Numbers

Michael Taylor

Contents

1. Peano arithmetic2. The integers3. Prime factorization and the fundamental theorem of arithmetic4. The rational numbers5. Sequences6. The real numbers7. Irrational numbers8. Cardinal numbers9. Metric properties of R

10. Complex numbers

1

2

Introduction

The basic goal of this work is to provide the reader with a solid understandingof the real number system. Texts on this subject vary on just how to achievethis. Some take an axiomatic approach. In such an approach, the set of realnumbers is hypothesized to have a number of properties, including various algebraicproperties satisfied by addition and multiplication, order axioms, and, crucially, thecompleteness property, sometimes expressed as the supremum property.

This is not the approach we will take. Rather, we will start with a small list ofaxioms for the natural numbers (i.e., the positive integers), and then build the restof the edifice logically, obtaining the basic properties of the real number system,particularly the completeness property, as theorems.

Sections 1–3 deal with the integers, starting in §1 with the set N of natural num-bers. The development proceeds from axioms of G. Peano. The main one is theprinciple of mathematical induction. We deduce basic results about integer arith-metic from these axioms. A high point is the fundamental theorem of arithmetic,presented in §3.

Section 4 discusses the set Q of rational numbers, deriving the basic algebraicproperties of these numbers from the results of §§1–3. Section 5 provides a bridgebetween §4 and §6. It deals with infinite sequences, including convergent sequencesand “Cauchy sequences.”

This prepares the way for §6, the main section of this chapter. Here we constructthe set R of real numbers, as “ideal limits” of rational numbers. We extend basicalgebraic results from Q to R. Furthermore, we establish the result that R is“complete,” i.e., Cauchy sequences always have limits in R. Section 7 providesexamples of irrational numbers, such as

√2,√

3,√

5,...Section 8 deals with cardinal numbers, an extension of the natural numbers N,

that can be used to “count” elements of a set, not necessarily finite. For example,N is a “countably” infinite set, and so is Q. We show that R “uncountable,” andhence much larger than N or Q.

Section 9 returns to the real number line R, and establishes further metric prop-erties of R and various subsets, with an emphasisis on the notion of compactness.The completeness property established in §6 plays a crucial role here.

Section 10 introduces the set C of complex numbers and establishes basic alge-braic and metric properties of C. While some introductory treatments of analysisavoid complex numbers, we embrace them, and consider their use in basic analysistoo precious to omit.

3

1. Peano arithmetic

In Peano arithmetic, we assume we have a set N (the natural numbers). Weassume given 0 /∈ N, and form N = N ∪ 0. We assume there is a map

(1.1) s : N −→ N,

which is bijective. That is to say, for each k ∈ N, there is a j ∈ N such that s(j) = k,so s is surjective; and furthermore, if s(j) = s(j′) then j = j′, so s is injective. Themap s plays the role of “addition by 1,” as we will see below. The only other axiomof Peano arithmetic is that the principle of mathematical induction holds. In otherwords, if S ⊂ N is a set with the properties

(1.2) 0 ∈ S, k ∈ S ⇒ s(k) ∈ S,

then S = N.Actually, applying the induction principle to S = 0 ∪ s(N), we see that it

suffices to assume that s in (1.1) is injective; the induction principle ensures thatit is surjective.

We define addition x + y, for x, y ∈ N, inductively on y, by

(1.3) x + 0 = x, x + s(y) = s(x + y).

Next, we define multiplication x · y, inductively on y, by

(1.4) x · 0 = 0, x · s(y) = x · y + x.

We also define

(1.5) 1 = s(0).

We now establish the basic laws of arithmetic.

Proposition 1.1. x + 1 = s(x).

Proof. x + s(0) = s(x + 0).

Proposition 1.2. 0 + x = x.

Proof. Use induction on x. First, 0 + 0 = 0. Now, assuming 0 + x = x, we have

0 + s(x) = s(0 + x) = s(x).

4

Proposition 1.3. s(y + x) = s(y) + x.

Proof. Use induction on x. First, s(y + 0) = s(y) = s(y) + 0. Next, we have

s(y + s(x)) = ss(y + x),

s(y) + s(x) = s(s(y) + x).

If s(y + x) = s(y) + x, the two right sides are equal, so the two left sides are equal,completing the induction.

Proposition 1.4. x + y = y + x.

Proof. Use induction on y. The case y = 0 follows from Proposition 1.2. Now,assuming x + y = y + x, for all x ∈ N, we must show s(y) has the same property.In fact,

x + s(y) = s(x + y) = s(y + x),

and by Proposition 1.3 the last quantity is equal to s(y) + x.

Proposition 1.5. (x + y) + z = x + (y + z).

Proof. Use induction on z. First, (x + y) + 0 = x + y = x + (y + 0). Now, assuming(x+y)+ z = x+(y + z), for all x, y ∈ N, we must show s(z) has the same property.In fact,

(x + y) + s(z) = s((x + y) + z),

x + (y + s(z)) = x + s(y + z) = s(x + (y + z)),

and we perceive the desired identity.

Proposition 1.6. x · 1 = x.

Proof. We havex · s(0) = x · 0 + x = 0 + x = x,

the last identity by Proposition 1.2.

Proposition 1.7. 0 · y = 0.

Proof. Use induction on y. First, 0 · 0 = 0. Next, assuming 0 · y = 0, we have0 · s(y) = 0 · y + 0 = 0 + 0 = 0.

Proposition 1.8. s(x) · y = x · y + y.

Proof. Use induction on y. First, s(x) · 0 = 0, while x · 0 + 0 = 0 + 0 = 0. Next,assuming s(x) · y = x · y + y, for all x, we must show that s(y) has this property.In fact,

s(x) · s(y) = s(x) · y + s(x) = (x · y + y) + (x + 1),

x · s(y) + s(y) = (x · y + x) + (y + 1),

and identity then follows via the commutative and associative laws of addition,Propositions 1.4 and 1.5.

5

Proposition 1.9. x · y = y · x.

Proof. Use induction on y. First, x ·0 = 0 = 0 ·x, the latter identity by Proposition1.7. Next, assuming x ·y = y ·x for all x ∈ N, we must show that s(y) has the sameproperty. In fact,

x · s(y) = x · y + x = y · x + x,

s(y) · x = y · x + x,

the last identity by Proposition 1.8.

Proposition 1.10. (x + y) · z = x · z + y · z.

Proof. Use induction on z. First, the identity clearly holds for z = 0. Next, assumingit holds for z (for all x, y ∈ N), we must show it holds for s(z). In fact,

(x + y) · s(z) = (x + y) · z + (x + y) = (x · z + y · z) + (x + y),

x · s(z) + y · s(z) = (x · z + x) + (y · z + y),

and the desired identity follows from the commutative and associative laws of ad-dition.

Proposition 1.11. (x · y) · z = x · (y · z).

Proof. Use induction on z. First, the identity clearly holds for z = 0. Next, assumingit holds for z (for all x, y ∈ N), we have

(x · y) · s(z) = (x · y) · z + x · y,

whilex · (y · s(z)) = x · (y · z + y) = x · (y · z) + x · y,

the last identity by Proposition 1.10. These observations yield the desired identity.

We next demonstrate the cancellation law of addition:

Proposition 1.12. Given x, y, z ∈ N,

(1.6) x + y = z + y =⇒ x = z.

Proof. Use induction on y. If y = 0, (1.6) obviously holds. Assuming (1.6) holdsfor y, we must show that

(1.7) x + s(y) = z + s(y)

implies x = z. In fact, (1.7) is equivalent to s(x + y) = s(z + y). Since the map s isassumed to be one-to-one, this implies that x + y = z + y, so we are done.

We next define an order relation on N. Given x, y ∈ N, we say

(1.8) x < y ⇐⇒ y = x + u, for some u ∈ N.

Similarly there is a definition of x ≤ y. We have x ≤ y if and only if y ∈ Rx, where

(1.9) Rx = x + u : u ∈ N.

6

Proposition 1.13. If x ≤ y and y ≤ x then x = y.

Proof. The hypotheses imply

(1.10) y = x + u, x = y + v, u, v ∈ N.

Hence x = x + u + v, so, by Proposition 1.12, u + v = 0. Now, if v 6= 0, thenv = s(w), so u + v = s(u + w) ∈ N. Thus v = 0, and u = 0.

Proposition 1.14. Given x, y ∈ N, either

(1.11) x < y, or x = y, or y < x,

and no two can hold.

Proof. That no two of (1.11) can hold follows from Proposition 1.13. To show thatone must hold, we want to show that

(1.12) y /∈ Rx =⇒ y < x.

To do this, use induction on y. If 0 /∈ Rx, then x 6= 0, so x ∈ N, and hence x = 0+xshows that 0 < x. Now, assuming that y has the property (1.12), we must showthat s(y) has this property.

So assume s(y) /∈ Rz. Since R0 = N, we deduce that z 6= 0, hence z ∈ N, hencez = s(x) for some x. But

s(y) /∈ Rs(x) ⇐⇒ y /∈ Rx.

The inductive hypothesis gives x = y + u, u ∈ N, hence s(x) = s(y) + u, and weare done.

We can now establish the cancellation law for multiplication.

Proposition 1.15. Given x, y, z ∈ N,

(1.13) x · y = x · z, x 6= 0 =⇒ y = z.

Proof. If y 6= z, then either y < z or z < y. Suppose y < z, i.e., z = y + u, u ∈ N.Then the hypotheses of (1.13) imply

x · y = x · y + x · u, x 6= 0,

hence, by Proposition 1.12,

(1.14) x · u = 0, x 6= 0.

We thus need to show that (1.14) implies u = 0. In fact, if not, then we can writeu = s(w), and x = s(a), with w, a ∈ N, and we have

(1.15) x · u = x · w + s(a) = s(x · w + a) ∈ N.

This contradicts (1.14), so we are done.

Remark. Note that (1.15) implies

(1.16) x, y ∈ N =⇒ x · y ∈ N.

We next establish the following variant of the principle of induction, called thewell-ordering property of N.

7

Proposition 1.16. If S ⊂ N is nonempty, then S contains a smallest element.

Proof. Suppose S contains no smallest element. Then 0 /∈ S. Let

(1.17) T = x ∈ N : x < y, ∀ y ∈ S.Then 0 ∈ T. We claim that

(1.18) x ∈ T =⇒ s(x) ∈ T.

Indeed, suppose x ∈ T, so x < y for all y ∈ S. If s(x) /∈ T, we have s(x) ≥ y0 forsome y0 ∈ S. Now, using Proposition 1.13, one can show that

(1.19) x < y0, s(x) ≥ y0 =⇒ s(x) = y0.

In turn, from this one can deduce that y0 must be the smallest element of S. Thus,if S has no smallest element, (1.18) must hold. The induction principle then impliesthat T = N, which implies S is empty.

Exercises

Given n ∈ N, we define∑n

k=1 ak inductively, as follows.

(1.20)1∑

k=1

ak = a1,

n+1∑

k=1

ak =( n∑

k=1

ak

)+ an+1.

Use the principle of induction to establish the following identities.

(1) 2n∑

k=1

k = n(n + 1).

(2) 6n∑

k=1

k2 = n(n + 1)(2n + 1).

(3) (a− 1)n∑

k=1

ak = an+1 − a, if a 6= 1.

In (3), we define an inductively by a1 = a, an+1 = an · a. We also set a0 = 1 ifa ∈ N, and

∑nk=0 ak = a0 +

∑nk=1 ak. Verify that

(4) (a− 1)n∑

k=0

ak = an+1 − 1, if a 6= 1.

5. Given k ∈ N, show that2k ≥ 2k,

with strict inequality for k > 1.

6. Show that, for x, x′, y, y′ ∈ N,

x < x′, y ≤ y′ ⇒ x · y < x′y′ and x + y < x′ + y′.

8

2. The integers

An integer is thought of as having the form x − a, with x, a ∈ N. To be moreformal, we will define an element of Z as an equivalence class of ordered pairs(x, a), x, a ∈ N, where we define

(2.1) (x, a) ∼ (y, b) ⇐⇒ x + b = y + a.

We claim (2.1) is an equivalence relation. In general, an equivalence relation on aset S is a specification s ∼ t for certain s, t ∈ S, which satisfies the following threeconditions.

(a) Reflexive. s ∼ s, ∀ s ∈ S.

(b) Symmetric. s ∼ t ⇐⇒ t ∼ s.

(c) Transitive. s ∼ t, t ∼ u =⇒ s ∼ u.

We will encounter various equivalence relations in this and subsequent sections.Generally, (a) and (b) are quite easy to verify, and we will be content with verifying(c).

Proposition 2.1. The relation (2.1) is an equivalence relation.

Proof. We need to check that

(2.2) (x, a) ∼ (y, b), (y, b) ∼ (z, c) =⇒ (x, a) ∼ (z, c),

i.e., that, for x, y, z, a, b, c ∈ N,

(2.3) x + b = y + a, y + c = z + b =⇒ x + c = z + a.

In fact, the hypotheses of (2.3), and the results of §1, imply

(x + c) + (y + b) = (z + a) + (y + b),

and the conclusion of (2.3) then follows from the cancellation property, Proposition1.12.

Let us denote the equivalence class containing (x, a) by [(x, a)]. We then defineaddition and multiplication in Z to satisfy

(2.4)[(x, a)] + [(y, b)] = [(x, a) + (y, b)], [(x, a)] · [(y, b)] = [(x, a) · (y, b)],

(x, a) + (y, b) = (x + y, a + b), (x, a) · (y, b) = (xy + ab, ay + xb).

To see that these operations are well defined, we need:

9

Proposition 2.2. If (x, a) ∼ (x′, a′) and (y, b) ∼ (y′, b′), then

(2.5) (x, a) + (y, b) ∼ (x′, a′) + (y′, b′),

and

(2.6) (x, a) · (y, b) ∼ (x′, a′) · (y′, b′).

Proof. The hypotheses say

(2.7) x + a′ = x′ + a, y + b′ = y′ + b.

The conclusions follow from results of §1. In more detail, adding the two identitiesin (2.7) gives

x + a′ + y + b′ = x′ + a + y′ + b,

and rearranging, using the commutative and associative laws of addition, yields

(x + y) + (a′ + b′) = (x′ + y′) + (a + b),

implying (2.5). The task of proving (2.6) is simplified by going through the inter-mediate step

(2.8) (x, a) · (y, b) ∼ (x′, a′) · (y, b).

If x′ > x, so x′ = x + u, u ∈ N, then also a′ = a + u, and our task is to prove

(xy + ab, ay + xb) ∼ (xy + uy + ab + ub, ay + uy + xb + ub),

which is readily done. Having (2.8), we apply similar reasoning to get

(x′, a′) · (y, b) ∼ (x′, a′) · (y′, b′),

and then (2.6) follows by transitivity.

Similarly, it is routine to verify the basic commutative, associative, etc. lawsincorporated in the next proposition. To formulate the results, set

(2.9) m = [(x, a)], n = [(y, b)], k = [(z, c)] ∈ Z.

Also, define

(2.10) 0 = [(0, 0)], 1 = [(1, 0)],

and

(2.11) −m = [(a, x)].

10

Proposition 2.3. We have

(2.12)

m + n = n + m,

(m + n) + k = m + (n + k),m + 0 = m,

m + (−m) = 0,

mn = nm,

m(nk) = (mn)k,

m · 1 = m,

m · 0 = 0,

m · (−1) = −m,

m · (n + k) = m · n + m · k.

To give an example of a demonstration of these results, the identity mn = nmis equivalent to

(xy + ab, ay + xb) ∼ (yx + ba, bx + ya).

In fact, commutative laws for addition and multiplication in N imply xy + ab =yx + ba and ay + xb = bx + ya. Verification of the other identities in (2.12) is leftto the reader.

We next establish the cancellation law for addition in Z.

Proposition 2.4. Given m, n, k ∈ Z,

(2.13) m + n = k + n =⇒ m = k.

Proof. We give two proofs. For one, we can add −n to both sides and use theresults of Proposition 2.3. Alternatively, we can write the hypotheses of (2.13) as

x + y + c + b = z + y + a + b

and use Proposition 1.12 to deduce that x + c = z + a.

Note that it is reasonable to set

(2.14) m− n = m + (−n).

This defines subtraction on Z.There is a natural injection

(2.15) N → Z, x 7→ [(x, 0)],

whose image we identify with N. Note that the map (2.10) preserves addition andmultiplication. There is also an injection x 7→ [(0, x)], whose image we identify with−N.

11

Proposition 2.5. We have a disjoint union:

(2.16) Z = N ∪ 0 ∪ (−N).

Proof. Suppose m ∈ Z; write m = [(x, a)]. By Proposition 1.14, either

a < x, or x = a, or x < a.

In these three cases,

x = a + u, u ∈ N, or x = a, or a = x + v, v ∈ N.

Then, either

(x, a) ∼ (u, 0), or (x, a) ∼ (0, 0), or (x, a) ∼ (0, v).

We define an order on Z by:

(2.17) m < n ⇐⇒ n−m ∈ N.

We then have:

Corollary 2.6. Given m, n ∈ Z, then either

(2.18) m < n, or m = n, or n < m,

and no two can hold.

The map (2.15) is seen to preserve order relations.Another consequence of (2.16) is the following.

Proposition 2.8. If m,n ∈ Z and m · n = 0, then either m = 0 or n = 0.

Proof. Suppose m 6= 0 and n 6= 0. We have four cases:

m > 0, n > 0 =⇒ mn > 0,

m < 0, n < 0 =⇒ mn = (−m)(−n) > 0,

m > 0, n < 0 =⇒ mn = −m(−n) < 0,

m < 0, n > 0 =⇒ mn = −(−m)n < 0,

the first by (1.16). This finishes the proof.

Using Proposition 2.7, we have the following cancellation law for multiplicationin Z.

12

Proposition 2.8. Given m, n, k ∈ Z,

(2.19) mk = nk, k 6= 0 =⇒ m = n.

Proof. First, mk = nk ⇒ mk − nk = 0. Now

mk − nk = (m− n)k.

See Exercise 3 below. Hence

mk = nk =⇒ (m− n)k = 0.

Given k 6= 0, Proposition 2.7 implies m− n = 0. Hence m = n.

Exercises

1. Verify Proposition 2.3.

2. We define∑n

k=1 ak as in (1.20), this time with ak ∈ Z. We also define ak

inductively as in (1.19), with a0 = 1 if a 6= 0. Use the principle of induction toestablish the identity

n∑

k=1

(−1)k−1k = −m if n = 2m,

m + 1 if n = 2m + 1.

3. Show that, if m,n, k ∈ Z,

−(nk) = (−n)k, and mk − nk = (m− n)k.

4. Deduce the following from Proposition 1.16. Let S ⊂ Z be nonempty and assumem ∈ Z such that m < n for all n ∈ S. Then S has a smallest element.

5. Show that Z has no smallest element.

13

3. Prime factorization and the fundamental theorem of arithmetic

Let x ∈ N. We say x is composite if one can write

(3.1) x = ab, a, b ∈ N,

with neither a nor b equal to 1. If x 6= 1 is not composite, it is said to be prime. If(3.1) holds, we say a|x (and that b|x), or that a is a divisor of x. Given x ∈ N, x > 1,set

(3.2) Dx = a ∈ N : a|x, a > 1.

Thus x ∈ Dx, so Dx is non-empty. By Proposition 1.16, Dx contains a smallestelement, say p1. Clearly p1 is a prime. Set

(3.3) x = p1x1, x1 ∈ N, x1 < x.

The same construction applies to x1, which is > 1 unless x = p1. Hence we haveeither x = p1 or

(3.4) x1 = p2x2, p2 prime , x2 < x1.

Continue this process, passing from xj to xj+1 as long as xj is not prime. The setS of such xj ∈ N has a smallest element, say xµ−1 = pµ, and we have

(3.5) x = p1p2 · · · pµ, pj prime.

This is part of the Fundamental Theorem of Arithmetic:

Theorem 3.1. Given x ∈ N, x 6= 1, there is a unique product expansion

(3.6) x = p1 · · · pµ,

where p1 ≤ · · · ≤ pµ are primes.

Only uniqueness remains to be established. This follows from:

Proposition 3.2. Assume a, b ∈ N, and p ∈ N is prime. Then

(3.7) p|ab =⇒ p|a or p|b.

We will deduce this from:

14

Proposition 3.3. If p ∈ N is prime and a ∈ N, is not a multiple of p, or moregenerally if p, a ∈ N have no common divisors > 1, then there exist m,n ∈ Z suchthat

(3.8) ma + np = 1.

Proof of Proposition 3.2. Assume p is a prime which does not divide a. Pick m,nsuch that (3.8) holds. Now, multiply (3.8) by b, to get

mab + npb = b.

Thus, if p|ab, i.e., ab = pk, we have

p(mk + nb) = b,

so p|b, as desired.

To prove Proposition 3.3, let us set

(3.9) Γ = ma + np : m,n ∈ Z.

Clearly Γ satisfies the following criterion:

Definition. A nonempty subset Γ ⊂ Z is a subgroup of Z provided

(3.10) a, b ∈ Γ =⇒ a + b, a− b ∈ Γ.

Proposition 3.4. If Γ ⊂ Z is a subgroup, then either Γ = 0, or there existsx ∈ N such that

(3.11) Γ = mx : m ∈ Z.

Proof. Note that n ∈ Γ ⇔ −n ∈ Γ, so, with Σ = Γ ∩ N, we have a disjoint union

Γ = Σ ∪ 0 ∪ (−Σ).

If Σ 6= ∅, let x be its smallest element. Then we want to establish (3.11), so setΓ0 = mx : m ∈ Z. Clearly Γ0 ⊂ Γ. Similarly, set Σ0 = mx : m ∈ N = Γ0 ∩ N.We want to show that Σ0 = Σ. If y ∈ Σ \ Σ0, then we can pick m0 ∈ N such that

m0x < y < (m0 + 1)x,

and hencey −m0x ∈ Σ

is smaller than x. This contradiction proves Proposition 3.4.

15

Proof of Proposition 3.3. Taking Γ as in (3.9), pick x ∈ N such that (3.11) holds.Since a ∈ Γ and p ∈ Γ, we have

a = m0x, p = m1x

for some mj ∈ Z. The assumption that a and p have no common divisor > 1 impliesx = 1. We conclude that 1 ∈ Γ, so (3.8) holds.

Exercises

1. Prove that there are infinitely many primes.Hint. If p1, . . . , pm is a complete list of primes, consider

x = p1 · · · pm + 1.

What are its prime factors?

2. Referring to (3.10), show that a nonempty subset Γ ⊂ Z is a subgroup of Zprovided

(3.12) a, b ∈ Γ =⇒ a− b ∈ Γ.

Hint. a ∈ Γ ⇒ 0 = a− a ∈ Γ ⇒ −a = 0− a ∈ Γ, given (3.12).

3. Let n ∈ N be a 12 digit integer. Show that if n is not prime, then it must bedivisible by a prime p < 106.

4. Determine whether the following number is prime:

(3.13) 201367.

Hint. This is for the student who can use a computer.

5. Find the smallest prime larger than the number in (3.13). Hint. Same as above.

16

4. The rational numbers

A rational number is thought of as having the form m/n, with m,n ∈ Z, n 6=0. Thus, we will define an element of Q as an equivalence class of ordered pairsm/n, m ∈ Z, n ∈ Z \ 0, where we define

(4.1) m/n ∼ a/b ⇐⇒ mb = an.

Proposition 4.1. This is an equivalence relation.

Proof. We need to check that

(4.2) m/n ∼ a/b, a/b ∼ c/d =⇒ m/n ∼ c/d,

i.e., that, for m, a, c ∈ Z, n, b, d ∈ Z \ 0,

(4.3) mb = an, ad = cb =⇒ md = cn.

Now the hypotheses of (4.3) imply (mb)(ad) = (an)(cb), hence

(md)(ab) = (cn)(ab).

We are assuming b 6= 0. If also a 6= 0, then ab 6= 0, and the conclusion of (4.3)follows from the cancellation property, Proposition 2.8. On the other hand, ifa = 0, then m/n ∼ a/b ⇒ mb = 0 ⇒ m = 0 (since b 6= 0), and similarlya/b ∼ c/d ⇒ cb = 0 ⇒ c = 0, so the desired implication also holds in that case.

We will (temporarily) denote the equivalence class containing m/n by [m/n]. Wethen define addition and multiplication in Q to satisfy

(4.4)[m/n] + [a/b] = [(m/n) + (a/b)], [m/n] · [a/b] = [(m/n) · (a/b)],

(m/n) + (a/b) = (mb + na)/(nb), (m/n) · (a/b) = (ma)/(nb).

To see that these operations are well defined, we need:

Proposition 4.2. If m/n ∼ m′/n′ and a/b ∼ a′/b′, then

(4.4A) (m/n) + (a/b) ∼ (m′/n′) + (a′/b′),

and

(4.4B) (m/n) · (a/b) ∼ (m′/n′) · (a′/b′).

17

Proof. The hypotheses say

(4.4C) mn′ = m′n, ab′ = a′b.

The conclusions follow from the results of §2. In more detail, multiplying the twoidentities in (4.4C) yields

man′b′ = m′a′nb,

which implies (4.4B). To prove (4.4A), it is convenient to establish the intermediatestep

(m/n) + (a/b) ∼ (m′/n′) + (a/b).

This is equivalent to

(mb + na)/nb ∼ (m′b + n′a)/(n′b),

hence to(mb + na)n′b = (m′b + n′a)nb,

or tomn′bb + nn′ab = m′nbb + n′nab.

This in turn follows readily from (4.4C). Having (4.4D), we can use a similar argu-ment to establish that

(m′/n′) + (a/b) ∼ (m′/n′) + (a′/b′),

and then (4.4A) follows by transitivity of ∼.

From now on, we drop the brackets, simply denoting the equivalence class ofm/n by m/n, and writing (4.1) as m/n = a/b. We also may denote an element ofQ by a single letter, e.g., x = m/n. There is an injection

(4.5) Z → Q, m 7→ m/1,

whose image we identify with Z. This map preserves addition and multiplication.We define

(4.6) −(m/n) = (−m)/n,

and, if x = m/n 6= 0, (i.e., m 6= 0 as well as n 6= 0), we define

(4.7) x−1 = n/m.

The results stated in the following proposition are routine consequences of theresults of §2.

18

Proposition 4.3. Given x, y, z ∈ Q, we have

x + y = y + x,

(x + y) + z = x + (y + z),x + 0 = x,

x + (−x) = 0,

x · y = y · x,

(x · y) · z = x · (y · z),x · 1 = x,

x · 0 = 0,

x · (−1) = −x,

x · (y + z) = x · y + x · z.

Furthermore,x 6= 0 =⇒ x · x−1 = 1.

For example, if x = m/n, y = a/b with m,n, a, b ∈ Z, n, b 6= 0, the identityx ·y = y ·x is equivalent to (ma)/(nb) ∼ (am)/(bn). In fact, the identities ma = amand nb = bn follow from Proposition 2.3. We leave the rest of Proposition 4.3 tothe reader.

We also have cancellation laws:

Proposition 4.4. Given x, y, z ∈ Q,

(4.8) x + y = z + y =⇒ x = z.

Also,

(4.9) xy = zy, y 6= 0 =⇒ x = z.

Proof. To get (4.8), add −y to both sides of x + y = z + y and use the results ofProposition 4.3. To get (4.9), multiply both sides of x · y = z · y by y−1.

It is natural to define

(4.10) x− y = x + (−y),

and, if y 6= 0,

(4.11) x/y = x · y−1.

We now define the order relation on Q. Set

(4.12) Q+ = m/n : mn > 0,

19

where, in (4.12), we use the order relation on Z, discussed in §2. This is welldefined. In fact, if m/n = m′/n′, then mn′ = m′n, hence (mn)(m′n′) = (mn′)2,and therefore mn > 0 ⇔ m′n′ > 0. Results of §2 imply that

(4.13) Q = Q+ ∪ 0 ∪ (−Q+)

is a disjoint union, where −Q+ = −x : x ∈ Q+. Also, clearly

(4.14) x, y ∈ Q+ =⇒ x + y, xy,x

y∈ Q+.

We define

(4.15) x < y ⇐⇒ y − x ∈ Q+,

and we have, for any x, y ∈ Q, either

(4.16) x < y, or x = y, or y < x,

and no two can hold. The map (4.5) is seen to preserve the order relations. In lightof (4.14), we see that

(4.17) given x, y > 0, x < y ⇔ x

y< 1 ⇔ 1

y<

1x

.

As usual, we say x ≤ y provided either x < y or x = y. Similarly there are naturaldefinitions of x > y and of x ≥ y.

The following result implies that Q has the Archimedean property.

Proposition 4.5. Given x ∈ Q, there exists k ∈ Z such that

(4.18) k − 1 < x ≤ k.

Proof. It suffices to prove (4.18) assuming x ∈ Q+; otherwise, work with −x (andmake a few minor adjustments). Say x = m/n, m, n ∈ N. Then

S = ` ∈ N : ` ≥ x

contains m, hence is nonempty. By Proposition 1.16, S has a smallest element; callit k. Then k ≥ x. We cannot have k − 1 ≥ x, for then k − 1 would belong to S.Hence (4.18) holds.

Exercises

1. Verify Proposition 4.3.

20

2. Look at the exercise set for §1, and verify (3) and (4) for a ∈ Q, a 6= 1, n ∈ N.

3. Here is another route to (4) of §1, i.e.,

(4.19)n∑

k=0

ak =an+1 − 1

a− 1, a 6= 1.

Denote the left side of (4.19) by Sn(a). Multiply by a and show that

aSn(a) = Sn(a) + an+1 − 1.

4. Given a ∈ Q, n ∈ N, define an as in Exercise 3 of §1. If a 6= 0, set a0 = 1 anda−n = (a−1)n, with a−1 defined as in (4.7). Show that, if a, b ∈ Q \ 0,

aj+k = ajak, ajk = (aj)k, (ab)j = ajbj , ∀ j, k ∈ Z.

5. Prove the following variant of Proposition 4.5.

Proposition 4.5A. Given ε ∈ Q, ε > 0, there exists n ∈ N such that

ε >1n

.

6. Work through the proof of the following.Assertion If x = m/n ∈ Q, then x2 6= 2.Hint. We can arrange that m and n have no common factors. Then

(m

n

)2

= 2 ⇒ m2 = 2n2 ⇒ m even (m = 2k)

⇒ 4k2 = 2n2

⇒ n2 = 2k2

⇒ n even.

Contradiction? (See Proposition 7.2 for a more general result.)

7. Given xj , yj ∈ Q, show that

x1 < x2, y1 ≤ y2 =⇒ x1 + y1 < x2 + y2.

Show that0 < x1 < x2, 0 < y1 ≤ y2 =⇒ x1y1 < x2y2.

21

5. Sequences

In this section, we discuss infinite sequences. For now, we deal with sequencesof rational numbers, but we will not explicitly state this restriction below. In fact,once the set of real numbers is constructed in §6, the results of this section will beseen to hold also for sequences of real numbers.

Definition. A sequence (aj) is said to converge to a limit a provided that, for anyn ∈ N, there exists K(n) such that

(5.1) j ≥ K(n) =⇒ |aj − a| < 1n

.

We write aj → a, or a = lim aj , or perhaps a = limj→∞ aj .

Here, we define the absolute value |x| of x by

(5.2)|x| = x if x ≥ 0,

−x if x < 0.

The absolute value function has various simple properties, such as |xy| = |x| ·|y|, which follow readily from the definition. One basic property is the triangleinequality:

(5.3) |x + y| ≤ |x|+ |y|.

In fact, if either x and y are both positive or they are both negative, one has equalityin (5.3). If x and y have opposite signs, then |x + y| ≤ max(|x|, |y|), which in turnis dominated by the right side of (5.3).

Proposition 5.1. If aj → a and bj → b, then

(5.4) aj + bj → a + b,

and

(5.5) ajbj → ab.

If furthermore, bj 6= 0 for all j and b 6= 0, then

(5.6) aj/bj → a/b.

Proof. To see (5.4), we have, by (5.3),

(5.7) |(aj + bj)− (a + b)| ≤ |aj − a|+ |bj − b|.

22

To get (5.5), we have

(5.8)|ajbj − ab| = |(ajbj − abj) + (abj − ab)|

≤ |bj | · |aj − a|+ |a| · |b− bj |.The hypotheses imply |bj | ≤ B, for some B, and hence the criterion for convergenceis readily verified. To get (5.6), we have

(5.9)∣∣∣aj

bj− a

b

∣∣∣ ≤ 1|b| · |bj |

|b| · |a− aj |+ |a| · |b− bj |.

The hypotheses imply 1/|bj | ≤ M for some M, so we also verify the criterion forconvergence in this case.

We next define the concept of a Cauchy sequence.

Definition. A sequence (aj) is a Cauchy sequence provided that, for any n ∈ N,there exists K(n) such that

(5.10) j, k ≥ K(n) =⇒ |aj − ak| ≤ 1n

.

It is clear that any convergent sequence is Cauchy. On the other hand, we have:

Proposition 5.2. Each Cauchy sequence is bounded.

Proof. Take n = 1 in the definition above. Thus, if (aj) is Cauchy, there is a Ksuch that j, k ≥ K ⇒ |aj − ak| ≤ 1. Hence, j ≥ K ⇒ |aj | ≤ |aK |+ 1, so, for all j,

|aj | ≤ M, M = max(|a1|, . . . , |aK−1|, |aK |+ 1

).

Now, the arguments proving Proposition 5.1 also establish:

Proposition 5.3. If (aj) and (bj) are Cauchy sequences, so are (aj + bj) and(ajbj). Furthermore, if, for all j, |bj | ≥ c for some c > 0, then (aj/bj) is Cauchy.

The following proposition is a bit deeper than the first three.

Proposition 5.4. If (aj) is bounded, i.e., |aj | ≤ M for all j, then it has a Cauchysubsequence.

Proof. We may as well assume M ∈ N. Now, either aj ∈ [0,M ] for infinitely manyj or aj ∈ [−M, 0] for infinitely many j. Let I1 be any one of these two intervalscontaining aj for infinitely many j, and pick j(1) such that aj(1) ∈ I1. Write I1 as theunion of two closed intervals, of equal length, sharing only the midpoint of I1. LetI2 be any one of them with the property that aj ∈ I2 for infinitely many j, and pickj(2) > j(1) such that aj(2) ∈ I2. Continue, picking Iν ⊂ Iν−1 ⊂ · · · ⊂ I1, of lengthM/2ν−1, containing aj for infinitely many j, and picking j(ν) > j(ν−1) > · · · > j(1)such that aj(ν) ∈ Iν . Setting bν = aj(ν), we see that (bν) is a Cauchy subsequenceof (aj), since, for all k ∈ N,

|bν+k − bν | ≤ M/2ν−1.

23

Proposition 5.5. Each bounded monotone sequence (aj) is Cauchy.

Proof. To say (aj) is monotone is to say that either (aj) is increasing, i.e., aj ≤ aj+1

for all j, or that (aj) is decreasing, i.e., aj ≥ aj+1 for all j. For the sake of argument,assume (aj) is increasing.

By Proposition 5.4, there is a subsequence (bν) = (aj(ν)) which is Cauchy. Thus,given n ∈ N, there exists K(n) such that

(5.11) µ, ν ≥ K(n) =⇒ |aj(ν) − aj(µ)| <1n

.

Now, if ν0 ≥ K(n) and k ≥ j ≥ j(ν0), pick ν1 such that j(ν1) ≥ k. Then

aj(ν0) ≤ aj ≤ ak ≤ aj(ν1),

so

(5.12) k ≥ j ≥ j(ν0) =⇒ |aj − ak| < 1n

.

We give a few simple but basic examples of convergent sequences.

Proposition 5.6. If |a| < 1, then aj → 0.

Proof. Set b = |a|; it suffices to show that bj → 0. Consider c = 1/b > 1, hencec = 1 + y, y > 0. We claim that

cj = (1 + y)j ≥ 1 + jy,

for all j ≥ 1. In fact, this clearly holds for j = 1, and if it holds for j = k, then

ck+1 ≥ (1 + y)(1 + ky) > 1 + (k + 1)y.

Hence, by induction, the estimate is established. Consequently,

bj <1jy

,

so the appropriate analogue of (5.1) holds, with K(n) = Kn, for any integer K >1/y.

Proposition 5.6 enables us to establish the following result on geometric series.

Proposition 5.7. If |x| < 1 and

aj = 1 + x + · · ·+ xj ,

thenaj → 1

1− x.

24

Proof. Note that xaj = x + x2 + · · ·+ xj+1, so (1− x)aj = 1− xj+1, i.e.,

aj =1− xj+1

1− x.

The conclusion follows from Proposition 5.6.

Note in particular that

(5.13) 0 < x < 1 =⇒ 1 + x + · · ·+ xj <1

1− x.

It is an important mathematical fact that not every Cauchy sequence of rationalnumbers has a rational number as limit. We give one example here. Consider thesequence

(5.14) aj =j∑

`=0

1`!

.

Then (aj) is increasing, and

an+j − an =n+j∑

`=n+1

1`!≤ 1

n!

( 1n + 1

+1

(n + 1)2+ · · ·+ 1

(n + 1)j

),

since (n + 1)(n + 2) · · · (n + j) ≥ (n + 1)j . Using (5.13), we have

(5.15) an+j − an ≤ 1(n + 1)!

11− 1

n+1

=1n!· 1n

.

Hence (aj) is Cauchy. Taking n = 2, we see that

(5.16) j > 2 =⇒ 212 < aj < 2 3

4 .

Proposition 5.8. The sequence (5.14) cannot converge to a rational number.

Proof. Assume aj → m/n with m,n ∈ N. By (5.16), we must have n > 2. Now,write

(5.17)m

n=

n∑

`=0

1`!

+ r, r = limj→∞

(an+j − an).

Multiplying both sides of (5.17) by n! gives

(5.18) m(n− 1)! = A + r · n!

25

where

(5.19) A =n∑

`=0

n!`!∈ N.

Thus the identity (5.17) forces r · n! ∈ N, while (5.15) implies

(5.20) 0 < r · n! ≤ 1/n.

This contradiction proves the proposition.

Exercises

1. Show thatlim

k→∞k

2k= 0,

and more generally for each n ∈ N,

limk→∞

kn

2k= 0.

2. Show that

limk→∞

2k

k!= 0,

and more generally for each n ∈ N,

limk→∞

2nk

k!= 0.

The following exercises discuss continued fractions. We assume

(5.21) aj ∈ Q, aj ≥ 1, j = 1, 2, 3, . . . ,

and set

(5.22) f1 = a1, f2 = a1 +1a2

, f3 = a1 +1

a2 + 1a3

, . . . .

Having fj , we obtain fj+1 by replacing aj by aj + 1/aj+1. In other words, with

(5.23) fj = ϕj(a1, . . . , aj),

given explicitly by (5.22) for j = 1, 2, 3, we have

(5.24) fj+1 = ϕj+1(a1, . . . , aj+1) = ϕj(a1, . . . , aj−1, aj + 1/aj+1).

26

3. Show thatf1 ≤ fj , ∀ j ≥ 2, and f2 ≥ fj , ∀ j ≥ 3.

Going further, show that

(5.25) f1 ≤ f3 ≤ f5 ≤ · · · ≤ f6 ≤ f4 ≤ f2.

4. If also aj+1 ∈ Q, aj+1 ≥ 1, show that

(5.26)ϕj+1(a1, . . . , aj , aj+1)− ϕj+1(a1, . . . , aj , aj+1)

= ϕj(a1, . . . , aj−1, bj)− ϕj(a1, . . . , aj−1, bj),

with

(5.27)bj = aj +

1aj+1

, bj = aj +1

aj+1,

bj − bj =1

aj+1− 1

aj+1=

aj+1 − aj+1

aj+1aj+1.

Note that bj , bj > 1. Iterating this, show that

(5.28) f2j − f2j+1 → 0, as j →∞.

Deduce that (fj) is a Cauchy sequence.

5. Suppose a sequence (aj) has the property that there exist

r < 1, K ∈ N

such thatj ≥ K =⇒

∣∣∣aj+1

aj

∣∣∣ ≤ r.

Show that aj → 0 as j →∞. How does this result apply to Exercises 1 and 2?

6. If (aj) satisfies the hypotheses of Exercise 5, show that there exists M < ∞ suchthat

k∑

j=1

|aj | ≤ M, ∀ k.

Remark. This yields the ratio test for infinite series.

27

6. The real numbers

We think of a real number as a quantity that can be specified by a process ofapproximation arbitrarily closely by rational numbers. Thus, we define an elementof R as an equivalence class of Cauchy sequences of rational numbers, where wedefine

(6.1) (aj) ∼ (bj) ⇐⇒ aj − bj → 0.

Proposition 6.1. This is an equivalence relation.

Proof. This is a straightforward consequence of Proposition 5.1. In particular, tosee that

(6.2) (aj) ∼ (bj), (bj) ∼ (cj) =⇒ (aj) ∼ (cj),

just use (5.4) of Proposition 5.1 to write

aj − bj → 0, bj − cj → 0 =⇒ aj − cj → 0.

We denote the equivalence class containing a Cauchy sequence (aj) by [(aj)]. Wethen define addition and multiplication on R to satisfy

(6.3)[(aj)] + [(bj)] = [(aj + bj)],

[(aj)] · [(bj)] = [(ajbj)].

Proposition 5.3 states that the sequences (aj + bj) and (ajbj) are Cauchy if (aj)and (bj) are. To conclude that the operations in (6.3) are well defined, we need:

Proposition 6.2. If Cauchy sequences of rational numbers are given which satisfy(aj) ∼ (a′j) and (bj) ∼ (b′j), then

(6.4) (aj + bj) ∼ (a′j + b′j),

and

(6.5) (ajbj) ∼ (a′jb′j).

The proof is a straightforward variant of the proof of parts (5.4)-(5.5) in Propo-sition 5.1, with due account taken of Proposition 5.2. For example, ajbj − a′jb

′j =

ajbj − ajb′j + ajb

′j − a′jb

′j , and there are uniform bounds |aj | ≤ A, |b′j | ≤ B, so

|ajbj − a′jb′j | ≤ |aj | · |bj − b′j |+ |aj − a′j | · |b′j |≤ A|bj − b′j |+ B|aj − a′j |.

28

There is a natural injection

(6.6) Q → R, a 7→ [(a, a, a, . . . )],

whose image we identify with Q. This map preserves addition and multiplication.If x = [(aj)], we define

(6.7) −x = [(−aj)].

For x 6= 0, we define x−1 as follows. First, to say x 6= 0 is to say there exists n ∈ Nsuch that |aj | ≥ 1/n for infinitely many j. Since (aj) is Cauchy, this implies thatthere exists K such that |aj | ≥ 1/2n for all j ≥ K. Now, if we set αj = ak+j , wehave (αj) ∼ (aj); we propose to set

(6.8) x−1 = [(α−1j )].

We claim that this is well defined. First, by Proposition 5.3, (α−1j ) is Cauchy.

Furthermore, if for such x we also have x = [(bj)], and we pick K so large that also|bj | ≥ 1/2n for all j ≥ K, and set βj = bK+j , we claim that

(6.9) (α−1j ) ∼ (β−1

j ).

Indeed, we have

(6.10) |α−1j − β−1

j | ≤ |βj − αj ||αj | · |βj | ≤ 4n2|βj − αj |,

so (6.9) holds.It is now a straightforward exercise to verify the basic algebraic properties of

addition and multiplication in R. We state the result.

Proposition 6.3. Given x, y, z ∈ R, all the algebraic properties stated in Proposi-tion 4.3 hold.

For example, if x = [(aj)] and y = [(bj)], the identity xy = yx is equivalentto (ajbj) ∼ (bjaj). In fact, the identity ajbj = bjaj for aj , bj ∈ Q, follows fromProposition 4.3. The rest of Proposition 6.3 is left to the reader.

As in (4.10)-(4.11), we define x− y = x + (−y) and, if y 6= 0, x/y = x · y−1.We now define an order relation on R. Take x ∈ R, x = [(aj)]. From the dis-

cussion above of x−1, we see that, if x 6= 0, then one and only one of the followingholds. Either, for some n,K ∈ N,

(6.11) j ≥ K =⇒ aj ≥ 12n

,

or, for some n,K ∈ N,

(6.12) j ≥ K =⇒ aj ≤ − 12n

.

29

If (aj) ∼ (bj) and (6.11) holds for aj , it also holds for bj (perhaps with different nand K), and ditto for (6.12). If (6.11) holds, we say x ∈ R+ (and we say x > 0),and if (6.12) holds we say x ∈ R− (and we say x < 0). Clearly x > 0 if and only if−x < 0. It is also clear that the map Q → R in (6.6) preserves the order relation.

Thus we have the disjoint union

(6.13) R = R+ ∪ 0 ∪ R−, R− = −R+.

Also, clearly

(6.14) x, y ∈ R+ =⇒ x + y, xy ∈ R+.

As in (4.15), we define

(6.15) x < y ⇐⇒ y − x ∈ R+.

The following results are straightforward.

Proposition 6.4. For elements of R, we have

(6.16) x1 < y1, x2 < y2 =⇒ x1 + x2 < y1 + y2,

(6.17) x < y ⇐⇒ −y < −x,

(6.18) 0 < x < y, a > 0 =⇒ 0 < ax < ay,

(6.19) 0 < x < y =⇒ 0 < y−1 < x−1.

Proof. The results (6.16) and (6.18) follow from (6.14); consider, for example, a(y−x). The result (6.17) follows from (6.13). To prove (6.19), first we see that x > 0implies x−1 > 0, as follows: if −x−1 > 0, the identity x · (−x−1) = −1 contradicts(6.14). As for the rest of (6.19), the hypotheses imply xy > 0, and multiplying bothsides of x < y by a = (xy)−1 gives the result, by (6.18).

As in (5.2), define |x| by

(6.20)|x| = x if x ≥ 0,

−x if x < 0.

It is straightforward to verify

(6.21) |xy| = |x| · |y|, |x + y| ≤ |x|+ |y|.

We now show that R has the Archimedean property.

30

Proposition 6.5. Given x ∈ R, there exists k ∈ Z such that

(6.22) k − 1 < x ≤ k.

Proof. It suffices to prove (6.22) assuming x ∈ R+. Otherwise, work with −x. Sayx = [(aj)] where (aj) is a Cauchy sequence of rational numbers. By Proposition5.2, there exists M ∈ Q such that |aj | ≤ M for all j. By Proposition 4.5, we haveM ≤ ` for some ` ∈ N. Hence the set S = ` ∈ N : ` ≥ x is nonempty. As in theproof of Proposition 4.5, taking k to be the smallest element of S gives (6.22).

Proposition 6.6. Given any real ε > 0, there exists n ∈ N such that ε > 1/n.

Proof. Using Proposition 6.5, pick n > 1/ε and apply (6.19). Alternatively, use thereasoning given above (6.8).

We are now ready to consider sequences of elements of R.

Definition. A sequence (xj) converges to x if and only if, for any n ∈ N, thereexists K(n) such that

(6.23) j ≥ K(n) =⇒ |xj − x| < 1n

.

In this case, we write xj → x, or x = lim xj .The sequence (xj) is Cauchy if and only if, for any n ∈ N, there exists K(n)

such that

(6.24) j, k ≥ K(n) =⇒ |xj − xk| < 1n

.

We note that it is typical to phrase the definition above in terms of picking anyreal ε > 0 and demanding that, e.g., |xj − x| < ε, for large j. The equivalence ofthe two definitions follows from Proposition 6.6.

As in Proposition 5.2, we have that every Cauchy sequence is bounded.It is clear that, if each xj ∈ Q, then the notion that (xj) is Cauchy given above

coincides with that in §5. If also x ∈ Q, the notion that xj → x also coincides withthat given in §5. Furthermore, if each aj ∈ Q, and x ∈ R, then

(6.25) aj → x ⇐⇒ x = [(aj)].

In fact, given x = [(aj)],

(6.26)(j, k ≥ K ⇒ |aj − ak| ≤ 1/n

)=⇒ (

j ≥ K ⇒ |aj − x| ≤ 1/n).

The proof of Proposition 5.1 extends to the present case, yielding:

31

Proposition 6.7. If xj → x and yj → y, then

(6.27) xj + yj → x + y,

and

(6.28) xjyj → xy.

If furthermore yj 6= 0 for all j and y 6= 0, then

(6.28) xj/yj → x/y.

So far, statements made about R have emphasized similarities of its propertieswith corresponding properties of Q. The crucial difference between these two setsof numbers is given by the following result, known as the completeness property.

Theorem 6.8. If (xj) is a Cauchy sequence of real numbers, then there existsx ∈ R such that xj → x.

Proof. Take xj = [(aj` : ` ∈ N)] with aj` ∈ Q. Using (6.26), take aj,`(j) = bj ∈ Qsuch that

(6.29) |xj − bj | ≤ 2−j .

Then (bj) is Cauchy, since |bj − bk| ≤ |xj − xk|+ 2−j + 2−k. Now, let

(6.30) x = [(bj)].

It follows that

(6.31) |xj − x| ≤ |xj − bj |+ |x− bj | ≤ 2−j + |x− bj |,and hence xj → x.

If we combine Theorem 6.8 with the argument behind Proposition 5.4, we obtainthe following important result, known as the Bolzano-Weierstrass Theorem.

Theorem 6.9. Each bounded sequence of real numbers has a convergent subse-quence.

Proof. If |xj | ≤ M, the proof of Proposition 5.4 applies without change to showthat (xj) has a Cauchy subsequence. By Theorem 6.8, that Cauchy subsequenceconverges.

Similarly, adding Theorem 6.8 to the argument behind Proposition 5.5 yields:

Proposition 6.10. Each bounded monotone sequence (xj) of real numbers con-verges.

A related property of R can be described in terms of the notion of the “supre-mum” of a set.

32

Definition. If S ⊂ R, one says that x ∈ R is an upper bound for S provided x ≥ sfor all s ∈ S, and one says

(6.32) x = sup S

provided x is an upper bound for S and further x ≤ x′ whenever x′ is an upperbound for S.

For some sets, such as S = Z, there is no x ∈ R satisfying (6.32). However, thereis the following result, known as the supremum property.

Proposition 6.11. If S is a nonempty subset of R that has an upper bound, thenthere is a real x = sup S.

Proof. We use an argument similar to the one in the proof of Proposition 5.3. Letx0 be an upper bound for S, pick s0 in S, and consider

I0 = [s0, x0] = y ∈ R : s0 ≤ y ≤ x0.If x0 = s0, then already x0 = sup S. Otherwise, I0 is an interval of nonzero length,L = x0 − s0. In that case, divide I0 into two equal intervals, having in commononly the midpoint; say I0 = I`

0 ∪ Ir0 , where Ir

0 lies to the right of I`0.

Let I1 = Ir0 if S ∩ Ir

0 6= ∅, and otherwise let I1 = I`0. Let x1 be the right endpoint

of I1, and pick s1 ∈ S ∩ I1. Note that x1 is also an upper bound for S.Continue, constructing

Iν ⊂ Iν−1 ⊂ · · · ⊂ I0,

where Iν has length 2−νL, such that the right endpoint xν of Iν satisfies

(6.33) xν ≥ s, ∀ s ∈ S,

and such that S ∩ Iν 6= ∅, so there exist sν ∈ S such that

(6.34) xν − sν ≤ 2−νL.

The sequence (xν) is bounded and monotone (decreasing) so, by Proposition 6.10,it converges; xν → x. By (6.33), we have x ≥ s for all s ∈ S, and by (6.34) we havex− sν ≤ 2−νL. Hence x satisfies (6.32).

We turn to infinite series∑∞

k=0 ak, with ak ∈ R. We say this series converges ifand only if the sequence of partial sums

(6.35) Sn =n∑

k=0

ak

converges:

(6.36)∞∑

k=0

ak = A ⇐⇒ Sn → A as n →∞.

The following is a useful condition guaranteeing convergence.

33

Proposition 6.12. The infinite series∑∞

k=0 ak converges provided

(6.37)∞∑

k=0

|ak| < ∞,

i.e., there exists B < ∞ such that∑n

k=0 |ak| ≤ B for all n.

Proof. The triangle inequality (the second part of (6.21)) gives, for ` ≥ 1,

(6.38)

|Sn+` − Sn| =∣∣∣

n+∑

k=n+1

ak

∣∣∣

≤n+∑

k=n+1

|ak|,

and we claim this tends to 0 as n →∞, uniformly in ` ≥ 1, provided (6.37) holds.In fact, if (6.37) holds, we can apply Proposition 6.10 to

(6.39) σn =n∑

k=0

|ak|, σn , σn ≤ B,

to deduce that there exists β ∈ R such that σn → β as n →∞. Then

(6.40) |Sn+` − Sn| ≤ σn+` − σn ≤ β − σn → 0 as n →∞.

Thus (6.37) ⇒ (Sn) is Cauchy. Convergence follows, by Theorem 6.8.

When (6.37) holds, we say the series∑∞

k=0 ak is absolutely convergent.The following result on alternating series gives another sufficient condition for

convergence.

Proposition 6.13. Assume ak > 0, ak 0. Then

(6.41)∞∑

k=0

(−1)kak

is convergent.

Proof. Denote the partial sums by Sn, n ≥ 0. We see that, for m ∈ N,

(6.42) S2m+1 ≤ S2m+3 ≤ S2m+2 ≤ S2m.

Iterating this, we have, as m →∞,

(6.43) S2m α, S2m+1 β, β ≤ α,

34

and

(6.44) S2m − S2m+1 = a2m+1,

hence α = β, and convergence is established.

Here is an example:

∞∑

k=0

(−1)k

k + 1= 1− 1

2+

13− 1

4+ · · · is convergent.

This series is not absolutely convergent (cf. Exercise 6 below). For an evaluation ofthis series, see exercises in §5 of Chapter 4.

Exercises

1. Verify Proposition 6.3.

2. If S ⊂ R, we say that x ∈ R is a lower bound for S provided x ≤ s for all s ∈ S,and we say

(6.45) x = inf S,

provided x is a lower bound for S and further x ≥ x′ whenever x′ is a lower boundfor S. Mirroring Proposition 6.11, show that if S ⊂ R is a nonempty set that hasa lower bound, then there is a real x = inf S.

3. Given a real number ξ ∈ (0, 1), show it has an infinite decimal expansion, i.e.,there exist bk ∈ 0, 1, . . . , 9 such that

(6.46) ξ =∞∑

k=1

bk · 10−k.

Hint. Start by breaking [0, 1] into ten subintervals of equal length, and picking oneto which ξ belongs.

4. Show that if 0 < x < 1,

∞∑

k=0

xk =1

1− x< ∞.

Hint. As in (4.19), we have

n∑

k=0

xk =1− xn+1

1− x, x 6= 1.

35

5. Assume ak > 0 and ak 0. Show that

(6.47)∞∑

k=1

ak < ∞⇐⇒∞∑

k=0

bk < ∞,

where

(6.48) bk = 2ka2k .

6. Deduce from Exercise 5 that the harmonic series 1 + 12 + 1

3 + 14 + · · · diverges,

i.e.,

(6.49)∞∑

k=1

1k

= ∞.

7. Deduce from Exercises 4–5 that

(6.50) p > 1 =⇒∞∑

k=1

1kp

< ∞.

For now, we take p ∈ N. We will see later that (6.50) is meaningful, and true, forp ∈ R, p > 1.

8. Given a, b ∈ R \ 0, k ∈ Z, define ak as in Exercise 4 of §4. Show that

aj+k = ajak, ajk = (aj)k, (ab)j = ajbj , ∀ j, k ∈ Z.

9. Given k ∈ N, show that, for xj ∈ R,

xj → x =⇒ xkj → xk.

Hint. Use Proposition 6.7.

10. Given xj , x, y ∈ R, show that

xj ≥ y ∀ j, xj → x =⇒ x ≥ y.

11. Given the alternating series∑

(−1)kak as in Proposition 6.13 (with ak 0),with sum S, show that, for each N ,

N∑

k=0

(−1)kak = S + rN , |rN | ≤ |aN+1|.

36

The following exercises deal with the sequence (fj) of partial fractions associatedto a sequence (aj) as in (5.21), via (5.22)–(5.24), leading to Exercises 3–4 of §5.

12. Deduce from (5.25) that there exist fo, fe ∈ R such that

f2k+1 fo, f2k fe, fo ≤ fe.

13. Deduce from (5.28) that fo = fe (= f , say), and hence

fj −→ f, as j →∞,

i.e., if (aj) satisfies (5.21),

ϕj(a1, . . . , aj) −→ f, as j →∞.

We denote the limit by ϕ(a1, . . . , aj , . . . ).

14. Show that ϕ(1, 1, . . . , 1, . . . ) = x solves x = 1 + 1/x, and hence

ϕ(1, 1, . . . , 1, . . . ) =1 +

√5

2.

Note. The existence of such x implies that 5 has a square root,√

5 ∈ R. SeeProposition 7.1 for a more general result.

37

7. Irrational numbers

There are real numbers that are not rational. One, called e, is given by the limitof the sequence (5.14); in standard notation,

(7.1) e =∞∑

`=0

1`!

Proposition 5.8 implies that e is not rational. One can approximate e to highaccuracy. In fact, as a consequence of (5.15), one has

(7.2) e−n∑

`=0

1`!≤ 1

n!· 1n

.

For example, one can verify that

(7.3) 120! > 6 · 10198,

and hence

(7.4) e−120∑

`=0

1`!

< 10−200.

In a fraction of a second, a personal computer with the right program can performa highly accurate approximation to such a sum, yielding

2.7182818284 5904523536 0287471352 6624977572 47093699959574966967 6277240766 3035354759 4571382178 52516642742746639193 2003059921 8174135966 2904357290 03342952605956307381 3232862794 3490763233 8298807531 · · ·

accurate to 190 places after the decimal point.A number in R \ Q is said to be irrational. We present some other common

examples of irrational numbers, such as√

2. To begin, one needs to show that√

2is a well defined real number. The following general result includes this fact.

Proposition 7.1. Given a ∈ R+, k ∈ N, there is a unique b ∈ R+ such thatbk = a.

Proof. Consider

(7.5) Sa,k = x ≥ 0 : xk ≤ a.

38

Then Sa,k is a nonempty bounded subset of R. Note that if y > 0 and yk > a theny is an upper bound for Sa,k. Take b = sup Sa,k. We claim that bk = a. In fact, ifbk < a, it follows from Exercise 9 of §6 that there exists b1 > b such that bk

1 < a,hence b1 ∈ Sa,k, so b < supSa,k. Similarly, if bk > a, there exists b0 < b such thatbk0 > a, hence b0 is an upper bound for Sa,k, so b > sup Sa,k.

We write

(7.6) b = a1/k.

Now for a list of some irrational numbers:

Proposition 7.2. Take a ∈ N, k ∈ N. If a1/k is not an integer, then a1/k isirrational.

Proof. Assume a1/k = m/n, with m,n ∈ N. Then

(7.7) mk = ank.

Using the Fundamental Theorem of Arithmetic, Theorem 3.1, write

(7.8) m = pµ11 · · · pµ`

` , n = pν11 · · · pν`

` , a = pα11 · · · pα`

` ,

with p1 < · · · < p` prime and µj , νj , αj ∈ N = N ∪ 0. The identity (7.7) implies

(7.9) pkµ11 · · · pkµ`

` = pα1+kν11 · · · pα`+kν`

` ,

and the uniqueness part of Theorem 3.1 then implies that kµj = αj+kνj , 1 ≤ j ≤ `,hence

(7.10) αj = kβj , βj ∈ N,

and hence

(7.11) a = bk, b = pβ11 · · · pβ`

` ∈ N.

Noting that 12 = 1, 22 = 4, 32 = 9, we have:

Corollary 7.3. The following numbers are irrational:√

2,√

3,√

5,√

6,√

7,√

8.

The real line is thick with both rational numbers and irrational numbers. By(6.25), given any x ∈ R, there exist aj ∈ Q such that aj → x. Also, given any x ∈ R,there exist irrational bj such that bj → x. To see this, just take aj ∈ Q, aj → x,

and set bj = aj + 2−j√

2.

39

In a sense that can be made precise, there are more irrational numbers thanrational numbers. Namely, Q is countable, while R is uncountable. See §8 for atreatment of this.

Perhaps the most intriguing irrational number is π. We recommend [Be] formaterial on this.

Exercises

1. Let ξ ∈ (0, 1) have a decimal expansion of the form (6.46), i.e.,

(7.12) ξ =∞∑

k=1

bk · 10−k, bk ∈ 0, 1, . . . , 9.

Show that ξ is rational if and only if (7.12) is eventually repeating, i.e., if and onlyif there exist N, m ∈ N such that

k ≥ N =⇒ bk+m = bk.

2. Show that ∞∑

k=1

10−k2is irrational.

3. Making use of Proposition 7.1, define ap for real a > 0, p = m/n ∈ Q. Showthat if also q ∈ Q,

apaq = ap+q.

Hint. You might start with am/n = (a1/n)m, given n ∈ N, m ∈ Z. Then you needto show that if k ∈ N,

(a1/nk)mk = (a1/n)m.

You can use the results of Exercise 8 in §6.

4. Show that, if a, b > 0 and p ∈ Q, then

(ab)p = apbp.

Hint. First show that (ab)1/n = a1/nb1/n.

5. Using Exercises 3 and 4, extend (6.50) to p ∈ Q, p > 1.Hint. If ak = k−p, then bk = 2ka2k = 2k(2k)−p = 2−(p−1)k = xk with x = 2−(p−1).

6. Show that√

2 +√

3 is irrational.Hint. Square it.

40

8. Cardinal numbers

We return to the natural numbers considered in §1 and make contact with thefact that these numbers are used to count objects in collections. Namely, let S besome set. If S is empty, we say 0 is the number of its elements. If S is not empty,pick an element out of S and count “1.” If there remain other elements of S, pickanother element and count “2.” Continue. If you pick a final element of S andcount “n,” then you say S has n elements. At least, that is a standard informaldescription of counting. We wish to restate this a little more formally, in the settingwhere we can apply the Peano axioms.

In order to do this, we consider the following subsets of N. Given n ∈ N, set

(8.1) In = j ∈ N : j ≤ n.While the following is quite obvious, it is worthwhile recording that it is a conse-quence of the Peano axioms and the material developed in §1.

Lemma 8.1. We have

(8.2) I1 = 1, In+1 = In ∪ n + 1.

Proof. Left to the reader.

Now we propose the following

Definition 8.1. A nonempty set S has n elements if and only if there exists abijective map ϕ : S → In.

A reasonable definition of counting should permit one to demonstrate that, if Shas n elements and it also has m elements, then m = n. The key to showing thisfrom the Peano postulates is the following.

Proposition 8.2. Assume m,n ∈ N. If there exists an injective map ϕ : Im → In,then m ≤ n.

Proof. Use induction on n. The case n = 1 is clear (by Lemma 8.1). Assume nowthat N ≥ 2 and that the result is true for n < N . Then let ϕ : Im → IN beinjective. Two cases arise: either there is an element j ∈ Im such that ϕ(j) = N ,or not. (Also, there is no loss of generality in assuming at this point that m ≥ 2.)

If there is such a j, define ψ : Im−1 → IN−1 by

ψ(`) = ϕ(`) for ` < j,

ϕ(` + 1) for j ≤ ` < m.

Then ψ is injective, so m− 1 ≤ N − 1, and hence m ≤ N .On the other hand, if there is no such j, then we already have an injective map

ϕ : Im → IN−1. The induction hypothesis implies m ≤ N−1, which in turn impliesm ≤ N .

41

Corollary 8.3. If there exists a bijective map ϕ : Im → In, then m = n.

Proof. We see that m ≤ n and n ≤ m, so Proposition 1.13 applies.

Corollary 8.4. If S is a set, m,n ∈ N, and there exist bijective maps ϕ : S →Im, ψ : S → In, then m = n.

Proof. Consider ψ ϕ−1.

Definition 8.2. If either S = ∅ or S has n elements for some n ∈ N, as inDefiniton 8.1, we say S is finite.

The next result implies that any subset of a finite set is finite.

Proposition 8.5. Assume n ∈ N. If S ⊂ In is nonempty, then there exists m ≤ nand a bijective map ϕ : S → Im.

Proof. Use induction on n. The case n = 1 is clear (by Lemma 8.1). Assume theresult is true for n < N . Then let S ⊂ IN . Two cases arise: either N ∈ S or N /∈ S.

If N ∈ S, consider S′ = S \N, so S = S′∪N and S′ ⊂ IN−1. The inductivehypothesis yields a bijective map ψ : S′ → Im (with m ≤ N − 1), and then weobtain ϕ : S′∪N → Im+1, equal to ψ on S′ and sending the element N to m+1.

If N /∈ S, then S ⊂ IN−1, and the inductive hypothesis directly yields the desiredbijective map.

Proposition 8.6. The set N is not finite.

Proof. If there were an n ∈ N and a bijective map ϕ : In → N, then, by restriction,there would be a bijective map ψ : S → In+1 for some subset S of In, hence bythe results above a bijective map ψ : Im → In+1 for some m ≤ n < n + 1. Thiscontradicts Corollary 8.3.

The next result says that, in a certain sense, N is a minimal set that is not finite.

Proposition 8.7. If S is not finite, then there exists an injective map Φ : N→ S.

Proof. We aim to show that there exists a family of injective maps ϕn : In → S,with the property that

(8.3) ϕn

∣∣Im

= ϕm, ∀m ≤ n.

We establish this by induction on n. For n = 1, just pick some element of S and callit ϕ1(1). Now assume this claim is true for all n < N . So we have ϕN−1 : IN−1 → Sinjective, but not surjective (since we assume S is not finite), and (8.3) holds forn ≤ N −1. Pick x ∈ S not in the range of ϕN−1. Then define ϕN : IN → S so that

(8.3A)ϕN (j) = ϕN−1(j), j ≤ N − 1,

ϕN (N) = x.

Having the family ϕn, we define Φ : N→ S by Φ(j) = ϕn(j) for any n ≥ j.

42

Two sets S and T are said to have the same cardinality if there exists a bijectivemap between them; we write Card(S) = Card(T ). If there exists an injective mapϕ : S → T , we write Card(S) ≤ Card(T ). The following result, known as theSchroeder-Bernstein theorem, implies that Card(S) = Card(T ) whenever one hasboth Card(S) ≤ Card(T ) and Card(T ) ≤ Card(S).

Theorem 8.8. Let S and T be sets. Suppose there exist injective maps ϕ : S → Tand ψ : T → S. Then there exists a bijective map Φ : S → T .

Proof. Let us say an element x ∈ T has a parent y ∈ S if ϕ(y) = x. Similarlythere is a notion of a parent of an element of S. Iterating this gives a sequence of“ancestors” of any element of S or T . For any element of S or T , there are threepossibilities:

a) The set of ancestors never terminates.b) The set of ancestors terminates at an element of S.c) The set of ancestors terminates at an element of T .

We denote by Sa, Ta the elements of S, T , respectively for which case a) holds.Similarly we have Sb, Tb and Sc, Tc. We have disjoint unions

S = Sa ∪ Sb ∪ Sc, T = Ta ∪ Tb ∪ Tc.

Now note thatϕ : Sa → Ta, ϕ : Sb → Tb, ψ : Tc → Sc

are all bijective. Thus we can set Φ equal to ϕ on Sa ∪ Sb and equal to ψ−1 on Sc,to get a desired bijection.

The terminology above suggests regarding Card(S) as an object (some sort ofnumber). Indeed, if S is finite we set Card(S) = n if S has n elements (as inDefinition 8.1). A set that is not finite is said to be infinite. We can also have anotion of cardinality of infinite sets. A standard notation for the cardinality of N is

(8.4) Card(N) = ℵ0.

Here are some other sets with the same cardinality:

Proposition 8.9. We have

(8.5) Card(Z) = Card(N× N) = Card(Q) = ℵ0.

Proof. We can define a bijection of N onto Z by ordering elements of Z as follows:

0, 1,−1, 2,−2, 3,−3, · · · .

We can define a bijection of N and N×N by ordering elements of N×N as follows:

(1, 1), (1, 2), (2, 1), (1, 3), (2, 2), (3, 1), · · · .

We leave it to the reader to produce a similar ordering of Q.

An infinite set that can be mapped bijectively onto N is called countably infinite.A set that is either finite or countably infinite is called countable. The followingresult is a natural extension of Proposition 8.5.

43

Proposition 8.10. If X is a countable set and S ⊂ X, then S is countable.

Proof. If X is finite, then Proposition 8.5 applies. Otherwise, we can assume X =N, and we are looking at S ⊂ N, so there is an injective map ϕ : S → N. If S isfinite, there is no problem. Otherwise, by Proposition 8.7, there is an injective mapψ : N → S, and then Theorem 8.8 implies the existence of a bijection between Sand N.

There are sets that are not countable; they are said to be uncountable.

Proposition 8.11. The set R of real numbers is uncountable.

Proof. We may as well show that (0, 1) = x ∈ R : 0 < x < 1 is uncountable. Ifit were countable, there would be a bijective map ϕ : N→ (0, 1). Expand the realnumber ϕ(j) in its infinite decimal expansion:

(8.6) ϕ(j) =∞∑

k=1

ajk · 10−k, ajk ∈ 0, 1, . . . 9.

Now set

(8.7)bk = 2 if akk 6= 2,

3 if akk = 2,

and consider

(8.8) ξ =∞∑

k=1

bk · 10−k, ξ ∈ (0, 1).

It is seen that ξ is not equal to ϕ(j) for any j ∈ N, contradicting the hypothesisthat ϕ : N→ (0, 1) is onto.

A common notation for the cardinality of R is

(8.9) Card(R) = c.

We leave it as an exercise to the reader to show that

(8.10) Card(R× R) = c.

Further development of the theory of cardinal numbers requires a formalizationof the notions of set theory. In these notes we have used set theoretical notionsrather informally. Our use of such notions has gotten somewhat heavier in thislast section. In particular, in the proof of Proposition 8.7, the innocent looking useof the phrase “pick x ∈ S . . . ” actually assumes a weak version of the Axiom ofChoice. For an introduction to the axiomatic treatment of set theory we refer to[Dev].

44

Exercises

1. What is the cardinality of the set P of prime numbers?

2. Let S be a nonempty set and let T be the set of all subsets of S. Adapt theproof of Proposition 8.11 to show that

Card(S) < Card(T ),

i.e., there is not a surjective map ϕ : S → T .Hint. There is a natural bijection of T and T , the set of functions f : S → 0, 1,via f ↔ x ∈ S : f(x) = 1. Given ϕ : S → T , describe a function g : S → 0, 1,not in the range of ϕ, taking a cue from the proof of Proposition 8.11.

3. Finish the proof of Proposition 8.9.

4. Use the map f(x) = x/(1 + x) to prove that

Card(R+) = Card((0, 1)).

5. Find a one-to-one map of R onto R+ and conclude that Card(R) = Card((0, 1)).

6. Use an interlacing of infinite decimal expansions to prove that

Card((0, 1)× (0, 1)) = Card((0, 1)).

7. Prove (8.10).

8. Let m ∈ Z, n ∈ N, and consider

Sm,n = k ∈ Z : m + 1 ≤ k ≤ m + n.Show that

Card Sm,n = n.

Hint. Produce a bijective map In → Sm,n.

9. Let S and T be sets. Assume

Card S = m, Card T = n, S ∩ T = ∅,with m,n ∈ N. Show that

CardS ∪ T = m + n.

Hint. Produce bijective maps S → Im and T → Sm,n, leading to a bijectionS ∪ T → Im+n.

45

9. Metric properties of R

We discuss a number of notions and results related to convergence in R. Recallthat a sequence of points (pj) in R converges to a limit p ∈ R (we write pj → p) ifand only if for every ε > 0 there exists N such that

(9.1) j ≥ N =⇒ |pj − p| < ε.

A set S ⊂ R is said to be closed if and only if

(9.2) pj ∈ S, pj → p =⇒ p ∈ S.

The complement R \ S of a closed set S is open. Alternatively, Ω ⊂ R is open ifand only if, given q ∈ Ω, there exists ε > 0 such that Bε(q) ⊂ Ω, where

(9.3) Bε(q) = p ∈ R : |p− q| < ε,

so q cannot be a limit of a sequence of points in R \ Ω.We define the closure S of a set S ⊂ R to consist of all points p ∈ R such that

Bε(p)∩S 6= ∅ for all ε > 0. Equivalently, p ∈ S if and only if there exists an infinitesequence (pj) of points in S such that pj → p.

An important property of R is completeness, which we recall is defined as follows.A sequence (pj) of points in R is called a Cauchy sequence if and only if

(9.4) |pj − pk| −→ 0, as j, k →∞.

It is easy to see that if pj → p for some p ∈ R, then (9.4) holds. The completenessproperty is the converse, given in Theorem 6.8, which we recall here.

Theorem 9.1. If (pj) is a Cauchy sequence in R, then it has a limit.

Completeness provides a path to the following key notion of compactness. Anonempty set K ⊂ R is said to be compact if and only if the following propertyholds.

(9.5)Each infinite sequence (pj) in K has a subsequencethat converges to a point in K.

It is clear that if K is compact, then it must be closed. It must also be bounded, i.e.,there exists R < ∞ such that K ⊂ BR(0). Indeed, if K is not bounded, there existpj ∈ K such that |pj+1| ≥ |pj |+ 1. In such a case, |pj − pk| ≥ 1 whenever j 6= k, so(pj) cannot have a convergent subsequence. The following converse statement is akey result.

46

Theorem 9.2. If a nonempty K ⊂ R is closed and bounded, then it is compact.

Clearly every nonempty closed subset of a compact set is compact, so Theorem9.2 is a consequence of:

Proposition 9.3. Each closed bounded interval I = [a, b] ⊂ R is compact.

Proof. This is a direct consequence of Theorem 6.9, the Bolzano-Weierstrass theo-rem.

Let K ⊂ R be compact. Since K is bounded from above and from below, wehave well defined real numbers

(9.6) b = sup K, a = inf K,

the first by Proposition 6.11, and the second by a similar argument (cf. Exercise 2of §6). Since a and b are limits of elements of K, we have a, b ∈ K. We use thenotation

(9.7) b = max K, a = min K.

We next discuss continuity. If S ⊂ R, a function

(9.8) f : S −→ Ris said to be continuous at p ∈ S provided

(9.9) pj ∈ S, pj → p =⇒ f(pj) → f(p).

If f is continuous at each p ∈ S, we say f is continuous on S.The following two results give important connections between continuity and

compactness.

Proposition 9.4. If K ⊂ R is compact and f : K → R is continuous, then f(K)is compact.

Proof. If (qk) is an infinite sequence of points in f(K), pick pk ∈ K such thatf(pk) = qk. If K is compact, we have a subsequence pkν → p in K, and thenqkν → f(p) in R.

This leads to the second connection.

Proposition 9.5. If K ⊂ R is compact and f : K → R is continuous, then thereexists p ∈ K such that

(9.10) f(p) = maxx∈K

f(x),

and there exists q ∈ K such that

(9.11) f(q) = minx∈K

f(x).

Proof. Since f(K) is compact, we have well defined numbers

(9.12) b = max f(K), a = min f(K), a, b ∈ f(K).

So take p, q ∈ K such that f(p) = b and f(q) = a.

The next result is called the intermediate value theorem.

47

Proposition 9.6. Take a, b, c ∈ R, a < b. Let f : [a, b] → R be continuous.Assume

(9.13) f(a) < c < f(b).

Then there exists x ∈ (a, b) such that f(x) = c.

Proof. Let

(9.14) S = y ∈ [a, b] : f(y) ≤ c.Then a ∈ S, so S is a nonempty, closed (hence compact) subset of [a, b]. Note hatb /∈ S. Take

(9.15) x = maxS.

Then a < x < b and f(x) ≤ c. If f(x) < c, then there exists ε > 0 such thata < x − ε < x + ε < b and f(y) < c for x − ε < y < x + ε. Thus x + ε ∈ S,contradicting (9.15).

Returning to the issue of compactness, we establish some further properties ofcompact sets K ⊂ R, leading to the important result, Proposition 9.10 below.

Proposition 9.7. Let K ⊂ R be compact. Assume X1 ⊃ X2 ⊃ X3 ⊃ · · · form adecreasing sequence of closed subsets of K. If each Xm 6= ∅, then ∩mXm 6= ∅.Proof. Pick xm ∈ Xm. If K is compact, (xm) has a convergent subsequence,xmk

→ y. Since xmk: k ≥ ` ⊂ Xm`

, which is closed, we have y ∈ ∩mXm.

Corolary 9.8. Let K ⊂ R be compact. Assume U1 ⊂ U2 ⊂ U3 ⊂ · · · form anincreasing sequence of open sets in R. If ∪mUm ⊃ K, then UM ⊃ K for some M .

Proof. Consider Xm = K \ Um.

Before getting to Proposition 9.10, we bring in the following. Let Q denote theset of rational numbers. The set Q ⊂ R has the following “denseness” property:given p ∈ R and ε > 0, there exists q ∈ Q such that |p− q| < ε. Let

(9.16) R = Brj (qj) : qj ∈ Q, rj ∈ Q ∩ (0,∞).Note that Q is countable, i.e., it can be put in one-to-one correspondence with N.Hence R is a countable collection of balls. The following lemma is left as an exercisefor the reader.

Lemma 9.9. Let Ω ⊂ R be a nonempty open set. Then

(9.17) Ω =⋃B : B ∈ R, B ⊂ Ω.

To state the next result, we say that a collection Uα : α ∈ A covers K ifK ⊂ ∪α∈AUα. If each Uα ⊂ R is open, it is called an open cover of K. If B ⊂ Aand K ⊂ ∪β∈BUβ , we say Uβ : β ∈ B is a subcover. This result is called theHeine-Borel theorem.

48

Proposition 9.10. If K ⊂ R is compact, then it has the following property.

(9.18) Every open cover Uα : α ∈ A of K has a finite subcover.

Proof. By Lemma 9.9, it suffices to prove the following.

(9.19)Every countable cover Bj : j ∈ N of K by open intervals

has a finite subcover.

For this, we set

(9.20) Um = B1 ∪ · · · ∪Bm

and apply Corollary 9.8.

Exercises

1. Consider a polynomial p(x) = xn + an−1xn−1 + · · · + a1x + a0. Assume each

aj ∈ R and n is odd. Use the intermediate value theorem to show that p(x) = 0 forsome x ∈ R.

We describe the construction of a Cantor set. Take a closed, bounded interval[a, b] = C0. Let C1 be obtained from C0 by deleting the open middle third interval,of length (b − a)/3. At the jth stage, Cj is a disjoint union of 2j closed intervals,each of length 3−j(b − a). Then Cj+1 is obtained from Cj by deleting the openmiddle third of each of these 2j intervals. We have C0 ⊃ C1 ⊃ · · · ⊃ Cj ⊃ · · · , eacha closed subset of [a, b].

2. Show that

(9.21) C =⋂

j≥0

Cj

is nonempty, and compact. This is the Cantor set.

3. Suppose C is formed as above, with [a, b] = [0, 1]. Show that points in C areprecisely those of the form

(9.22) ξ =∞∑

j=0

bj 3−j , bj ∈ 0, 2.

4. If p, q ∈ C (and p < q), show that the interval [p, q] must contain points not inC. One says C is totally disconnected.

49

5. If p ∈ C, ε > 0, show that (p − ε, p + ε) contains infinitely many points in C.Given that C is closed, one says C is perfect.

6. Show that Card(C) = Card(R).Hint. With ξ as in (9.22) show that

ξ 7→ η =∞∑

j=0

(bj

2

)2−j

maps C onto [0, 1].

7. Show that Proposition 9.6 implies Proposition 7.1.

50

10. Complex numbers

A complex number is a number of the form

(10.1) z = x + iy, x, y ∈ R,

where the new object i has the property

(10.2) i2 = −1.

We denote the set of complex numbers by C. We have R → C, identifying x ∈ Rwith x + i0 ∈ C.

We define addition and multiplication in C as follows. Suppose w = a + ib,a, b ∈ R. We set

(10.3)z + w = (x + a) + i(y + b),

zw = (xa− yb) + i(xb + ya).

It is routine to verify various commutative, associative, and distributive laws, par-allel to those in Proposition 4.3. If z 6= 0, i.e., either x 6= 0 or y 6= 0, we canset

(10.4) z−1 =1z

=x

x2 + y2− i

y

x2 + y2,

and verify that zz−1 = 1.For some more notation, for z ∈ C of the form (10.1), we set

(10.5) z = x− iy, Re z = x, Im z = y.

We say z is the complex conjugate of z, Re z is the real part of z, and Im z is theimaginary part of z.

We next discuss the concept of the magnitude (or absolute value) of an elementz ∈ C. If z has the form (10.1), we take a cue from the Pythagorean theorem,giving the Euclidean distance from z to 0, and set

(10.6) |z| =√

x2 + y2.

Note that

(10.7) |z|2 = z z.

51

With this notation, (10.4) takes the compact (and clear) form

(10.8) z−1 =z

|z|2 .

We have

(10.9) |zw| = |z| · |w|,for z, w ∈ C, as a consequence of the identity (readily verified from the definition(10.5))

(10.10) zw = z · w.

In fact, |zw|2 = (zw)(zw) = z w z w = zzww = |z|2|w|2. This extends the first partof (6.21) from R to C. The extension of the second part also holds, but it requiresa little more work. The following is the triangle inequality in C.

Proposition 10.1. Given z, w ∈ C,

(10.11) |z + w| ≤ |z|+ |w|.

Proof. We compare the squares of each side of (10.11). First,

(10.12)

|z + w|2 = (z + w)(z + w)

= |z|2 + |w|2 + wz + zw

= |z|2 + |w|2 + 2 Re zw.

Now, for any ζ ∈ C, Re ζ ≤ |ζ|, so Re zw ≤ |zw| = |z| · |w|, so (10.12) is

(10.13) ≤ |z|2 + |w|2 + 2|z| · |w| = (|z|+ |w|)2,and we have (10.11).

We now discuss matters related to convergence in C. Parallel to the real case,we say a sequence (zj) in C converges to a limit z ∈ C (and write zj → z) if andonly if for each ε > 0 there exists N such that

(10.14) j ≥ N =⇒ |zj − z| < ε.

Equivalently,

(10.13) zj → z ⇐⇒ |zj − z| → 0.

It is easily seen that

(10.16) zj → z ⇐⇒ Re zj → Re z and Im zj → Im z.

The set C also has the completeness property, given as follows. A sequence (zj)in C is said to be a Cauchy sequence if and only if

(10.17) |zj − zk| → 0, as j, k →∞.

It is easy to see (using the triangle inequality) that if zj → z for some z ∈ C, then(10.17) holds. Here is the converse:

52

Proposition 10.2. If (zj) is a Cauchy sequence in C, then it has a limit.

Proof. If (zj) is Cauchy in C, then (Re zj) and (Im zj) are Cauchy in R, so, byTheorem 6.8, they have limits.

We turn to infinite series∑∞

k=0 ak, with ak ∈ C. We say this converges if andonly if the sequence of partial sums

(10.18) Sn =n∑

k=0

ak

converges:

(10.19)∞∑

k=0

ak = A ⇐⇒ Sn → A as n →∞.

The following is a useful condition guaranteeing convergence. Compare Proposition6.12.

Proposition 10.3. The infinite series∑∞

k=0 ak converges provided

(10.20)∞∑

k=0

|ak| < ∞,

i.e., there exists B < ∞ such that∑n

k=0 |ak| ≤ B for all n.

Proof. The triangle inequality gives, for ` ≥ 1,

(10.21)

|Sn+` − Sn| =∣∣∣

n+∑

k=n+1

ak

∣∣∣

≤n+∑

k=n+1

|ak|,

which tends to 0 as n →∞, uniformly in ` ≥ 1, provided (10.20) holds (cf. (6.39)–(6.40)). Hence (10.20) ⇒ (Sn) is Cauchy. Convergence then follows, by Proposition10.2.

As in the real case, if (10.20) holds, we say the infinite series∑∞

k=0 ak is absolutelyconvergent.

An example to which Proposition 10.3 applies is the following power series, givingthe exponential function ez:

(10.22) ez =∞∑

k=0

zk

k!, z ∈ C.

53

We turn to a discussion of polar coordinates on C. Given a nonzero z ∈ C, wecan write

(10.23) z = rω, r = |z|, ω =z

|z| .

Then ω has unit distance from 0. If the ray from 0 to ω makes an angle θ with thepositive real axis, we have

(10.24) Re ω = cos θ, Im ω = sin θ,

by definition of the trigonometric functions cos and sin. Hence

(10.25) z = r cis θ,

where

(10.26) cis θ = cos θ + i sin θ.

If also

(10.27) w = ρ cis ϕ, ρ = |w|,

then

(10.28) zw = rρ cis(θ + ϕ),

as a consequence of the identity

(10.29) cis(θ + ϕ) = (cis θ)(cis ϕ),

which in turn is equivalent to the pair of trigonometric identities

(10.30)cos(θ + ϕ) = cos θ cosϕ− sin θ sin ϕ,

sin(θ + ϕ) = cos θ sinϕ + sin θ cos ϕ.

There is another way to write (10.25), using the classical Euler identity

(10.31) eiθ = cos θ + i sin θ.

Then (10.25) becomes

(10.32) z = r eiθ.

The identity (10.29) is equivalent to

(10.33) ei(θ+ϕ) = eiθeiϕ.

54

For a self-contained derivation of (10.31) (and also of (10.30) and (10.33)), werecommend [T], Chapter 4, especially §5.

We next define closed and open subsets of C, and discuss the notion of compact-ness. A set S ⊂ C is said to be closed if and only if

(10.34) zj ∈ S, zj → z =⇒ z ∈ S.

The complement C \ S of a closed set S is open. Alternatively, Ω ⊂ C is open ifand only if, given q ∈ Ω, there exists ε > 0 such that Bε(q) ⊂ Ω, where

(10.35) Bε(q) = z ∈ C : |z − q| < ε,

so q cannot be a limit of a sequence of points in C \ Ω. We define the closure S ofa set S ⊂ C to consist of all points p ∈ C such that Bε(p) ∩ S 6= ∅ for all ε > 0.Equivalently, p ∈ S if and only if there exists an infinite sequence (pj) of points inS such that pj → p.

Parallel to (9.5), we say a nonempty set K ⊂ C is compact if and only if thefollowing property holds.

(10.36)Each infinite sequence (pj) in K has a subsequencethat converges to a point in K.

As in §9, if K ⊂ C is compact, it must be closed and bounded. Parallel to Theorem9.2, we have the converse.

Proposition 10.4. If a nonempty K ⊂ C is closed and bounded, then it is compact.

Proof. Let (zj) be a sequence in K. Then (Re zj) and (Im zj) are bounded, soTheorem 6.9 implies the existence of a subsequence such that Re zjν and Im zjν

converge. Hence the subsequence (zjν ) converges in C. Since K is closed, the limitmust belong to K.

If S ⊂ C, a function

(10.37) f : S −→ C

is said to be continuous at p ∈ S provided

(10.38) pj ∈ S, pj → p =⇒ f(pj) → f(p).

If f is continuous at each p ∈ S, we say f is continuous on S. The following resulthas the same proof as Proposition 9.4.

Proposition 10.5. If K ⊂ C is compact and f : K → C is continuous, then f(K)is compact.

Then the following variant of Proposition 9.5 is straightforward.

55

Proposition 10.6. If K ⊂ C is compact and f : K → C is continuous, then thereexists p ∈ K such that

(10.39) |f(p)| = maxz∈K

|f(z)|,

and there exists q ∈ K such that

(10.40) |f(q)| = minz∈K

|f(z)|.

There are also straightforward extensions to K ⊂ C of Propositions 9.7–9.10.We omit the details.

Exercises

1. Use (10.25)–(10.28) in conjunction with Proposition 7.1 to prove the following:

Given a ∈ C, a 6= 0, n ∈ N, there exist z1, . . . , zn ∈ Csuch that zn

j = a.

2. Compute (12

+√

32

i)3

,

and verify that

(10.41) cosπ

3=

12, sin

π

3=√

32

.

3. Find z1, . . . , zn such that

(10.42) znj = 1,

explicitly in the form a + ib (not simply as cis(2πj/n)), in case

(10.43) n = 3, 4, 6, 8.

Hint. Use (10.41), and also the fact that the equation u2j = i has solutions

(10.44) u1 =1√2

+i√2, u2 = −u1.

4. Take the following path to finding the 5 solutions to

(10.45) z5j = 1.

56

One solution is z1 = 1. Since z5− 1 = (z− 1)(z4 + z3 + z2 + z + 1), we need to find4 solutions to z4 + z3 + z2 + z + 1 = 0. Write this as

(10.46) z2 + z + 1 +1z

+1z2

= 0,

which, for

(10.47) w = z +1z,

becomes

(10.48) w2 + w − 1 = 0.

Use the quadratic formula to find 2 solutions to (10.48). Then solve (10.47), i.e.,z2 − wz + 1 = 0, for z.

5. Take the following path to explicitly finding the real and imaginary parts of asolution to

z2 = a + ib.

Namely, with x = Re z, y = Im z, we have

x2 − y2 = a, 2xy = b,

and alsox2 + y2 = ρ =

√a2 + b2,

hence

x =

√ρ + a

2, y =

b

2x,

as long as a + ib 6= −|a|.

6. Taking a cue from Exercise 4 of §6, show that

(10.49)1

1− z=

∞∑

k=0

zk, for z ∈ C, |z| < 1.

7. Show that1

1− z2=

∞∑

k=0

z2k, for z ∈ C, |z| < 1.

8. Produce a power series series expansion in z, valid for |z| < 1, for

11 + z2

.

57

References

[BS] R. Bartle and D. Sherbert, Introduction to Real Analysis, J. Wiley, Newyork, 1992.

[Be] P. Beckmann, A History of π, St. Martin’s Press, New York, 1971.[Dev] K. Devlin, The Joy of Sets: Fundamentals of Contemporary Set Theory,

Springer-Verlag, New York, 1993.[T] M. Taylor, Introduction to Analysis in One Variable, Preprint, 2014.