Math 140A - Sketch Notes

Neil Donaldson

Fall 2019

1 Introduction

The primary purpose of 140A/B is to organize the logical background of calculus. Calculus is excep- tionally useful, yet at its heart lies an ancient philosophical debate dating at least to Zeno’s paradoxes of the 5thC BC: how do we calculate with infinitesimally small quantities, and is it even legitimate to do so? Definite integration, for instance, requires the summation of the areas of vanishingly small rectangles. Is it legitimate to use such a method to ‘compute’ the area under a curve? Long after its pioneers1 had applied calculus to many important problems in the sciences, philosophers were ar- guing over its validity, and mathematicians were struggling to put calculus on a firm logical footing. Modern descriptions of calculus rely heavily on the notion of limit, though it took mathematicians over a century to realize the importance of this concept and how to formulate it clearly. The ‘modern’ e-δ definition that we shall explore was developed in the early 1800’s, courtesy of Bolzano, Cauchy and Weierstrass (amongst others), all of whom feature prominently in 140A/B. The climax of 140B is a thorough description of the Riemann integral, which did not appear until the mid 1800’s. Elemen- tary Analysis is therefore a history of one of the great mathematical triumphs of the 16–1800’s.

Most of the work on calculus will have to wait until 140B. In this course we primarily consider limits and continuity. We start, however, with something even more basic: the structure of the real numbers, without which the smooth, continuous changes of calculus cannot be described. While you will certainly be familiar with many examples of real numbers, they are, in general, significantly harder to describe than the integers or the rationals. It is perhaps surprising, though it is certainly a fact you’ve encountered before, that the set of irrational numbers has greater cardinality than the set of rationals.2 However, beyond specific examples such as π, e and simple radicals like √2, it is rather difficult to exhibit explicit examples of irrational numbers. The reason is simple: the integers are intuitive and you’ve been using them since infancy. The natural way to construct more numbers is to start with integers and build from there. . .

1.1 The Set N of Natural Numbers

Here are one version of the standard axioms of the natural numbers, known as Peano’s Axioms. While reading them, think of the challenge mathematicians faced in getting here: what is the essential nature of the natural numbers, expressible without a prior understanding of number, and what distinguishes the natural numbers from other sets?

1In particular Newton and Leibniz in the late 1600’s. 2Q is countable while R (and consequently) R Q is uncountable. \ Axioms 1.1 (Peano). The natural numbers are a set N satisfying the following properties:

1. (Non-emptiness) N is non-empty.

2. (Successor function) There exists a function f : N N. This is usually denoted ‘+1’ so that we may → write,

n N, n + 1 N ∀ ∈ ∈ 3. (Initial element) f is not surjective. Otherwise said, there exists an element 1 range( f ) which is not 6∈ the successor of any element.3)

4. (Unique predecessor/order) f is injective. If m and n have the same successor, then m = n.

5. (Induction) Suppose that A N has the following properties: 1 A and n A = n + 1 A. ⊆ ∈ ∈ ⇒ ∈ Then A = N.

Axioms 1–4 should be obvious and can be summarized by saying that there exists a set N with an injective, non-surjective function f : N N. Axiom 5 is trickier, though it should remind you of → proof by induction: it is essentially the same as the principle of mathematical induction and the notion of well-ordering. We won’t dwell on it, but you should review induction yourself. Here is an example of a simple induction argument written in the language of axiom 5.

Example Prove that 7n 4n is divisible by 3 for all n N. − ∈ Let A be the set of natural numbers for which 3 7n 4n. It is required to prove that A = N. | − • If n = 1, then 71 41 = 3, so 1 A. − ∈ • Suppose that n A. Then 7n 4n = 3λ for some λ N. Hence ∈ − ∈ 7n+1 4n+1 = 7 7n 4n+1 = 7(3λ + 4n) 4n+1 = 3 7λ + (7 4) 4n − · − − · − · which is divisible by 3. Hence n + 1 A. ∈ Appealing to axiom 5, we see that A = N, hence result. Note that the two bulleted arguments are what you likely called the base case and the induction step in a previous course.

What is different about the integers? It should be clear that the integers satisfy axioms 1, 2 and 4, but not 3 and 5. In particular:

• 1 has a predecessor (namely zero).

• A = N is a proper subset of Z satisfying axiom 5, but A = Z. 6 We can provide an explicit construction of the integers from the natural numbers however: simply extend the function f so that 0 is the unique predecessor of 1, 1 the unique predecessor of 2, etc. − 3It is purely a convention to denote the ﬁrst natural number by 1: we could use 0, x, α, or any symbol you wish! In particular, you should not assume any additional properties of the element 1 beyond those described in the axioms.

2 1.2 The Set Q of Rational Numbers There are several ways to deﬁne the rational numbers from the integers. For example, we could consider the set of relatively prime ordered pairs Q = (p, q) : p Z, q N, gcd(p, q) = 1 Z N { ∈ ∈ } ⊆ × Writing p instead of (p, q) and adopting the convention that λp = p for any non-zero λ Z makes q λq q ∈ things more familiar. It is easy to deﬁne the usual operations (+, , etc.) consistently with those for · the integers.

An alternative, historical, approach involves equations: new sets of numbers can be ‘invented’ for the purpose of solving equations.4 The simplest situation involves considering linear equations qx p = 0 − where p, q Z and q = 0. Each such equation essentially describes a rational number! For example ∈ 6 27 13x + 27 = 0 ! x = −13 Extending this process further, we might consider higher degree polynomials. Deﬁnition 1.2. A number x is algebraic if it satisﬁes an equation of the form n n 1 anx + an 1x − + + a1x + a0 = 0 ( ) − ··· ∗ for some integers a0,..., an. (You should be alarmed by this! We have given up on constructing new numbers and instead are simply describing their properties. No matter, a construction of the real numbers will come later.)

Examples 1. √2 is algebraic, since it satisﬁes the equation x2 2 = 0. − p 2. 5 √3 + 7 is also algebraic. This is slightly harder to check: clearly x5 7 = √3 = (x5 7)2 = 3 = x10 14x5 + 46 = 0 − ⇒ − ⇒ − It is far from clear, though is it true, that there exist non-algebraic (or transcendental numbers): e and π are classic examples. These satisfy no polynomial equation with integer coefﬁcients, though demon- strating this is somewhat tricky and beyond this course.

Deciding on the rationality or irrationality of an algebraic number is often an easy application of the next result. Theorem 1.3 (Rational Roots). Suppose that a ,..., a Z and that r Q satisﬁes ( ). If r = p in lowest 0 n ∈ ∈ ∗ q terms, then p a and q a . | 0 | n Proof. Simply calculate n p p n n 1 n 1 n an + + a1 + a0 = 0 = an p + an 1 p − q + + a1 pq − + a0q = 0 q ··· q ⇒ − ··· All terms except the last contain a factor of p, whence p a qn. Since gcd(p, q) = 1 it follows that | 0 p a . The result for q is almost identical. | 0 4Take a class on abstract algebra/ring theory for more of this approach.

3 Examples

1. √2 is irrational. Clearly √2 satisﬁes x2 2 = 0. If √2 = p were rational in lowest terms, then − q the rational roots theorem says that

p 2 and q 1 = √2 1, 2 | | ⇒ ∈ {± ± } This is clearly a contradiction since none of the numbers 1, 2 satisfy x2 2 = 0. ± ± − You should compare this argument to the standard proof of the irrationality of √2 as seen in a previous course. Note how easy it is to extend this argument to √3, √29, √3 2, √5 8 etc.

2. (√3 1)1/3 is irrational. It satisﬁes (x3 + 1)2 = 3, from which − x6 + 2x3 2 = 0 − By the Theorem, if x = p we’d necessarily have p 2 and q 1, whence x = 1, 2, none of q | | ± ± which satisﬁes (x3 + 1)2 = 3.

1/2 3. 4+√3 is irrational. It satisﬁes 5x2 4 = √3, from which 5 − 25x4 40x2 + 13 = 0 − whence, if x = p we’d have p 13 and q 25. There are twelve possibilities in all: q | | 1 13 1 13 x = 1, 13, , , , ± ± ±5 ± 5 ±25 ±25 It is tedious to check all cases, but none satisfy the required polynomial. In fact it is much easier with this example to bypass the Theorem entirely: note that if x Q ∈ then √3 = 5x2 4 Q immediately gives the desired irrationality! − ∈ The rational roots theorem can also be use to help ﬁnd rational solutions to high-degree polynomials and aid in factorization. For example, suppose you want to solve the equation

3x3 + x2 + x 2 = 0 − You might try guessing, but this likely won’t get you far. To search for a rational solution, assume that x = p solves the equation where p, q are relatively prime. Then p 2 and q 3 which gives q | | several possibilities:

1 2 x 1, 2, , ∈ ± ± ±3 ±3

2 It does not take long to try these and observe that x = 3 is the only rational solution. Thus the polynomial has a factor of 3x 2 which we can extract by long division to obtain − 3x3 + x2 + x 2 = (3x 2)(x2 + x + 1) − − All solutions can now be found using the quadratic formula.

4 We ﬁnish with a nice corollary which allows us to decide precisely which square-roots of integers are rational.

Corollary 1.4. Let n N. Then √n is rational if and only if it is an integer. ∈ Otherwise said, √n is irrational unless n is a perfect square.

Proof. √n satisﬁes x2 n = 0. If √n = p with gcd(p, q) = 1, the rational roots theorem implies that − q q 1. It follows that x = √n = p is itself an integer. |

1.3 The Set R of Real Numbers Thusfar, we have formally constructed the natural numbers and used them to (loosely) build the integers and rationals. It is a signiﬁcantly greater challenge to construct the real numbers. We start by thinking about ordered ﬁelds, of which both Q and R are examples.

Axioms 1.5 (Field). Consider a set F with two binary operations5 + and . We say that F is an ﬁeld if, for · all a, b, c F, ∈ A0a + b F (closure) ∈ A1a + (b + c) = (a + b) + c (associativity)

A2a + b = b + a (commutativity)

A3 0 F such that a + 0 = a (identity) ∃ ∈ A4 Given a F, a F such that a + ( a) = 0 (inverse) ∈ ∃ − ∈ − M0 ab F (closure) ∈ M1a (bc) = (ab)c (associativity)

M2 ab = ba (commutativity)

M3 1 F such that a 1 = a (identity) ∃ ∈ · M4 Given a F 0 , a 1 F such that aa 1 = 1 (inverse) ∈ \{ } ∃ − ∈ − Da (b + c) = ab + ac (distributivity)

For those with some algebraic training:

• The Addition Axioms A0–A4 say that (F, +) is an abelian group. • The Multiplication Axioms M0–M4 say that (F 0 , ) is an abelian group. \{ } · • The Distributive Axiom describes how addition and multiplication interact.

In addition to these axioms, we can introduce a notion of ordering, based on a binary relation6 . The ≤ symbol < is used in the usual way: x < y x y and x = y. ⇐⇒ ≤ 6 5We write multiplication as juxtaposition unless it is helpful for clarity: thus a b = ab. · · 6Thus x y is a statement which is true or false. ≤

5 Axioms 1.6 (Ordered Field). An ordered ﬁeld is a ﬁeld F together with a binary relation such that, for ≤ all a, b, c F, ∈ O1a b or b a ≤ ≤ O2a b and b a = a = b ≤ ≤ ⇒ O3a b and b c = a c ≤ ≤ ⇒ ≤ O4a b = a + c b + c ≤ ⇒ ≤ O5a b and 0 c = ac bc ≤ ≤ ⇒ ≤ You should mentally run through each of the axioms. In particular, you should consider the following:

• Recall that we deﬁned Q from Z and N, in which ordering is natural. We can therefore deﬁne the ordering of two rational numbers via p r ps qr q ≤ s ⇐⇒ ≤

You should be able to prove that every one of the axioms holds for Q, using only basic facts about multiplication, addition and ordering within the integers. For instance,

p s M2 Given a = q and b = t rational, we have ps sp ab = = = ba qt qt

since multiplication of integers is commutative. O3 Suppose p r and r t . Then q ≤ s s ≤ u ps qr and ru st ≤ ≤ whence p ps t psu qru qst = = ≤ ≤ ⇒ q qs ≤ u

• Make sure you believe that each axiom holds for your current understanding of the real numbers. You can prove anything, since the real numbers haven’t been defined yet! • Z is not an ordered field: which of the axioms are not satisfied? • The complex numbers C form a field. Consider the lexicographic ordering of C defined by ( x < p or x + iy p + iq ≤ ⇐⇒ x = p and y q ≤ Does this ordering satisfy axioms O1–O5?

6 Theorem 1.7. If F is any ordered ﬁeld with at least two elements 0 = 1 6 (a)a + c = b + c = a = b ⇒ (b)a 0 = 0 · (c) ( a)b = (ab) − − (d) ( a)( b) = ab − − (e) ac = bc and c = 0 = a = b 6 ⇒ (f) ab = 0 = a = 0 or b = 0 ⇒ (g)a b = b a ≤ ⇒ − ≤ − (h)a b and c 0 = bc ac ≤ ≤ ⇒ ≤ (i) 0 a and 0 b = 0 ab ≤ ≤ ⇒ ≤ (j) 0 a2 (a2 is a shorthand for a a) ≤ · (k) 0 < 1

(l) 0 < a = 0 < a 1 ⇒ − (m) 0 < a < b = 0 < b 1 < a 1 ⇒ − − All of these statements should be intuitive for the ﬁelds Q and R. Try abstractly proving a few, using only the axioms! We now deﬁne the last ingredient before we analyzing the difference between R and Q.

Deﬁnition 1.8. If F is an ordered ﬁeld, then the absolute value of an element a F is ∈ ( a if a 0 a = ≥ | | a if a < 0 − Theorem 1.9. 1. a 0 | | ≥ 2. ab = a b | | | | · | | 3. a + b a + b ( -inequality7) | | ≤ | | | | 4 Proof. For (a) and (b) just check the cases. For part (c), observe that ± a b a + b a + b − | | − | | ≤ ≤ | | | | which is obvious by thinking about the cases. The result is immediate. ± 7The name comes from the corresponding result for vectors a + b a + b . If a triangle has two edges described ≤ vectors a and b then the third edge a + b has length no more than| the| sum| of| the| other| two sides.

7 1.4 The Completeness Axiom & 1.5 The Symbols ∞ ± Everything in the previous section is as valid for Q as it is for R: both are ordered fields. Distinguish- ing between them requires the introduction of the completeness axiom or least upper bound principle: every bounded above subset of R has a least upper bound. To flesh this out we first need some terminology.

Deﬁnition 1.10 (Maxima, Minima & Boundedness). Let S R be non-empty. ⊆ (a) S is bounded above if it has an upper bound M: that is,

M R such that s S, s M ∃ ∈ ∀ ∈ ≤

(b) We write M = max S, the maximum of S, if M is an upper bound for S and M S. ∈ (c) The concepts of S being bounded below, having a lower bound m, and of having a minimum min S are similar.

(d) We say that S is bounded if it is bounded above and below. We say that S is bounded by if M s S, s ∀ ∈ | | ≤ M

The following should be immediate: try formally proving them yourself.

Lemma 1.11. 1. If M is an upper bound for S, so is M + ε for any ε 0. ≥ 2. If max S exists, then it is unique.

3. A set is bounded if and only if it is bounded above and below. In particular, if m, M are lower/upper bounds, then

s S, s max( m , M ) ∀ ∈ | | ≤ | | | |

Examples

1. If S is a ﬁnite set, then it is bounded and has both a maximum and a minimum.

2. Intervals should be familiar. For example: (2, 4] = x R : 2 < x 4 in bounded, with { ∈ ≤ } maximum 4 but no minimum.

3. N has a minimum, but no maximum. Z and Q have neither: both are unbounded.

4. S = x Q : x 4 has min S = 4 and max S = 4 . { ∈ | | ≤ 3 } − 3 3 5. S = Q [0, 3) has neither a maximum nor a minimum; it is bounded, for example by = 5. ∩ M 6. S = x Q : 0 x < √2 is bounded but has neither a maximum nor a minimum. { ∈ ≤ } It is worth checking examples 5 and 6 explicitly.

8 1 Example 5 Suppose x = max S exists and deﬁne s = 2 (x + 3). We have: •0 x < 3. This is immediate since x S. ≤ ∈ • s S. This follows since 3 s = 1 (3 x) > 0 = s < 3 and because s is clearly rational and ∈ − 2 − ⇒ positive.

• s < x. This follows since s x = 1 (3 x) > 0. − 2 − There exists an element s S which is larger than x, whence x is not the maximum: contradiction. ∈ The goal of the argument should be clear from the picture: given a hypothetical maximum M, ﬁnd an element s S between x and 3. ∈ x+3 s = 2

0 x 3

Example 6 The goal is the same as in the previous example: given a hypothetical maximum x, ﬁnd an element s S between x and √2. The difﬁculty is that we can’t simply use the average 1 (x + √2): ∈ 2 this isn’t rational and so doesn’t lie in S!

To ﬁx this, we informally invoke sequences: this might seem quite hard at the moment, but we will properly discuss sequences later. The rough idea is to construct a sequence (sn) of elements of S which increases to √2. Eventually one of these must be larger than x.

1 n 8 Deﬁne a sequence of rational numbers s , s , s , . . . by s = n 10 √2 . This sequence recovers the 0 1 2 n 10 b c decimal expansion of √2 to n decimal places: 14 141 1414 s = 1, s = 1.4 = , s = 1.41 = , s = 1.414 = ,... 0 1 10 2 100 3 1000 The sequence has the following properties: (a)0 < s < √2 whence s S. n n ∈ (b) √2 s < 10 n (since 10n√2 10n√2 < 1). − n − − b c Now suppose x = max S exists. Clearly x < √2. Choose n N large enough9 so that √2 x > 10 n. ∈ − − But then sn is an element of S and

√2 s < √2 x = x < s − n − ⇒ n The hypothetical maximum x is therefore not an upper bound for S: contradiction.

s0 sn 1 sn − ≈ ≈ 01 x √2

8 y is the ﬂoor of y and returns the greatest integer less than or equal to y. b c 9Any n > log (√2 x) will do the trick! − 10 −

9 Suprema and Inﬁma We can now generalize the idea of maximum and minimum values for bounded sets. For example, [2, 5) has least upper bound 5: among all upper bounds, 5 is the smallest.

Deﬁnition 1.12. Let S R be non-empty. ⊆ 1. If S is bounded above, deﬁne its supremum sup S to be its least upper bound. Otherwise said,

(a) sup S is an upper bound: s S, s sup S, ∀ ∈ ≤ (b) sup S is the least such: if M is an upper bound, then sup S M. ≤ 2. If S is bounded below, we similarly deﬁne its inﬁmum inf S to be its greatest lower bound. Equiv- alently,

(a) inf S is a lower bound: s S, s inf S, ∀ ∈ ≥ (b) inf S is the greatest such: if m is a lower bound, then m inf S. ≤ Of course we are assuming something here. . .

Axiom 1.13 (Completeness of R). If S R is non-empty and bounded above, then sup S exists (and is a ⊆ real number!). It is this property that distinguishes the real numbers from the rationals. In particular, every bounded set S of rational numbers has a supremum, the issue is that sup S might not be rational! In a moment we shall think more abstractly and deal with the issue of inﬁma. First some examples.

Examples

1. If a set has a maximum, then max S = sup S (similarly min S = inf S if a minimum exists).

2. sup S and inf S can exist even when the maximum and minimum do not. For example, the interval S = ( 1, 4) has sup S = 4: note here that sup S S. − 6∈ 3. S = 1 : n N has sup S = max S = 1, inf S = 0, no minimum. { n ∈ } ∞ T 1 1 4. S = [ n , 1 + n ) has inf S = 1 = sup S since S = 1 . n=1 { } 5. sup Q [0, 3) = 3. ∩ 6. sup Q [0, √2) = √2. ∩

A Useful Contrapositive The contrapositives of parts (b) of Deﬁnition 1.12 read as follows:

Lemma 1.14. 1. (b) If x < sup S, then s S such that s > x. ∃ ∈ 2. (b) If x > inf S, then s S such that s < x. ∃ ∈ Recall Examples 5 and 6 on page 9. In both cases we showed that any given x < sup S could not be a maximum (indeed upper bound), precisely as in the Lemma. This formulation is very useful in proofs and will be used repeatedly: the existence of s gives us something to work with. . .

10 Theorem 1.15 (Existence of Inﬁma). If S R non-empty and bounded below, then inf S R exists. ⊆ ∈ Proof. The picture summarizes the proof: sup T inf S = sup T −

m 0 m T − S Let m be a lower bound for S. Deﬁne T = t R : t S by reﬂecting S across zero. Clearly m { ∈ − ∈ } − is an upper bound for T:

t T, m t = t m ∀ ∈ ≤ − ⇒ ≤ − Completeness (Axiom 1.13) says that sup T exists and is a real number. We claim that inf S = sup T, − and check by verifying Deﬁnition 1.12:

(a) sup T is a lower bound for S: If s S, then − ∈ s T = s sup T = s sup T − ∈ ⇒ − ≤ ⇒ ≥ − (b) sup T is the greatest lower bound: Suppose x > sup T is also a lower bound for S. Then − − x > sup T is an upper bound for T: contradiction.10 − sup T inf S = sup T −

x mx 0 m T − − S

In particular, sup T = inf S, as required. −

Unbounded sets In order that every subset of the real numbers have both a supremum and an inﬁmum, we extend Deﬁnition 1.12.

Deﬁnition 1.16. 1. If S is unbounded above, we write sup S = ∞. Otherwise said

x R, s S such that s > x ( ) ∀ ∈ ∃ ∈ ∗ 2. If S is unbounded below, we write inf S = ∞. − The symbols ∞ have no other meaning (as yet): in particular, they are not numbers! If one is willing ± to abuse notation and write x < ∞ for any real number x, then the conclusions of Lemma 1.14 still hold for unbounded sets: indeed ( ) is precisely this statement! ∗ 10We could have done this using Lemma 1.14, though it would be a little excessive!

x > sup T = x < sup T = t T such that t > x − ⇒ − ⇒ ∃ ∈ − But then s := t S satisﬁes s > x and so x cannot be a lower bound for S. − ∈

11 Aside: A Synthetic Construction of R We have only observed that completeness is a property of R. It is, in fact, the deﬁning difference between R and Q. Explicitly:

Theorem 1.17. Up to relabeling (isomorphism of ﬁelds), there is precisely one set R which satisﬁes the following properties:

• R is an ordered ﬁeld such that 0 = 1 (Axioms 1.5 and 1.6) 6 • R has the completeness property (Axiom 1.13)

The proof is too difﬁcult and far too algebraic for us! To unpack things slightly, one must show that if (R, 0R, 1R, +R, R, R) and (S, 0S, 1S, +S, S, S) are two models satisfying the axioms, then there · ≤ · ≤ exists an isomorphism φ : R S: that is, → 1. φ is bijective

2. φ(x +R y) = φ(x) +S φ(y) and φ(x R y) = φ(x) S φ(y) for all x, y R · · ∈ 3. x R y φ(x) S φ(y) ≤ ⇐⇒ ≤ This approach will likely only satisfy you if you are addicted to abstract algebra!

The Archimidean Property and the Density of the Rationals We ﬁnish this section with an important discussion regarding how the rationals are distributed among the reals. Indeed, you should think carefully about where exactly we used this property in our discussion on page 9.

Theorem 1.18 (Archimidean Property). If b > 0 is a real number, then n N such that n > b. ∃ ∈ Equivalently: a, b > 0 = n N such that an > b. ⇒ ∃ ∈ It is important to note that in this result we are assuming nothing about R beyond the fact that it is an ordered ﬁeld satisfying the completeness axiom. The natural numbers in this context are deﬁned to be the set

N = 1, 1 + 1, 1 + 1 + 1, . . . R { } ⊆ Strictly speaking Peano’s axioms for this set N are themselves a theorem. In the next section we’ll brieﬂy think about how the Archimidean property is essentially trivial if one has an explicit construction of R built upon N and Q.

Proof. Suppose the result were false. Then b > 0 such that n b for all n N. Otherwise said, N ∃ ≤ ∈ is bounded above. The completeness axiom says that M := sup N exists. We now essentially do induction:

0 < 1 = M < M + 1 = M 1 < M = sup N ⇒ ⇒ − By Lemma 1.14, n N such that n > M 1. But then M < n + 1 which is clearly a natural number. ∃ ∈ − M cannot be an upper bound for N: contradiction.

12 Note particularly how we used the completeness axiom: indeed it can in fact be shown that there exist non-Archimidean ordered ﬁelds!

The Archimedean principle leads to a deep connection between Q and R.

Theorem 1.19 (Density of Q in R). Between any two real numbers, there exists a rational number. < 1 > Proof. WLOG suppose 0 a b. The Archimedean property applied to b a 0 says ≤ − 1 n N such that n > from which bn > 1 + an ∃ ∈ b a − A second application says that k N such that k > an. Now consider ∃ ∈ J := j N : an < j k { ∈ ≤ } and deﬁne m = min J: this exists since J is a ﬁnite non-empty set of natural numbers.11 Clearly m > an > m 1, since m = min J. But then m an + 1 < bn. We conclude that − ≤ m an < m < bn = a < < b ⇒ n as required.

The idea of the proof is straightforward even if the argument might seem abstract:

• Stretch the interval (a, b) by a factor of n until (an, bn) has width greater than 1, and thus contains an integer m

• Squash: divide by n so that m (a, b). n ∈ The Archimedean property is used purely to guarantee the existence of m, n.

Squash m J n ≈ m k 0 ab an bn

Stretch

1.6 A Development of R: non-examinable While we could use the synthetic approach, it is nice to be able to provide an explicit construction. The following approach uses Dedekind cuts. Here is a summary.

1. (Define N, Z and Q) Use Peano’s Axioms and proceed as in sections 1 and 2. One now carefully defines +, and , first on N and then for Z and Q building on the concepts for the integers. · ≤ 11This part of the argument is needed because, in this context, we haven’t established the well-ordering property of N (equivalent to Peano’s fifth axiom).

13 2. (Deﬁne R) Let α be a non-empty proper subset of Q with the following properties: • If r α and s Q with s < r then s α. ∈ ∈ ∈ • α has no largest rational number (max α makes no sense). Such sets α are called Dedekind cuts. Now deﬁne R to be the set of Dedekind cuts! This construction probably seems weird, though at least it is explicit. For example, the Dedekind cut α = x Q : x < 4 { ∈ } corresponds to the real number 4. The real number √2 corresponds to the Dedekind cut β = x Q : x < 0 or x2 < 2 { ∈ } Colloquially, the real number α corresponds to the Dedekind cut ‘all rationals less than α.’

It remains to prove all the axioms of a complete ordered field. We omit the details, though here is a rough overview. • Define the ordering of Dedekind cuts by α β α β ≤ ⇐⇒ ⊆ One can now prove axioms O1–O3 and that the ordering corresponds to that of Q. • Define addition by α + β := a + b : a α, b β { ∈ ∈ } Now prove axioms A0–A4 and O4 (with careful definition of α: this is quite hard. . . ). − • Multiplication is horrible: if α, β 0 then ≥ αβ := ab : a 0, a α, b 0, b β q Q : q < 0 { ≥ ∈ ≥ ∈ } ∪ { ∈ } which is then extended to a definition when at least one of α, β < 0. Once can then prove axioms M0–M4, O5 and D. • The completeness axiom must be proved, but it comes essentially for free! If A R (so that A ⊆ is a set of Dedekind cuts!), then the supremum of A is [ sup A = α α A ∈ An alternative approach using sequences of rational numbers will be given later.

Revisiting the Archimidean Property In this context, the Archimidean property is close to trivial, since we already have the unboundedness of N and Q built in! • First observe that every Dedekind cut β is bounded above by some rational number (if not then β = Q; contradiction. . . ). • If β > 0 is a positive Dedekind cut bounded above by p with p, q N, then n = p + 1 satisﬁes q ∈ n∗ > β, where

n∗ = x Q : x < n { ∈ } is the Dedekind cut corresponding to n.