. $\{n + \frac{(-1)^n}{n} : n \in \mathbb{N}\}$
12. $\{r \in \mathbb{Q} : r < 2\}$
13. $\{r \in \mathbb{Q} : r^2 < 4\}$
14. $\{r \in \mathbb{Q} : r^2 < 2\}$
15. $\{x \in \mathbb{R} : x < 0\}$
16. $\{1, \frac\pi3, \pi^2, 10\}$
17. $\{0, 1, 2, 4, 8, 16\}$
18. $\bigcap_{n=1}^\infty (1 - \frac1n, 1 + \frac1n)$
19. $\{\frac1n : n \in \mathbb{N},\ n \text{ prime}\}$
20. $\{x \in \mathbb{R} : x^3 < 8\}$
21. $\{x^2 : x \in \mathbb{R}\}$
22. $\{\cos\frac{n\pi}{3} : n \in \mathbb{N}\}$
23. $\{\sin\frac{n\pi}{3} : n \in \mathbb{N}\}$

Solution

Any number at least as large as the supremum (Exercise 4.3) is an upper bound. For example, for (a) use $1, 2, 3$. The sets that are not bounded above are (h), (k), (u). For (r), the intersection is $\{1\}$, and for (i) it is $[0, 1]$.

Tags: Warm-upComputationSup & inf

Used in Exercise 4.2, Exercise 4.3, Exercise 5.3

Exercise 4.2

Repeat Exercise 4.1 for lower bounds.

Solution

The sets that are not bounded below are (l), (o) and (t). Every other set is bounded below by its infimum (Exercise 4.4) and anything smaller.

Tags: Warm-upComputationSup & inf

Exercise 4.3

For each set in Exercise 4.1, give its supremum if it has one. Otherwise write “NO sup.”

Solution

(a) 1 (b) 1 (c) 7 (d) $\pi$ (e) 1 (f) 0 (g) 3 (h) NO sup (i) 1 (j) 1 (k) NO sup (l) 2 (m) 2 (n) $\sqrt2$ (o) 0 (p) 10 (q) 16 (r) 1 (s) $\frac12$ (t) 2 (u) NO sup (v) 1 (w) $\frac{\sqrt3}{2}$

Tags: Warm-upComputationSup & inf

Used in Exercise 4.1, Exercise 4.4

Exercise 4.4

Repeat Exercise 4.3 for infima (the plural of infimum).

Solution

(a) 0 (b) 0 (c) 2 (d) $e$ (e) 0 (f) 0 (g) 0 (h) 2 (i) 0 (j) $\frac23$ (k) 0 (l) NO inf (m) $-2$ (n) $-\sqrt2$ (o) NO inf (p) $\frac\pi3$ (q) 0 (r) 1 (s) 0 (t) NO inf (u) 0 (v) $-1$ (w) $-\frac{\sqrt3}2$

Tags: Warm-upComputationSup & inf

Used in Exercise 4.2

Exercise 4.5

Let $S$ be a nonempty subset of $\mathbb{R}$ that is bounded above. Prove that if $\sup S$ belongs to $S$, then $\sup S = \max S$. (Your proof should be very short.)

Solution

$\sup S$ is an upper bound for $S$ and belongs to $S$, so it is the largest element of $S$.

Tags: Warm-upSup & inf

Exercise 4.6

Let $S$ be a nonempty bounded subset of $\mathbb{R}$.

1. Prove $\inf S \le \sup S$. (This is almost obvious; your proof should be short.)
2. What can you say about $S$ if $\inf S = \sup S$?

Solution

(a) Pick $s \in S$. Then $\inf S \le s \le \sup S$. (b) Every $s \in S$ satisfies $\inf S \le s \le \sup S = \inf S$, so $S$ consists of exactly one point.

Tags: StandardSup & inf

Used in Exercise 5.5

Exercise 4.7

Let $S$ and $T$ be nonempty bounded subsets of $\mathbb{R}$.

1. Prove that if $S \subseteq T$, then $\inf T \le \inf S \le \sup S \le \sup T$.
2. Prove $\sup(S \cup T) = \max\{\sup S, \sup T\}$. (Do not assume $S \subseteq T$.)

Hint

(a) $\sup T$ is an upper bound for $T$, hence for $S$. (b) Show $\max\{\sup S, \sup T\}$ is an upper bound for $S \cup T$, then use (a) for “$\ge$”.

Solution

1. $\sup T$ is an upper bound of $T$, hence of $S \subseteq T$, so the least upper bound satisfies $\sup S \le \sup T$. Likewise $\inf T$ is a lower bound of $S$, so $\inf T \le \inf S$. Finally, pick $s \in S$; then $\inf S \le s \le \sup S$.
2. Let $m = \max\{\sup S, \sup T\}$. Every $x \in S \cup T$ lies in $S$ or in $T$, so $x \le \sup S \le m$ or $x \le \sup T \le m$. Thus $m$ is an upper bound of $S \cup T$, which is nonempty and bounded, and $\sup(S \cup T) \le m$. Conversely $S \subseteq S \cup T$ and $T \subseteq S \cup T$, so by (a) $\sup S \le \sup(S \cup T)$ and $\sup T \le \sup(S \cup T)$, whence $m \le \sup(S \cup T)$.

Tags: StandardSup & inf

Used in Exercise 5.6, §10

Exercise 4.8

Let $S$ and $T$ be nonempty subsets of $\mathbb{R}$ such that $s \le t$ for all $s \in S$ and $t \in T$.

1. Observe $S$ is bounded above and $T$ is bounded below.
2. Prove $\sup S \le \inf T$.
3. Give an example of such sets $S$ and $T$ where $S \cap T$ is nonempty.
4. Give an example of sets $S$ and $T$ where $\sup S = \inf T$ and $S \cap T$ is empty.

Solution

(b) Each $t \in T$ is an upper bound for $S$, so $\sup S \le t$. Thus $\sup S$ is a lower bound for $T$, and $\sup S \le \inf T$. (c) $S = [0, 1]$, $T = [1, 2]$. (d) $S = [0, 1)$, $T = [1, 2]$.

Tags: StandardSup & inf

Exercise 4.9

Complete the proof that $\inf S = -\sup(-S)$ in Corollary 4.5 by proving (1) and (2).

Hint

(1): for $s \in S$, $-s \in -S$, so $-s \le s_0$. (2): if $t \le s$ for all $s \in S$, then $-t$ is an upper bound for $-S$, so $s_0 \le -t$.

Solution

Let $s_0 = \sup(-S)$.

(1) If $s \in S$ then $-s \in -S$, so $-s \le s_0$, i.e. $-s_0 \le s$.

(2) Suppose $t \le s$ for all $s \in S$. Then $-t \ge -s$ for all $s \in S$, so $-t$ is an upper bound of $-S$ (every element of $-S$ has the form $-s$). Hence $s_0 \le -t$, i.e. $t \le -s_0$.

Together, (1) says $-s_0$ is a lower bound of $S$ and (2) says it is the greatest one, so $\inf S = -s_0 = -\sup(-S)$.

Tags: StandardSup & inf

Used in Corollary 4.5, Exercise 5.4

Exercise 4.10

Prove that if $a > 0$, then there exists $n \in \mathbb{N}$ such that $\frac1n < a < n$.

Hint

Use $(*)$ and $(**)$ to get $n_1$ and $n_2$, then take $n = \max\{n_1, n_2\}$.

Solution

By $(*)$ applied to $a$ there is $n_1 \in \mathbb{N}$ with $\frac1{n_1} < a$, and by $(**)$ there is $n_2$ with $a < n_2$. Let $n = \max\{n_1, n_2\}$. Then $\frac1n \le \frac1{n_1} < a < n_2 \le n$.

Tags: StandardSup & inf

Exercise 4.11

Consider $a, b \in \mathbb{R}$ with $a < b$. Use Denseness of $\mathbb{Q}$ 4.7 to show there are infinitely many rationals between $a$ and $b$.

Hint

Get $r_1 \in (a, b)$, then $r_2 \in (a, r_1)$, then $r_3 \in (a, r_2)$, and so on.

Solution

By Denseness of $\mathbb{Q}$ choose a rational $r_1$ with $a < r_1 < b$. Having chosen $r_k \in (a,b)$, apply denseness to $a < r_k$ to get a rational $r_{k+1}$ with $a < r_{k+1} < r_k$. Then $r_1 > r_2 > r_3 > \cdots$ are distinct rationals, all in $(a, b)$. So there are infinitely many.

Tags: StandardSup & inf

Used in §11

Exercise 4.12

Let $\mathbb{I}$ be the set of real numbers that are not rational; these are the irrational numbers. Prove that if $a < b$, then there exists $x \in \mathbb{I}$ with $a < x < b$.

Hint

First show $\{r + \sqrt2 : r \in \mathbb{Q}\} \subseteq \mathbb{I}$. Then apply 4.7 to $a - \sqrt2 < b - \sqrt2$.

Solution

$\sqrt2$ is irrational (Corollary 2.3). For rational $r$, $r + \sqrt2$ is irrational: if $r + \sqrt2 = q \in \mathbb{Q}$ then $\sqrt2 = q - r \in \mathbb{Q}$, a contradiction. Now $a - \sqrt2 < b - \sqrt2$, so Denseness of $\mathbb{Q}$ gives a rational $r$ with $a - \sqrt2 < r < b - \sqrt2$. Then $x = r + \sqrt2$ is irrational and $a < x < b$.

Tags: StandardSup & infContradiction

Used in Exercise 13.10

Exercise 4.13

Prove the following are equivalent for real numbers $a, b, c$ (either all of them hold or none does):

1. $|a - b| < c$,
2. $b - c < a < b + c$,
3. $a \in (b - c, b + c)$.

Hint

Use Exercise 3.7(b).

Solution

(i) $\Leftrightarrow$ (ii) is Exercise 3.7(b). (ii) $\Leftrightarrow$ (iii) holds because $(b - c, b + c)$ is, by definition, the set of $a$ with $b - c < a < b + c$. So the three statements are equivalent.

Tags: Warm-upInequalities

Exercise 4.14

Let $A$ and $B$ be nonempty bounded subsets of $\mathbb{R}$, and let $A + B = \{a + b : a \in A,\ b \in B\}$.

1. Prove $\sup(A + B) = \sup A + \sup B$.
2. Prove $\inf(A + B) = \inf A + \inf B$.

Hint

For $\sup A + \sup B \le \sup(A+B)$: for each $b \in B$, $\sup(A+B) - b$ is an upper bound for $A$, so $\sup A \le \sup(A+B) - b$. Then $\sup(A+B) - \sup A$ is an upper bound for $B$.

Solution

1. $\le$: for $a \in A$, $b \in B$ we have $a + b \le \sup A + \sup B$, so $\sup(A+B) \le \sup A + \sup B$. $\ge$: let $s = \sup(A+B)$. For fixed $b \in B$ and any $a \in A$, $a + b \le s$, so $a \le s - b$. So $s - b$ is an upper bound of $A$, and $\sup A \le s - b$, i.e. $b \le s - \sup A$. This holds for every $b \in B$, so $s - \sup A$ is an upper bound of $B$ and $\sup B \le s - \sup A$. Thus $\sup A + \sup B \le s$.
2. Note $-(A + B) = (-A) + (-B)$. By Corollary 4.5 and part (a), $\inf(A+B) = -\sup\bigl((-A) + (-B)\bigr) = -\bigl[\sup(-A) + \sup(-B)\bigr] = \inf A + \inf B .$

Tags: ChallengeSup & inf

Used in §5

Exercise 4.15

Let $a, b \in \mathbb{R}$. Show that if $a \le b + \frac1n$ for all $n \in \mathbb{N}$, then $a \le b$. Compare Exercise 3.8.

Hint

If $a > b$, use the Archimedean property to find $n$ with $\frac1n < a - b$.

Solution

Suppose instead $a > b$. By the Archimedean property $(*)$ there is $n$ with $\frac1n < a - b$, i.e. $b + \frac1n < a$. That contradicts the hypothesis $a \le b + \frac1n$. Hence $a \le b$.

Tags: StandardInequalitiesContradiction

Exercise 4.16

Show $\sup\{r \in \mathbb{Q} : r < a\} = a$ for each $a \in \mathbb{R}$.

Hint

$a$ is clearly an upper bound. If $M_1 < a$, Denseness of $\mathbb{Q}$ gives a rational $r$ with $M_1 < r < a$.

Solution

Let $T = \{r \in \mathbb{Q} : r < a\}$. It is nonempty (by denseness there is a rational in $(a-1, a)$) and bounded above by $a$, so $\sup T \le a$. If $\sup T < a$, denseness gives a rational $r$ with $\sup T < r < a$; then $r \in T$ but $r > \sup T$, which is impossible. Hence $\sup T = a$.

Tags: StandardSup & inf

Used in §6