Math LM (24543) Lectures 01

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1 Math LM (24543) Lectures 01 Ethan Akin Office: NAC 6/287 Phone: Spring, 2018

2 Contents Introduction, Ross Chapter 1 and Appendix The Natural Numbers N and The Rational Numbers Q The Real Numbers R Sequences and Metric Spaces, Ross Chapter 2 Limits of Sequences and Subsequences Metric Space Topology Open and Closed Sets Compactness

3 Introduction The book is Kenneth A. Ross Elementary Analysis, The Theory of Calculus (2 nd Ed). Keep up with the homework. Ask questions. The course information sheet with homework assignments is posted on my site: http : /math.sci.ccny.cuny.edu//peoplename = EthanAkin I will be posting there the pdf I am using here and later on review questions for the tests. Office: NAC 6/287. Hours: Tuesday 12-1:50pm. Phone: , ethanakin@earthlink.net

4 Sets and Quantifiers, Ross Appendix We will begin by reviewing some notation and properties about sets which we will be using. A set is a collection of objects: numbers, sets, whales etc.the particular set is described by a well-defined property P(x). This means that there is no ambiguity about whether x satisfies P(x) or not. The property x is blue. is too vague. Blue shades into green. Is this blue or green? We won t look at properties like that. On the other hand the answer can be unambiguous but unknown. A prime number of the form 2 k + 1 is called a Fermat prime. There are only five known. Suppose P(x) is x is the largest Fermat prime. If there are infinitely many Fermat primes then this is always false. If there are only finitely many then it is true for exactly one number but until we know what they all are, we don t know which one it is.

5 The set associated with a property P(x) is written S = {x : P(x)}. So x S, meaning x is an element of the set S, iff (= if and only if) P(x) is true. That is, x S P(x) is true. This association of sets and properties can lead to difficulties. Consider the property of sets P(x) is x x, that is, the set is not an element of itself. So for a set A, P(A) means A is not an element of itself. Exercise A1.1: Let R = {x : P(x) is true } = {x : x x}. Prove the following: (i) If R R, then P(R) is false and so R R. (ii) If R R, then R R. This is called Russell s Paradox.

6 The set theory paradoxes like this are avoided by staying inside some big set. Thus, our sets will be of the form S = {x U : P(x)}. Often we will just write {x : P(x)} because the big set U is understood. That is, we will be looking at sets of numbers or functions or subsets of some big set. This gets us to quantifiers and. ( x)p(x) means for all x in U, P(x) is true, while ( x)p(x) means for some x in U, P(x) is true. The property P(x) is true when P(x) is false. We will frequently use the negation properties: [( x)p(x)] is exactly ( x) P(x) and [( x)p(x)] is exactly ( x) P(x).

7 Lists or Indexed Sets Often we will be looking a set indexed by a set I. This is just a function defined on I. The example to keep in mind is a sequence of numbers {a n } which can be written a 1, a 2,.... We are using subscript notation a n instead of the function notation a(n). Notice the difference between the sequence and its set of values, the image of the function. Think of the sequence a n = ( 1) n with only two values. For an indexed collection {A i } of subsets of U or a subset A of U we define the union, the intersection and the complement by x i {A i} ( i)x A i x i {A i} ( i)x A i x U \ A (x A) x A.

8 Exercise A1.2:(a) Use quantifier negation to prove DeMorgan s Laws: U \ ( i {A i }) = i {U \ A i }. U \ ( i {A i }) = i {U \ A i }. (b) For A U, prove the Distributive Laws: A ( i {A i }) = i {A A i }. A ( i {A i }) = i {A A i }.

9 The Natural Numbers N and The Rational Numbers Q, Ross Chap. 1, Secs. 1, 2 For the natural numbers N the first of the two things we need to review is the Uniqueness of the Prime Factorization. It follows that if c = ab then the prime factorization of c is the product of the factorizations of a and b. We write a c (read as a divides c) when c/a is a natural number. This is the relation you learned in elementary school as gazinta. Natural numbers a and b are relatively prime when 1 is the only number which divides both of them, i.e. 1 is the only common factor of a and b. Exercise A1.3: Show that if a bc and a and b are relatively prime then a c (Hint: Factor).

10 The other thing we need to look at is Mathematical Induction: If a set A of natural numbers contains 1 and contains n + 1 whenever it contains n then A is all of N. In practise we use this to prove for a property P(n) that ( n N)P(n) is true. There are two steps. Initial Step: Check that P(1) is true. The property P holds for 1. Inductive Step: Show that P(n) P(n + 1). This means that you can assume P is true at level n and use it to prove it holds at level n + 1. It even suffices to show that P(1),..., P(n) imply P(n + 1). This feels like a circular argument, but it is really spiralling upward.

11 The equation n = 1 n(n + 1) is used on Ross 2 page 3 [R 1 page 2] to illustrate induction. Proofs which avoid induction are often more illuminating. There is a classic one due to Gauss (when he was in elementary school) and also a geometric one. HW Exercises: Ross p. 5 [R 1 p. 4] 1.1, 1.3, 1.6, Now we will use induction to prove: All horses are the same color.

12 Call a set of horses equicolored when all the horses in it are the same color. We prove P(n) : Every set of n horses is equicolored. Observe that P(1) is true. Now suppose we have a set of n + 1 horses. Line them up in stalls and put horse n + 1 in the corral. The remaining horses are a set of size n and so by inductive hypothesis they are all the same color. Now bring in horse n + 1 and throw horse 1 in the corral. Again we have a set of size n which is equicolored by the inductive hypothesis again. Bring back horse 1. The two horses on the end are each the same color as the horses in the middle and so all n + 1 are the same color. This can t be right. What is the problem?

13 Rational Zeroes Theorem For rational numbers recall that ac = c by cancelling a. So we ad d can cancel common factors and write any fraction as c with c d and d relatively prime and d > 0. Now suppose that r = c satisfies the polynomial equation d c n x n + c n 1 x n c 1 x + c 0 = 0 with c 0,..., c n integers such that c 0 0 and c n 0. Substitute x = r and multiply by d n to get c n c n + c n 1 c n 1 d + + c 1 cd n 1 + c 0 d n = 0. Solve to get c 0 d n = c[ c 1 d n 1 c n 1 c n 2 d c n c n 1 ]. So c c 0 d n and so, by Exercise 1.3, c c 0. Solve to get c n c n = d[ c 0 d n 1 c 1 cd n 2 c n 1 c n 1 ]. So d c n.

14 We can use this to see that x 2 3 has no rational roots, i.e. 3 is not rational. HW Exercises: Ross p. 13 [R 1 p. 9] 2.3, 2.5, 2.7*(b). Let us look at Exercise 2.7(a). Let x = = (x + 3) 2 = x x = x (x 1) = (x 1)[x ]. Cancel x 1 and we see that x = which is not rational because 3 is not rational. But we were supposed to show that x is rational. What is the problem?

15 The Real Numbers R, Ross Chap. 1, Secs. 3, 4 In addition to the standard algebra and order properties of the real numbers, we focus our review on the absolute value function. { a if a 0, We define the absolute value of a by a = a if a 0. From this we see, for example, that a = a 2. Alternatively, we see that a = max(a, a). So we consider the properties of max and min applied to a set.

16 If a nonempty subset S R has a largest element then it is the maximum of S, written max S. The smallest element, if it exists, is the minimum, min S. So s = max S iff s S and ( x S)x s. If S has a maximum then max S b iff ( x S)x b. Similarly, s = min S iff s S and ( x S)x s. Exercise A1.4: (a)prove that max(max{a 1,..., a n }, a n+1 ) = max{a 1,..., a n, a n+1 }. Use this to prove, by induction, that every nonempty, finite subset of R has a maximum. (b) Prove that if S has a maximum then S = { s : s S} has a minimum and min S = max S. (c) If S has a maximum then max S > b iff ( x S)x > b (Hint: Use quantifier negation). (d) Use induction to prove the well-ordering property for N: Any nonempty subset S of N has a minimum element.

17 Back to absolute values. b a iff b a and b a. Of course, this requires b 0 since always a 0. Ross Theorem 3.5 (i): a 0 and a = 0 only when a = 0. Also obvious, a = a. (ii): ab = a b. Proof: Always ab = ±( a b ) (iii): a + b a + b. Proof: a a, a, b b, b. So a + b a + b, ( a) + ( b) = (a + b). Ross Exercise 3.5 (a): b a iff b a, b a iff b a, b a iff a b a. For Ross Exercise 3.7 (a) replace by < and by >.

18 The absolute value is especially important because it is used to measure distance between points of R. For a, b R the distance between them, d(a, b) = a b (The book uses dist(a, b) instead of d(a, b)). From the properties of absolute value we obtain the properties of the metric d: (i)(positivity): d(a, b) 0 and d(a, b) = 0 only when a = b. (ii) (Symmetry): d(a, b) = d(b, a). (iii) (Triangle Inequality): d(a, c) d(a, b) + d(b, c). These are general properties of metrics. Special for R is: (iv) d(a, b) c iff b c a b + c iff a [b c, b + c]. Also, d(a, b) < c iff b c < a < b + c iff a (b c, b + c). You will see this a lot with ɛ written for c. HW Exercises: Ross p. 19 [R 1 p. 14] 3.5, 3.6, 3.7, 3.8.

19 Completeness, Ross Chap. 1, Secs. 4, 5 For a nonempty subset S of R we call M an upper bound for S when ( x S)x M, or S (, M]. We call m a lower bound for S when ( x S)x m, or S [m, ). Completeness Axiom for R: If S is a nonempty subset of R which is bounded above then the set of upper bounds has a minimum element. The least upper bound (or lub) of S is called the supremum of S and is denoted sup S. If S is a nonempty subset of R which is bounded below then the set of lower bounds has a maximum element. The greatest lower bound (or glb) of S is called the infimum of S and is denoted inf S. It is only necessary to assume one of these as the other follows from inf S = sup S.

20 If S is not bounded above, we write + = sup S and if S is not bounded below, we write = inf S. In general, we have b sup S iff b is an upper bound for S iff ( x S)b x. Be careful: ( x S)b > x does not imply b > sup S (Example?). By negating quantifiers we have (and this is important) b < sup S iff b is not an upper bound for S iff ( x S)b < x. HW Exercises: Ross p [R 1 p ] 4.6, 4.7, 4.8, 4.9, 4.14.

21 Archimedean Property: Assume a > 0 and b > 0. There exists n such that b < na. Proof: If not then b is an upper bound for {na : n N}, then the supremum s 0 = sup{na : n N} exists. Since a < 0, s 0 a < s 0 and so s 0 a is not an upper bound for the set. So there exists n 0 N such that s 0 a < n 0 a and so s 0 < (n 0 + 1)a and so s 0 is not an upper bound for the set. This contradiction proves that b is not an upper bound for the set. If b R then there exists an integer k Z such that k b < k + 1. The integer k is denoted [b]. Proof: If b > 0 then by the Archimedean Property and Well-Ordering of N: There is a smallest n N such that b < n. So k = n 1 b.

22 If b = 0 let k = 0. If b < 0 then there is an n N such that n 1 b < n and so n < b n + 1. Use k = n or if b = n + 1, use b = n + 1. Density of the Rationals: If a < b then there exists r Q such that a < r < b. Proof: There exists n N such that n(b a) > 2. That is, na + 2 < nb. With k = [na] we have k na < k + 1 and so na < k + 1 < na + 2 < nb. Let m = k + 1 and r = m/n. HW Exercises: Ross p [R 1 p ] 4.10, 4.11, 4.12, 4.15, 4.16.

23 Notice that these two results only required the Archimedean Property, not the full power of Completeness. Conversely, from the density of the rationals we can obtain the Archimedean Property. Proof: Given 0 < a < b. By density of the rationals, there exist rationals r 1, r 2 such that 0 < r 1 < a < b < r 2 < b + 1 Write the two rationals as fractions with a common, positive denominator. So r 1 = m c and r 2 = M c with 1 m. So M < M m. b < M c < M m c < M a.

24 Ross Exercise 3.8: If a b + ɛ for all ɛ > 0 then a b. Proof: If a > b then a > 1(b + a) = b + 1 (b a). Similarly, if 2 2 a b ɛ for all ɛ > 0 then a b. Ross Exercise 4.7 (a): If S T then sup T is an upper bound for S and so sup T sup S. Similarly, inf T inf S. Always inf S sup S. Therefore, inf S sup S sup S sup T. (b:) S, T S T. By (a) sup S, sup T sup(s T ). Therefore, max(sup S, sup T ) sup(s T ). On the other hand, max(sup S, sup T ) is an upper bound for S T and so max(sup S, sup T ) sup(s T ). 4.8 (b): If s S, t T imply that s t then sup S inf T. Proof: Each t T is an upper bound for S and so sup S t for all t T. Thus, sup S is a lower bound for T and so sup S inf T.

25 4.14 (a): sup A + sup B is an upper bound for A + B and so sup A + sup B sup(a + B). If ɛ > 0 then sup A ɛ is not an upper bound for A and so there exists a A with a > sup A ɛ and similarly there exists b B such that b > sup B ɛ. Hence, sup(a + B) a + b > sup A + sup B 2ɛ. This is true for all ɛ > 0 and so sup(a + B) sup A + sup B.

26 Sequences and Subsequences, Ross Chap. 2, Secs. 7, 11 A sequence {s n } in a set S is a function from N (or sometimes N {0} to S). In the language I introduced earlier, it is an indexed set with index N. Instead of writing s(n) we use the subscript notation s n. Two important bits of language which we will repeatedly use for a property P(s) which it true or false about elements of S. Eventually P(s n ) is true means ( N)( n N)P(s n ) is true. That is, P(s n ) is true from some time N on. It is true for all but finitely many n s. Frequently P(s n ) is true means ( N)( n N)P(s n ) is true. That is, P(s n ) keeps happening. It is true for infinitely many n s.

27 For any N N the set T N = {s n : n N} is called a tail of the sequence. So Eventually P(s n ) is true says that {s n : P(s n ) is true } contains some tail of the sequence. On the other hand, Frequently P(s n ) is true says that {s n : P(s n ) is true } meets every tail of the sequence. That is, its intersection with every tail is non-empty.

28 For example, s n = ( 1) n is frequently equal to 1, but not eventually equal to 1. The sequence s n = n is eventually larger than K for any positive number K, but notice that if you change K then the time it takes to get bigger than K changes. Exercise A2.1: Use negation of quantifiers to show that the negation of Eventually P(s n ) is true is not Eventually P(s n ) is false, but is, instead, Frequently P(s n ) is false. If {s n } is a sequence of then we say that {s n } is increasing when s n s n+1 for all n N and strictly increasing when the inequalities are all strict: s n < s n+1 for all n N. We say that {s n } is decreasing when s n s n+1 for all n N and strictly decreasing when the inequalities are all strict: s n > s n+1 for all n N.

29 Instead of the pairing: increasing/strictly increasing, you will sometimes see non-decreasing/increasing. A sequence of real numbers is called monotonic when it is either increasing or decreasing. If K is any infinite subset of N then it is the set of values of a unique strictly increasing sequence: K = {k 1, k 2,... } with k 1 the minimum element of K, k 2 the next element, i.e. the minimum element of K \ {k 1 } and so forth. This is just a list of the set, labeling the elements in order of size. A subsequence {s ki } of a sequence {s n } just the composition of the function s : N S with a strictly increasing function k : N N. We can instead think of it as restricting the function s to the infinite subset which is the image of k.

30 A useful example is Ross Theorem 11.4: Every sequence of real numbers has a monotone subsequence. Proof: For a sequence {s n } call n a dominant index if s n > s m for all m > n. Let K be the set of dominant indices. Case 1: K is infinite an so defines a subsequence s ki. Since k i+1 > k i and k i is dominant, s ki > s ki+1 and so the subsequence is strictly decreasing. Case 2: K is finite and so there exists n 1 so that all the dominant terms are smaller than n 1. We use an inductive construction to build an increasing sequence. Since n 1 is not dominant there exists n 2 > n 1 such that s n2 s n1. Since n 2 is not dominant there exists n 3 > n 2 such that s n3 s n2.

31 Assume that n 1, n 2,..., n k have been defined. Since n k is not dominant there exists n k+1 > n k such that s nk+1 s nk. By induction the sequence is defined for all k N and so it is an increasing subsequence. Notice that for s n = ( 1) n n the even values of n give an increasing subsequence and the odd values give a decreasing subsequence.

32 Limits of Sequences and Subsequences, Ross Chap. 2, Secs. 7, 10, 11 For a sequence {s n } of real numbers and a real number s, we say that: {s n } converges to s, or The limit of {s n } is s, written lim n {s n } = s or just lim{s n } = s, or {s n } approaches s, written s n s. when ( ɛ > 0)( N N)( n N) s n s < ɛ. All of these mean the same thing. That is, for every ɛ > 0 eventually the sequence gets ɛ close to s, or, eventually the sequence is arbitrarily close to s. Recall that s n s < ɛ iff s ɛ < s n < s + ɛ. Thus, lim{s n } = s means that eventually the sequence is in the interval (s ɛ, s + ɛ) for any ɛ > 0.

33 There is a kind of sloppiness which we will allow ourselves in proofs about limits. Suppose you start with an arbitrary ɛ > 0 and for a particular sequence you find an N so that when n N we have s n s 17 ɛ. That is, you have shown ( ɛ > 0)( N N)( n N) s n s 17 ɛ. It is important to understand that the quantified variables are dummy variables like the variables which are integrated away in a definite integral. That is, you have proved ( δ > 0)( M N)( n M) s n s 17 δ. You want to prove the original statement, but you are free to pick any δ. Starting with ɛ > 0, you pick δ > 0 so that 17 δ < ɛ and so that δ < ɛ 2 /17. You then use you N(ɛ) to be the M(δ) for this δ.

34 We will repeatedly use this idea that if something is true for all δ > 0, then it is true for any δ which you pick for the convenience of the rest of your argument. HW Exercises: Ross p [R 1 p ] 8.4, 8.5, 8.6, 8.9, Infinite limits are defined by: s n + ( {s n } diverges to plus infinity) when ( K)( N)n N s n > K. That is, eventually {s n } is arbitrarily large. Similarly, s n when ( K)( N)n N s n < K. Ross Theorem 11.3: If s R {± } and s n s, then for any subsequence s nk s. Proof: As Ross observes n k k. So if P(s n ) is true when n N then P(s nk ) is true when k N.

35 Ross Theorems 10.2, 10.4: Assume {s n } is an increasing sequence. (a) If {s n } is bounded, then {s n } converges to sup{s n }. (b) If {s n } is not bounded, then {s n } diverges to +. Proof: (a) Let s = sup{s n } and let ɛ > 0. Because s ɛ is not an upper bound, there exists N such that s ɛ < s N. For any n N, s ɛ < s N s n and since s is an upper bound, s n s < s + ɛ. (b) s 1 = min{s n } and so the sequence is bounded below. If it is not bounded, and so not bounded above, then any K is not an upper bound. As before, there exists N such that K < s N. For any n N, K < s N s n.

36 Convergence of bounded monotone sequences is equivalent to completeness. That is, if we assume that any bounded increasing sequence has a limit, then any nonempty, bounded subset S has a least upper bound. Proof: Let s 1 S and let t 1 be an upper bound for S. Let D = t 1 s 1 0. Notice that the midpoint m 1 = s 1+t 1 lies 2 halfway between the two points. If m 1 is an upper bound then let s 2 = s 1 and t 2 = m 1. If m 1 is not an upper bound then choose a point of S s 2 > m 1. Let t 2 = t 1. Notice that in either case, s 1 s 2 in S and t 1 t 2, upper bounds for S and that t 2 s 2 D/2 1.

37 Assume, inductively, that we have constructed s 1 s 2... s k in S and t 1 t 2... t k upper bounds for S with t k s k D/2 k. Let m k = s k+t k 2. If m k is an upper bound, let t k+1 = m k and s k+1 = s k. If m k is not an upper bound then choose a point of S s k+1 > m k. Let t k+1 = t k. In either case, t k+1 s k (t k s k ) D/2 k+1. We have an increasing sequence {s n } in S and a decreasing sequence {t n } of upper bounds, with t n s n D/2 n. By assumption {s n } has a limit s. We will show that s = sup S. Because 0 t n s n D/2 n, it follows that lim{t n s n } = 0. It follows that lim{t n } = lim{s n } + lim{t n s n } = s. That is, the sequence {t n } has limit s, as well. This uses some algebraic properties of limits which we will prove later.

38 For any ɛ > 0 both sequences are eventually in (s ɛ, s + ɛ). That is, there exists s n > s ɛ and so s ɛ is not an upper bound. There exists t n < s + ɛ and so s + ɛ is an upper bound. This means that there is no element of S larger than s and no upper bound smaller than s. That is, s = sup S. For a sequence {s n }, a point s R is a cluster point of the sequence when ( ɛ > 0)( N N)( n N) s n s < ɛ. That is, for every ɛ > 0 the sequence is frequently in the interval (s ɛ, s + ɛ). Exercise A2.2: Use negation of quantifiers to explain when s is not the limit of {s n } and when s is not a cluster point of {s n }.

39 Theorem: If {s n } converges to s then s is the unique cluster point of the sequence. In particular, a sequence converges to at most one point. Proof: It should be clear that a limit is a cluster point. (Eventually true is stronger than frequently true.) Let t s so that 2ɛ = t s > 0 (This is the definition of ɛ). There exists N so that for all n N, s n (s ɛ, s + ɛ) which is disjoint from (t ɛ, t + ɛ). (Check this using the triangle inequality). Thus, there does not exist n N such that s n (t ɛ, t + ɛ). So t is not a cluster point.

40 Ross Theorem 11.2 (i): For a sequence {s n }, a point s R is a cluster point iff s is the limit of some subsequence of {s n }. Proof: If a subsequence {s nk } converges to s and ɛ > 0 then there exists N 1 so that k N 1 implies s nk s < ɛ. For any N N let k max(n 1, N). Then n k k N and since k N 1, s nk s < ɛ. Thus, s is a cluster point. Now assume that s is a cluster point. We construct, inductively, a subsequence {s nk } so that s nk s < 1/k. (Show that this implies that {s nk } converges to s.) First, since s n s < 1 infinitely often, we can choose n 1 so that s n1 s < 1. This is the initial step. Now assume that n 1 < n 2 <... n k have been chosen. Since s n s < 1/(k + 1) infinitely often we can choose n k+1 > n k so that s nk+1 s < 1/(k + 1). By induction we obtain the required subsequence.

41 Ross Theorem 11.5 (Bolzano-Weierstrass Theorem): Any bounded sequence {s n } in R has a convergent subsequence, or, equivalently, has a cluster point in R. Proof: Any sequence has a monotone subsequence which is bounded if the original sequence was. A bounded increasing sequence converges to its supremum and a bounded decreasing sequence converges to its infimum. A limit of a subsequence is a cluster point. Ross Exercise 8.5(a) (Squeeze Theorem): Assume a n s n b n for all n. If lim{a n } = lim{b n } = s, then lim{s n } = s. Proof: If ɛ > 0 then there exist N 1, N 2 such that n N 1 implies a n (s ɛ, s + ɛ) and n N 2 implies b n (s ɛ, s + ɛ). If n max(n 1, N 2 ), then s ɛ < a n s n b n < s + ɛ.

42 Ross Exercises 8.10, 8.9: Assume {s n } converges to s and a R. (a) If s < a, then eventually s n < a. (b) If frequently s n a, then s a. Proof: (a) Let ɛ = a s. Since, ɛ > 0, eventually, s n < s + ɛ = a. (b) This is the contrapositive of (a). Theorem: If s n s and t n t, then s n t n for all n implies s t. Proof: s n t n 0 and so by (b) s t 0.

43 Algebraic Properties of Limits,Ross Chap. 2, Sec. 9 Ross Theorem 9.1: A convergent sequence is bounded. Proof: If s n s then there exists N such that s n s 1 and so s n s + 1 for n N. If M = max( s 1,..., s N, s + 1) then the sequence is contained in [ M, M]. Ross Theorems 9.3, 9.4: If s n s and t n t in R then s n + t n s + t and s n t n s t. Proof: Assume that s n, t n M for all n.

44 (s n + t n ) (s + t) s n s + t n t. s n t n s t = s n (t n t) + t(s n s) M( s n s + t n t ). Corollary: If s n s and t is a cluster point of {t n } in R then s + t is a cluster point of {s n + t n } and s t is a cluster point of {s n t n }. Proof: Choose a subsequence t nk s nk s. t and use the fact that Exercise A2.3: Give an example where s is a cluster point of {s n } and t is a cluster point of {t n }, but s + t is not a cluster point of {s n + t n } and s t is not a cluster point of {s n t n }.

45 Ross Exercise 8.6: If s n s then s n s. If s n 0 then s n 0. Proof: s n s s n s and s n 0 = s n 0. Ross Example 8.6: If s n 0 for all n and lim{s n } = s 0 then inf{ s n } > 0. Proof: Let ɛ = s /2. There exists N so that s /2 = s s /2 < s n for n N. Let m = min( s 1,..., s N, s /2).

46 Ross Theorem 9.6: If s n s and t n t in R, and s n, s 0 for all n then t n /s n t s. Proof: It suffices to show that 1/s n 0 < m s, s n for all n. 1/s. Assume (1/s n ) (1/s) = s s n / s s n s s n /m 2. Look at the proofs of Ross Theorem 9.7 (Basic Examples). Recall the Binomial Theorem (a + b) n = a n + na n 1 b + n (n 1) 2 a n 2 b

47 Lim Sup and Lim Inf, Ross Chap. 2, Secs. 10, 11 Assume that {s n } is a bounded sequence in R. Recall that the set T n = {s i : i n} is called a tail of the sequence. We let s n = sup T n and s n = inf T n. Clearly, s n s n s n for all n. Since T n+1 T n, Exercise 4.7 implies that s n s n+1 s n+1 s n. Thus, {s n } is an increasing sequence and {s n } is a decreasing sequence. As each is bounded, each has a limit.

48 We define lim sup{s n } = lim{s n } = inf{s n }. lim inf{s n } = lim{s n } = sup{s n }. All of the s n s are less than or equal to the s m s and so Ross Exercise 4.8 implies lim inf{s n } lim sup{s n }.

49 Ross Theorem 11.7 If {s n } is a bounded sequence, then lim inf{s n } and lim sup{s n } are cluster points of {s n }. Proof: We do the result for lim sup. Let s = lim sup{s n } and let ɛ > 0. There exists N 1 such that for all n N 1, s n = sup T n (s ɛ, s + ɛ). For any N let n max(n, N 1 ) and so s ɛ is not an upper bound for T n. This means there exists i n N such that s ɛ < s i s n < s + ɛ. Thus, {s n } is frequently in (s ɛ, s + ɛ).

50 Ross Theorem 11.8 (c) A bounded sequence {s n } converges iff lim inf{s n } = lim sup{s n }, in which case the common value is lim{s n }. Proof: Because s n s n s n for all n, lim inf{s n } = lim sup{s n } implies that common value is lim{s n } by the Squeeze Theorem. If lim inf{s n } lim sup{s n }, then {s n } has at least two distinct cluster points and so does not converge. Exercise A2.3: If {s nk } is a subsequence of {s n }, show that lim inf{s n } lim inf{s nk } lim sup{s nk } lim sup{s n }. Conclude that every cluster point of {s n } is contained in the interval [lim inf{s n }, lim sup{s n }].

51 Cauchy Sequences in R, Ross Chap. 2, Sec. 10 A sequence {s n } in R is called a Cauchy sequence when ( ɛ > 0)( N N)( n, m N) s n s m < ɛ. An equivalent statement is ( ɛ > 0)( N N)( n N) s n s N < ɛ, because then n, m N imply that s n s m s n s N + s m s N < 2ɛ.

52 Ross Lemmas 10.9, 10.10: A convergent sequence is Cauchy and a Cauchy sequence is bounded. Proof: If s n s, then given ɛ > 0 there exists N so that n, m N imply s n s, s m s < ɛ. By the triangle inequality, s n s m s n s + s m s < 2ɛ. If {s n } is Cauchy, then there exists N such that s n s N < 1 and so s n < s N + 1 for all n N. So for all n, s n max( s 1,..., s N 1, s N + 1).

53 Lemma: If {s n } is a Cauchy sequence and s is a cluster point of {s n } then {s n } converges to s. Proof: If ɛ > 0 there exists N such that s n s m < ɛ for all n, m N ({s n } is Cauchy). There exists m 0 N such that s m0 s < ɛ (s is a cluster point). Hence, for all n N, s n s s n s m0 + s m0 s < 2ɛ. Theorem: A sequence {s n } converges iff it is Cauchy. Proof: A Cauchy sequence is bounded and so has cluster points, e.g. the lim inf and lim sup. So it converges. On the other hand we saw above that a convergent sequence is Cauchy.

54 As a side note, we show that from the Archimedean Property it follows, without Completeness, that a bounded monotone sequence is Cauchy. If a sequence {s n } is not Cauchy, then ( ɛ > 0)( N)( n > N) s n s N ɛ. We construct a subsequence {s nk } such that s nk+1 s nk ɛ. If {s n } is increasing then s nk+1 s nk telescoping sum ɛ and so we have the s nk+1 s n1 = (s nk+1 s nk ) + (s nk s nk 1 ) + + (s n2 s n1 ) kɛ. So s nk+1 s n1 + kɛ. By the Archimedean Property, the sequence is unbounded. The contrapositive statement is that a bounded increasing sequence is Cauchy.

55 So we see that the Archimedean Property and convergence of Cauchy sequences implies that bounded monotone sequences converge and so that any nonempty bounded subset has a supremum and infimum. HW Exercises: Ross p. 55 [R 1 p. 41] 9.12; p. 65 [R 1 p. 48] 10.8; p.77 [R 1 p. 56]

56 Series, Ross Chap. 2, Sec. 14 The expression n=1 a n = s means that s is the limit of the sequence of partial sums {s n } with s n = n k=1 a k. We say that the series converges when the limit of partial sums exists. Often we cannot compute the limit when we know it exists. An important exception is the geometric series: Ross Example 14.1: If r < 1, then n=1 ar n = a. 1 r Ross Corollary 14.5 (N th term test): If n=1 a n converges then lim a n = 0. Proof: a n = s n s n 1 for n > 1. Both sequences {s n } and {s n 1 } have the same limit s and so the difference has limit s s = 0.

57 Convergence or divergence is not affected by changing a finite number of terms. Replacing the terms up to N by 0 replaces, for n > N, s n = n k=1 a k by n k=n+1 a k = s n N k=1 a k and so the limit of the new series is s N k=1 a k. Most important, and easiest, is the case when the terms of the series are positive (actually non-negative) a n 0. This is exactly the case when {s n } is an increasing, positive sequence. When {s n } is bounded, it converges to sup{s n }. When it is unbounded, it diverges to infinity. Ross Theorem 14.6 (Comparison Test) If 0 a n b n, and n=1 b n converges, then n=1 a n converges. Proof: The partial sums of the b n series remain bounded and are larger than the partial sums of the a n series. So the a n partial sums are bounded.

58 In any case, a series converges iff {s n } is a Cauchy sequence. Notice that if m > n, then s m s n = m k=n+1 a k. Ross Theorem 14.4: The series n=1 a n converges iff for every ɛ > 0 there exists N N so that if m > n N, m k=n+1 a k < ɛ. The series n=1 a n is called absolutely convergent when the series of positive terms n=1 a n is convergent. Ross Corollary 14.7: An absolutely convergent series is convergent. Proof: m k=n+1 a k m k=n+1 a k.

59 Lemma: (a) If L > 0 and either lim sup( a n ) 1/n < L or lim sup( a n+1 / a n ) < L, then there exists M > 0 such that eventually, a n < M L n. (b) If l > 0 and either lim sup( a n ) 1/n > l or lim inf( a n+1 / a n ) > l, then there exists m > 0 such that frequently, a n > m l n. Proof: (a) Let A n = sup{( a k ) 1/k : k n}. The sequence {A n } decreases with limit lim sup( a n ) 1/n. If for some N, A N < L, then a n 1/n < L and so a n < L n for all n N. If for some N, sup{( a n+1 / a n ) : n N} < L, then a n+1 < L a n for all n N and, by induction, a n ( a N /L N ) L n.

60 (b) If A n > l, then for some k n, a k 1/k > l and so a k > l k. These suprema decrease with n. So if lim sup( a n ) 1/n > l, then A n > l for all n N. It follows that for every n there exists k n such that a k > l k. If for some N, inf{( a n+1 / a n ) : n N} > l, then a n+1 > l a n for all n N and, by induction, a n > ( a N /l N ) l n for all n N.

61 Ross Theorem Root Test: Let α = lim sup( a n ) 1/n. If α < 1, then the series converges absolutely. If α > 1, then the series diverges. Ross Theorem Ratio Test: If lim sup( a n+1 / a n ) < 1, then the series converges absolutely. If lim inf( a n+1 / a n ) > 1, then the series diverges. Proof: If lim sup( a n ) 1/n < 1 or lim sup( a n+1 / a n ) < 1, then we can choose a positive L < 1 such that lim sup( a n ) 1/n < L or lim sup( a n+1 / a n ) < L. From the lemma, there exists N N and M > 0 so that a n < M L n for all n N. By omitting the terms up to N, we can use the comparison test with the geometric series to see that n=1 a n converges.

62 If lim sup( a n ) 1/n > 1 or lim inf( a n+1 / a n ) > 1, then we can choose a positive l > 1 such that lim sup( a n ) 1/n > l or lim inf( a n+1 / a n ) > l. From the lemma, there exists m > 0 so that a n > m l n for infinitely many n N. Since l > 1, m l n and so infinitely often a n > 1. Since infinitely often a n > 1, the N th term test implies that the series diverges.

63 Metric Spaces: Completeness and Compactness, Ross Chap. 2, Sec. 10 A metric space X is a set on which distance is measured by a metric d which satisfies for all x, y X : Positivity d(x, y) 0 and = 0 iff x = y. Symmetry d(x, y) = d(y, x). Triangle Inequality d(x, y) d(x, z) + d(z, y) for all z X. For a sequence {x n } in a metric space X the following definitions generalize those on R. {x n } is convergent with limit x X when ( ɛ > 0)( N N)( n N) d(x n, x) < ɛ. x X is a cluster point of {x n } when ( ɛ > 0)( N N)( n N) d(x n, x) < ɛ. {x n } is a Cauchy sequence when ( ɛ > 0)( N N)( n, m N) d(x n, x m ) < ɛ.

64 Exercise A2.3: For a sequence {x n } in a metric space X, prove each of the following by generalizing the corresponding proofs for R. (a) A point x is a cluster point of {x n } iff there is a subsequence of {x n } which converges to x. (b) If {x n } is convergent with limit x, then x is the unique cluster point of {x n }. (c) If {x n } is convergent then {x n } is Cauchy. (d) If {x n } is Cauchy and x is a cluster point of {x n }, then {x n } converges to x. The metric space X is called complete when every Cauchy sequence is convergent. Two metrics d 1, d 2 on X are called equivalent when there exists M 1 such that M 1 d 1 (x, y) d 2 (x, y) Md 2 (x, y).

65 Exercise A2.4: Show that the properties {x n } converges to x, x is a cluster point of {x n }, {x n } is Cauchy and X is complete, each remain unchanged when the metric is replaced by an equivalent metric. Our first example of a metric space is R k the space of k-tuples x = x 1,..., x k. The usual Euclidean metric is given by d(x, y) = ( k i=1 (x i y i ) 2 ) 1/2. We will not check the Triangle Inequality which uses a bit of linear algebra. Instead, we define the absolute value x = max( x 1,..., x n ) and use as metric applied to the difference of the two vectors x and y. That is, the distance from x to y is x y = max( x 1 y 1,..., x n y n ). (Check the Triangle Inequality). We call a subset S R k bounded when there exists M R such that x M for all x S. Observe that for x, y R k, x y d(x, y) k x y and so the metrics are equivalent.

66 Because of the coordinate subscripts, we write {x (n) } for a sequence in R k. For each i = 1,..., k, the sequence {x (n) i } in R is called the i th coordinate sequence. Ross Lemma 13.3: For a sequence {x (n) } in R k, (a) {x (n) } is convergent with limit x, iff for all i = 1,... k, {x (n) i } is convergent with limit x i. (b) {x (n) } is Cauchy iff for all i = 1,... k, {x (n) i } is Cauchy. (c) {x (n) } is bounded, iff for all i = 1,... k, {x (n) i } is bounded. Proof: x y < ɛ iff for all i, x i y i < ɛ. Ross Theorem 13.4: R k is complete.

67 The Bolzano-Weierstrass Property also extends to R k but is a bit trickier. Exercise A2.5: Show that the sequence with x (n) = n(( 1) n 1), n(( 1) n + 1) in R 2 has no convergent subsequence although 0 is a cluster point for each coordinate sequence. What we do, following Ross, to take successive subsequences, using the fact that a subsequence of a subsequence is a subsequence. This is easier to see if you think of a subsequence as defined by an infinite subset of N. Ross Theorem 13.5 Bolzano-Weierstrass Theorem: Every bounded sequence in R k has a convergent subsequence.

68 Proof: Repeatedly apply the property for R. If {x (n) } is a bounded sequence in R k then {x (n) 1 } is a bounded sequence and by restricting to some infinite subset S 1 N we define a subsequence with convergent first coordinate sequence. {x (n) 2 } on S 1 is a bounded sequence and so by restricting to some infinite subset S 2 S 1 we define a subsequence with convergent second coordinate sequence. On S 2 the first coordinate sequence still converges, and to the same limit. Continue to define infinite subsets S k S k 1 S 2 S 1 N so that for the subsequence defined on S k all of the coordinate subsequences converge.

69 Open and Closed Sets We now define some general topology concepts for a metric space X. Given r > 0, the set B r (x) = {y X : d(x, y) < r} is called the open ball centered at x. For s R the ball B r (s) is the open interval (s r, s + r). For E X a point x is interior to E if for some r > 0, the ball B r (x) E. That is, all points close enough to x also lie in E. The interior E is the set of all interior points of E. E is called an open set if E = E. Lemma: An open ball B r (x) is an open set. Proof: This is not as obvious as the name suggests. For y B r (x) we have to find an r 1 so that the ball B r1 (y) is contained in B r (x).

70 By definition, d(x, y) < r and so r 1 = r d(x, y) > 0. If z B r1 (y) then d(x, z) d(x, y) + d(y, z) < r. Thus, a set is open exactly when it is a union of open balls. Thus, the interior E of any subset E is open. It is the largest open set contained in E. It also follows that any union of any family of open sets is open. If B r1 (x) E 1 and B r2 (x) E 2 then with r = min(r 1, r 2 ), B r (x) E 1 E 2. So the intersection of any two open sets is open and, by induction, the intersection of any finite number of open sets is open. A subset F is closed when its complement X \ F is open.

71 Notice that x E iff ( r > 0) B r (x) E. Furthermore, A B iff A meets the complement of B. That is, A (X \ B). For a subset F, the closure of F, denoted F, is X \ (X \ F ), the complement of the interior of the complement. Thus, x is in the closure of F if for every ɛ > 0 there are points of F ɛ close to x. A set F is closed iff F = F. By DeMorgan s Laws the intersection of any family of closed sets is closed and the finite union of closed sets is closed. The closure of F is the smallest closed subset containing F.

72 Theorem: x F iff there is a sequence {x n } in F which converges to x. In particular, if F is a closed set and {x n } is a sequence in F with limit x then x F. Proof: If {x n } is a sequence in F with limit x and ɛ > 0 then eventually x n B ɛ (x) and so in B ɛ (x) F. Now assume that x F. For every n N choose x n B 1/n (x) F. {x n } is a sequence in F with d(x n, x) < 1/n and so x n x. HW Exercises: Ross p. 94 [R 1 p. 67] 13.9,

73 Open and Closed Set Summary 1- x E means r > 0, B r (x) E. 2- E is open means E = E. That is, for every x E, r > 0, B r (x) E. 3- x F means r > 0, B r (x) E. 4- F is closed means F = F. That is, if r > 0, B r (x) E then x F. 5- F is closed iff the complement X \ F is open. 6- x F iff there is a sequence in F with limit x.

74 7- Any union and any finite intersection of open sets is open. 8- Any intersection and any finite union of closed sets is closed. 9- E is the largest open set contained in E, the union of all open sets contained in E. 10- F is the smallest closed set containing F, the intersection of all closed sets containing F.

75 Theorem: For a sequence {x n } in X the tail set T n = {x i : i n}. The sequence {T n } is a decreasing sequence of closed sets. The intersection T = n T n is the set of cluster points of the sequence {x n }. In particular, the set of cluster points of a sequence is always a closed set. Proof: If x is a cluster point, then for every ɛ > 0 and every N N, there exists n N such that d(x n, x) < ɛ. That is, B ɛ (x) T N. Since this is true for all ɛ > 0, x T N. Since this is true for all N N, x T. Furthermore, this argument is completely reversible, showing that if x T, then x is a cluster point.

76 For a nonempty subset F we define the diameter diam(f ) = sup{d(x, y) : x, y F } +. Notice that the diameter of a nonempty set is 0 exactly when the set consists of a single point. Exercise A2.6: Show that diam(f ) = diam(f ). (Hint: Show that diam(f ) diam(f ) + 2ɛ for any ɛ > 0.) We will say that X has the Cantor Intersection Property if whenever {F n } is a decreasing sequence of nonempty closed sets such that lim diam(f n ) = 0 then F = n F n is nonempty. Since F F n for all n it is clear that the diameter of F is 0 and so it consists of a single point when it is nonempty.

77 Theorem: A metric space X is complete iff it has the Cantor Intersection Property. Proof: For {F n } a decreasing sequence of nonempty closed sets we can choose x n F n. For any N N x n, x m F N for all n, m N. So if lim diam(f N ) = 0, this is a Cauchy sequence. If X is complete, then it converges to some point x. The tail subsequence T N = {x n : n N} lies in F N and has limit x. Since F N is closed, x F N. Since this is true for all N, x F = N F N. Thus, a complete space has the Cantor Intersection Property.

78 Now assume that {x n } is a Cauchy sequence. Let F n = T n, the closure of the n th tail set. We saw before that this is a decreasing sequence of closed sets whose intersection is the set of cluster points. Now given ɛ > 0 there exists N so that for all n, m N, d(x n, x m ) < ɛ. This implies that for all k N, diam(f k ) = diam(t k ) ɛ. Hence, lim diam(f k ) = 0. If the Cantor Intersection Property holds, then there exists a point x in the intersection T n. That is, x is a cluster point of {x n }. Since {x n } is Cauchy, it converges to x. A family U of open sets is called a cover of a subset E when E {U : U U}. That is, every point of E is contained in some member of U. A subfamily U 0 U is called a subcover when it still covers U. We are interested when a finite subcover exists. Example:If E is an unbounded subset of R k, then U = {{x : x < n} : n N} has no finite subcover of E.

79 Compactness We call a subset E of X compact when every open cover of E has a finite subcover. This is also called the Heine-Borel Property. For a family U of open sets, we let F = {X \ U : U U} be the family of complementary closed sets. Notice that U does not cover E iff E ( {F : F F}) is nonempty. That is, iff there is a point x E which is in the complement of every member of U. It follows that E is compact when for a family of closed sets F, E meets the intersection of F when it meets the intersection of each finite subfamily. A subset E is called totally bounded if for every ɛ > 0 the open cover {B ɛ (x) : x E} has a finite subcover. This says that for every ɛ > 0 there is a finite list x 1,..., x k of points in E such that for every x E, d(x, x i ) < ɛ for some i = 1,..., k.

80 Theorem: For a subset E of a metric space X the following three conditions are equivalent. (i) E is compact. That is, E satisfies the Heine-Borel Property. (ii) Every sequence in E has a cluster point in E. That is, E satisfies the Bolzano-Weierstrass Property. (iii) E is totally bounded and every Cauchy sequence in E has a limit in E. That is, E is complete and totally bounded. Proof: (i) implies (ii) - For a sequence {x n } in E, F n = Tn is a decreasing sequence of closed sets. Since it is decreasing, we have E ( N n=1 F n) = E F N T N. Assuming compactness it follows that E ( n=1 F n) is nonempty. These are the cluster points of the sequence which lie in E. (ii) implies (iii) - We show the contrapositive: If either E is not totally bounded or not complete, then the Bolzano-Weierstrass Property fails.

81 If E is not complete then there is a Cauchy sequence {x n } in E which does not converge to a point of E. Since {x n } is Cauchy this is the same as saying that the sequence has no cluster point in E contradicting Bolzano-Weierstrass. If E is not totally bounded then there is an ɛ > 0 such that {B ɛ (x) : x E} does not have a finite subcover. Choose x 1 E. There exists x 2 E \ B ɛ (x 1 ) and so d(x 1, x 2 ) ɛ. Inductively, we construct a sequence {x n } in E such that m < n implies d(x n, x m ) ɛ. Assume x 1,..., x k have been chosen. We may choose x k+1 E \ (B ɛ (x 1 ) B ɛ (x k )). This sequence has no cluster points. In fact, for any x X the open ball B ɛ/2 (x) contains at most one term of the sequence. Again this contradicts Bolzano-Weierstrass. (iii) implies (i)- We assume that E is complete and totally bounded and that U is an open cover of E with no finite subcover. We then derive a contradiction, proving the result indirectly.

82 We begin by using total boundedness to choose for each n N a list B 1,n,..., B kn,n of balls of radius 1/n, centered at points of E so that for each n, E k n i=1 B i,n. Each F i,n = B i,n is a closed set with diameter at most 2/n. As they are subsets of E, each E F i,n is covered by U. Now if U had a finite subcover for each E F i,1 then by putting them all together we would have a finite subcover for E. So there is some F i,1 whose intersection with E cannot be covered by a finite subfamily of U. Relabel this F i.1 as K 1. Again if U had a finite subcover for each E K 1 F i,2 then we would be able to get a finite subcover for E K 1. So there is a K 2 = K 1 F i,2 such that E K 2 has no finite subcover.

83 We define a decreasing sequence of closed sets {K n } each with diameter at most 2/n and such that U has no finite subcover of E K n. Inductively, from K n we choose K n+1 to be K n F i,n+1 for some i = 1,..., k n+1. Since diam(k n ) 0, we can choose x n E K n and, as in the proof of the Cantor Intersection Theorem, we obtain a Cauchy sequence in E. By completeness, it converges to x E so that {x} = n K n. Now there exists U U with x U. Since U is open, there exists ɛ > 0 such that B ɛ (x) U. For sufficiently large n, diam(k n ) < ɛ. Since x K n, we have K n B ɛ (x) U. But this means that {U} is a finite subcover of K n. However, K n was built to have no finite subcover of U. This contradiction completes the proof.

84 Corollary: If E is a compact subset of a metric space X, then E is closed. Proof: If x E then there is a sequence {x n } in E which converges to x. Since convergence implies Cauchy, {x n } is a Cauchy sequence in E. So completeness of E implies that the limit x lies in E. Thus, E = E. Thus, a subset E of R k is compact iff it is closed and bounded. HW Exercises: Ross p. 94 [R 1 p. 67] 13.12,

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