Real Analysis Andrew Kobin Fall 2012 Contents Contents Contents 1 Introduction 1 1.1 The Natural Numbers . .1 1.2 The Rational Numbers . .1 1.3 The Real Numbers . .6 1.4 A Note About Infinity . .8 2 Sequences and Series 9 2.1 Sequences . .9 2.2 Basic Limit Theorems . 12 2.3 Monotone Sequences . 16 2.4 Subsequences . 19 2.5 lim inf and lim sup . 22 2.6 Series . 23 2.7 The Integral Test . 28 2.8 Alternating Series . 30 3 Functions 32 3.1 Continuous Functions . 32 3.2 Properties of Continuous Functions . 35 3.3 Uniform Continuity . 37 3.4 Limits of Functions . 41 3.5 Power Series . 44 3.6 Uniform Convergence . 46 3.7 Applications of Uniform Convergence . 49 4 Calculus 52 4.1 Differentiation and Integration of Power Series . 52 4.2 The Derivative . 54 4.3 The Mean Value Theorem . 57 4.4 Taylor's Theorem . 61 4.5 The Integral . 63 4.6 Properties of Integrals . 67 i 1 Introduction 1 Introduction These notes were compiled from a semester of lectures at Wake Forest University by Dr. Sarah Raynor. The primary text for the course is Ross's Elementary Analysis: The Theory of Calculus. In the introduction, we develop an axiomatic presentation for the real numbers. Subsequent chapters explore sequences, continuity, functions and finally a rigorous study of single-variable calculus. 1.1 The Natural Numbers The following notation is standard. N = f1; 2; 3; 4;:::g the natural numbers Z = f:::; −3; −2; −1; 0; 1; 2; 3;:::g the integers n p o Q = q j p; q 2 Z; q 6= 0 the rational numbers R ? the real numbers Note that N ⊂ Z ⊂ Q ⊂ R. We begin with a short axiomatic presentation of the natural numbers N. The five axioms below are called the Peano Axioms. (1) There is a special element 1 2 N, so N is nonempty. (2) Every element of N has a successor which is also in N. For n 2 N, its successor is denoted n + 1. (3) 1 is not the successor of any element of N. (4) Suppose S ⊆ N is a subset. If 1 2 S and for all n 2 S, n + 1 2 S as well, then S = N. This is called the Axiom of Induction. (5) If the successors of some m; n 2 N are the same, then m and n are the same. In other words, m + 1 = n + 1 =) m = n. Remark. Any set that satisfies all five Peano Axioms is isomorphic to the natural numbers. 1.2 The Rational Numbers Q is the arithmetic completion of N, i.e. the rationals are closed under +; −; × and ÷. 2 However, there are issues withpQ. For example, try finding roots of the polynomial x −2 = 0. The only solutions are x = ± 2, but it turns out that these are not rational numbers: p Theorem 1.2.1. 2 is irrational. 2 p Proof. Suppose x = 2 and x 2 Q. Then x = q for some p; q 2 Z, q 6= 0. We may assume p p and q have no common factors, i.e. q is reduced. Consider p2 x2 = = 2 =) p2 = 2q2: q2 1 1.2 The Rational Numbers 1 Introduction Then p2 is even, so p must be even, i.e. there is some integer k such that p = 2k. This means 2 2 2 2 2 2 p = 4k = 2q =) q = 2k . Thus q is even, so q is even.p This shows that 2 is a common factor of p and q, which contradicts our assumption. Hence 2 is irrational. So it's clear that the rational numbers are missing some ‘stuff' that should be there. Definition. A number x is algebaic if there are integers a0; a1; : : : ; an, with an 6= 0 and n n−1 n ≥ 1, such that anx + an−1x + ::: + a1x + a0 = 0. p p p Example 1.2.2. 2 is algebraic since it is a root of x2 − 2 = 0. Is 2 + 2 algebraic? To p p see that it is, let x = 2 + 2 and consider p x2 = 2 + 2 (x2 − 2)2 = 2 x4 − 4x2 + 2 = 0: So a4 = 1, a2 = −4 and a0 = 2 shows that x is algebraic. Proposition 1.2.3. Every rational number is algebraic. p Proof. Let x = q for p; q 2 Z and q 6= 0. We can rewrite this as qx − p = 0 so x is algebraic. p The converse is false: note that every algebraic number is rational (see 2 above). Definition. The algebraic completion of Q is the set Q of all algebraic numbers. p Theorem 1.2.4 (Rational Roots Theorem). Let r 2 Q with r = q such that p and q have n no common factors. If anr + ::: + a1r + a0 = 0 with n ≥ 1, an 6= 0 and a0 6= 0, then q j an and p j a0. Proof. Plug in: pn p a + ::: + a + a = 0 n q 1 q 0 n n−1 n n anp + ::: + a1pq + a0q = 0 · q = 0 n n−1 n−1 n anp = −an−1p q − ::: − a1pq − a0q : n Since q divides everything on the right, q must be a factor of anp . But q and p share no common factors, so q divides an. Likewise p divides a0. Example 1.2.5. The rational roots theorem is particularly effective for showing that a 4 2 polynomial has no rational roots. For example, consider x − 4x + 2 = 0. Note that a0 = 2 and a4 = 1, so all possible rational roots are ±1; ±2. But none of these are roots, so all roots of x4 − 4x2 + 2 are irrational. 2 1.2 The Rational Numbers 1 Introduction Are all real numbers algebraic? In other words, does Q = R? It turns out that the answer is no. Examples include π, e, sin(1), etc. but these are hard to prove. In fact, it wasn't until 1844 that Liouville proved the existence of such numbers, called transcendental numbers. It has since been shown that Q is a subset of the complex numbers C. This means that not all real numbers are algebraic and not all algebraic numbers are real. Next we detail an axiomatic presentation of the rationals, similar to the Peano Axioms in Section 1.1. First, there are special numbers 0; 1 2 Q with 0 6= 1, so Q is nonempty. Define the binary operations +; · : Q × Q ! Q that satisfy the following axioms for all a; b; c 2 Q: (1)( a + b) + c = a + (b + c) (additive associativity). (2) a + b = b + a (additive commutativity). (3) a + 0 = a (additive identity). (4) For every a there exists an element −a such that a + (−a) = 0 (additive inverse). (5) a(bc) = (ab)c (multiplicative associativity). (6) ab = ba (multiplicative commutativity). (7) a · 1 = a (multiplicative identity). 1 1 (8) For every a there exists an element a such that a · a = 1 (multiplicative inverse). (9) a(b + c) = ab + ac (distribution). Any set that satisfies (1) { (9) is called a field. Q is an example of a field, however this set of axioms is not a unique presentation of Q. Proposition 1.2.6. Some algebraic rules regarding a; b; c 2 Q are: (a) If a + c = b + c then a = b. (b) a · 0 = 0. (c) (−a)b = −(ab). (d) (−a)(−b) = ab. (e) If ac = bc and c 6= 0 then a = b. (f) If ab = 0 then either a = 0 or b = 0. Proof. All of these may be proven from axioms (1) { (9). We will prove (a), (b) and (e) and leave the rest for exercise. (a) If a + c = b + c for a; b; c 2 Q then (a + c) + (−c) = (b + c) + (−c) a + (c + −c) = b + (c + −c) by axiom (1) a + 0 = b + 0 by axiom (4) a = b by axiom (3). 3 1.2 The Rational Numbers 1 Introduction (b) Take a 2 Q and consider a · 0. By axiom (3) we can write 0 + 0 = 0, so a · 0 = a(0 + 0) =) a · 0 = a · 0 + a · 0 by axiom (9). By axiom (4), there exists an element −(a · 0) 2 Q such that (a · 0) + −(a · 0) = 0, so 0 = a · 0 + −(a · 0) = (a · 0 + a · 0) + −(a · 0) = a · 0 + (a · 0 + −(a · 0)) by association = a · 0 + 0 by additive inverse = a · 0 by additive identity: 1 (e) Let a; b; c 2 Q with c 6= 0 and suppose ac = bc. Then there exists some element c 2 Q 1 such that c · c = 1. Then 1 1 (ac) · = (bc) · c c 1 1 a c · = b c · by association c c a · 1 = b · 1 by multiplicative inverse a = b by multiplicative identity. To further develop the rational numbers, we state the five Order Axioms: for a; b; c 2 Q, we have (10) Either a ≤ b or b ≤ a (comparability). (11) If a ≤ b and b ≤ a then a = b (antisymmetry). (12) If a ≤ b and b ≤ c then a ≤ c (transitivity). (13) If a ≤ b then a + c ≤ b + c. (14) If a ≤ b and c ≥ 0 then ac ≤ bc. Note that (10) { (12) endow Q with a total ordering, so that the 14 axioms together mean that Q is a totally ordered field. There are other examples of ordered fields, such as R. ? Example 1.2.7. C is not an ordered field: 3 + i ≤ 2 + 2i | these can't be ordered.