Let s Learn Limits or Basics of Analysis By Howard G. Tucker Professor Emeritus of Mathematics University of California, Irvine

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1 by Let s Learn Limits or Basics of Analysis By Howard G. Tucker Professor Emeritus of Mathematics University of California, Irvine I. A word about absolute values -6. Definition. If x is any real number, we define its absolute value, x, ½ x if x 0 x = x if x< Proposition. If x is a real number, and if >0, then x < if and only if <x<. Proof. Only if part: Assume x < ;toprove <x<. Case (i) x 0. In this case, <0 x<,so <x<. Case (ii) x<0. In this case, x = x, so0 < x < or <x<0 <. If part. Assume <x< ;toprove x <. Again there are cases. Case (i) x 0. In this case, 0 x<,sobyourdefinition of absolute value, x <. Case (ii) x<0. Inthiscase, <x<0, or x <. But in this case, x = x. So x <. Q.E.D. -4. Corollary. If x a <,thena <x<a+. Proof: By Proposition -5, wehave <x a <.Addinga throughout, we obtain the conclusion. -3. Proposition. For all real numbers, a and b, ab = a b. Proof. If both numbers are nonnegative, then so is ab and thus ab = ab = a b. In the case where a<0 and if b>0, thenab < 0 so ab = ab. But a = a and b = b, so in this case, ab = ab = a b. Incasea<and b<0, thenab > 0 so ab = ab. Also,a = a and b = b, andab 0, so ab = a b in this case too. -. Lemma. If x 0 and if y 0, andifx <y,thenx<y. Proof. Since x <y,theny x > 0. But y x =((y + x)(y x). Since y + x>0, therefore y x>0, which in turn implies y>x. -1. Proposition: For all real numbers a and b, a + b a + b. Proof. Observe that, by Proposition -, a + b =(a + b) = a +ab + b a + a b + b ( a + b ). 1

2 Applying Lemma - gives us our result. 0. Flavor Theorem. If x /0, andifforevery >0, the inequality x< is true, then x =0. Proof: Suppose not: Then x > 0. Let = x. Then x = >, contradicting the hypothesis. II. Limits of Sequences. 1. Definition. A sequence of numbers, {a 1,a,,a n, },or{a n }, is said to converge to a number a, called its limit, if for every >0,(no matter how small might be,) there exists a natural number N (which possibly depends on ) suchthatforalln>n the inequality a n a < / is satisfied.in suchacasewewrite a n a as n, or lim a n = a. n. Theorem. If a n a as n,andifk is any real number, then Ka n Ka as n. Proof: There are two cases. Case 1: K =0. In this case, let >0 be arbitrary, and let N denote any natural number. Then for every n > N, it follows that Ka n Ka = 0 0 =0<, so the theorem is true in this case. In the case K 6= 0, again let >0 be arbitrary. Then > 0, sobythedefinition of a K n a as n,there exists a natural number N such that for all n>n, a n a < K. Thissaysthesamethingas K a n a < for all n>n, which, by proposition -3, is the same as Ka n Ka < for all n>n, which proves the theorem in this case.

3 3. Theorem. If a n a as n,andifb n b as n,then a n + b n a + b as n. Proof: Let >0 be arbitrary. We must prove that there exists a natural number N such that for all n>n, the inequality (a n + b n ) (a + b) <. Indeed, since >0, thenalso > 0, so we know that there exists N 1 and N such that a n a < for all n>n 1 and b n b < for all n>n, Let N be any natural number larger than N 1 and N. Then for every n>n,both inequalities are true, so, by Proposition -1, we have (a n + b n ) (a + b) = (a n a)+(b n b) (a n a) + (b n b) < + =, which proves the theorem. (Look again at this proof and see why I had the foresight to use the s at the beginning of the proof.) 4. Lemma. If a n a as n,andifa>0, thenthereexistsn N such that for all n>n, it follows that a n > a and a n < 3a Proof. Since by hypothesis a>0, then it follows that a > 0. Therefore by the definition of a n a as n, it follows that there exists N N such that for all n>n, a n a < a. This is the same as stating: for all n>n, the inequalities a <a n a< a are true. By the first inequality we get a n >a a = a and a n <a+ a = 3 a for all n>n. Q.E.D. 5. Theorem. If a n a as n,andifa n a>0for all n, then 1 a n 1 a as n. Proof: Let >0 be arbitrary; we wish to find N N such that for all n>n, it follows that 1 1 a n a <. Since by hypothesis, a n a>0for all n, it follows that the a n s and a are either all positive or are all negative. We consider only the case when all are positive. Then we know that there exits N 1 N such that for all n>n, a n a < a. Also, by lemma 4, there exists N N such that for all n>n, a n > a,or 1 <. So let N equal the larger of N a 1 and N. Then for all n>n, a n 1 1 a n a = a n a <. Q.E.D. a n a 6. Definition. A sequence {a n } is said to be bounded if there exist numbers u<vsuch that u<a n <vfor all n. 3

4 7. Lemma. If a n a as n, then the sequence {a n } is bounded. Proof: This lemma is just a slight extension of lemma 4. Let >0 be arbitrary.. Then by definition of a n a as n, there exists N N such that for all n > N, a n a <, or, what amounts to the same, a <a n <a+. Now let u in Definition 6 be the smallest number in the set {a, a 1 1,,a N 1}, andletv be the largest number in {a+, a 1 +1,,a N +1}. Fromthisitfollowsthatu<a n <vfor all n N. Q.E.D. 8. Theorem. If a n a as n,and if b n b as n,then a n b n ab as n. Proof: Let >0bearbitrary; we shall show there exists a natural number N such that for all n>n,itfollowsthat a n b n ab <. A consequence ofthehypothesesisthatbylemma7thereexistsapositivenumberk such that b n <Kfor all n N. We also know that there exists N 1 N and N N such that for all n>n 1, a n a < 1 and for all n>n K, b n b < 1.LetN =max{n a 1,N }.Thenforalln>N, a n b n ab = (a n b n ab n )+(ab n ab) a n b n ab n + ab n ab = a m a b n + b n b a < 1 K + 1 a =. Q.E.D. K a 9. Theorem. If a n a as n,ifb n b as n and if b n b>0 for all n, then a n bn a as n. b 1 Proof. By hypothesis and Theorem 5, b n 1 as n. Since by b hypothesis a n a as n, it follows by Theorem 8 that a n bn a as b n. Q.E.D. 10. Corollary. If a n a as n,thena n a as n, Proof. In theorem 8, replace b by a everywhere in the proof you see it. Pre-11. Behold the following tautology: p, q, r, (( q) (r ( r))) (p q). 11. Theorem. If a n a as n and if a n b as n,then a = b. (Thus uniqueness.) Proof. Suppose the conclusion is not true. In this case, assume a<b. Then let = 1 (b a). Clearly >0. By hypothesis there exists N N such that for all n>n, a n a < and a n b <. By the first inequality, 4

5 a n < 1(b a) for all n>nandbythesecondinequality,a n 1 (b a). Now apply the above tautology. 1. Proposition.If a n a as n,andifb n b as n, and if a n b n for all n N, thena b. Proof. Suppose to the contrary that a>b. Then a + a a + b, so a >a+b or a> a+b a+b. Similarly, b+b a+b, sob a+b or b. Since a a+b = a b > 0, we know that there exists N 1 N such that for all n>n 1, a n > a b. Similarly, b + a b = a+b > 0, so there exists N. N such that for all n>n, the inequality b n < a+b is true.hence for all n>max{n 1,N }, we obtain the inequality a n > a+b >b n, yielding a contradiction. III Limits of Sets. 13. Definition: If S is a nonempty set of real numbers, then we define arealnumbers as a supremum of S, and denote it by sup S or lub(s) if for every >0, (i) s is an upper bound of S, thatis,foreveryt S, t s, and (ii) no number less that s is an upper bound if S, thatis,forevery >0 there exist t S such that t>s. 14. Completeness Axiom. If S is a set of real numbers that satisfies (i) S 6=, and(ii) S is bounded above, then sup S exists. (This is frequently referred to as an axiom. Actually, it is a theorem, way down the road from the Peano Axioms that I gave in Math 13.) 15. Definition. A sequence of real numbers {a n } is said to be a Cauchy sequence if for every >0, there exists a natural number N such that for all n>n and for all m>n,then a n a m <. 16. Theorem. Ifa n a as n,then{a n } is a Cauchy sequence. Proof. Let >0 be arbitrary. Then > 0, sothereexistsn such that for all n>n and for all m>n, a n a < and a m a <. Thus for all n>n and for all m>n,wehave a n a m = (a n a)+(a a m ) a n a + a a m < + =. Q.E.D. 17. Lemma. If {a n } is a Cauchy sequence, then it is bounded. (In other words, there exist real numbers K<Lsuch that for all n, K<a n <L.) Proof: By the definition of a Cauchy sequence, there exists a natural number N such that for all m>nand all m>n, a n a m < 1. By 5

6 Corollary -1, the inequality is true for all n N +1.Solet and let a N+1 1 <a m <a N+1 +1, K =min{a 1,,a N,a N+1 1}, L =max{a 1,,a N,a N+1 +1}, and we may conclude that for all n, K a n L. 18. Theorem. If {a n } is a Cauchy sequence, then there exists a real number a such that a n a as n. Proof: Let S denote the set of all real numbers x such that only a finite number of terms of the sequence are less than x. This set is nonempty by lemma 16. Also by lemma 17, this set is bounded above, that is, there exists an upper bound of S. By the completeness axiom given above, there exists a least upper bound of S, sup Ṡ. Let >0 be arbitrary. Then there exists N such that for all n>n 1 and all m>n 1, a n a m <.Bythedefinition of sup S there are only a finite number of terms of the sequence that are equal to or less than sup S. We denote these terms by a k1,,a kr, where k 1 <k < <k r.letn =max{k r,n 1 }.Thenforalln>N, a n sup S <. Thus, a n sup S as n. Q.E.D. 19. Theorem. If {a n } is a bounded sequence, then sup{a n : n 1} exists. Proof. This follows from the Completeness Axiom. 0. Definition. A sequence {a n } is said to be a nondecreasing sequence if a n a n+1 for all n Theorem. If {a n } is a a nondecreasing and bounded sequence, then there exists a real number a such that a n a as a,anda =sup{a n : n 1}. Proof. Since the sequence is bounded, we know that sup{a n.n 1} exists. Denote this number by a. Let >0 be arbitrary. Then by definition of sup, 6

7 there exists N that satisfies a N >a. Since by hypothesis, {a n } is a nondecreasing sequence, we have for all n>n, a <a N a n a<a+, which implies: for all n>n, a <a n <a+, or a n a <.. Theorem. If A B aresetsinr 1,ifA 6= and if B is bounded, then sup A and sup B exist and sup A sup B. Proof. Since sup B is an upper bound of A, then by the definition of sup A, it follows that sup A sup B. 3. Definition. If a set S of real numbers is nonempty, then we define inf S as a number that satisfies: (i) infs is a lower bound of S, thatis, inf S s for all s S, and(ii) for every >0, there exists an s S such that s<inf S Theorem. If S is a nonempty set that is bounded below, then inf S exists. Proof. Let K denote a lower bound of S. Thismeansthatforalls S, s K. Let S be defined by S = { s : s S}. Then the set S is a set that is bounded above., so sup( S) exists by theorem. Easily, inf S = sup( S). 5. Definition. A sequence {a n } is said to be nonincreasing if a n a n+1 for all n Theorem. If {a n } is a nonincreasing sequence that is bounded (below), then there exists a number a such that a n a as n,and a =inf{a n }. Proof: Replace {a n } by { a n } and apply theorem Definition. If {a n } is a sequence that is bounded above, then we define lim sup n a n as a number a that satisfies: for every >0, then (i) a n >a+ foratmostafinite number of values of n, and (ii) a n >a for infinitelymanyvaluesofn. 8. Theorem. Suppose {a n } is a sequence that is bounded above and for which there exists a number K such that infinitely many c n s are greater than K. If the sequence {c n } is defined by c n =sup{a k : k n}, 7

8 then there exists a number, c, suchthatc n c, which is denoted by c = lim sup a n. n Proof: The hypotheses imply that each c n exists. Let >0 be arbitrary. We wish to show that there exists a real number c such that c n c. Note that {a k : k n +1} {a k : k n}, and hence c n+1 =sup{a k : k n +1} sup{a k : k n} = c n for all natural numbers n. By theorem 6, we obtain the conclusion. 9. Theorem. If {a n } is a sequence that is bounded from below and is such that there exists a real number K such that infinitely many a n s are less than K, and if the sequence {c n } is defined by c n =inf{a k : k n}, then there exists a number, c, suchthatc n c, which is denoted by c = lim inf n a n. Proof. The hypotheses imply that each c n exists. We wish to show that there exists a real number, c, such that c n c as n.notethat and hence {a k : k n +1} {a k : k n}, c n+1 =inf{a k : k n +1} inf{a k : k n} = c n for all natural numbers n. By theorem 1, we obtain the conclusion 30. Theorem. If {a n } and {b n } are sequences that converge to a and b respectively as n,then max{a n,b n } max{a, b} as n and min{a n b n } min{a, b} as n. 8

9 Proof. Case (i) a = b. Let c = a = b, andlet >0 be arbitrary. By hypothesis, a n c as n, and so there exists N 1 such that for all n>n 1, c <a n <c+. Also, by hypothesis, b n c as n,and so there exists N such that for all n>n, c <a n <c+. Now let N =max{n 1,N }. Then, for all n>n,wehave c <max{a n,b n } <c+. which is the conclusion in case (i). Case (ii).a < b. Let = 1 (b a). Note that >0. So by hypothesis there exists a natural number N 1 such that for all n>n 1, b n >b = 1(b a), and there exists a natural number N such that for all n>n, a n <a+ = 1(a + b). NowletN =max{n 1,N }.Then for all n>n, a n < 1(a + b) <b n.thusforalln>n, max{a n,b n } = b n b =max{a, b} as n. Case (iii) a>b. The proof in this case is the same as that of case (ii), provided each a is replaced by a b, andeachh is replaced by an a. 31. The Bolzano-Weierstrass Theorem. If {a n,n 1} is a bounded sequence, then there exists a subsequence {a kn,n 1} of {a n,n 1} that converges (to some number, a.). Proof: We firstneedtoproveaclaim. Claim 1. The sequence {b n } defined by b n = 1 for all n 1 satisfies: n b n 0 as n. Proof of Claim 1. Let >0 be arbitrary. Let N denote 1+[ 1],where [ 1] denotes the largest integer equal to or less than 1, Then, for all n>n, 1 0 = 1 < 1.ButsinceN> 1 or > 1, it follows that for all n>n, n n N N 1 0 <,thusprovingtheclaim. n Let B denote the set of all numbers b such that only a finite number of numbers in the sequence {a n } are equal to or less than b. We know that B 6= since by hypothesis the sequence is bounded.. By hypothesis, B is bounded above by any upper bound of the sequence {a n }. Hence by the completeness axiom, sup B exists. Then define k(1) to be any number such that sup B 1 <a k(1) < sup B +1. By the claim, such a number exists. We may define k() so that it satisfies k() >k(1) and sup B 1 <a k() < sup B + 1 ; this is possible since infinitely many a n s satisfy the inequality. In general, having defined k(1), k(),, k(n 1), wedefine k(n) so that it satisfies 9

10 k(n) >k(n 1) and sup B 1 n <a k(n) < sup B + 1 n. So let >0 be arbitrary. If N is as defined in Claim 1 for this, it follows by the last displayed inequality that for all n>n,wehave a n sup B <, i.e., a n sup B as n. Q.E.D. 3. Theorem. If {a n } is a bounded sequence, then it converges to a number a if and only if every subsequence converges to that same number. Proof: Suppose {a n,n 1} is a bounded sequence. The only if part of the proof is straightforward. Now suppose every subsequence converges to the same limit, and we wish to prove that the sequence converges to a. Suppose to the contrary that the sequence does not converge to a. Then there exists an >0 such that for infinitely many values of n, a n> a + or a n a. Whichever case it might be, say the first, that subsequence, {a kn,n 1} is bounded, and so by the Bolzano-Weierstrass theorem, there exists b a +, such that a kn b as n. This contradicts the hypothesis that every subsequence converses to the same limit a. Q.E.D. 3A. Corollary. If 0 <p<1, thenlim n p n =0. Proof. We first prove two small claims. Claim 1. For all natural numbers n, 0 <p n+1 <p n < 1. This follows from the fact that 0 <p<1 and for all natural numbers, n, p n =(p +(1 p))p n = p n+1 +(1 p)p n >p n+1 for all n. Claim. There exists a number c 0 such that p n c as n. In order to prove this, consider the set S = {p n : n 1}. In order to prove this, then by claim (i), 1 >p n >p +1 > 0. Hence the sequence is monotone decreasing and bounded. By theorem 1, this sequence converges to some number c [0, 1). Since p n inf S as n, then, by theorem 3, every subsequence of {p n : n 1} converges to the same limit. Thus p n sup S as n.but p n = p n p n,andp n inf S as n. By theorem 11, we have inf S =(infs). Hence inf S is 0 or 1. Butsinceinf S<1, it follows that inf S =0. 10

11 1 3B. Corollary. If p>0, thenlim n =0. n p Proof: Since n +1>n,then(n +1) p >n p,or 1 n > 1 p (n +1), p so the sequence is decreasing and bounded below by 0. So the sequence converges to some number, c, where1 >c 0. By theorem 3, we know that every subsequence of a converging sequenc converges and converges toi thesamenumnber.so 1 c as n. (n ) p But since 1 (n ) = 1 1 p n p n c c = p c, which implies again by theorem 11 that c = c. The only solutions for C in this equation are 0 and 1. Sincebytheabove,c<1, it follows that c =0. 3C. Corollary. If p>0, then p 1/n 1 as n. Proof: We shall prove this in two cases. Case (i) 0 <p<1. First note in this case that for every n, p 1/n < 1. Also, 0 <p n+1 <p n,. Also, 0 < (p n+1 ) 1/n(n+1) < (p n ) 1/n(n+1) < 1, or 0 <p 1/n <p 1/(n+1) < 1. Thus the sequence {p 1/n } is an increasing sequence, bounded above by 1. Thus there exists a positive number c such that p 1/n c as n. By theorem 3, every subsequence of a converginfg sequence converges and converges to the same limit, we have p 1/n c as n. But since p 1/n =(p 1/n ), 11

12 it follows again by theorem 11 that c = c. Sincec>0, it follows that c =1, thus proving the theorem in the case when p<1. Case (ii) p>1. First note in this case that for every n, p 1/n > 1. Also, Also, or 1 <p n <p n+1,. 1 < (p n ) 1/n(n+1) < (p n+1 ) 1/n(n+1), 1 <p 1/(n+1) <p 1/n. Thus the sequence {p 1/n } is a decreasing sequence, bounded below by 1. Thus there exists a positive number c 1 such that p 1/n c as n. By theorem 3, every subsequence of a converginfg sequence converges and converges to the same limit, we have But since p 1/n c as n. p 1/n =(p 1/n ), it follows again by theorem 11 that c = c. Sincec 1, it follows that c =1, thus proving the theorem in the case when p>1. iv. Some Topology of Real Numbers. 33. Notation. If a and b are real numbers where a<b,wedefine the four kinds of intervals by [a, b] [a, b) (a, b] (a, b) = {x R 1 : a x b}, = {x R 1 : a x<b}, = {x R 1 : a<x b}, and = {x R 1 : a<x<b} Note: If a b, then(a, b) is empty, and the other three intervals defined above could be empty. An interval of the form (a, b) is called an open interval, and an interval of the form [a, b] is called a closed interval. 34. Definition. AsubsetS of R 1 is said to be an open set if for every x S thereexistsan x > 0 such that (x x,x+ x ) S. (Notethatthe value of x depends on x.) 1

13 35. Theorem. If {S λ : λ Λ} is any collection of open subsets of R 1, then [ {Sξ : λ Λ}} is an open set. Proof. Let x [ {S ξ : λ Λ}}. Then there exists λ Λ such that x S λ.thusthereexists >0 such that (x, x + ) [ {S μ : μ Λ}} thus establishing that it is an open set. 36. Proposition. The empty set of real numbers,, is an open set. Proof. A proof of this relies on our definition that, for sentences p and q> the sentence p q is true if both p and q are false sentences. In the proof of this proposition, p is the sentence x,andq is the sentence there exists >0 such that (x, x + ). Since both hypothesis and conclusion are false, it follows that the implication is true. 37. Theorem. If A and B are open subsets of R 1,thenA B is an open set.. Proof: If A B =, thenitisopenbyproposition36. IfA B 6=, then let x A B be arbitrary. Since x A, thenthereexists 1 > 0 such that (x 1,x+ 1 ) A. Also,sincex B, it follows that there exists > 0 such that (x,x+ ) B. Nowlet =min{ 1, }.Then >0, so and (x, x + ) (x 1,x+ 1 ) A (x, x + ) (x,x+ ) B, so (x, x + ) A B, which establishes the conclusion. 38. Lemma. If x is a real number, then n=1(x 1 n,x+ 1 )=[x.x] ={x}. n Proof. The right hand side is clearly a subset of the left hand side. Now if I can show that every number unequal to x is not in the left side of the above equation, then I should know that the only number in the left side is x. So let y 6= x. Therefore x y > 0. Now let n be any natural number 13

14 1 greater than. Then 1 < x y. If y>x,then x y = x + y so x y n y>x+ 1,andincasey<x,then x y = x y, soy<x 1.Ineither n n case, y/ (x 1,x+ 1 ), and therefore not in the left hand side. This is all n n we needed to establish. 39. Theorem. If A is a nonempty open set, then it is an open interval if and only if for every pair of numbers a and b in A, wherea b then [a, b] A. Proof. Suppose A is open, and suppose that it is an interval, say, (x, y), and let a and b belong to A so that a b. Then if a z /b, x<a z b<y, so by transitivity, x<z<yfor all such z, i.e.,[a, b] A. Now, suppose A is an open set and is such that for every a and b in A, [a, b] A. We must show that A is an open interval. Case (i) A is bounded. Since by hypothesis, A 6=, there exist numbers x =infa and y =supa. Notethat x / A and b / A, but that A (x, y). Now let z (x, y). Then, since x<z<y,andbythedefinitions of sup A and inf A, thereexistu and v such that x<u<z<v<y. Then [u, v] A, soz A. Thus the definition of the open set being an open interval is satisfied. 40 Lemma. If (a, b /) is a nonempty open interval, then there exists a rational number q in (a, b). Proof. Without loss of generality we may assume that a > 0. Letn be a natural number so large that n>max{ 1 1 }. It follows that if we define a b a S = {k N : k a} 6=, thens 6=, andsowemaydefine r by r =maxs. n Thus r n a< r n + 1 <a+(b a) =b. n This implies a< r +1 n <b, which gives us our conclusion. 41. Theorem.If A is an open set, then A can be represented in a unique way as a union of a countable number of disjoint open intervals. Proof. If x and y are points in A, we shall write x y if [x, y] [y, x] A is true. It is observed that the symbol,,is an equivalence relation. This means: (i) for all x A, x x, (ii) if x y, theny x, and (iii) if x y and if y z, thenx z. Thus by the fundamental theorem of equivalence 14

15 relations, A is a union of disjoint equivalence classes. By the definition of an equivalence class here, and by theorem 39, an equivalence class is an open interval. By lemma 40, each equivalence class contains a rational number. Since the rationals are countable, we may conclude that A is a disjoint union of open intervals. 4. Definition. A subset S of R 1 is said to be a closed set if its complement S c = R 1 \S is an open set..43. Theorem. If S R 1,thenS is closed if and only if every convergent sequence in S converges to a number in S. Proof: Only if part. Suppose S is closed, and let {x n } be any convergent sequence in S. We wish to prove that x 0 S, wherex 0 = lim n x n. Suppose not. Then x 0 R 1 \S, which,byourdefinition of a closed set, is an open set. So there exists >0 such that (x 0, x 0 + ) R 1 \S. But since x n x 0 as n,thereexistsann such that for all n>n,the inequalities x 0 <x n <x 0 + aretrue.butthismeansthat{x n,n>n} is a subset of R 1 \S, which contradicts the hypothesis that the sequence {x x } is a convergent sequence in S. This proves the only if part. In order to prove the if part, we assume that every convergent sequence in S converges to a number in S. We wish to prove that S is closed. We shall do this by proving that R 1 \S is open. Suppose to the contrary that R 1 \S is not open. Then there exists a number x 0 R 1 \S such that for every >0, (x 0, x 0 + ) S 6=. Hence for every natural number n, thereexists x n (x 0 1,x n ),wherex n n S. Hence x n x 0 as n,thus violating the hyothesis that every convergent sequence in S converges to a number in S 44. Heine-Borel Theorem. If O is a collection of open sets, and if the closed, bounded interval [a, b] is a subset of the union of the open sets in O, then there exists a finite number of open sets in O, callthemo 1,,O n, such that n[ [a, b] O i. Proof. Let S denote the set of numbers x in [a, b] such that the theorem is true for [a, x) when x is substituted for b. This set is not empty, since a S. This set is also bounded above, by b Hence by the completeness axiom, the least upper bound, y, ofs exists. We first prove that. y S.. Since y [a, b], thereisan >0 and an O O such that (y, y + ) O. If y <x<y, then we know that [a, x] isasubsetofaunionofonly i=1 15

16 a finite number of sets in O. Hence [a, y] a subset of just the union of this finite union and just one more open set in O. Butify<b, then there exists z (y, min{b, y + }) such that [a, z] is a subset of the union of that same finite number of open sets in O. Consequently, y = b, and the theorem is proved.. V. Limits of Functions. 45. Definition. Let f :(a, b) R 1 be a function, and let x 0 (a, b). We say that f(x) L as x x 0,orwrite lim f(x) =L, x x 0 if for every >0, (no matter how small might be), there exists δ > 0 such that for all x (a, b) that satisfy 0 < x x 0 <δ, then the inequality f(x) L < is true. 46. Theorem. Let f :(a, b) R 1 be a function, and let x 0 (a, b). If f(x) L as x x 0,andifK R 1,thenKf(x) KL as x x 0. Proof. Let >0bearbitrary. We must show that there exists δ > 0 such that Kf(x) KL < for all values of x (a, b) that satisfy 0 < x x 0 <δ. If K =0, then the condition is satisfied for any δ > 0 we might choose. If K 6= 0, then, since K > 0, there exists a positive number δ such that f(x) L < K for all values of x (a, b) that satisfy 0 <x x 0 <δ. But the displayed inequality is the same as stating Kf(x) KL <. This proves our theorem. 47. Theorem. Let f and g be functions over (a, b), andletx 0 (a, b). If f(x) L as x x 0 and if g(x) M as x x 0,then lim (f(x)+g(x)) = L + M. x x 0 Proof. Let >0 be arbitrary. We must show that there exists δ > 0 such that (f(x)+g(x)) (L + M) < for all values of x that satisfy 0 < x x 0 <δ. Observe that > 0. Thus there exists a positive number δ 0 > 0 such that for all values of x in (a, b) that 16

17 satisfy 0 < x x 0 <δ 0,then f(x) L <. Also, there exists a positive number δ 00 > 0 such that for all values of x that satisfy 0 < x x 0 <δ 00, g(x) M <. Now let δ =min{δ 0,δ 00 }. Note that δ > 0. Then for all values of x (a, b) that satisfy 0 < x x 0 <δ, it follows that (f(x)+g(x)) (L + M) = (f(x) L)+(g(x)) M) f(x) L + g(x) M + =. 48. Theorem. Let f :(a, b) R 1 be a function, and let x 0 (a, b). Then f(x) L as x x 0 if and only f(x n ) L as n for every sequence {x n } in (a, b) that satisfies x n x 0 as n. Proof. Suppose f(x) L as x x 0.Let >0 be arbitrary. Then there exists δ>0such that f(x) L < for all values of x (a, b) that satisfy 0 < x x 0 <δ. But δ>0, so by the definition of a sequence {x n } that satisfies x n x 0 as n, we know that there exists N such that for all n>n, x n x 0 <δ. Therefore, for all n>n, it follows that f(x n ) L <, i.e., f(x n ) L as n. Conversely, suppose that for every sequence {x n } in (a, b) that satisfies x n x 0 as n also satisfies f(x n ) L as n. Suppose to the contrary that the conclusion, f(x) L as x x 0,isnottrue. Then there exists a sequence {x n } that satisfies x n x 0 as n,butsuch that f(x n ) L is not true.. This means that there exists an >0such that for every δ>0, there exists a number y that satisfies 0 < y x 0 <δand such that f(y) L. If we replace δ by δ n = 1 and the corresponding n value of y by y n,weobtainthaty n x 0 but that the sequence {f(y n )} does not converge to x 0, thus giving us a contradiction, the contradiction being that we have found a sequence {y n },suchthaty n x 0 as n but that {f(y n ) /} does not converge to L. And so... The 48 theorems, definitions and lemmas given so far constitute a reasonable background for the theory of differentiation and integration. Mission (hopefully) accomplished. The End 17