1 Sets Functions and Relations Mathematical Induction Equivalence of Sets and Countability The Real Numbers...

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1 Contents 1 Sets Functions nd Reltions Mthemticl Induction Equivlence of Sets nd Countbility The Rel Numbers Sequences nd Convergence The Arithmetic of Sequences Subsequences Convergence of Sequences nd Series Open sets, Closed sets Limits of Functions 29 4 Continuitiy Uniform Continuity Properties of Continuous Functions Differentition The Men Vlue Theorem L Hôpitl s Rule The Riemnn Integrl Integrbility i

2 ii CONTENTS 6.2 Arithmetic of Integrls The Fundmentl Theorem

3 0 CONTENTS

4 Chpter 1 Sets A set is colletion of mthemticl objects. Sets, themselves, re mthemticl objects. One wy to think of sets is s jrs or continers tht hve mthemticl objects, possibly including other sets, inside of them. If X is set nd x is some mthemticl object, we write x X to denote the fct tht x is in the collection X. We red this s * x is n element of X or s x is in X. If this is not the cse, we write, insted, x X nd red it s x is not in X. We sy two sets re equl if they hve exctly the sme elements. Tht is, X = Y if nd only if for every mthemticl object x, we hve x X if nd only if x Y. There re two stndrd wys of describing sets. In the first, we simply list the elements of the set between brces ({}). So, for exmple, the empty set = {}. This is the set tht hs no elements in it. We could lso spek of the set A = {1, 2, 3, 4} where 1 A, but 5 A. Some cre is needed since, for exmple, { } is set with exctly one element in it (nmely, the empty set), so { }. In this wy, we hve the set N = {1, 2, 3, } of ll nturl numbers nd the set Z = {, 2, 1, 0, 1, 2, } of ll integers. The second wy to describe set is by considering ll x tht hve some property P (x). For this we write the set A = {x P (x)}. Then x A if nd only if P (x) is true. Now, in this P (x) should be sttement tht depends on x nd 1

5 2 CHAPTER 1. SETS is either true or flse for ech mthemticl object x. So, nother description of the emptyset is = {x x x}. In this, x x is our property P (x) nd this is flse for ll objects x. In this wy, we cn define Q = {x : there re m, n Z with x = m n nd n 0}, the set of ll rtionl numbers. Then 1 3 Q, for exmple. Definition 1.1 Let A nd B be sets. We sy A is subset of B nd write A B if for ll objects x, x A implies tht x B. In other words, everything in A is in B. Directly from the definitions of equlity of sets nd of subsets, we hve tht A = B if nd only if both A B nd B A. This is very common wy to how tht two sets re equl. Some cution with lnguge is good thing t this point. If A nd B re sets, the phrse A is contined in B could men either A B or A B. These re very different concepts nd it should lwys be mde cler which of these two possibilities is ment. For exmple {1} {1, 2}, but {1} {1, 2}. If X is set, we define the power set of X by P(X) = {A A X}. Hence, P(X) is the collection of ll subsets of X. Notice tht P(X) nd X P(X). As n exmple, if X = {1, 2}, then P(X) = {, {1}, {2}, {1, 2}}. Notice tht, here, we hve {1} P(X), but 1 P(X). Next if A nd B re sets, we define A B = {x x A nd x B} nd A B = {x x A or x B}. We cll these sets the intersection nd the union of A nd B, respectively. We hve the following: Proposition 1.2 Let A, B, d C be sets. then 1. A B = B A 2. A B = B A 3. (A B) C = A (B C) 4. (A B) C = A (B C) 5. A (B C) = (A B) (A C)

6 1.1. FUNCTIONS AND RELATIONS 3 6. A (B C) = (A B) (A C) Next, if A nd B re sets, we define the complement of B in A by A \ B = {x x A nd x B}. We then hve the following: Proposition 1.3 Let A, B, nd C be sets. Then, 1. A \ (A \ B) = A B 2. A \ (B C) = (A \ B) (A \ C) 3. A \ (B C) = (A \ B) (A \ C) 4. (A B) \ C = (A \ C) (B \ C) 5. (A B) \ C = (A \ C) (B \ C) 1.1 Functions nd Reltions If x nd y re mthemticl objects, we define the ordered pir (x, y) = {{x}, {x, y}}. The only property of (x, y) tht is required is the following: Proposition 1.4 We hve tht (x, y) = (, b) if nd only if x = nd y = b. A funtion is defined to be set f of ordered pirs such tht whenever (x, y) f nd (x, z) f, we my conclude tht y = z. In other words, there is t most one second component for ech first component for elements of f. In this cse, if we hve (x, y) f, we write y = f(x). If f is function, we define the domin of f by dom f = {x there is y such tht (x, y) f} nd the imge of f by im f = {y there is n x such tht (x, y) f}. If, now, A = dom f nd im f B, we sy tht f is function from the set A to the set B nd write f : A B. Notice tht the domin of f must be the

7 4 CHAPTER 1. SETS set A, but the imge does not hve to be the set B. The imge need only be subset of B. Now, let A nd B be sets, nd define the cross product A B = {(x, y) : x A, y B} = {z : prticulr, if f : A B, then f A B. there re x A nd y B such tht z = (x, y)}. In Now suppose tht Λ is set nd f is function with Λ = dom f such tht f(λ) is set for ech λ Λ. Let s write A λ for f(λ). Then, we sy tht {A λ } λ Λ is n indexed fmily of sets with index set Λ. For exmple, we might tke Λ = N nd define A n = { m n : m Z}. Then A n consists of those frctions whose demonintors re n. It is very common for the index set of fmily of sets to be either N or some subset of N. As nother exmple, we might hve Λ = Z nd define A n = {x Z : x > n}. Then A 0 = N, for exmple. Now, suppose tht {A λ } λ Λ is n indexed fmily of sets. union nd intersection of this fmily by nd A λ = {x there exists λ Λ with x A λ } λ Λ A λ = {x x A λ for ll λ Λ}. λ Λ We define the As n exmple, suppose tht Λ = N nd A n = { m n m Z}. We clim tht n N A n = Q nd n N A n = Z. As nothr exmple, if Λ = Z nd A n = {x Z x > n}, then n Z A n = Z nd n Z A n =. In generl, we write n=1 A n = n N A n whenever {A n } n N is fmily of sets indexed by N with similr nottion for intersections. We now hve: Proposition 1.5 Suppose tht X is set nd tht {A λ } λ Λ is n indexed fmily of sets. Then ) X \ ( λ Λ A λ) = λ Λ (X \ A λ) b) X \ ( λ Λ A λ) = λ Λ (X \ A λ)

8 1.2. MATHEMATICAL INDUCTION 5 Definition 1.6 Suppose tht f : X Y is function. We sy tht f is one-toone if the only wy for f(x 1 ) = f(x 2 ) is for x 1 = x 2. Alterntively, function is ne-to-one when x 1 x 2 implies tht f(x 1 ) f(x 2 ). Next, we sy tht f is onto if whenever y Y, there is n x X with y = f(x). In other words, f : X Y is onto when im f = Y. Finlly, we sy tht f is one-to-one correspondence (or n equivlence, or bijection) if f is both one-to-one nd onto. Definition 1.7 Suppose tht f : X Y nd g : Y Z re functions. We define the composition, g f : X Z by setting g f(x) = g(f(x)) for ll x X. Definition 1.8 A reltion is ny set of ordered pirs. If R is reltion, sy R X Y, then we cn define the opposite reltion ˆR = {(x, y) (y, x) R}. Notice tht ˆR = R. Proposition 1.9 A function f : X Y is one-to-one exctly when ˆf is function. In this cse, dom ˆf = im f nd im ˆf = dom f. In prticulr, if f : X Y is bijection, then ˆf : Y X is lso bijection. In this cse, we write f 1 for ˆf. Proposition 1.10 Suppose tht f : X Y nd g : Y Z. then we hve ) If both f nd g re one-to-one, so is g f. b) If both f nd g re onto, so is g f. c) If both f nd g re bijections, so is g f. d) If g f is onto, so is g. e) If g f is one-to-one, so is f. 1.2 Mthemticl Induction We ssume property of the nturl numbers clled well-ordering. It sys tht if A N is non-empty, then there is smllest element of A. In other words, there is n A such tht x for every x A.

9 6 CHAPTER 1. SETS Theorem 1.11 Suppose tht P (n) is property tht nturl number cn either hve or not hve. Suppose 1. P (0) is true. 2. For every n N, if P (n) is true, then P (n + 1) follows. Then P (n) is true for ll n N. Proof: Let A = {n N : P (n) is true}. Then 0 A, so A is not empty. We hve to show tht A = N. Suppose this is not the cse, so N \ A. Let be the smllest element of N \ A. Then 0, so 1 N. But then 1 A, so P ( 1). By the second hypothesis, we cn conclude tht P (( 1) + 1) is true. This show tht A, contrry to the definition of. 1.3 Equivlence of Sets nd Countbility Definition 1.12 We sy two sets X nd Y re equivlent if there exists bijection f : X Y. In this cse, we write X Y. Theorem 1.13 Let X, Y, nd Z be sets. Then ) X X. b) If X Y, then Y X. c) If X Y nd Y Z, then X Z. Definition 1.14 We sy set A is finite if it is either empty or there is n n N such tht A {1, 2,, n}. We sy tht A is countbly infinite if A N nd tht A is countble if it is either finite or countbly infinite. Proposition 1.15 A set A is countble if nd only if there is subset B N with A B. Proof:

10 1.3. EQUIVALENCE OF SETS AND COUNTABILITY 7 The forwrd direction is cler from the defintion of countbility. So now suppose tht B N nd tht A B. To prove the result, it is enough to show tht eitherb =, B {1,, n} for some n N or tht B N. If B is not empty, then there is smllest element of B. We cll this element b 1. Now, if B = {b 1 }, then B {1}, so we re done. Otherwise, there is smllest element b 2 of B \ {b 1 }. If B = {b 1, b 2 }, then B {1, 2}, so we re done. We proceed by induction. Either this process stops t some point, in which cse B {1,, n} for some n N, or B = {b 1, b 2, }, so B N. Proposition 1.16 Subsets of countble sets re countble. Proposition 1.17 N N is countble. Proof: Define f : N N N by f(m, n) = 2 m 3 n. By unique fctoriztion, f is one-to-one. Thus N N im f N, so N N is countble. Theorem 1.18 A set A is countble if nd only if there is one-to-one mp f : A N. f : N A. Similrly, A is countble if nd only if there is n onto mp Proof: If f : A N is one-to-one, then A im f N, so A is countble. On the other hnd, if f : N A is onto, then for ech A, there is n n N with f(n ) =. Define g : A N by g() = n. Then g is one-to-one, so A is countble. Conversely, if A is countble, the definition provides f : A N which is one-to-one. If A is countbly infinite, then the definition lso gives n f tht is onto. Now, if A is finite, choose f : {1,, n} A which is one-to-one nd onto. Then define F : N A by F (k) = f(k) if k n nd F (k) = f(1) if k > n. Then F is onto. Theorem 1.19 Supose tht {A n } n=1 is fmily of countble sets indexed by N. Then A = n=1 A n is countble.

11 8 CHAPTER 1. SETS Proof: Choose, for ech n, n onto function f n : N A n. Now define f : N N A by f(n, m) = f n (m). Then f is onto, so A is countble. It is good question whether every set is countble. Tht this is not the cse follows from the following. Proposition 1.20 Let X be set. Then X P(X). Proof: Suppose tht f : X P(X) is ny mp. We will show tht f cnnot be onto. Hence there is no bijection from X to P(X). Let A = {x X : x f(x)}. Then A X, so A P(X). Suppose, though tht f(y) = A. There re two choices for y. Either y A, in which cse the definition of A shows tht y f(y) = A, which is contrdiction, or y A = f(y), n which cse y A. Since either cse leds to contrdiction, no such y cn exist, so f is not onto. We will lter show tht the set of rel numbers, R is uncountble. 1.4 The Rel Numbers We ssume the existence of set R nd two opertions + : R R R nd : R R R. We usully write x+y nd x y for +(x, y) nd (x, y), resp. We lso ssume there is reltion < R R where we write x < y for (x, y) <. We further ssume the following properties: 1. For ll x, y, z R, we hve (x + y) + z = x + (y + z). 2. For ll x, y R, we hve x + y = y + x. 3. There is n element 0 R such tht x + 0 = x for ll x R. 4. For ech x R, there is y R such tht x + y = For ll x, y, z R, we hve (x y) z = x (y z). 6. For ll x, y R, x y = y x. 7. There is n element 1 R with 0 1 nd x 1 = x for ll x R.

12 1.4. THE REAL NUMBERS 9 8. For ech x 0 in R, there is y R with x y = For ll x, y, z R, we hve x (y + z) = x y + x z. 10. If x < y nd z R, then x + z < y + z. 11. If 0 < x nd y < z in R, then x y < x z. 12. If x, y, z R nd x < y nd y < z, then x < z. 13. If x, y R, exctly one of the following is true: x < y,y < x, or x = y. We write x y if x < y or x = y. Similrly, we write x > y is y < x nd x y if x > y or x = y. Notice tht ll the xioms so fr re stisied by Q with the usul opertions. This mens tht the finl xiom is our only wy to distinguish Q from R. Proposition 1.21 ) The element 0 bove is unique. b) For ech x, the element y such tht x + y = 0 is unique. We cll it x. c) The element 1 boce is unique. d) For ech x 0, the element y with x y = 1 is unique. We cll it 1 x. e) If x < y, then y < x. f) If 0 < x < y, then 0 < 1 y < 1 x. g) 0 < 1. h) If x R, then 0 x 2. Definition 1.22 For < b in R, define [, b] = {x R : x b}. Similrly, define (, b), [, b), nd (, b]. Also, define [, ) = {x R : x}. Similrly define (, ), (, ), nd (, ]. To understnd our finl xiom, we need some terminology. Suppose tht A R. We sy tht A is bounded bove if there is n x R such tht x for ll A. We lso sy tht x is n upper bound for A in this cse.

13 10 CHAPTER 1. SETS We sy tht x 0 R is lest upper bound for A if it is n upper bound for A nd if x is ny upper bound for A, we hve x 0 x. In other words, x 0 is the smllest possible upper bound for A. Our xiom is then Lest Upper Bound Property: Every non-empty subset of R with n upper bound in R hs lest upper bound in R. Proposition 1.23 For ech x R, there is n n N with x n. Proof: Assume not. Then there is n x R with n < x for ll n N. In other words, x is n upper bound for N. By our xiom, there is lest upper bound for N. Cll it x 0. Then n x 0 for ll n N. But, since x 0 1 < x 0, x 0 is NOT n upper bound for N. This mens tht there is n n 0 N with x 0 1 < n 0. But then x 0 < n 0 + 1, which contrdicts tht x 0 is n upper bound for N. Proposition 1.24 If x R, there is n n Z such tht n x < n + 1. Proof: First ssume tht x 0. By the previous result, there is n n N with x < n. By the fct tht N is well-ordered, there is smllest n 0 N such tht x < n 0. But then n 0 1 x, so n = n 0 1 works in the result. Now suppose tht x < 0. Then x > 0, so there is n n 0 Z with n 0 x < n But then n 0 1 < x n 0. If x n 0, let n = n 0 1. Otherwise, let n = n 0. Proposition 1.25 Let < b in R. Then, there is n r Q with < r < b. Proof: First, b > 0, so 1/(b ) > 0, so there is n n N with 1/(b ) < n. now find m Z such tht m < n b m + 1. Then we hve 1 < n (b ) = nb n, so n < nb 1 m < nb, so < m/n < b. let r = m/n. Proposition 1.26 There is no x Q with x 2 = 2 (here, 2 = by definition). There is n x R with x 2 = 2. In fct, for ech 0, there is n x 0 with x 2 =. Proof: First ssume tht 1. Consider the set A = {x R : x > 0, x 2 }. Notice tht 1 A, so A is non-empty. Also, suppose tht x R. Then either

14 1.4. THE REAL NUMBERS 11 x <, x =, or x >. In the lst cse, though x 2 = x x > x > 1 =, so x A. In other words, x A implies tht x, so is n upper bound for A. Using our xiom, let x 0 be the lest upper bound for A. The clim is tht x 2 0 =. To show this, we note tht the only other possibilities re tht x 2 0 < or x 2 0 >. We eliminte these two possibilities. Cse I: x 2 0 <. Let ε = min{1, ( x 2 0)/(2x 0 + 1)}. Then ε > 0,so x 0 + ε > x 0. But (x 0 + ε) 2 = x x 0 ε + ε 2 x x 0 ε + ε x (2x 0 + 1)ε x ( x 2 0) =. This mens tht x 0 + ε A, which contrdicts the fct tht x 0 is n upper bound for A. Cse II: x 2 0 >. In this cse, let ε = (x 2 0 )/(2x 0. Then (x 0 ε) 2 = x 2 0 2x 0 ε + ε 2 > x 2 0 2x 0 ε = x 2 0 (x 2 0 ) =. Now, if x > x 0 ε, we hve tht x 2 > (x 0 ε) 2 >, so x A. in other words, if x A, x x 0 ε. This shows tht x 0 ε is n upper bound for A. But, x 0 ε < x 0, which contrdicts tht x 0 is the lest upper bound for A. The only other possibility is tht x 2 0 =. This is our result for 1. Now, if = 0, x = 0 works for x 2 =. If 0 < < 1, then 1/ > 1, so by the previous cse there is n x 0 with x 2 0 = 1/. But now (1/x 0 ) 2 =, so we hve the result in this cse lso. Proposition 1.27 Let x < y in R. x < z < y. Then there is z R \ Q such tht Proof: Since x < y, x 2 < y 2. By previous result, there is r Q with x 2 < r < y 2. Then x < r + 2 < y. Then, with z = r + 2, we hve the result. Finly, we mke definition. Definition 1.28 For x R, we define x to be x if x 0 nd to be x if x < 0. Notice tht x 0 in every cse nd tht x x x. Proposition 1.29 We hve the folowing:

15 12 CHAPTER 1. SETS ) xy = x y b) x + y x + y c) x < is equivlent to < x <. d) If b R, then x b < is equivlent to b < x < b +.

16 Chpter 2 Sequences nd Convergence Definition 2.1 A sequence is function with domin N. We will be concerned with rel-vlued sequences, which re functions : N R. Insted of (n), it is trditionl to write n nd to write { n } n=1 for the sequence. Typiclly, formul is given for n in terms of n N. For exmple, if we let = { 1 n } n=1, we hve n = 1 n, so for exmple 3 = 1 3. As nother exmple, let b = {1 + ( 1) n } n=1. Then b 1 = 0, b 2 = 2, b 3 = 0,. Definition 2.2 Let { n } n=1 be sequence nd A R. We sy tht { n } n=1 converges to A if for every ε > 0, there is n N N such tht whenever n N, we hve n A < ε. This will be written s n A. As n exmple, we show tht { 1 n } n=1 converges to 0. Suppose tht ε > 0. Choose N N such tht 1 ε < N. Then, whenever n N, we hve 1 n 0 = 1 n 1 N < ε. On the other hnd, we show tht { 1 n } n=1 does NOT converge to 1. To do this, we need to negte the definition of convergence. Hence, we need to show tht there is n ε > 0 such tht for ll N N, there is n n N with 1 n 1 ε. I tke ε = 1 2. For ny N N, pick n mx{2, N}. Then 1 n 1 = 1 1 n = ε. In this definition, ε represents how close we wnt the terms ot he sequence to be to A nd N is how fr out in the sequence we hve to go to gurntee 13

17 14 CHAPTER 2. SEQUENCES AND CONVERGENCE tht the sequence is within this degree of closeness to A. It is cler tht if N works for given ε, ny lrger nturl number will lso. The first thing we wnt to do is show tht sequence cnnot converge to two different rel numbers simultneously. Proposition 2.3 Suppose tht the sequence { n } n=1 converges to both A nd B. Then A = B. Proof: Suppose not, in other words, suppose tht A B. Then ε = A B /3 > 0. Since { n } n=1 converges to A, there is n N 1 N such tht whenever n geqn 1 we hve n A < ε. SImilrly, there is n N 2 N such tht whenever n N 2, we hve n B < ε. Now, to obtin our contrdiction, let n N be chosen so tht n N 1, N 2. We then hve tht A B A n + n B ε + ε = 2ε = 2 A B. 3 This is contrdiction since A B > 0. Definition 2.4 We sy sequence { n } n=1 converges if there exists n A R such tht { n } n=1 converges to A. By the previous result, if such n A exists, it is unique. As nother exmple, we show tht the sequence { n = 1 + ( 1) n } n=1 does not converge. In fct, suppose tht n A nd let ε = 1. Then, there is n N N such tht whenever n N we hve n A < ε. Let n be ny odd integer with n N. Then n + 1 N lso, so we hve n A, n+1 A < ε. But n = 0 nd n+1 = 2, so A < ε nd 2 A < ε. But this shows tht 2 = 2 = 2 A + A < ε + ε = = 2, contrdiction. We will find mny more cses of sequences tht fil to converge in this chpter. 2.1 The Arithmetic of Sequences Proposition 2.5 Suppose tht n A nd b n B. Then n + b n A + B.

18 2.1. THE ARITHMETIC OF SEQUENCES 15 Proof: Let ε > 0. Choose N 1 N such tht whenever n N 1, we hve n A < ε/2. Similrly, choose N 2 N such tht whenever n N 2, we hve b n B < ε/2. Let N mx{n 1, N 2 }. Then, if n geqn, we hve ( n + b n ) (A + B) = n A + b n B n A + b n B < ε/2 + ε/2 = ε. It should be pointed out tht essentilly the sme proof shows tht n b n A B. Now let s consider wht it will tke to show tht n b n AB. Clerly, we hve to consider the expression n b n AB. The quetion is how to mke this smll (i.e., less thn ε) given the knowledge tht n A nd b n B re both smll. After some plying round, nd remembering trick used in the proof of the product rule for derivtives, we dd nd subtrct n B nd rgue tht n b n AB = n b n n B + n B AB n b n B + B n A. Now, in both terms we hve some expressions tht re smll. In the second, the expression is multiplied by B, so we cn del with it by using ε/ B insted of ε in the defintion of convergence of n. But the first term hs n, which depends on n nd so cnnot simply be divided off the ε. But, if we cn show tht n never gets too lrge, we cn still mke things work. We give definition first. Definition 2.6 We sy sequence { n } n=1 is bounded if there exists M R such tht n M for ll n N. As n exmple, the sequence {( 1) n } n=1 is bounded (we cn tke M = 1), but it not convergent. However, we do hve the following positive result: Proposition 2.7 Suppose tht { n } n=1 converges. Then it is bounded. Proof: Suppose tht n A. In the definition of convergence, tke ε = 1 nd find N N such tht whenever n N, we hve n A < ε = 1. Then, whenever n N, we lso hve n = n A + A n A + A < 1 + A. Now, if we let M = mx{ 1, 2,, N 1, 1 + A }, we hve tht n M for ll n N. Now we cn prove Proposition 2.8 Suppose tht n A nd b n B. Then n b n AB.

19 16 CHAPTER 2. SEQUENCES AND CONVERGENCE Proof: Let ε > 0. Since { n } n=1 converges (to A), it is bounded. Let M R with n M for ll n N. We my ssume M 0 (if it ws, M = 1 works just s well). Now, find N 1 N such tht whenever n N 1 we hve n A < ε/2( B + 1). We use B + 1 insted of B just in cse B = 0. Also, pick N 2 N such tht whenever n N 2, we hve b n B < ε/2m. Let N = mx{n 1, N 2 }. Then, for ny n N, we hve n b n AB = n b n n B + n B AB n b n B + B n A < M ε/2m + B ε/2( B + 1) < ε/2 + ε/2 = ε nd we re done. If we try to pply this result to the sequence { ( 1)n n } n=1, by writing n = ( 1) n nd b n = 1 n, we find tht it doesn t work becuse {( 1)n } n=1 doesn t converge. Nonetheless, we cn show tht { ( 1)n n } n=1 converges to 0 s follows: Proposition 2.9 Suppose tht { n } n=1 is bounded nd b n 0. Then n b n 0. Proof: Let ε > 0 nd M R such tht n M for ll n N. We my ssume M 0. Choose N N such tht whenever n N, we hve b n 0 < eps/m. Then, for these sme n, we hve n b n 0 = n b n M b n 0 < M ε/m = ε. If { n } n=1 is not bounded, this result fils. For exmple, tke n = n nd b n = 1/n. Then b n 0, yet n b n 1. Of course, our next step is to show tht division works for limits of sequences. We hve to be sure both tht the sequence we divide by is never 0 nd tht the limit is not 0 just to be sure everything we wnt is defined. Our first step is to show tht reciprocls work well. So, suppose tht n 0 for ll n nd tht n A with A 0. How do we show tht 1 n 1 A? We hve to consider 1 n 1 A = A n n A. Now, the numertor converges to 0 by ssumption, so if we 1 cn show tht n A is bounded, we will be done by the following result. Proposition 2.10 Let { n } n=1 be sequence. Then n A if nd only if n A 0.

20 2.1. THE ARITHMETIC OF SEQUENCES 17 This is n ssigned homework problem. Another useful, if esy result: Proposition 2.11 Suppose tht n b n for ll n N nd suppose tht b n 0. Then n 0. Proof: Suppose ε > 0 nd choose N N such tht n N gives b n 0 < ε. Then n 0 = n b n < ε. Lemm 2.12 Suppose tht n 0 for ll n N nd n A with A 0. Then { 1 n } n=1 is bounded. Proof: In the definition of convergence, choose ε = A /2 > 0 nd find N N so tht n N implies tht n A < ε = A /2. Then, for these n, we hve A = A n + n n A + n < A /2+ n, so n > A A /2 = A /2. But then 1 n < 2/ A. Now, let M = mx{ 1 1,, 1 N 1, 2/ A } to se tht 1 n M for ll n N. Proposition 2.13 Suppose tht n 0 for ll n N nd tht n A with A 0. Then 1 n 1 A. Proof: We need to show tht 1 n 1 1 A 0. So, n 1 A = A n = 1 n A A 1 n n A. Since { 1 n } n=1 is bounded nd 1 A is constnt, nd n A 0, we obtin our result. Proposition 2.14 Suppose tht n 0 for ll n N nd n A 0. Suppose lso tht b n B. Then bn n B A. Proof: b n n = b n 1 n B 1 A = B A. It is often the cse tht none of our results so fr pplies to given sequence, but the sequence cn be mnipulted so tht these results pply. Exmple: Let n = 2n+1 3n+2. Since neither {2n + 1} n=1 nor {3n + 2} n=1 converge, the result on division of convergent sequences does not pply. However, if we multiply nd divide by 1 2n+1 n, we see tht 3n+2 = 2+ 1 n 3+ 2 n. SInce 1 n 0, we see

21 18 CHAPTER 2. SEQUENCES AND CONVERGENCE tht n = 2 nd n = 3, so we find tht 2n+1 3n Much more complicted results cn be obtined in this wy. Finlly, we give result showing how convergence of sequences reltes to the order properties of R. Proposition 2.15 Suppose tht n A, b n B nd n b n for ll n N. Then A B. Proof: Suppose not, tht is B < A. Let ε = (A B)/2 > 0. Find N N such tht whenever n N, we hve both n A < ε nd b n B < ε. Then, for every n N, we hve b n < B + ε = B + (A B)/2 = (A + B)/2 = A (A B)/2 = A ε < n, in contrdiction to our ssumption. It should be pointed out tht the previous result is flse if is replced by < throughout. In fct, let n = 1 1 n nd b n = n. Then n < b n for every n. But, n 1 nd b n 1, so we hve equlity of the limits. 2.2 Subsequences Definition 2.16 Suppose tht { n } n=1 is sequence nd n 1 < n 2 < is sequence of nturl numbers. sequence { n } n=1. We sy tht { nk } k=1 is subsequence of the For exmple, suppose tht n = ( 1) n. If we choose n k = 2k, we get tht nk = ( 1) 2k = 1 for ll k N. Alterntively, if we chose m k = 2k 1, then mk = 1 for ll k N. So, even though the originl sequence fils to converge, these prticulr subsequences converge. Since the originl sequence is subsequence of itself, it is cler tht some subsequences fil to converge. A positive result in this direction is the following. Proposition 2.17 Suppose tht n A. Then for ny subsequence, nk A. Proof: First notice tht since n 1 N, we hve 1 n 1. Also, if k n k for some k N, we hve tht k + 1 n k + 1 n k+1. Thus, indiction shows tht k n k for every k N.

22 2.2. SUBSEQUENCES 19 Now, let ε > 0 nd find N N such tht whenever n N, we hve n A < ε. Then, if k N, we hve n k k N, so nk A < ε. Thus nk A. This is ctully very hndy wy to show tht some sequences fil to converge. If we cn find two subsequences tht converge to different limits, the whole sequence cnnot converge t ll. For exmple, suppose tht n = ( 1) n (1 + 1 n ). Then, it is esy to see tht 2k = k 2k+1 = ( k+1 ) 1. Hence { n} n=1 cnnot converge. 1 while Definition 2.18 Let { n } n=1 be sequence nd A R. We sy tht A is cluster point of { n } n=1 if for ll ε > 0, the set {n N : n A < ε} is infinite. For exmple, 1 is cluster point of the sequence {( 1) n } n=1. On the other hnd, the sequence {n} n=1 hs no cluster points t ll. More interestingly, we know tht Q is countble set, so there is n onto function : N Q. This function is, by definition sequence. Furthermore, if A R is ny rel number nd ε > 0, we know tht there re infinitely mny rtionl numbers in the open intervl (A ε, A + ε), so the set {n N : n A < ε} is infinite. Hence every rel number is cluster point of this sequence! Theorem 2.19 A point A is cluster point of the sequence { n } n=1 if nd only if there is subsequence { nk } k=1 such tht n k A. Proof: First ssume tht there is subsequence with nk A. Let ε > 0 nd choose N so tht k N implies tht nk A < ε. Then every n k with k N is in the set {n N : n A < ε}, so this set is infinite. This A is cluster point of the sequence { n } n=1. Conversely, suppose tht A is cluster point of { n } n=1. We need to find subsequence which converges to A. First, tke ε = 1 in the definition of cluster point to see tht there is n n 1 such tht n1 A < 1 (in fct, there re infinitely mny!). Now, suppose tht n 1 < n 2 < n k hve been chosen. Choose ε = 1 k+1 in the definition of cluster point to see tht the set {n N : n A < 1 k+1 } is

23 20 CHAPTER 2. SEQUENCES AND CONVERGENCE infinite. In prticulr, there is n n k+1 in this set nd with n k < n k+1. In this wy, we inductively find n 1 < n 2 < such tht nk A < 1 k for ll k N. Since 1 k 0, we see tht n k A 0, so nk A. Theorem 2.20 (Bolzno-Weierstrss Theorem) Every bounded sequence hs cluster point. Proof: Suppose tht { n } n=1 is bounded sequence with n M for ll n N. Let x 0 = M nd y 0 = M. Then x 0 n y 0 for ll n N. In prticulr, the set {n N : x 0 n y 0 } is infinite. Let z 0 = (x 0 + y 0 )/2 nd notice tht {n N : x 0 n y 0 } = {n N : x 0 n z 0 } {n N : z 0 n y 0 }. Hence t lest one of the two sets on the right is infinite. If the first one is, let x 1 = x 0 nd y 1 = z 0. If not, let x 1 = z 0 nd y 1 = y 0. Then, we hve tht {n N : x 1 n y 1 } is infinite nd tht y 1 x 1 = (y 0 x 0 )/2. If we ssume tht we hve chosen x 0 x k < y k y 0 such tht {n N : x k n y k } is infintie nd such tht y k x k = (y 0 x 0 )/2 k, then let z k = (x k +y k )/2 so tht z k x k = y k z k = (y k x k )/2 = (y 0 x 0 )/2 k+1. Now {n N : x k n y k } = {n N : x k n z k } {n N : z k n y k }, so t lest one of the sets on the right is infinite. if it is the first, let x k+1 = x k nd y k+1 = z k, nd otherwise let x k+1 = z k nd y k+1 = y k. In this wy, the induction continues. In this wy, we obtin sequences {x n } n=1 nd {y n } n=1 such tht x 0 x 1 x k y k y 1 y 0 such tht the set {n N : x k n y k } is infinite nd such tht y k x k = (y 0 x 0 )/2 k for ech k N. There re now two things to notice: the first is tht the set {x k : k N} is bounded bove (by y 0, for exmple). The second is tht y k x k = (y 0 x 0 )/2 k 0. Let A = sup{x k : k N}. The clim is tht A is cluster point of the originl sequence { n } n=1. To see this, let ε > 0. Then A ε is not n upper

24 2.3. CONVERGENCE OF SEQUENCES AND SERIES 21 bound for {x k : k N}, so there is K 1 N with A ε < x K1 A. Notice tht for ll k K 1, we hve A ε < x K1 leqx k A. Next, there is K 2 such tht k K 2 implies tht y k x k < ε. Choose ny k K 1, K 2. Then, we hve A ε < x k < y k = x k + (y k x k ) < x k + ε A + ε. In other words, [x k, y k ] (A ε, A + ε). But then, {n N : x k n y k } {n N : n A < ε}. Since the first set is infinite, so is the second. This proves tht A is cluster point, proving the result. Corollry 2.21 (Sequentil comleteness of bounded sets) Every bounded sequence hs convergent subsequence. Theorem 2.22 A bounded sequence converges if nd only if every convergent subsequence converges to the sme vlue. Proof: One direction is esy since if { n } n=1 converges to A, every subsequence converges to A lso. Hence every convergent subsequence converges to A. Now, suppose tht { n } n=1 is bounded sequence nd tht every convergent subsequence converges to A. We need to show tht n A. Assume not. Then, there is n ε > 0 such tht for every N N, there is n N with n A ε. Pick n 1 such tht n1 A ε. Next, pick n 2 n with n2 A ε. Continue in the wy to find n 1 < n 2 < with nk A ε for ll k N. The problem is tht the sequence { nk } k=1 probbly does not converge. However, it is bounded sequence, so it hs subsequence { nkj } which does converge. But this is subsequence of { n } n=1 which does not converge to A, which violtes our ssumption. 2.3 Convergence of Sequences nd Series Definition 2.23 We sy sequence { n } n=1 is non-decresing if n n+1 for ll n N. Similrly, we sy the sequence is non-incresing if n+1 n for ll n N. Finlly, we sy { n } n=1 is monotone if it is either non-decresing or non-incresing.

25 22 CHAPTER 2. SEQUENCES AND CONVERGENCE Theorem 2.24 A non-decresing sequence is convergent if nd only if it is bounded. Proof: We know tht every convergent sequence is bounded, so one direction is lredy known. For the converse, ssume tht { n } n=1 is non-decresing nd bounded. Consider the set { n : n N}. It is non-empty nd bounded bove, so it hs lest upper bound. Let A = sup{ n : n N}. We clim tht n A. In fct, suppose tht ε > 0. Since A ε is not n upper bound, there is N N such tht A ε < N A. But then, for ech n N, we hve A ε < N n A < A + ε, so n A < ε. As n exmple, let n = n. Using induction, it is esy to see tht n = n. Hence, n 2. But now, consider b n = n!. Then {b n } n=1 is clerly non-decresing nd b n n < 2, so it is bounded lso. Hence {b n } n=1 converges. This is very useful wy to show sequences correspoonding to infinite sums re convergent. Definition 2.25 We sy tht sequence { n } n=1 is Cuchy if whenever ε > 0, there is N N such tht whenever n, m N, we hve n m < ε. The difference between convergence to point nd being Cuchy is tht for n A, we know the limit point (A), nd the terms of the sequence re getting close to A. For Cuchy sequences, we only know tht the terms re getting closer to ech other. Proposition 2.26 Every convergent sequence is Cuchy. Proof: Suppose tht n A nd tht ε > 0. Choose N N such tht for ll n N we hve n A < ε/2. Then, whenever m, n N, we hve n m = ( n A) ( m A) n A + m A < ε/2 + ε/2 = ε. Proposition 2.27 A Cuchy sequence with convergent subsequence is, itself, convergent.

26 2.3. CONVERGENCE OF SEQUENCES AND SERIES 23 Proof: Suppose tht { n } n=1 is Cuchy nd tht nk A. We show tht n A. Let ε > 0. Choose N 1 N such tht m, n N implies tht n m < ε/2. Also, choose N 2 N such tht k N 2 implies tht nk A < ε/2. Now, let n N be ny nturl number. Choose ny k N, N 2. Then n k N so n A n nk + nk A < ε/2 + ε/2 = ε nd we re done. Proposition 2.28 Every Cuchy sequence is bounded. Proof: Suppose tht { n } n=1 is Cuchy equence. Choose ε = 1 in the definition of Cuchy to find N N such tht whenever m, n N we hve n m < 1. Then, for ech n N we hve n N + N n < N +1. Now let M = mx{ 1,, N 1, N +1} to see tht n M for ll n N. Theorem 2.29 A sequence is convergent if nd only if it is Cuchy. Proof: We hve shown tht convergent sequences re Cuchy. Conversely, A Cuchy sequence is bounded, nd so hs convergent subsequence. But then, the originl sequence must converge. While the previus proof ws esy, it required the very difficult Bolzno- Weiestrss theorem for its proof. This mkes it mjor result. Definition 2.30 Let { n } n=1 be sequence. We cn construct nother sequence {s n } n=1 of prtil sums by setting s n = n. We sy the series n=1 n converges if nd only if {s n } n=1 converges. Furthermore, if s n A, we write n=1 n = A. Proposition 2.31 If n=1 n converges, then n 0. Proof: Let {s n } n=1 be the sequence of prtil sums nd ssume tht s n A. Then n = s n s n 1 A A = 0. Proposition 2.32 Suppose tht n 0 for ll n N. Then n=1 n converges if nd only if the sequence of prtil sums is bounded.

27 24 CHAPTER 2. SEQUENCES AND CONVERGENCE Proof: Notice tht the sequence of prtil sums {s n } n=1 is non-decresing since n 0. Hence, it converges if nd only if it is bounded. Proposition 2.33 Supose tht n b n nd tht n=1 b n converges. Then n=1 n converges. Proof: Let {s n } n=1 be the sequence of prtil sums for n=1 n nd {t n } n=1 the sequence of prtil sums for n=1 b n. Then, since {t n } n=1 converges, it is Cuchy. We show tht {s n } n=1 is lso Cuchy, so it, in turn, converges. Let ε > 0. Choose N so tht m, n N implies tht t n t m < ε. We my ssume m < n. But then s n s m = m+1 + n m+1 + n b m+1 + b n = t n t m < ε nd we re done. Theorem 2.34 (Alternting Series Test) Suppose tht { n } n=1 is non-incresing with n 0 for ll n. Then n=1 ( 1)n+1 n converges if nd only if n 0. Proof: Let {s n } n=1 be the sequence of prtil sums nd let x n = s 2n 1 nd y n = s 2n. Then, x n+1 = s 2n+1 = S 2n 1 2n + 2n+1 x n nd y n+1 = s 2n+2 = s 2n + 2n+1 2n+2 s 2n = y n since n is non-incresing. Thus, {y n } n=1 is non-decresing nd {x n } n=1 is non-incresing. Next, y n = s 2n = n = 1 ( 2 3 ) ( 4 5 ) 2n 1, so {y n } n=1 is bounded bove. Similrly, {x n } n=1 is bounded below. Hence, both {x n } n=1 nd {y n } n=1 converge. Let x n A nd y n B. Then x n y n = 2n 0, so A B = 0, so A = B. Thus, s 2n nd s 2n 1 both convege to A, so s n A. In other words, our series converges. As n exmple, ( 1) n+1 n=1 n converrges even though n=1 1 n does not. 2.4 Open sets, Closed sets Definition 2.35 We sy tht subset U R is open if, for every x U, there is n ε > 0 such tht (x ε, x + ε) U. We sy subset C R is closed if R \ C is open.

28 2.4. OPEN SETS, CLOSED SETS 25 For exmple, let R nd set U = (, + ). We show tht U is open. In fct, if x U, then < x. Let ε = x > 0. Then (x ε, x+ε) = (, 2x ) (, + ) = U. Similrly, (, ) is lso open. Proposition 2.36 Suppose tht U nd V re open sets. Then U V is open. Proof: Let x U V. Then, there is n ε 1 > 0 such tht (x ε 1, x + ε 2 ) U nd n ε 2 > 0 such tht (x ε 2, x + ε 2 ) V. Let ε = min{ε 1, ε 2 }. Then (x ε, x + ε) U V. In prticulr, if < b, then (, b) = (, + ) (, b) is open. Hence open intervls re open. Proposition 2.37 Suppose tht {U α } α I Then α I U α is open. is n indexed fmily of open sets. Proof: Suppose tht x U = U α. Then there is n α I with x U α. Since U α is open, ther eis n ε > 0 such tht (x ε, x + ε) U α U. Thus, every union of open intervls is n open set. Next, if R, then [, + ) is closed becuse R\[, + ) = (, ) is open. Also, (, ] is closed. We hve the following using DeMorgn s theorem. Proposition 2.38 Suppose tht C nd D re closed sets. Then C D is closed. Proposition 2.39 Supose tht {C α } is n indexed fmily of closed sets. Then Cα is closed. The investigtion of open nd closed sets is prt of the subject mtter of topology. It turns out to be very useful to use convergence of sequences to determine when set is open or closed. The following provides route to do this. Theorem 2.40 Let C R. The following re equivlent: ) C is closed. b) Whenever { n } n=1 is sequence with n C for ll n N nd n x, then we cn conclude tht x C.

29 26 CHAPTER 2. SEQUENCES AND CONVERGENCE Proof: Suppose tht C is closed nd tht { n } n=1 is sequence with n C nd n x. Suppose lso tht x C. Then x R \ C, which is n open set. Hence, there is n ε > 0 with (x ε, x + ε) R \ C. Since n x, there is n N N such tht whenever n N, we hve n x < ε, which implies tht x ε < n < x + ε. But this implies tht n R \ C for ll n N, in contrdiction to our ssumption. For the converse, suppose tht C is not closed. Then R \ C is not open, which mens there is x R\C such tht whenever ε > 0, the set (x ε, x+ε) is not contined in R \ C. But this mens tht for ech ε > 0, there is n C with x < ε. Now, for ech n N, choose n C with n x < 1 n. Then, n x nd n C for ll n N but x C. In other words, if C is not closed, the second condition fils lso. Definition 2.41 Let A R nd let x R. We sy tht x is n ccumultion point of A if for ech ε > 0, the set A (x ε, x + ε) is infinite. As n exmple 1 is n ccumultion point of the set (0, 1) since (0, 1) (1 ε, 1 + ε) = (1 ε, 1) is infinite for every ε > 0. Similrly, every rel number is n ccumultion point of the set Q nd of the set R \ Q. Theorem 2.42 Let A R nd x R. Then the following re equivlent: ) x is n ccumultion point of A. b) For ech ε > 0, there is n element of A (x ε, x + ε) different from x. c) There is sequence { n } n=1 of distinct points in A with n x. Proof: ) implies b):suppose tht x is n ccumultion point of A. Then for ech ε > 0, the set A (x ε, x + ε) is infinite, so hs t lest one point different from x. c) implies ): If { n } n=1 is sequence of distinct points in A with n x, let ε > 0. Then, there is n N N such tht n x < ε for ll n N. In other

30 2.4. OPEN SETS, CLOSED SETS 27 words, the infinite set { n : n N} A (x ε, x + ε). Thus A (x ε, x + ε) is infinite for every ε > 0, so x is n ccumultion point of A. b) implies c): Let ε 1 = 1. There is n 1 A with 1 x nd 1 x < 1. Now, let ε 2 = min{ 1 2, 1 x }. Then ε 2 > 0 nd so there is n 2 x with 2 x < ε 2. Notice tht 1 2 by construction. Inductively, suppose we hve 1, k A ll distinct nd different from x with j x < 1 j for 1 j k. Set ε k+1 = min{ 1 k+1, 1 x,, k x }. Then ε k+1 > 0 so there is n k+1 A with k+1 x nd k+1 x < ε k+1. Then k+1 is different thn ll the previous j nd k+1 x < 1 k+1, so the induction continues. This gives us sequence { n } n=1 of distinct terms with n A for ll n nd n x. An importnt consequence of this is the following: Proposition 2.43 Assume A R is bounded bove. Then sup A is n element of A or n ccumultion point of A. Proof: Suppose tht x = sup A A. Then, for ech ε > 0, there is n A with x ε < < x < x + ε. Hence, x is n ccumultion point of A. Theorem 2.44 (Bolzno-Weierstrss Theorem) Every bounded infinie set hs n ccumultion point. Proof: Let A [ M, M] be infinite. Inductuvely, pick sequence { n } n=1 of distinct terms with n A for ll n. This sequence is then bounded, so it hs convergent subsequence. This subsequence lso hs distinct terms, so its limit is n ccumultion point of A. Theorem 2.45 Let A R nd let A be the set of ccumultion points of A. Then A A is closed. Proof: So, let A R nd set B = A A. We show tht R \ B is open. So, suppose tht x R \ B. Then x A, so there is n ε > 0 so tht A (x ε, x + ε) contins no point other thn x. But, since x A, we must

31 28 CHAPTER 2. SEQUENCES AND CONVERGENCE hve A (x ε, x + ε) =. But then, no point of (x ε, x + ε) cn be n ccumultion point of A, so A (x ε, x + ε) = lso. In other words, (x ε, x + ε) B =, so (x ε, x + ε) R \ B. This shows tht R \ B is open, so B is closed. Corollry 2.46 Let A R. Then A is closed if nd only if A A. Proof: If A is closed nd x A, then there is sequence { n } n=1 in A of distinct terms with n x. By our chrcteriztion of closed sets vi sequences, x A. Hence A A. Conversely, if A A, then A = A A is closed. Corollry 2.47 Suppose A is closed nd bounded bove. Then sup A A. Exercises: 1. Suppose tht 0 < r < 1. Show tht the sequence {r n } n=1 converges. Then find its limit. Wht hppens if 1 < r < 0? 2. Suppose tht 1 < r. Show tht {r n } n=1 is unbounded. Hint: if, not, show it converges. Wht would it hve to converge to? 3. Suppose tht r 1. Show tht 1 + r + r 2 + r r n 1 = 1 rn 1 r. 4. Suppose tht r < 1. Show tht n=1 rn 1 = 1 1 r. 5. Suppose tht 0 <, b. Show tht lim n ( n + b n ) 1/n = mx{, b}. ( 6. Let x 1 = 1 nd x n+1 = 1 2 x n + 2 x n ). Show tht x 2 n x2 n 2 2. From this, show tht x n 2.

32 Chpter 3 Limits of Functions In this chpter, we wnt to consder limits of functions. In prticulr, we wnt to justify clcultions like those done in elementry clsses such s x 2 4 lim x 2 x 2 = lim x 2 (x 2)(x + 2) x 2 = lim x 2 x + 2 = 4. Sevrl things should be pointed out in this clcultion. First, nd most significntly, the function involved, f(x) = (x 2 4)/(x 2) is not defined t the plce we re pproching in the limit (i.e. x = 2). However, the point x = 2 is n ccumultion point of the domin of f, i.e. {x R : x 2}. Next, the second inequlity depends on the fct tht the x-vlues we tke re not equl to 2, while the lst limit seems to plug in x = 2. These chrcteristics of limits mde the forml definition reltively lte occurence. We give it here. Definition 3.1 Let f : D R be function with D R nd suppose tht is n ccumultion point of D. We sy tht f hs limit L R s x pproches nd write lim x f(x) = L if for ll ε > 0, there is δ > 0 such tht whenever x D with 0 < x < δ, we hve f(x) L < ε. We sy tht the limit of f s x pproches exists if there is n L R such tht the limit of f s x pproches is L. 29

33 30 CHAPTER 3. LIMITS OF FUNCTIONS Another thing to point out is tht since we use 0 < x < ε, the vlue of f() is irrelevnt for the existence of the limit even when D. As n exmple, we show tht lim x 2 x 2 = 4. Let ε > 0 nd choose δ = min{1, ε/5}. Then δ > 0. Implicitly in this exmple, we hve D = R, so 2 is n ccumultion point of D. So, suppose tht x R with 0 < x 2 < δ. Then, first, x x < = 3, so x = 5. Then, we hve f(x) L = x 2 4 = (x 2)(x + 2) x 2 5 < ε s required. Proposition 3.2 Suppose tht the limit of f s x pproches is both L 1 nd L 2. Then L 1 = L 2. In other words, limits re unique. We hve implicitly used this when we write lim x f(x) = L, so some cre is required here. Proof: Suppose L 1 L 2 nd set ε = L 1 L 2 /2. Then pick δ 1 > 0 such tht whenever x D with 0 < x < δ 1, we hve f(x) L 1 < ε. Smilrly, pick δ 2 > 0 such tht whenever x D with 0 < x < δ 2, we hve f(x) L 2 < ε. Let δ = min{δ 1, δ 2 }. Since is n ccumultion point of D, there is n x D with x nd x < δ. For this x, we hve f(x) L 1 < ε nd f(x) L 2 < ε. But this implies tht L 1 L 2 = (f(x) L 2 ) (f(x) L 1 ) < ε + ε = 2ε = L 1 L 2, contrdction. Exmples: ) We lwys hve tht lim x x =. In fct, if ε > 0, simply choose δ = ε. Then, if x D nd 0 < x < δ, we hve x < ε. b) let D = R. We show tht lim x 3 x 2 + x = 12. In fct, let ε > 0. Choose δ = min{1, ε/8}. Then, if x D with 0 < x 3 < δ, we hve first of ll tht x + 4 = (x 3) + 7 x < = 8. Now, this shows tht (x 2 + x) 12 = (x 3)(x + 4) = x 3 x + 4 < 8 x 3 < ε. c) Let D = R \ {0} nd let f(x) = cos( 1 x ). Notice tht 0 is n ccumultion point of D. We clim tht f does not hve limit s x pproches 0. In fct, suppose to the contrry tht lim x 0 f(x) = L nd choose ε = 1. Then, there would be δ > 0 such tht whenever 0 < x 0 < δ, we

34 31 hve f(x) L < ε. But, we hve tht the sequence 1 πn 0, so there is n N N such tht whenever n N, we hve 1 πn 0 < δ. But then one of N nd N + 1 is odd nd the other is even. Thus one of cos(πn) nd cos(π(n + 1)) is 1 nd the other is 1. letting x = 1 πn 1 nd x = π(n+1) shows tht L 1 < ε nd L ( 1) < ε. But then, 2 = 1 ( 1) 1 L + L ( 1) < 2ε = 2. This is contrdiction. d) Let D = (0, 1) nd define f(x) = 0 if x is irrtionl nd f(x) = 1 q if x = p q is rtionl nd in lowest terms. We clim tht lim x f(x) = 0 for ll [0, 1]. In fct, let [0, 1] nd let ε > 0. There is n N N such tht 1 N < ε. Now consider the set of frctions of the form p q where q N. There re only finitely mny such frctions. Let δ = min{ p q : q N, p q } be the smllest distnce from to one of these frctions. Then δ > 0. Now suppose tht x D nd 0 < x < δ. Either x is irrtionl, in which cse f(x) 0 = 0 0 = 0 < ε, or x is rtionl, but x = p q where q > N. In the ltter cse, f(x) 0 = 1 q 0 = 1 q < 1 N < ε. At this point, we cn prove ll of the lgebric properties of limits for functions in wy tht is similr to how we showed them for sequences in the lst chpter. The following result is n exmple of this technique. Proposition 3.3 Let f, g : D R nd n ccumultion point of D. Suppose tht lim x f(x) = L 1 nd lim x g(x) = L 2. Then lim x f(x) + g(x) = L 1 + L 2. Proof: Let ε > 0. Choose δ 1 > 0 such tht whenever x D with 0 < x < δ 1 we hve f(x) L 1 < ε/2. Similrly, find δ 2 > 0 such tht whenever x D with 0 < x < δ 2 we hve g(x) L 2 < ε/2. let δ = min{δ 1, δ 2 }, so tht δ > 0. Now, if x D nd 0 < x < δ, we hve f(x) + g(x) (L 1 + L 2 ) f(x) L 1 + g(x) L 2 < ε/2 + ε/2 = ε, s required. Just s for sequences, we cn define boundedness of functions: Definition 3.4 Let f : D R. We sy tht f is bounded if there is n M R such tht f(x) M for ll x D.

35 32 CHAPTER 3. LIMITS OF FUNCTIONS Proposition 3.5 Suppose tht f : D R nd is n ccumultion point of D where lim x f(x) exists. Then, there is n δ > 0 such tht f is bounded on the set D ( δ, + δ). It is possible to prove ll the results for the lgebric properties of limits of functions vi techniques similr to those used for the properties of limits of sequences. However, it turns out to be more helpful to use the results from sequences to do this. For this we need the following, very importnt result. Theorem 3.6 Let f : D R nd nd ccumultion point of D. The following re equivlent. ) The limit lim x f(x) = L. b) Whenever { n } n=1 is sequence in D with n for ll n N nd n, we hve tht f( n ) L. Proof: ) implies b): Suppose tht lim x f(x) = L. Also, suppose tht n D, n, nd n. We need to show tht f( n ) L. So let ε > 0. The, there is δ > 0 such tht whenever x D with 0 < x < δ, we hve f(x) L < ε. But then, there is n N N such tht whenever n N, we hve n < δ. But now, for these n, we hve tht n, so 0 < n < δ, which implies tht f( n ) L < ε. In other words, f( n ) L. b) implies ): Suppose tht f does not hve limit of L s x D pproches. Then, there is n ε > 0 such tht for every δ > 0, there is n x D with 0 < x < δ yet f(x) L ε. Let δ = 1 n nd find x n D with 0 < x < 1 n but such tht f(x n) L ε. Then x n with x n for ll n, but {f(x n )} n=1 does not converge to L. This contrdicts our ssumption nd shows tht lim x f(x) = L. As n exmple of the use of this result, we show tht limits of products work the right wy. Proposition 3.7 Let f, g : D R nd n ccumultion point of D. Assume tht lim x f(x) = L nd lim x g(x) = M. Then lim x f(x)g(x) = LM. Similrly, if g(x) 0 for ll x D nd M 0, then lim x f(x) g(x) = L M.

36 33 Proof: Suppose tht { n } n=1 in D with n for ll n. Then, f( n ) L nd g( n ) M, so by the result on products of sequences, f( n )g( n ) LM. Since this hppens for ll sequences { n } n=1, we hve tht lim x f(x)g(x) exists nd equls LM. The result for quotients follows similrly. The next result is bit more subtle. Insted of ssuming convergence to specific point, it merely ssumes convergence. Theorem 3.8 Assume f : D R nd is n ccumultion point of D. The following re equivlent: ) The limit of f s x pproches exists. b) Whenever { n } n=1 is sequence in D with n nd n, then the sequence {f( n )} n=1 converges. Proof: ) implies b): This follows directly from Theorem 3.6. b) implies ): The difficulty here is tht we only ssume tht the sequences {f( n )} n=1 converge, not tht they ll converge to the sme thing. Since is n ccumultion point of D, there is t lest one sequence { n } n=1 in D with n for ll n nd n. By ssumption, {f( n )} n=1 converges, sy f( n ) L. Now, ssume tht {b n } n=1 is ny other sequence in D with b n for ll n nd b n. We need to show tht f(b n ) L lso (we only know this sequence converges, NOT tht it converges to L). To do this, let {c n } n=1 be defined by c 2n 1 = n nd c 2n = b n, so c n lterntes between n nd b n. Then c n D nd c n neqx for ll n. Also, c n. By ssumption {f(c n )} n=1 converges, sy f(c n ) L. But now, {f( n )} n=1 is subsequence of {f(c n )} n=1, so f( n ) L lso. Hence, L = L. But now, {f(b n )} n=1 is lso subsequence of {f(c n )} n=1, so f(b n ) L = L. Now we cn pply Theorem 3.6 gin to see tht lim x f(x) = L. Exercises: 1. Show tht lim x 3 x 2 9 x 3 = 6 using the ε, δ definition. 2. Show tht lim x 2 x 2 + 4x + 5 = 17 using the ε, δ definition.

37 34 CHAPTER 3. LIMITS OF FUNCTIONS 3. Suppose tht f, g : D R nd is n ccumultion point of D. Suppose tht g is bounded nd tht lim x f(x) = 0. Show tht lim x f(x)g(x) = 0 using the ε, δ definition. 4. Suppose tht f, g, h : D R nd is n ccumultion point of D. Suppose tht f(x) g(x) h(x) for ll x D nd tht lim x f(x) = lim x h(x) = L. Show tht lim x g(x) = L. 5. Suppose tht f : D R nd is n ccumultion point of D. Suppose lso tht for ech ε > 0 there is δ > 0 such tht whenever x, y D with x y nd x y < δ, then f(x) f(y) < ε. Show tht lim x f(x) exists. Hint: Suppose tht n D with n. Show tht {f( n )} is Cuchy.