MATH 117 LECTURE NOTES

Size: px
Start display at page:

Download "MATH 117 LECTURE NOTES"

Transcription

1 MATH 117 LECTURE NOTES XIN ZHOU Abstract. This is the set of lecture notes for Math 117 during Fall quarter of 2017 at UC Santa Barbara. The lectures follow closely the textbook [1]. Contents 1. The set N of natural numbers 2 2. The set Q of rational numbers 4 3. Ordered fields 7 4. Real numbers R and the Completeness Axiom and Limits of sequences Proofs Problem session Limit theorems for sequences Monotone sequences and Cauchy sequences Subsequences lim sup and lim inf revisited Series Alternating series and Integral Tests Continuous functions Properties of continuous functions 55 References 59 Date: December 5,

2 2 XIN ZHOU 1. The set N of natural numbers We introduce basic properties for the set of natural numbers. N = {1, 2, 3, } denotes the set of all natural numbers, or all positive integers; each n N 1 has a successor n + 1. For example, the successor of 5 is 6. Proposition 1.1. The set N satisfies the following properties: N1. 1 N; N2. if n N, then its successor n + 1 N; N3. 1 is not the successor of any element in N; N4. if n and m have the same successor, then n = m; N5. suppose S is a subset of N satisfying: 1 N and if n S then n+1 S, then S = N. Property (N5) is called the Peano Axiom or Peano Postulate. Property (N5) is the basis for mathematical induction: Let P 1, P 2, be a list of statements of propositions that may or may not be true; The Principle of Mathematical Induction asserts that: All of P 1, P 2, are true provided: P 1 is true; (this is called the basis of induction) if P n is true, then P n+1 is true for all n N. (this is called induction step) Example 1.2. All numbers of form 7 n 2 n (n N) is divisible by 5. Proof. The n-th proposition is: P n : 7 n 2 n is divisible by 5. Basis of induction P 1 : = 5 is divisible by 5, so P 1 is true; Induction step: Suppose P n is true, then 7 n 2 n = 5m for some m N. To verify P n+1, we have 7 n+1 2 n+1 = 7 n n n 2 n+1 = 7 n (7 2) + 2(7 n 2 n ) = 5 7 n + 2(5m) = 5(7 n + 2m). 1 The symbol n N means n is an element of the set N

3 MATH 117 FALL 2017 UCSB 3 That is, 7 n+1 2 n+1 is also divisible by 5. Therefore P n implies P n+1, and the induction step holds. Example 1.3. Show that sin(nx) n sin(x) for all n N and all x R. Proof. The n-th proposition is: P n : sin(nx) n sin(x) for all x R. Basis of induction: sin(x) = 1 sin(x), hence P 1 is true; Induction step: Suppose P n is true, that is sin(nx) n sin(x) for all x R. To verify P n+1, we have sin((n + 1)x) = sin(nx + x) = sin(nx) cos(x) + sin(x) cos(nx) sin(nx) cos(x) + sin(x) cos(nx) ( sin(x), cos(nx) 1) sin(nx) + sin(x) (by P n ) n sin(x) + sin(x) = (n + 1) sin(x). That is, P n+1 is true. Therefore P n implies P n+1, and the induction step holds.

4 4 XIN ZHOU 2. The set Q of rational numbers We introduce basic properties for the set of integers and rational numbers. Z = {0, 1, 1, 2, 2, } denotes the set of all integers; Q = { m : m, n Z, n 0} denotes the set of all rational numbers. n Q is a good algebraic system: one can do basic operations like additions, multiplication, subtraction, division over Q; Q is inadequate when solving algebraic equations like: x 2 2 = 0. Actually, the solutions of x 2 2 = 0 makes sense: (1) by the Pythagorean Theorem, if d is the length of the diagonal of a square of edge length 1, then d 2 = = 2, and hence d is a solution; (2) the graph of the function y = x 2 2 will pass through the x-axis at two points which are both solutions of x 2 2 = 0. This means that: there are gaps in Q, as 2 Q. Moreover, there are more exotic numbers like π, and e : π appears when studying the perimeters of circles and spheres; e appears when studying the infinite sums: n N Definition 2.1. A number is called an algebraic number if it satisfies a polynomial equation: a n x n + a n 1 x n a 1 x + a 0 = 0, where a i Z for all i = 0,, n, and a n 0. Example 2.2. We know that: all rational numbers are algebraic numbers: if r = m, m, n Z, n 0, then it satisfies nx m = 0; n square root, cubic root 3, etc, are algebraic numbers; ordinary algebraic operations (including addition, multiplication, subtraction, division, taking roots) of algebraic numbers are also algebraic numbers. Example /3, (2+5 1/3 ) 1/2, [ ( /2 )/7 ] 1/2 are all algebraic numbers. Proof. (1) 17 1/3 is a solution of x 3 17 = 0; (2) Let a = ( /3 ) 1/2, = a 2 = /3, = (a 2 2) 3 = 5, = a 6 2a 4 + 4a 2 8 = 5, = a 6 2a 4 + 4a 2 13 = 0; 1. n!

5 MATH 117 FALL 2017 UCSB 5 (3) Let b = [ ( /2 )/7 ] 1/2, = 7b 2 = /2, = 7b 2 4 = 2 3 1/2, = (7b 2 4) 2 = 12, = 49b 4 56b = 12, = 49b 4 56b 2 +4 = 0. Recall that an integer k is a factor of another integer m, or k divides m, if m/k is also an integer; an integer p 2 is a prime provided that the only positive factors of p is 1 and p. Proposition is not a rational number. Proof. We prove by contradiction argument. Suppose the statement is not true, that is 2 Q, = 2 = p, where p, q are both integers with no q common factor and q 0. Then p 2 q 2 = 2, = p2 = 2q 2. Lemma 2.5. If p 2 is an even number (p Z), then p is also even. Proof. Using contradiction argument again, suppose not, then p is odd, = p = 2m + 1 for some m Z. Then p 2 = (2m + 1) 2 = 4m 2 + 4m + 1 = 2(2m 2 + 2) + 1, so p 2 is odd, and this is a contradiction. Let us go back to the proof of Proposition 2.4. Since p 2 is even, by the above lemma p is also even, = p = 2m, hence p 2 = 4m 2 = 2q 2, = q 2 = 2m 2. So q 2 is even, and hence q is even by the above lemma. But this is a contradiction to the assumption that p, q has no common factor. Theorem 2.6 (Rational Zero Theorem). Suppose a 0, a 1,, a n are all integers, and r is a rational number satisfying the polynomial equation: where n 1, a n 0, a 0 0. Write a n x n + + a 1 x + a 0 = 0, r = p q, where p, q Z have no common factor and q 0. Then q divides a n, and p divides a 0.

6 6 XIN ZHOU Proof. Plug in r = p q into the equation: Multiply by q n, = Re-arrange the equation: a n ( p q )n + a n 1 ( p q )n a 1 ( p q ) + a 0 = 0. a n p n + a n 1 p n 1 q + + a 1 pq n 1 + a 0 q n = 0. a n p n = q(a n 1 p n a 1 pq n 2 + a 0 q n 1 ), = q divides a n p n. Since p, q has no common factor, q must divide a n. We can prove p divides a 0 in a similar manner. In particular, a 0 q n = p(a n p n 1 + a n 1 p n a 1 q n 1 ), = p divides a 0 q n, and hence p must divide a 0. Example 2.7. Use the above theorem to prove that 2 is not a rational number. Proof. Suppose the equation x 2 2 = 0 has a rational root p, where p, q Z q has no common factor and q 0. Then by Theorem 2.6, q divides 1, and p divides 2. There p can only be ±1 or ±2, and q can only be ±1, = p must be of the q forms ±1, ±2. By substituting back to x 2 2 = 0, we can see that none of them is a solution. This is a contradiction. Example 2.8. Prove that a = ( /3 ) 1/2 is not a rational number. Proof. By the above example, we know that a satisfies: a 6 2a 4 + 4a 2 13 = 0. If a is a rational number, then write a = p, where p, q Z has no common q factor and q 0. By the Rational Zero Theorem, q divides 1, and p divides 13. There p can only be ±1 or ±13, and q can only be ±1, = p must be of q the forms ±1, ±13. By substituting back, we can see that none of them is a solution. This is a contradiction.

7 MATH 117 FALL 2017 UCSB 7 3. Ordered fields The basic algebraic operations on Q are + and. The following properties hold in this algebraic system (Q, +, ): Properties for + : A1. a + (b + c) = (a + b) + c, a, b, c Q 2 associative law; A2. a + b = b + a, a, b Q commutative law; A3. a + 0 = a, a Q existence of 0; A4. For each a Q, there exists an element a Q, such that a+( a) = 0 existence of reverse; properties for : M1. a (b c) = (a b) c, a, b, c Q associative law; M2. a b = b a, a, b Q commutative law; M3. a 1 = a, a Q existence of 1; M4. For each a Q, a 0, there exists an element a 1 Q, such that a (a 1 ) = 1 existence of reciprocal; a property relating and : DL. a (b + c) = a b + a c, a, b, c Q distributive law. Definition 3.1. A set F is a field if it has two operations + and, and elements 0 and 1 satisfying all the properties (A1)...(A4) (M1)...(M4) (DL) (with Q changed to F ). Example 3.2. There are two examples which are not fields: N (the natural numbers ) fails properties (A4) and (M4); Z (the integers) fails (M4). Q also has an order structure satisfying the properties: O1. given a, b Q, either a b or b a; O2. if a b and b a, then a = b; O3. if a b and b c, then a c transitive law; O4. if a b, then a + c b + c; O5. if a b and 0 c, then a c b c. Definition 3.3. A field F is called an ordered field if it has an order structure satisfying properties (O1)...(O5). 2 Here the symbol means for all.

8 8 XIN ZHOU Remark 3.4. a < b means that a b and a b. Properties of ordered fields: based on the defining properties, we can deduce more properties for a field or an ordered field. Theorem 3.5. Suppose (F, +, ) is a field with elements 0, 1, and a, b, c denote arbitrary elements in F, then: (i) a + c = b + c, = a = b; (ii) a 0 = a, a; (iii) ( a) b = (a b); (iv) ( a) ( b) = a b; (v) a c = b c, and c 0, = a = b; (vi) a b = 0, = either a = 0 or b = 0. Proof. (i) (a + c) + ( c) = a + (c + ( c)) = a + 0 = a = (b + c) + ( c) = b + (c + ( c)) = b + 0 = b. (ii) a 0 = a (0 + 0) = a 0 + a 0, = (by (i)) a 0 = 0. (iii) ( a) b + a b = ( a + a) b = 0 b = 0 (by (ii)), so ( a) b = (a b). (iv) By (iii), ( a) ( b) = a ( b) = ( a b) = ab. (v) Homework. (vi) Homework. Theorem 3.6. Suppose F is an ordered field with order structure, and a, b, c denotes arbitrary elements in F, then: (i) if a b, then b a; (ii) if a b and c 0, then bc ac; (iii) if 0 a and 0 b, then 0 ab; (iv) 0 a 2 (= a a), a; (v) 0 < 1; (vi) if 0 < a, then 0 < a 1 ; (vii) if 0 < a < b, then 0 < b 1 < a 1. Proof. (i) By (O4), a + ( b) b + ( b), = a + ( b) 0; by (O4), a + [a + ( b)] a + 0 = a; by (A1), b a.

9 MATH 117 FALL 2017 UCSB 9 (ii) By (i), 0 c, so by (O5), a ( c) b ( c); by Theorem 3.5(iii), ac bc; by (i), bc ac. (iii) By(ii). (iv) If 0 a, then it follows from (iii); if a 0, then by (i), 0 ( a); by (iii), 0 ( a) ( a) = a 2 (by Theorem 3.5)(iv). (v) Since 1 2 = 1, by (v), 0 1; also as 0 1, we have 0 < 1. (vi) Homework. (vii) Homework. Definition 3.7. Let F be an ordered field. Given an element a F, we can define the absolute value a of a as: a := a if a 0, and a := a if a < 0. Here a 0 is an equivalent way to write 0 a. The absolute value gives rise to the notion of distance. Definition 3.8. Given a, b F, the distance between them is defined as: dist(a, b) := a b. Theorem 3.9. Let F be an ordered field. For all a, b F, (i) a 0; (ii) a b = a b ; (iii) a + b a + b. Proof. (i) If a 0, then by definition a = a 0; if a < 0, then a = a 0 by Theorem 3.6(i). (ii) We can compare both sides in four cases: (1) a 0, b 0, (2) a 0, b < 0, (3) a < 0, b 0, (4) a < 0, b < 0. For case (1): a b 0 by Theorem 3.6(iii), so ab = ab; on the other hand, a = a, b = b, so a b = ab. We will leave other cases as exercises. (iii) By definition, since either a = a or a = a, we have Using (O4), a a a, and b b b. a + ( b ) a + b a + b a + b a + b.

10 10 XIN ZHOU Therefore ( a + b ) a + b a + b ( a + b ) (a + b) a + b } = a + b a + b. Corollary We have the triangle inequality: dist(a, c) dist(a, b) + dist(b, c). Proof. dist(a, c) = a c = a b + b c a b + b c = dist(a, b) + dist(b, c).

11 MATH 117 FALL 2017 UCSB Real numbers R and the Completeness Axiom Heuristically, the set of real numbers R is the order field F containing Q with no gaps. We will make rigorous what does no gap mean. Definition 4.1. Let S F be a subset of an ordered field F, and S. (a) If S contains a largest element s 0, [that is to say, s 0 S, and s s 0, s S], we call s 0 the maximum of S, and write s 0 = max S; (b) If S contains a smallest element s 0, [that is to say, s 0 S, and s 0 s, s S], we call s 0 the minimum of S, and write s 0 = min S. Example 4.2. (1) Every finite subset of Q has a maximum and a minimum, i.e. max{1, 2, 2.6, 3, 4} = 4, and min{1, 2, 2.6, 3, 4} = 1. (2) Given a, b F, with a < b, we denote by Among them: [a, b] = {x F : a x b} closed interval, (a, b) = {x F : a < x < b} open interval, [a, b) = {x F : a x < b} half-open interval, (a, b] = {x F : a < x b} half-open interval. max[a, b] = b, min[a, b] = a, (a, b) has no maximum and minimum in general. (3) The set {r Q : 0 r 2} has no maximum. These examples say that maximums and minimums may not always exist. Nevertheless, we can generate the concepts to upper and lower bounds. Definition 4.3. Let S F be a subset and S. (a) If M F and s M s S, then M is called an upper bound of S, and S is said to be bounded from above; (b) If m F and m s s S, then m is called an lower bound of S, and S is said to be bounded from below; (c) S is said to be bounded if it is bounded above and below, and this is equivalent to say S [m, M] for some m, M F. Example 4.4. (1) If S has the maximum max S, then max S is an upper bound of S; similarly min S is a lower bound of S if it exists. (2) b is an upper bound of the intervals [a, b], (a, b), (a, b] and [a, b).

12 12 XIN ZHOU (3) 2 is an upper bound of {r Q : 0 r 2}; 2 is the least upper bound. Now we summarize the definition for least upper bound and greatest lower bound. Definition 4.5. Let S F be a subset and S. (a) If S is bounded from above and S has a least upper bound, then we call it the supreme of S and denote it by sup S. (b) If S is bounded from below and S has a greatest lower bound, then we call it the infimum of S and denote it by inf S. We have the following criterion for least upper bound: M = sup S if and only if (i)s M, s S, (ii) if M 1 < M, then s 1 S, such that s 1 > M. Example 4.6. (a) If S has a maximum, then max S = sup S, and similarly for the minimum and infimum of S. (b) sup[a, b] = sup[a, b) = sup(a, b] = sup(a, b) = b. (c) Let A = {r Q : 0 r 2}, then sup A = 2 / A. Let us elaborate a bit about this example using decimal expansions: first it is obvious that 2 is an upper bound, and we will explain why it is the least one; the decimal expansion for 2 is ; if M 1 is any number less than 2, even if it is very close to 2, there should exist a first decimal of M 1 that is less than the corresponding one for 2; for instance, M 1 = , or M 1 = , or M 1 = ; and for these numbers we can find s 1 = , s 1 = , s 1 = that is less than 2 but greater than M 1. Remark 4.7. By case (c) above, the least upper bound may not belong to S. This is a way to say that there exist gaps in F. Completeness Axiom: every nonempty subset S R that is bounded above has a least upper bound, sup S exists and is a real number. Definition 4.8. The set of real numbers R is an ordered field containing Q that satisfies the Completeness Axiom.

13 MATH 117 FALL 2017 UCSB 13 Remark 4.9. The example A = {r Q : 0 r 2} shows that Q does not satisfy the Completeness Axiom. Corollary Every nonempty subset S R which is bounded from below has a greatest upper bound. Proof. Consider the subset, denoted by S = { s : s S}. Since S is bounded from below, m R, such that m s, s S, = s m, s S. Thus S is bounded from above. By the Completeness Axiom, the supreme of S exists and sup( S) R. Let s 0 = sup( S), and we claim that s 0 = inf S. In particular, (1) s s 0, s S, = s 0 s, s S, and hence s 0 is a lower bound of S; (2) To show that s 0 is the greatest lower bound, consider any other lower bound t R, such that t s, s S, = s t, s S; thus t is an upper bound of S, so s 0 t, = t s 0 ; therefore s 0 is the greatest lower bound. There are two important corollaries of the completeness axiom. Theorem 4.11 (Archimedean Property). Given two real numbers a, b R, if a > 0 and b > 0, then for some positive integer n N, we have na > b. Proof. We use the contradiction argument. Suppose the conclusion is not true, which means that na b, n N. Denote the set S as S = {na : n N}. Then b is an upper bound of S. By the completeness axiom: the supreme of S exists and s 0 = sup S R. Since a > 0, then s 0 < s 0 + a, or equivalently s 0 a < s 0. By the definition of supreme, we know that s 0 a is no longer an upper bound of S, so there exists an element n 0 a, for some n 0 N, such that s 0 a < n 0 a. = s 0 < n 0 a + a = (n 0 + 1)a S. There is a contradiction to the fact that s 0 is an upper bound of S. Corollary (1) if a > 0, then 1/n(= n 1 ) < a for some positive integer n N; (2) if b > 0, then b < n for some positive integer n N.

14 14 XIN ZHOU Proof. (1) follows from the Archimedean Property by letting b = 1, and (2) by letting a = 1. Theorem 4.13 (Denseness of Q). Given two real numbers a, b R, if a < b, then there exists a rational number r Q, such that a < r < b. Proof. We need to find some integers m, n Z, n > 0, such that a < m n < b, and this is equivalent to na < m < nb. Since a < b = (b a) > 0, by the Archimedean property, there exists n N, such that n(b a) > 1. By the Archimedean property again, there exists k N, such that Therefore, k > max{ na, nb }. k < na < nb < k. Consider the set {j Z : k < j k, and na < j}, and it is a finite and nonempty set (since k belongs to it). As it is finite, we pick the minimum = min{j Z : k < j k, and na < j}. Then na < m, and moreover, m 1 na. Therefore na < m = (m 1) + 1 na + 1 < na + (nb na) = nb. So the pair (m, n) satisfies the requirement.

15 MATH 117 FALL 2017 UCSB and We add in two elements + and into R so that R {+, } is still a good ordered system, and it makes taking supreme and infimum easier in many cases. + and denote plus infinity and minus infinity; we will write + as ; we can add the ordering to R {+, } so that a R {+, }, a + ; this ordering system satisfies O1, O2, O3; +, do not represent any real numbers, and there is no algebraic structure like + and on R {+, }; we can introduce new notations: [a, ) = {x R : a x} unbounded closed interval, (a, ) = {x R : a < x <} unbounded open interval, (, b] = {x R : x b} unbounded closed interval, (, b) = {x R : x < b} unbounded open interval, (, + ) = R; given an nonempty subset S R, denote sup S = +, inf S =, therefore, if S R and S, then if S is not bounded from above, if S is not bounded from below; sup S and inf S always make sense in R {+, }.

16 16 XIN ZHOU 6. Limits of sequences We introduce the concept of sequences and their limits in this section. A sequence is a function, with domain a set of natural numbers of the form {n Z : n m}, where m is usually chosen to be 1 or 0, with ranges in R. Denote a sequence by (s n ) n=m, s n R, or (s m, s m+1, s m+2, ). If m = 2, write (s n ) n N. Example 6.1. a) Let s n = 1, then the sequence is (1, 1, 1, 1, ) = n (1, 1, 1, 1, ). The set of values is {1, 1, 1, 1, } b) Let a n = ( 1) n for n 0, then the sequence is (1, 1, 1, 1, ), and the set of values is {1, 1}. Remark 6.2. Note the difference between a sequence and its set of values: we use (,, ) to denote a sequence, and {,, } to denote the set of values. In particular, the sequence (a n = ( 1) n ) has infinite number of terms, while its set of values {( 1) n } consists of only two numbers. c) Consider the sequence ( cos( nπ)). It can be written as 3 n N ( 1, 1, 1, 1, 1, 1, 1, ). The set of values is {1, 1, 1, 1} d) Let s n = n 1 n for n N, then the sequence is (1, 2, 3 1/3, ). We will show that n 1/n is close to 1 for very large n. e) Let ( s n = (1 + 1 )n). n n N The limit of a sequence (s n ) is intuitively a real number for which the values s n accumulate to. To be more precise, Definition 6.3. A sequence (s n ) of real numbers is said to converge to the real number s, provided that (6.1) for each ɛ > 0, there exists a number N, such that n > N implies s n s < ɛ. If (s n ) converges to s, we will write lim n s n = s, or s n s. s is called the limit of the sequence (s n ). A sequence that does not converge to some real number is said to diverge. Remark 6.4. (1) ɛ, δ represent small numbers; (2) (6.1) is an infinite number of statements. We sometime denote N = N(ɛ) to show the dependence of N on ɛ, and in principle N(ɛ) is large when ɛ is small;

17 MATH 117 FALL 2017 UCSB 17 (3) N can be any large real or rational numbers by the Archimedean property. Example a) lim n = 0, and this will be proved soon. n 2 b) The sequence (a n = ( 1) n ) diverges: Proof. Assume by contradiction that the sequence converges to some real number s R, lim( 1) n = s. Since (6.1) is assumed to be true for any ɛ > 0, we choose ɛ = 1. By (6.1), there exists a number N > 0, such that when n > N, 2 we have ( 1) n s < 1 2 = s 1 2 < ( 1)n < s In particular, we can let n = 2N or n = 2N + 1 which are both greater than n, and then { When n = 2N, s 1 2 < 1 < s = 1 2 < s < 3 2, When n = 2N + 1, s 1 2 < 1 < s = 3 2 < s < 1 2. This is a contradiction, so we finish the proof. c) The sequence ( cos( nπ)) diverges, and this is an exercise. 3 n N d) lim n 1 n = 1, and this will be proved later. e) lim s n = (1 + 1 n )n = e, and this will be proved later. Using the definition, we can show that the limit, if it exists, is uniquely determined. Proposition 6.6. Let (s n ) be a sequence, and assume that lim s n = s and lim s n = t for some real numbers s, t R, then s = t. Proof. By (6.1) and the assumptions, for any ɛ > 0, there exists N 1 > 0, such that n > N 1 = s n s < ɛ 2 ; and there exists N 2 > 0, such that So for n > max{n 1, N 2 }, Therefore s = t. contradiction.) n > N 2 = s n t < ɛ 2. s t s s n + s n t < ɛ. (If this were not true, then take ɛ = 1 s t to get a 2

18 18 XIN ZHOU 7. Proofs The main purpose of this section is to introduce how to write rigorous proofs for limits. Example 7.1. lim n 1 n 2 = 0. Discussion: given ɛ > 0, we want to find N > 0, such that if n > N, then 1 n 0 2 < ɛ. Algebra: 1 n 2 0 < ɛ 1 n 2 < ɛ n 2 > 1 ɛ, or n > 1 ɛ. Take N = 1 ɛ, then it should serve our purpose. Proof. For any ɛ > 0, let N = 1 ɛ. If n > N, then n > 1 ɛ, so n 2 > 1, hence ɛ ɛ > 1. n 2 Thus n > N implies 1 0 n < ɛ, and this proves 1 2 limn = 0. n 2 Example 7.2. lim n 3n+1 7n 4 = 3 7. Discussion: given ɛ > 0, we want to find N > 0, such that if n > N, then 3n + 1 7n < ɛ. Algebra: 3n+1 3 7n 4 7 = 19 7(7n 4). 3n < ɛ 7n 4 7 7(7n 4) < ɛ < ɛ < 7n 4 7(7n 4) 7ɛ 7n > n > So put N = ɛ 49ɛ 7 49ɛ 7 Proof. For any ɛ > 0, let N = 19 above to show that 3n+1 < ɛ. 7n ɛ Example 7.3. lim n 4n 3 +3n n 3 6 = 4.. If n > N, then we can reverse all steps Discussion: given ɛ > 0, we want to find N > 0, such that if n > N, then (7.1) 4n 3 + 3n n < ɛ, 3n + 24 n 3 6 < ɛ. There is no way to easily solve this inequality for n by ɛ, so we need to do some rough estimates: (1) 3n n + 24n 27n; (2) n 3 6 > n 3 /2 provide n is large, actually this is equivalent to n 3 > 12, which holds true when n > 2.

19 By what we have above, we have MATH 117 FALL 2017 UCSB 19 (7.1) = 27n n 3 /2 < ɛ, and n > 2; 54 < ɛ, and n > 2; n2 54 n >, and n > 2. ɛ Proof. For any ɛ > 0, let N = max{ all steps above to (7.1). 54, 2}. If n > N, then we can reverse ɛ Example 7.4. Let (s n ) be a sequence of nonnegative numbers, i.e. s n 0. Suppose that lim n s n = s. (Show that s 0 as an exercise). Prove that lim s n = s. Discussion: given ɛ > 0, we want to find N > 0, such that if n > N, then Note that s n s < ɛ. sn s = We will divide the discussion into two cases: (1) if s > 0, then s n + s s, hence s n s sn + s. s n s s n s s. By the assumption lim n s n = s, we can select N > 0, such that n > N = s n s < sɛ, and this will serve our purpose. (2) The case when s = 0 will be left as homework. Proof. We only discuss the case when s > 0 here. For any ɛ > 0, since lim n s n = s, there exists N > 0, such that n > N = s n s < sɛ. This implies that s n s s n s < ɛ, s and this finishes the proof. Example 7.5. Let (s n ) n N be a sequence of real numbers such that s n 0 for all n N and lim s n = s 0. Prove that inf{ s n : n N} > 0.

20 20 XIN ZHOU Discussion: Let ɛ = s /2 in the definition (6.1). For N = N(ɛ) in (6.1), if n > N, then we have s n s < ɛ, = s n (s ɛ, s + ɛ). Proof. Let ɛ = s /2. Since lim s n = s, there exists N > 0, such that So n > N implies that n > N, = s n s < ɛ = s 2. s n = s + s n s s s 2 = s 2. Now set m = min{ s 2, s 1, s 2,, s N }, then s n m, for all nn, and this finishes the proof.

21 MATH 117 FALL 2017 UCSB Problem session (1) (4.15) Let a, b R. Show if a b + 1 for all n N, then a b. n Proof. Suppose not. Then a > b, or a b > 0. By the Archimedean Property, there exists n N, such that 1 n < a b. And this is equivalent to a > b + 1, but this is a contradiction. n (2) Show that the Archimedean Property of the real numbers holds if and only if the set of natural numbers N is not bounded above. Proof. First assume that the Archimedean Property holds. Then for any positive number M > 0, there exists a n N, such that M < n. This means that N is not bounded above. Now assume that N is not bounded above, and this means that for any N > 0, there exists n N, such that N < n. Now given any two numbers a, b R, a > 0, b > 0. Consider the number N = b/a, so there exists n N, such that N = b/a < n, which is equivalent to na > b. (3) (HW3.2) Show that if S Z is bounded above then S has a maximum, i.e., sups S. (4) Let x R. We define the Greatest Integer function of x as the largest integer m Z such that m x; it is denoted by [x] and satisfies the inequality [x] x < [x] + 1. For instance, [0.1] = 0, [1.1] = 1, [ 2.1] = 3, [5] = 5, [ 4] = 4, etc. Show that for any real number x R, [x] exists. Hint: Consider the set S = {n Z: n x}. Then consider two cases depending on whether x 0 or x < 0. In the case when x < 0, use the Archimedean property of R to show that S is nonempty. Then apply the completeness axiom and finally show that sups = [x], i.e., show that sups is an integer (this follows from Problem 3) and it satisfies sups x < sups + 1.

22 22 XIN ZHOU Proof. Let us only consider the case when x 0, then 0 S, hence S. Since S is bounded from above, (x is an upper bound), by the Completeness Axiom, sup S exists. By Problem 3, sup S S, and sup S = max S. Then the only thing we need to prove is x < sup S + 1. Assume by the contrary that x sup S + 1. Then since sup S is an integer, and hence sup S + 1 is an integer, so by the definition of S, we have sup S + 1 S. This is a contradiction to the fact that sup S = max S.

23 MATH 117 FALL 2017 UCSB Limit theorems for sequences Definition 9.1. A sequence (s n ) of real numbers is said to be bounded if the set of values {s n : n N} is bounded. This is equivalent to saying that: M R, such that s n M for all n N. Theorem 9.2. Convergent sequences are bounded. Proof. Let (s n ) be a convergent sequence, and let s be the limit, or equivalently s = lim s n. Apply the definition of convergence with ɛ = 1, then we get some large number N N, such that if n > N, then s n s < 1. By the triangle inequality, s n s < 1 = Define Then we have s n s + s n s s + s n s s + 1. M = max{ s 1, s 2,, s N, s + 1}. { s n M if n N s + 1 M if n > N. Therefore s n M for all n N, and so {s n } is bounded. Remark 9.3. In the proof above, we only apply the defining condition (6.1) for a single ɛ = 1. Actually, we can take ɛ to be any finite positive number in the proof, and this is left as an exercise. Theorem 9.4. If s sequence (s n ) converges to s, and k R, then the sequence (ks n ) converges to ks. That is, lim(ks n ) = k lim s n. Proof. If k = 0, the proof is trivial as ks n 0, and lim(0 s n ) = 0 = 0 lim s n. Assume k 0 now. Let ɛ > 0 be any positive real number, and we need to show that ks n ks < ɛ, for large n N. Note that the above inequality is equivalent to saying that s n s < ɛ, for large n N. k

24 24 XIN ZHOU Since lim s n = s, for ɛ k > 0, by (6.1) there exists N > 0, such that if n N, then s n s < ɛ k. Therefore for n > N, we have ks n ks < ɛ. Theorem 9.5. If (s n ) converges to s and (t n ) converges to t, then (s n + t n ) converges to s + t. That is to say, lim(s n + t n ) = lim s n + lim t n. Proof. Let ɛ > 0 be any positive real number, and we want to show that Note that (s n + t n ) (s + t) < ɛ, for all large n N. (s n + t n ) (s + t) s n s + t n t. (Hint: so we can try to show s n s < ɛ/2 and t n t < ɛ/2.) Since lim s n = s, N 1 N, such that if n N 1, then s n s < ɛ/2. Similarly by lim t n = t, N 2 N, such that Let Then if n > N, we have if n N 2, then t n t < ɛ/2. N = max{n 1, N 2 }. (s n + t n ) (s + t) s n s + t n t < ɛ. Theorem 9.6. If (s n ) converges to s and (t n ) converges to t, then (s n t n ) converges to s t. That is to say, lim(s n t n ) = (lim s n ) (lim t n ). Discussion: s n t n st = s n t n s n t + s n t st s n t n t + t s n s M t n t + t s n s. In the above inequality, t n t and s n s can be very small for large n; s n is bounded by Theorem 9.2, i.e. s n M.

25 MATH 117 FALL 2017 UCSB 25 Proof. Let ɛ > 0 be any positive real number. By Theorem 9.2, M > 0, such that s n M, for all n N. Since lim t n = t, N 1 N, such that if n N 1, then t n t < ɛ/(2m). Similarly by lim s n = s, N 2 N, such that Let Then if n > N, we have if n N 2, then s n s < N = max{n 1, N 2 }. ɛ 2( t + 1). s n t n st s n t n t + t s n s ɛ M 2M + t ɛ 2( t + 1) < ɛ. Lemma 9.7. If (s n ) converges to s and if s n 0 for all n N, and if s 0, then Discussion: lim 1 s n = 1 s. 1 1 s n s = s n s s n s s n s. m s We know that s n s is very small for n very large. To show the right hand side is small, we need the denominator s n s to be bounded away from 0. By Example 7.5, we do have that m = inf{ s n : n N} > 0. Proof. Let ɛ > 0 be any positive real number, and m be defined as above. Since lim s n = s, there exists N > 0, such that if n N, then s n s < ɛ m s. Therefore if n N, then 1 1 s n s ɛ m s s n s < ɛ.

26 26 XIN ZHOU Theorem 9.8. Suppose lim s n = s and lim t n = t. If s 0 and s n 0 for all n N, then ( ) tn lim = t s. Proof. By the previous lemma, lim 1 s n A few basic examples: s n = 1 s. Then lim t n s n = (lim t n )(lim 1 s n ) = t s. a) lim 1 = 0 for p > 0 (assume p N at this moment); n p b) lim a n = 0 if a < 1; c) lim n 1/n = 1; d) lim a 1/n = 1 for a > 0. Proof. a). Let ɛ > 0 be any positive real number. (We want to show 1 < ɛ, n p which is equivalent to n > ( ) 1 1/p.) ɛ Let N = ( ) 1 1/p, ɛ (Exercise: use the completeness axiom to show that: for any n N, and a > 0, there exists x > 0, such that x n = a, and we write x = a 1/n.) then if n > N, we have 1 < ɛ. Therefore n p 1 n 0 p = 1 n < ɛ. p b). If a = 0, then a n = 0, and the proof is trivial. If a 0 and a < 1, then a = 1 for some b > 0. Using the binomial 1+b expansion, (1 + b) n 1 + nb > nb. So a n = 0 = a n 1 = (1 + b) < 1 n nb. Let ɛ > 0 be any positive real number, and let N = 1. Then if n > N, we ɛb have a n = 0 < 1 nb < ɛ. c). Let s n = n 1/n 1. Note that s n 0. It suffices to show lim s n = 0. Note that 1 + s n = n 1/n (1 + s n ) n = n.

27 By the binomial expansion: MATH 117 FALL 2017 UCSB 27 n = (1 + s n ) n = 1 + n s n n(n + 1)s2 n > frac12n(n + 1)s 2 n. So s 2 n < 2 2 n 1, = s n < n 1. Exercise: use the above formula to show lim s n = 0. d). First assume that a 1. Then for n a, we have 1 a 1/n n 1/n. Exercise: use the fact lim n 1/n = 1 to prove lim a 1/n = 1. (Hint: let ɛ > 0 be any positive real number, there exists N > 0, such that n > N = n 1/n 1 < ɛ = a 1/n 1 < ɛ.) Suppose 0 < a < 1, then 1 > 1. So lim ( 1 1/n a a) = 1. By Lemma 9.7, we can get lim a 1/n = 1. Definition 9.9. A sequence (s n ) is said to diverges to +, provided: (9.1) for each M > 0, N > 0, such that n > N, = s n > M. We write lim s n = +. Similarly we write lim s n =, provided (9.2) for each M < 0, N > 0, such that n > N, = s n < M. Example Prove that lim( n + 5) = +. Discussion: consider an arbitrarily large M > 0, we need to find N > 0, such that n > N = n + 5 > M n > (M 5) 2. Proof. Given M > 0, let N = (M 5) 2, then n > N implies n + 5 > M. Theorem Assume that lim s n = + and lim t n > 0, then lim(t n s n ) = +. Discussion: consider an arbitrarily large M > 0, we want (s n t n ) > M. Since lim s n = +, s n can be as large as we want when n is large, so we need to show that t n s are bounded away from 0; Choose 0 < m < lim t n, and observe that t n > m for n large; Then we only need s n > M/m for large n.

28 28 XIN ZHOU Proof. Let M > 0 be an arbitrarily large number. Select m > 0, such that 0 < m < lim t n. By Problem 8.10 in textbook, N n > 0, such that n > N 1, = t n > m. Since lim s n = +, N 2 > 0, such that n > N 2, = s n > M m. Let N = max{n 1, N 2 }, then n > N, = s n t n > M m m = M. Theorem Let (s n ) be a sequence of positive numbers, then ( ) 1 lim s n = +, lim = 0. Proof. (1) Assume lim s n = +. Given any ɛ > 0, let M = 1 ɛ N 1, such that s n > 0, then n > N 1, = s n > M = 1 ɛ, = 1 < ɛ. s n ( ) 1 (2) Assume lim s n = 0. Given M > 0, let ɛ = 1, then N M 2, such that n > N 2, = 1 > M = 1 s n ɛ, = s n > M.

29 MATH 117 FALL 2017 UCSB Monotone sequences and Cauchy sequences Definition Let (s n ) be a sequence. (s n ) is called a non-decreasing sequence if s n s n+1 for all n N. (s n ) is called a non-increasing sequence if s n s n+1 for all n N. Such sequences are called monotone sequences. Example a n = 1 1, b n n = n 3, c n = ( 1 + n) 1 n are non-decreasing sequences; d n = 1 is a non-increasing sequence; n 2 s n = ( 1) n, t n = cos ( ) nπ 3, un = ( 1) n n, v n = ( 1)n are not monotone n sequences; x n = n 1/n is not monotone. Theorem All bounded monotone sequences converge. Proof. Let (s n ) be a bounded non-decreasing sequence. Consider the set of values: S = {s n : n N}. Since (s n ) is bounded, the set S is also bounded. By the Completeness Axiom, we can take u = sup S. Claim: lim s n = u. Given ɛ > 0, since u is the supremum, there exists N N, such that Since (s n ) is non-decreasing, s N > u ɛ. n > N, s n s N > u ɛ. On the other hand, since u is an upper bound for S, we have n N, s n u. Combining the two inequalities above, n > N, u ɛ < s n u, = s n u < ɛ. If (s n ) is a bounded non-increasing sequence, the proof is similar and will be left as an exercise. (Hint: let u = inf S.) Decimals:

30 30 XIN ZHOU Different decimal expansions can represent the same real number; Focus on non-negative decimals expansions and non-negative real numbers. Given a decimal: Let k.d 1 d 2 d 3 d 4, k Z +, d j {0, 1, 2,, 9}. s n = k + d d d n 2 10, n then (s n ) is a non-decreasing sequence, and (s n ) is bounded from above by k + 1. By Theorem 10.3, (s n ) converges and we denote the limit by s = k.d 1 d 2 d 3 d 4. Theorem (1) If (s n ) is an unbounded non-decreasing sequence, then lim s n = + ; (2) If (s n ) is an unbounded non-increasing sequence, then lim s n =. Proof. (1) Since (s n ) is unbounded, and s n s 1, (s n ) is unbounded from above. This implies that: for any M > 0,, N > 0, such that s N > M. Since (s n ) is non-decreasing, (2) Exercise. n > N, s n s N > M. Corollary If (s n ) is a monotone sequence, then it either converges, diverges to +, or diverges to. Therefore lim s n always is meaningful. lim sup and lim inf: Let (s n ) be a bounded sequence. It may or may not converge. The limiting behavior of (s n ) depends only on the terms {s n : n > N} for N large. Let N N be an arbitrary natural number: Claim 1: if lim s n exists, then lim s n [u N, v N ], where u N = inf{s n : n > N}, v N = sup{s n : s > N}.

31 MATH 117 FALL 2017 UCSB 31 Claim 2: we have: u 1 u 2 u 3 ; v 1 v 2 v 3. Note that as N increases, the sets S N = {s n : n > N} get smaller: S 1 S 2 S 3. The above inequalities follow by Problem 4.7 in textbook. By Theorem 10.3, the following limits exist: u = lim u N, v = lim v N, and u v. Moreover, if lim s n exists, = u lim s n v. Definition ( ) lim sup s n := lim sup{s n : n > N} ; n ( ) lim inf s n := lim inf{s n : n > N}. n Remark In the above definition, the sequence (s n ) is not restricted to be bounded. If sup{s n : n > N} = +, then lim sup s n = + ; If inf{s n : n > N} =, then lim inf s n =. Theorem Let (s n ) be a sequence. (i) If lim s n exists (in R, or is +, or ), then lim inf s n = lim s n = lim sup s n. (ii) If lim inf s n = lim sup s n, then lim s n exists, and Proof. As in the definition, denote lim s n = lim inf s n = lim sup s n. u N = inf{s n : n > N}, v N = sup{s n : s > N}, and u = lim inf s n = lim u N, v = lim sup s n = lim v N.

32 32 XIN ZHOU (i) Case 1: lim s n = +. Then M > 0, N > 0, such that n > N, = s n > M. Then u N M, = if m > N, then u m u N M, so lim u N = +. Therefore lim sup s n = lim v N lim u N = +. Case 2: lim s n = can be proven similarly. Case 3: assume lim s n = s. Then ɛ > 0, N > 0, such that We have s n s < ɛ, for n > N. = s n < s + ɛ, for n > N; = v N = sup{s n : n > N} s + ɛ; if m > N, = v m v N s + ɛ; = lim sup s n = lim v m s. Similarly, we have = s n > s ɛ, for n > N; = u N = inf{s n : n > N} s ɛ; if m > N, = u m u N s ɛ; = lim inf s n = lim u m s. Since we know lim inf s n lim sup s n, they must all equal to s. (ii) The proof when lim sup s n = lim inf s n = ± is left as exercise. Assume lim sup s n = lim inf s n = s. Given ɛ > 0, since lim v N = s, there exists N 1 > 0, such that if n > N 1 v N s = sup{s n : n > N} s < ɛ. Note that {v N } is monotone non-increasing, so we have v N s, and hence sup{s n : n > N} s < ɛ, = s n < s + ɛ, if n > N 1. Similarly, since since lim u N = s, there exists N 2 > 0, such that if n > N 2 u N s = inf{s n : n > N} s < ɛ. Note that {u N } is monotone non-decreasing, so we have u N s, and hence s inf{s n : n > N} < ɛ,

33 MATH 117 FALL 2017 UCSB 33 = s n > s ɛ, if n > N 2. Let N = max{n 1, N 2 }, then if n > N, s ɛ < s n < s + ɛ. This implies that lim s n = s. Definition A sequence (s n ) is called a Cauchy sequence if (10.1) ɛ > 0, N > 0, such that m, n > M, = s m s n < ɛ. Lemma Convergent sequences are Cauchy sequences. Proof. Assume that lim s n = s. Let ɛ > 0, then there exists N > 0, such that n > N, = s n s < ɛ 2. Therefore, if n, m > N, s n s m s n s + s s m < ɛ 2 + ɛ 2 = ɛ. Lemma Cauchy sequences are bounded. Proof. Apply (10.1) with ɛ = 1. Then there exists N > 0, such that m, n > N, = s n s m < 1. In particular, s n s N+1 < 1, and so s n < s N Let M = max{ s N+1 + 1, s 1,, s N }. Then s n M. Theorem A sequence converges if and only if it is a Cauchy sequence. Proof. According Lemma 10.10, we only need to show that Cauchy sequence must converge. Assume (s n ) is Cauchy. By Lemma 10.11, (s n ) is bounded, so by Theorem 10.8 we only need to show that lim inf s n = lim sup s n. Given ɛ > 0, there exists N > 0, such that if m, n > N, = s m s n < ɛ.

34 34 XIN ZHOU = s n < s m + ɛ; = v N = sup s n : n > N s m + ɛ, m > N; = v N ɛ s m,, m > N; = v N ɛ inf{s m : m > N} = u N ; = lim sup s n v N u N + ɛ lim inf s N + ɛ.

35 MATH 117 FALL 2017 UCSB Subsequences Definition Let (s n ) be a sequence. A subsequence of (s n ) is a sequence (t k ) k N, such that t k = s nk, with n 1 < n 2 < n 3 < < n k < n k+1 <. Example Consider s n = n 2 ( 1) n. The positive terms form a subsequence: Here t k = s 2k. (t k ) = (4, 16, 36, ). Example Consider a n = sin ( nπ 3 ( ,, 0, 0, 2 2, ). The nonnegative terms are 3 2, ) Example Q can be listed as s sequence (r n ). Proposition Given any real number a, there exists a subsequence (r nk ) of (r n ) that converges to a. Discussion: we want to construct r nk, such that r nk a < 1, for all k N. k Proof. We use Definition/construction by induction. (1) Select n 1, such that r n1 a < 1, and this follows from the Denseness of Q; (2) Suppose n 1,, n k have been selected such that n 1 < n 2 < < n k, and r nj a < 1 for j = 1, 2,, k. j We want to find r nk+1, with n k+1 > n k and r nk+1 a < 1, which k+1 is equivalent to r nk+1 ( a 1, a + 1 k+1 k+1). By Problem 4.11 in the textbook, there are infinitely many rationals in ( a 1, a + 1 k+1 k+1). Therefore, there exists nk+1 > n k, such that r nk+1 ( a 1, a + 1 k+1 k+1). By Mathematical Induction, there exists a subsequence (r nk ), with lim r nk = a.

36 36 XIN ZHOU Example Let (s n ) be a sequence of positive numbers, i.e. s n > 0, such that inf{s n : n N} = 0. Prove that a subsequence of (s n ) converges monotonically to 0. Discussion: we want to construct s nk, such that s 1 > s 2 > > s k ; and s k < 1, for all k N. k Proof. We use Definition/construction by induction. (1) Since inf{s n : n N} = 0, there exists n 1 N, such that s 1 < 1; (2) Suppose n 1,, n k have been selected such that and s nj+1 < min{s nj, n 1 < n 2 < < n k, 1 }, for j = 1, 2,, k 1. j We want to find s nk+1, such that s nk+1 < min{s nk, }. Note that k+1 min{s n : 1 n n k } > 0, = inf{s n : n > n k } = 0. Therefore, there exists n k+1 > n k, such that s nk+1 < min{s nk, 1 k + 1 }. By Mathematical Induction, there exists a decreasing subsequence (s nk ), with lim s nk = 0. Theorem If a sequence (s n ) converges, then every subsequence converges to the same limit. Proof. Let lim s n = s, and (s nk ) be a subsequence of (s n ). Note that n k k for all k. Given ɛ > 0, there exists N > 0, such that n > N, = s n s < ɛ. If k > N, then n k > N, so s nk s < ɛ. This implies that lim s nk = s. Theorem Every sequence (s n ) has a monotone subsequence. Proof. Say that the n-term s n is dominant if s n > s m for all m > n. We divide the proof into two cases:

37 MATH 117 FALL 2017 UCSB 37 Case 1: there exists infinitely many dominant terms (s nk ), then s nk+1 < s nk, so (s nk ) is a decreasing sequence. Case 2: suppose there are only finitely many dominant terms, and we only focus on this case in the following. Select n 1, such that s n1 is larger than all dominant terms. Then (11.1) For any N n 1, there exists m > N, such that s m s N. We are going to use inductive construction to find a monotone sequence. (1) Applying (11.1) with N = n 1 gives n 2 > n 1, such that s n2 s n1 ; (2) Suppose n 1,, n k have been selected such that and n 1 < n 2 < < n k, s n1 s n2 s nk. Applying (11.1) with N = n k gives n k+1 > n k, such that s nk+1 s nk. By Mathematical Induction, (s nk ) is a nondecreasing subsequence. Corollary Let (s n ) be a sequence, then there exists a monotone subsequence with limit lim sup s n, and a monotone subsequence with limit lim inf s n. Proof. We only prove the first statement for lim sup s n. Let v N = sup{s n : s > N}, N N, then v = lim v N = lim sup s n. If v =, then lim s n =, so any monotone sequence will converge to lim sup s n, and we finished. If v, select an arbitrary monotone increasing sequence (t N ), such that lim t N = v: if v is finite, let t N = v 1 N ; if v = +, let t N = N. Now we discuss the two cases as in the proof of the above theorem. Case 1: by assumption, s nk = sup{s m : m n k } = v nk 1, so {s nk } is decreasing and lim k s n k = lim v N = lim sup s n. Case 2: Given N > n 1 as above, by (11.1), there exists m 1 > N, such that s m1 s N. Since t N < v v N = sup{s m : m > N},

38 38 XIN ZHOU there exists m 2 > N, such that s m2 > t N. { either sm1 s m2, = s m1 s N and s m1 > t N ; or s m2 s m1, = s m2 s N and s m2 > t N. Then we have (11.2) Given N n 1, there exists m > N, such that s m s N and s m > t N. We are going to use inductive construction to find the monotone subsequence. (1) Applying (11.2) with N = n 1 gives n 2 > n 1, such that s n2 s n1 and s n2 > t n1 ; (2) Suppose n 1,, n k have been selected such that n 1 < n 2 < < n k, and s n1 s n2 s nk, s nj+1 > t nj, for all 1 j k 1. Applying (11.2) with N = n k gives n k+1 > n k, such that s nk+1 and s nk+1 > t nk. Moreover, s nk1 sup{s m : m > n k } = v nk, so s nk t nk < s nk+1 v nk. By Mathematical Induction, (s nk ) is a nondecreasing sequence and lim s nk = lim sup s n. Definition Let (s n ) be a sequence. A subsequential limit is any real number, or +, or, that is the limit of some subsequence. Example If lim s n = s, then the set of subsequential limit is {s}. Example Let (r n ) be the list of rational numbers, then the set of subsequential limits are R {+, }. Theorem Let (s n ) be a sequence, and S be the set of subsequential limits of (s n ). Then (i) S ; (ii) sup S = lim sup s n, inf S = lim inf s n ; (iii) lim s n exists if and only if S consists of only one element, i.e. S = {s}.

39 MATH 117 FALL 2017 UCSB 39 Proof. (i) follows from the above corollary since lim sup s n, lim inf s n S, and (iii) follows from (ii). To prove (ii), let t = lim s nk for some subsequence (s nk ), then Note that So Therefore, So t = lim sup s nk = lim inf s nk. {s nk : k > N} {s n : n > N}. sup{s nk : k > N} sup{s n : n > N}; inf{s nk : k > N} inf{s n : n > N}. lim inf s n lim inf s nk = t = lim sup s nk lim sup s n. lim inf s n inf S sup S lim sup s n. Also lim sup s n, lim inf s n S, so we finish the proof. Theorem Let S be the set of subsequential limits of (s n ). If t n S R and lim t n = t, then t S. Proof. We use construction by induction. (1) Since t 1 = lim s nk for some subsequence (s nk ), there exists n 1, such that s n1 t 1 < 1; (2) Assume n 1 < < n k have been selected with s nj t j < 1, for j = 1,, k. j Since t k+1 = lim s ni for some other subsequence (s ni ), there exists n k+1 > n k, such that s nk+1 t k+1 < 1 k + 1. By induction, we found a subsequence (s nk ) such that s nk t k < 1. It is then k an easy exercise to show that t = lim s nk. Definition A subset S R is called closed, if for any convergent sequence (t n ) where t n S, then the limit t = lim t n S.

40 40 XIN ZHOU 12. lim sup and lim inf revisited Theorem If (s n ) is a sequence with lim s n = s and s > 0, and (t n ) is an arbitrary sequence, then lim sup(s n t n ) = s lim sup t n. Proof. First we will show lim sup(s n t n ) s lim sup t n. Denote β = lim sup t n. Case 1: β R. There exists a subsequence (t nk ) of (t n ), such that lim t nk = lim sup t n = β. Then lim(s nk t nk ) = sβ lim sup(s n t n ). Case 2: β = +. There exists a subsequence (t nk ) of (t n ), such that lim t nk = +. Since lim s nk = s > 0, we have lim(s nk t nk ) = + by Theorem Hence Case 3: β =. Exercise. lim sup(s n t n ) = +. To show the reverse direction: assume s n 0 by ignore the first few terms. Then [ lim sup t n = lim sup ( 1 ] ( ) 1 ) (s n t n ) lim lim sup(s n t n ) = 1 s n s n 2 lim sup(s nt n ). This finishes the proof. Theorem Let (s n ) be a sequence with s n 0. Then lim inf s n+1 s n lim inf s n 1/n lim sup s n 1/n lim sup s n+1 s n. Proof. Let α = lim sup s n 1/n, and L = lim sup s n+1. s n If L = +, then we are done. Assume L < +. It suffices to show α L 1 for any L 1 > L. Since ( { }) s n+1 L = lim sup N : n > N < L 1, there exists N > 0, such that { } s n+1 sup : n N < L 1. s n s n

41 Then by iteration, for n > N, s n = Let a = L N 1 s N > 0, then As lim a 1/n = 1, we have MATH 117 FALL 2017 UCSB 41 s n s n 1 s n 1 s n 2 s N+1 s N s N < L n N 1 s N = L n 1 s N. L N 1 s n < a L n 1 = s n 1/n < L 1 a 1/n. α = lim sup s n 1/n L 1. The other part lim inf s n+1 s n lim inf sn 1/n is left as an exercise.

42 42 XIN ZHOU Denote 13. Series n a k = a m + a m a n. k=m We are interested in infinite series n=m a n. Let Definition s n = a m + a m a n = n a k. (1) The infinite series n=m a n is said to converge provided the sequence of partial sums (s n ) converges to a real number s R, and we define a n = s. n=m (2) A series that does not converge is said to diverge. (3) Say n=m a n = diverges to +, and write n=m a n = +, provided lim s n = +. Similarly we can define n=m a n =. (4) If we do not care the initial index m, just write a n. (5) If a n 0, then (s n ) is nondecreasing, so a n either converges or diverges to +. Hence for any series a n, the sum of absolute values an is meaningful. (6) a n is said to converge absolutely or be absolutely convergent if a n converges. Example A series n=0 arn for constants a and r is called a geometric series. For r 1, s n = n k=0 For r < 1, lim n r n+1 = 0, we have And k=m ar k = a 1 rn+1 1 r. lim s n = n ar n = n=0 a 1 r. a 1 r. If a 0, and r 1, the series does not converge.

43 Example The series n=1 be proved later.) MATH 117 FALL 2017 UCSB 43 1 n p converges if and only if p > 1. (This will Definition We say a series a n satisfies the Cauchy criterion if the sequence of partial sums (s n ) is a Cauchy sequence: (13.1) for each ɛ > 0, N > 0, such that if n, m > N, = s n s m < ɛ. And this is equivalent to (13.2) for each ɛ > 0, N > 0, such that n m > N, = s n s m 1 < ɛ. This is equivalent to (13.3) for each ɛ > 0, N > 0, such that n m > N, = As a direct corollary of Theorem 10.12, we have, n a k < ɛ. Theorem A series converges if and only if it satisfies the Cauchy criterion. Corollary If a n converges, then lim a n = 0. Proof. For any ɛ > 0, by the Cauchy criterion, there exists N > 0, such that if n m > N, then n k=m a k < ɛ. We can take n = m, then we have a m < ɛ. This implies that lim a m = 0. Remark The converse of the above result is not true. 1 = +, although lim 1 = 0. n n k=m For instance, Theorem 13.8 (Comparison Test). Consider a series a n, with a n 0 for all n N. (i) If a n converges, and b n a n, then b n converges; (ii) If a n = +, and b n a n, then b n = +. Proof. (i) For any n m, by the triangle inequality, n n n b k b k a k. k=m k=m Therefore a n satisfies the Cauchy criterion implies that b k also satisfies the Cauchy criterion. (ii) Denote t n = n k=0 b k and s n = n k=0 a k. Then t n s n. So lim s n = + implies lim t n = +. k=m

44 44 XIN ZHOU Corollary Absolutely convergent series are convergent. Proof. This is an easy corollary of the comparison test. Theorem (Ratio Test). Consider a series a n of nonzero terms (a n 0 for all n N). (i) a n converges absolutely, if lim sup a n+1 a n < 1; (ii) a n diverges If lim inf a n+1 a n > 1; (iii) otherwise if lim inf a n+1 a n 1 lim sup a n+1, a n then the test gives no information. We will use the following root test to prove the ratio test. Theorem (Root Test). Let a n be a series, and α = lim sup a n 1/n. The the series a n (i) converges absolutely, if α < 1; (ii) diverges if α > 1; (iii) otherwise if α = 1, then the test gives no information. Proof. (i) Suppose α < 1, and select β such that α < β < 1. By the definition of lim sup, there exists N > 0, such that sup{ a n 1/n : n > N} < β. In particular, a n 1/n < β for all n > N, and hence a n < β n, for all n > N. Since 0 < β < 1, the series β n converges. By the comparison test, an converges. (ii) If α > 1, then a subsequence of a n 1/n will converge to α > 1, hence a n > 1 for infinitely many choices of n. By Corollary 13.6, a n cannot converge. (iii) For 1 and 1, α turns out to be 1, but 1 diverges, and 1 n n 2 n n 2 converges. Proof of Ratio Test. Let α = lim sup a n 1/n. By Theorem 12.2:

2.2 Some Consequences of the Completeness Axiom

2.2 Some Consequences of the Completeness Axiom 60 CHAPTER 2. IMPORTANT PROPERTIES OF R 2.2 Some Consequences of the Completeness Axiom In this section, we use the fact that R is complete to establish some important results. First, we will prove that

More information

In N we can do addition, but in order to do subtraction we need to extend N to the integers

In N we can do addition, but in order to do subtraction we need to extend N to the integers Chapter The Real Numbers.. Some Preliminaries Discussion: The Irrationality of 2. We begin with the natural numbers N = {, 2, 3, }. In N we can do addition, but in order to do subtraction we need to extend

More information

Sequences. Limits of Sequences. Definition. A real-valued sequence s is any function s : N R.

Sequences. Limits of Sequences. Definition. A real-valued sequence s is any function s : N R. Sequences Limits of Sequences. Definition. A real-valued sequence s is any function s : N R. Usually, instead of using the notation s(n), we write s n for the value of this function calculated at n. We

More information

Sequences. Chapter 3. n + 1 3n + 2 sin n n. 3. lim (ln(n + 1) ln n) 1. lim. 2. lim. 4. lim (1 + n)1/n. Answers: 1. 1/3; 2. 0; 3. 0; 4. 1.

Sequences. Chapter 3. n + 1 3n + 2 sin n n. 3. lim (ln(n + 1) ln n) 1. lim. 2. lim. 4. lim (1 + n)1/n. Answers: 1. 1/3; 2. 0; 3. 0; 4. 1. Chapter 3 Sequences Both the main elements of calculus (differentiation and integration) require the notion of a limit. Sequences will play a central role when we work with limits. Definition 3.. A Sequence

More information

In N we can do addition, but in order to do subtraction we need to extend N to the integers

In N we can do addition, but in order to do subtraction we need to extend N to the integers Chapter 1 The Real Numbers 1.1. Some Preliminaries Discussion: The Irrationality of 2. We begin with the natural numbers N = {1, 2, 3, }. In N we can do addition, but in order to do subtraction we need

More information

Undergraduate Notes in Mathematics. Arkansas Tech University Department of Mathematics

Undergraduate Notes in Mathematics. Arkansas Tech University Department of Mathematics Undergraduate Notes in Mathematics Arkansas Tech University Department of Mathematics An Introductory Single Variable Real Analysis: A Learning Approach through Problem Solving Marcel B. Finan c All Rights

More information

Structure of R. Chapter Algebraic and Order Properties of R

Structure of R. Chapter Algebraic and Order Properties of R Chapter Structure of R We will re-assemble calculus by first making assumptions about the real numbers. All subsequent results will be rigorously derived from these assumptions. Most of the assumptions

More information

Real Analysis Math 131AH Rudin, Chapter #1. Dominique Abdi

Real Analysis Math 131AH Rudin, Chapter #1. Dominique Abdi Real Analysis Math 3AH Rudin, Chapter # Dominique Abdi.. If r is rational (r 0) and x is irrational, prove that r + x and rx are irrational. Solution. Assume the contrary, that r+x and rx are rational.

More information

Chapter 1 The Real Numbers

Chapter 1 The Real Numbers Chapter 1 The Real Numbers In a beginning course in calculus, the emphasis is on introducing the techniques of the subject;i.e., differentiation and integration and their applications. An advanced calculus

More information

Chapter 2. Real Numbers. 1. Rational Numbers

Chapter 2. Real Numbers. 1. Rational Numbers Chapter 2. Real Numbers 1. Rational Numbers A commutative ring is called a field if its nonzero elements form a group under multiplication. Let (F, +, ) be a filed with 0 as its additive identity element

More information

1 The Real Number System

1 The Real Number System 1 The Real Number System The rational numbers are beautiful, but are not big enough for various purposes, and the set R of real numbers was constructed in the late nineteenth century, as a kind of an envelope

More information

Homework 4, 5, 6 Solutions. > 0, and so a n 0 = n + 1 n = ( n+1 n)( n+1+ n) 1 if n is odd 1/n if n is even diverges.

Homework 4, 5, 6 Solutions. > 0, and so a n 0 = n + 1 n = ( n+1 n)( n+1+ n) 1 if n is odd 1/n if n is even diverges. 2..2(a) lim a n = 0. Homework 4, 5, 6 Solutions Proof. Let ɛ > 0. Then for n n = 2+ 2ɛ we have 2n 3 4+ ɛ 3 > ɛ > 0, so 0 < 2n 3 < ɛ, and thus a n 0 = 2n 3 < ɛ. 2..2(g) lim ( n + n) = 0. Proof. Let ɛ >

More information

Solutions for Homework Assignment 2

Solutions for Homework Assignment 2 Solutions for Homework Assignment 2 Problem 1. If a,b R, then a+b a + b. This fact is called the Triangle Inequality. By using the Triangle Inequality, prove that a b a b for all a,b R. Solution. To prove

More information

Advanced Calculus: MATH 410 Real Numbers Professor David Levermore 5 December 2010

Advanced Calculus: MATH 410 Real Numbers Professor David Levermore 5 December 2010 Advanced Calculus: MATH 410 Real Numbers Professor David Levermore 5 December 2010 1. Real Number System 1.1. Introduction. Numbers are at the heart of mathematics. By now you must be fairly familiar with

More information

Real Analysis - Notes and After Notes Fall 2008

Real Analysis - Notes and After Notes Fall 2008 Real Analysis - Notes and After Notes Fall 2008 October 29, 2008 1 Introduction into proof August 20, 2008 First we will go through some simple proofs to learn how one writes a rigorous proof. Let start

More information

Chapter One. The Real Number System

Chapter One. The Real Number System Chapter One. The Real Number System We shall give a quick introduction to the real number system. It is imperative that we know how the set of real numbers behaves in the way that its completeness and

More information

FINAL EXAM Math 25 Temple-F06

FINAL EXAM Math 25 Temple-F06 FINAL EXAM Math 25 Temple-F06 Write solutions on the paper provided. Put your name on this exam sheet, and staple it to the front of your finished exam. Do Not Write On This Exam Sheet. Problem 1. (Short

More information

Postulate 2 [Order Axioms] in WRW the usual rules for inequalities

Postulate 2 [Order Axioms] in WRW the usual rules for inequalities Number Systems N 1,2,3,... the positive integers Z 3, 2, 1,0,1,2,3,... the integers Q p q : p,q Z with q 0 the rational numbers R {numbers expressible by finite or unending decimal expansions} makes sense

More information

1. Supremum and Infimum Remark: In this sections, all the subsets of R are assumed to be nonempty.

1. Supremum and Infimum Remark: In this sections, all the subsets of R are assumed to be nonempty. 1. Supremum and Infimum Remark: In this sections, all the subsets of R are assumed to be nonempty. Let E be a subset of R. We say that E is bounded above if there exists a real number U such that x U for

More information

Math LM (24543) Lectures 01

Math LM (24543) Lectures 01 Math 32300 LM (24543) Lectures 01 Ethan Akin Office: NAC 6/287 Phone: 650-5136 Email: ethanakin@earthlink.net Spring, 2018 Contents Introduction, Ross Chapter 1 and Appendix The Natural Numbers N and The

More information

CHAPTER 8: EXPLORING R

CHAPTER 8: EXPLORING R CHAPTER 8: EXPLORING R LECTURE NOTES FOR MATH 378 (CSUSM, SPRING 2009). WAYNE AITKEN In the previous chapter we discussed the need for a complete ordered field. The field Q is not complete, so we constructed

More information

1. Theorem. (Archimedean Property) Let x be any real number. There exists a positive integer n greater than x.

1. Theorem. (Archimedean Property) Let x be any real number. There exists a positive integer n greater than x. Advanced Calculus I, Dr. Block, Chapter 2 notes. Theorem. (Archimedean Property) Let x be any real number. There exists a positive integer n greater than x. 2. Definition. A sequence is a real-valued function

More information

Math 117: Infinite Sequences

Math 117: Infinite Sequences Math 7: Infinite Sequences John Douglas Moore November, 008 The three main theorems in the theory of infinite sequences are the Monotone Convergence Theorem, the Cauchy Sequence Theorem and the Subsequence

More information

The Real Number System

The Real Number System MATH 337 The Real Number System Sets of Numbers Dr. Neal, WKU A set S is a well-defined collection of objects, with well-defined meaning that there is a specific description from which we can tell precisely

More information

Midterm Review Math 311, Spring 2016

Midterm Review Math 311, Spring 2016 Midterm Review Math 3, Spring 206 Material Review Preliminaries and Chapter Chapter 2. Set theory (DeMorgan s laws, infinite collections of sets, nested sets, cardinality) 2. Functions (image, preimage,

More information

Chapter 1. Sets and Numbers

Chapter 1. Sets and Numbers Chapter 1. Sets and Numbers 1. Sets A set is considered to be a collection of objects (elements). If A is a set and x is an element of the set A, we say x is a member of A or x belongs to A, and we write

More information

means is a subset of. So we say A B for sets A and B if x A we have x B holds. BY CONTRAST, a S means that a is a member of S.

means is a subset of. So we say A B for sets A and B if x A we have x B holds. BY CONTRAST, a S means that a is a member of S. 1 Notation For those unfamiliar, we have := means equal by definition, N := {0, 1,... } or {1, 2,... } depending on context. (i.e. N is the set or collection of counting numbers.) In addition, means for

More information

Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.

Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall. .1 Limits of Sequences. CHAPTER.1.0. a) True. If converges, then there is an M > 0 such that M. Choose by Archimedes an N N such that N > M/ε. Then n N implies /n M/n M/N < ε. b) False. = n does not converge,

More information

Continuity. Chapter 4

Continuity. Chapter 4 Chapter 4 Continuity Throughout this chapter D is a nonempty subset of the real numbers. We recall the definition of a function. Definition 4.1. A function from D into R, denoted f : D R, is a subset of

More information

MATH 1A, Complete Lecture Notes. Fedor Duzhin

MATH 1A, Complete Lecture Notes. Fedor Duzhin MATH 1A, Complete Lecture Notes Fedor Duzhin 2007 Contents I Limit 6 1 Sets and Functions 7 1.1 Sets................................. 7 1.2 Functions.............................. 8 1.3 How to define a

More information

Advanced Calculus: MATH 410 Real Numbers Professor David Levermore 1 November 2017

Advanced Calculus: MATH 410 Real Numbers Professor David Levermore 1 November 2017 Advanced Calculus: MATH 410 Real Numbers Professor David Levermore 1 November 2017 1. Real Number System 1.1. Introduction. Numbers are at the heart of mathematics. By now you must be fairly familiar with

More information

MATH 102 INTRODUCTION TO MATHEMATICAL ANALYSIS. 1. Some Fundamentals

MATH 102 INTRODUCTION TO MATHEMATICAL ANALYSIS. 1. Some Fundamentals MATH 02 INTRODUCTION TO MATHEMATICAL ANALYSIS Properties of Real Numbers Some Fundamentals The whole course will be based entirely on the study of sequence of numbers and functions defined on the real

More information

= 1 2x. x 2 a ) 0 (mod p n ), (x 2 + 2a + a2. x a ) 2

= 1 2x. x 2 a ) 0 (mod p n ), (x 2 + 2a + a2. x a ) 2 8. p-adic numbers 8.1. Motivation: Solving x 2 a (mod p n ). Take an odd prime p, and ( an) integer a coprime to p. Then, as we know, x 2 a (mod p) has a solution x Z iff = 1. In this case we can suppose

More information

2.1 Convergence of Sequences

2.1 Convergence of Sequences Chapter 2 Sequences 2. Convergence of Sequences A sequence is a function f : N R. We write f) = a, f2) = a 2, and in general fn) = a n. We usually identify the sequence with the range of f, which is written

More information

Suppose R is an ordered ring with positive elements P.

Suppose R is an ordered ring with positive elements P. 1. The real numbers. 1.1. Ordered rings. Definition 1.1. By an ordered commutative ring with unity we mean an ordered sextuple (R, +, 0,, 1, P ) such that (R, +, 0,, 1) is a commutative ring with unity

More information

Theorems. Theorem 1.11: Greatest-Lower-Bound Property. Theorem 1.20: The Archimedean property of. Theorem 1.21: -th Root of Real Numbers

Theorems. Theorem 1.11: Greatest-Lower-Bound Property. Theorem 1.20: The Archimedean property of. Theorem 1.21: -th Root of Real Numbers Page 1 Theorems Wednesday, May 9, 2018 12:53 AM Theorem 1.11: Greatest-Lower-Bound Property Suppose is an ordered set with the least-upper-bound property Suppose, and is bounded below be the set of lower

More information

. As the binomial coefficients are integers we have that. 2 n(n 1).

. As the binomial coefficients are integers we have that. 2 n(n 1). Math 580 Homework. 1. Divisibility. Definition 1. Let a, b be integers with a 0. Then b divides b iff there is an integer k such that b = ka. In the case we write a b. In this case we also say a is a factor

More information

We are going to discuss what it means for a sequence to converge in three stages: First, we define what it means for a sequence to converge to zero

We are going to discuss what it means for a sequence to converge in three stages: First, we define what it means for a sequence to converge to zero Chapter Limits of Sequences Calculus Student: lim s n = 0 means the s n are getting closer and closer to zero but never gets there. Instructor: ARGHHHHH! Exercise. Think of a better response for the instructor.

More information

Numerical Sequences and Series

Numerical Sequences and Series Numerical Sequences and Series Written by Men-Gen Tsai email: b89902089@ntu.edu.tw. Prove that the convergence of {s n } implies convergence of { s n }. Is the converse true? Solution: Since {s n } is

More information

Continuity. Chapter 4

Continuity. Chapter 4 Chapter 4 Continuity Throughout this chapter D is a nonempty subset of the real numbers. We recall the definition of a function. Definition 4.1. A function from D into R, denoted f : D R, is a subset of

More information

Solutions Manual for Homework Sets Math 401. Dr Vignon S. Oussa

Solutions Manual for Homework Sets Math 401. Dr Vignon S. Oussa 1 Solutions Manual for Homework Sets Math 401 Dr Vignon S. Oussa Solutions Homework Set 0 Math 401 Fall 2015 1. (Direct Proof) Assume that x and y are odd integers. Then there exist integers u and v such

More information

Math 118B Solutions. Charles Martin. March 6, d i (x i, y i ) + d i (y i, z i ) = d(x, y) + d(y, z). i=1

Math 118B Solutions. Charles Martin. March 6, d i (x i, y i ) + d i (y i, z i ) = d(x, y) + d(y, z). i=1 Math 8B Solutions Charles Martin March 6, Homework Problems. Let (X i, d i ), i n, be finitely many metric spaces. Construct a metric on the product space X = X X n. Proof. Denote points in X as x = (x,

More information

MATH 131A: REAL ANALYSIS

MATH 131A: REAL ANALYSIS MATH 131A: REAL ANALYSIS NICKOLAS ANDERSEN The textbook for the course is Ross, Elementary Analysis [2], but in these notes I have also borrowed from Tao, Analysis I [3], and Abbott, Understanding Analysis

More information

Principle of Mathematical Induction

Principle of Mathematical Induction Advanced Calculus I. Math 451, Fall 2016, Prof. Vershynin Principle of Mathematical Induction 1. Prove that 1 + 2 + + n = 1 n(n + 1) for all n N. 2 2. Prove that 1 2 + 2 2 + + n 2 = 1 n(n + 1)(2n + 1)

More information

MA103 Introduction to Abstract Mathematics Second part, Analysis and Algebra

MA103 Introduction to Abstract Mathematics Second part, Analysis and Algebra 206/7 MA03 Introduction to Abstract Mathematics Second part, Analysis and Algebra Amol Sasane Revised by Jozef Skokan, Konrad Swanepoel, and Graham Brightwell Copyright c London School of Economics 206

More information

We want to show P (n) is true for all integers

We want to show P (n) is true for all integers Generalized Induction Proof: Let P (n) be the proposition 1 + 2 + 2 2 + + 2 n = 2 n+1 1. We want to show P (n) is true for all integers n 0. Generalized Induction Example: Use generalized induction to

More information

Solutions, Project on p-adic Numbers, Real Analysis I, Fall, 2010.

Solutions, Project on p-adic Numbers, Real Analysis I, Fall, 2010. Solutions, Project on p-adic Numbers, Real Analysis I, Fall, 2010. (24) The p-adic numbers. Let p {2, 3, 5, 7, 11,... } be a prime number. (a) For x Q, define { 0 for x = 0, x p = p n for x = p n (a/b),

More information

Limits and Continuity

Limits and Continuity Chapter Limits and Continuity. Limits of Sequences.. The Concept of Limit and Its Properties A sequence { } is an ordered infinite list x,x,...,,... The n-th term of the sequence is, and n is the index

More information

Scalar multiplication and addition of sequences 9

Scalar multiplication and addition of sequences 9 8 Sequences 1.2.7. Proposition. Every subsequence of a convergent sequence (a n ) n N converges to lim n a n. Proof. If (a nk ) k N is a subsequence of (a n ) n N, then n k k for every k. Hence if ε >

More information

Homework 1 Solutions

Homework 1 Solutions MATH 171 Spring 2016 Problem 1 Homework 1 Solutions (If you find any errors, please send an e-mail to farana at stanford dot edu) Presenting your arguments in steps, using only axioms of an ordered field,

More information

MATH 101, FALL 2018: SUPPLEMENTARY NOTES ON THE REAL LINE

MATH 101, FALL 2018: SUPPLEMENTARY NOTES ON THE REAL LINE MATH 101, FALL 2018: SUPPLEMENTARY NOTES ON THE REAL LINE SEBASTIEN VASEY These notes describe the material for November 26, 2018 (while similar content is in Abbott s book, the presentation here is different).

More information

Sequences of Real Numbers

Sequences of Real Numbers Chapter 8 Sequences of Real Numbers In this chapter, we assume the existence of the ordered field of real numbers, though we do not yet discuss or use the completeness of the real numbers. In the next

More information

Supremum and Infimum

Supremum and Infimum Supremum and Infimum UBC M0 Lecture Notes by Philip D. Loewen The Real Number System. Work hard to construct from the axioms a set R with special elements O and I, and a subset P R, and mappings A: R R

More information

Introductory Analysis I Fall 2014 Homework #9 Due: Wednesday, November 19

Introductory Analysis I Fall 2014 Homework #9 Due: Wednesday, November 19 Introductory Analysis I Fall 204 Homework #9 Due: Wednesday, November 9 Here is an easy one, to serve as warmup Assume M is a compact metric space and N is a metric space Assume that f n : M N for each

More information

Seunghee Ye Ma 8: Week 2 Oct 6

Seunghee Ye Ma 8: Week 2 Oct 6 Week 2 Summary This week, we will learn about sequences and real numbers. We first define what we mean by a sequence and discuss several properties of sequences. Then, we will talk about what it means

More information

1 Homework. Recommended Reading:

1 Homework. Recommended Reading: Analysis MT43C Notes/Problems/Homework Recommended Reading: R. G. Bartle, D. R. Sherbert Introduction to real analysis, principal reference M. Spivak Calculus W. Rudin Principles of mathematical analysis

More information

Week 2: Sequences and Series

Week 2: Sequences and Series QF0: Quantitative Finance August 29, 207 Week 2: Sequences and Series Facilitator: Christopher Ting AY 207/208 Mathematicians have tried in vain to this day to discover some order in the sequence of prime

More information

Principles of Real Analysis I Fall I. The Real Number System

Principles of Real Analysis I Fall I. The Real Number System 21-355 Principles of Real Analysis I Fall 2004 I. The Real Number System The main goal of this course is to develop the theory of real-valued functions of one real variable in a systematic and rigorous

More information

DR.RUPNATHJI( DR.RUPAK NATH )

DR.RUPNATHJI( DR.RUPAK NATH ) Contents 1 Sets 1 2 The Real Numbers 9 3 Sequences 29 4 Series 59 5 Functions 81 6 Power Series 105 7 The elementary functions 111 Chapter 1 Sets It is very convenient to introduce some notation and terminology

More information

Contents Ordered Fields... 2 Ordered sets and fields... 2 Construction of the Reals 1: Dedekind Cuts... 2 Metric Spaces... 3

Contents Ordered Fields... 2 Ordered sets and fields... 2 Construction of the Reals 1: Dedekind Cuts... 2 Metric Spaces... 3 Analysis Math Notes Study Guide Real Analysis Contents Ordered Fields 2 Ordered sets and fields 2 Construction of the Reals 1: Dedekind Cuts 2 Metric Spaces 3 Metric Spaces 3 Definitions 4 Separability

More information

Contribution of Problems

Contribution of Problems Exam topics 1. Basic structures: sets, lists, functions (a) Sets { }: write all elements, or define by condition (b) Set operations: A B, A B, A\B, A c (c) Lists ( ): Cartesian product A B (d) Functions

More information

2.7 Subsequences. Definition Suppose that (s n ) n N is a sequence. Let (n 1, n 2, n 3,... ) be a sequence of natural numbers such that

2.7 Subsequences. Definition Suppose that (s n ) n N is a sequence. Let (n 1, n 2, n 3,... ) be a sequence of natural numbers such that 2.7 Subsequences Definition 2.7.1. Suppose that (s n ) n N is a sequence. Let (n 1, n 2, n 3,... ) be a sequence of natural numbers such that n 1 < n 2 < < n k < n k+1

More information

Metric Spaces and Topology

Metric Spaces and Topology Chapter 2 Metric Spaces and Topology From an engineering perspective, the most important way to construct a topology on a set is to define the topology in terms of a metric on the set. This approach underlies

More information

CHAPTER 3. Sequences. 1. Basic Properties

CHAPTER 3. Sequences. 1. Basic Properties CHAPTER 3 Sequences We begin our study of analysis with sequences. There are several reasons for starting here. First, sequences are the simplest way to introduce limits, the central idea of calculus.

More information

Chapter 8. P-adic numbers. 8.1 Absolute values

Chapter 8. P-adic numbers. 8.1 Absolute values Chapter 8 P-adic numbers Literature: N. Koblitz, p-adic Numbers, p-adic Analysis, and Zeta-Functions, 2nd edition, Graduate Texts in Mathematics 58, Springer Verlag 1984, corrected 2nd printing 1996, Chap.

More information

M17 MAT25-21 HOMEWORK 6

M17 MAT25-21 HOMEWORK 6 M17 MAT25-21 HOMEWORK 6 DUE 10:00AM WEDNESDAY SEPTEMBER 13TH 1. To Hand In Double Series. The exercises in this section will guide you to complete the proof of the following theorem: Theorem 1: Absolute

More information

A lower bound for X is an element z F such that

A lower bound for X is an element z F such that Math 316, Intro to Analysis Completeness. Definition 1 (Upper bounds). Let F be an ordered field. For a subset X F an upper bound for X is an element y F such that A lower bound for X is an element z F

More information

Math 140: Foundations of Real Analysis. Todd Kemp

Math 140: Foundations of Real Analysis. Todd Kemp Math 140: Foundations of Real Analysis Todd Kemp Contents Part 1. Math 140A 5 Chapter 1. Ordered Sets, Ordered Fields, and Completeness 7 1. Lecture 1: January 5, 2016 7 2. Lecture 2: January 7, 2016

More information

Sequences CHAPTER 3. Definition. A sequence is a function f : N R.

Sequences CHAPTER 3. Definition. A sequence is a function f : N R. CHAPTER 3 Sequences 1. Limits and the Archimedean Property Our first basic object for investigating real numbers is the sequence. Before we give the precise definition of a sequence, we will give the intuitive

More information

C.7. Numerical series. Pag. 147 Proof of the converging criteria for series. Theorem 5.29 (Comparison test) Let a k and b k be positive-term series

C.7. Numerical series. Pag. 147 Proof of the converging criteria for series. Theorem 5.29 (Comparison test) Let a k and b k be positive-term series C.7 Numerical series Pag. 147 Proof of the converging criteria for series Theorem 5.29 (Comparison test) Let and be positive-term series such that 0, for any k 0. i) If the series converges, then also

More information

A NEW SET THEORY FOR ANALYSIS

A NEW SET THEORY FOR ANALYSIS Article A NEW SET THEORY FOR ANALYSIS Juan Pablo Ramírez 0000-0002-4912-2952 Abstract: We present the real number system as a generalization of the natural numbers. First, we prove the co-finite topology,

More information

MATH 324 Summer 2011 Elementary Number Theory. Notes on Mathematical Induction. Recall the following axiom for the set of integers.

MATH 324 Summer 2011 Elementary Number Theory. Notes on Mathematical Induction. Recall the following axiom for the set of integers. MATH 4 Summer 011 Elementary Number Theory Notes on Mathematical Induction Principle of Mathematical Induction Recall the following axiom for the set of integers. Well-Ordering Axiom for the Integers If

More information

MAT 570 REAL ANALYSIS LECTURE NOTES. Contents. 1. Sets Functions Countability Axiom of choice Equivalence relations 9

MAT 570 REAL ANALYSIS LECTURE NOTES. Contents. 1. Sets Functions Countability Axiom of choice Equivalence relations 9 MAT 570 REAL ANALYSIS LECTURE NOTES PROFESSOR: JOHN QUIGG SEMESTER: FALL 204 Contents. Sets 2 2. Functions 5 3. Countability 7 4. Axiom of choice 8 5. Equivalence relations 9 6. Real numbers 9 7. Extended

More information

MATH 140B - HW 5 SOLUTIONS

MATH 140B - HW 5 SOLUTIONS MATH 140B - HW 5 SOLUTIONS Problem 1 (WR Ch 7 #8). If I (x) = { 0 (x 0), 1 (x > 0), if {x n } is a sequence of distinct points of (a,b), and if c n converges, prove that the series f (x) = c n I (x x n

More information

MATH202 Introduction to Analysis (2007 Fall and 2008 Spring) Tutorial Note #7

MATH202 Introduction to Analysis (2007 Fall and 2008 Spring) Tutorial Note #7 MATH202 Introduction to Analysis (2007 Fall and 2008 Spring) Tutorial Note #7 Real Number Summary of terminology and theorems: Definition: (Supremum & infimum) A supremum (or least upper bound) of a non-empty

More information

h(x) lim H(x) = lim Since h is nondecreasing then h(x) 0 for all x, and if h is discontinuous at a point x then H(x) > 0. Denote

h(x) lim H(x) = lim Since h is nondecreasing then h(x) 0 for all x, and if h is discontinuous at a point x then H(x) > 0. Denote Real Variables, Fall 4 Problem set 4 Solution suggestions Exercise. Let f be of bounded variation on [a, b]. Show that for each c (a, b), lim x c f(x) and lim x c f(x) exist. Prove that a monotone function

More information

Mathematics-I Prof. S.K. Ray Department of Mathematics and Statistics Indian Institute of Technology, Kanpur. Lecture 1 Real Numbers

Mathematics-I Prof. S.K. Ray Department of Mathematics and Statistics Indian Institute of Technology, Kanpur. Lecture 1 Real Numbers Mathematics-I Prof. S.K. Ray Department of Mathematics and Statistics Indian Institute of Technology, Kanpur Lecture 1 Real Numbers In these lectures, we are going to study a branch of mathematics called

More information

Mathematics 242 Principles of Analysis Solutions for Problem Set 5 Due: March 15, 2013

Mathematics 242 Principles of Analysis Solutions for Problem Set 5 Due: March 15, 2013 Mathematics Principles of Analysis Solutions for Problem Set 5 Due: March 15, 013 A Section 1. For each of the following sequences, determine three different subsequences, each converging to a different

More information

MATH 409 Advanced Calculus I Lecture 7: Monotone sequences. The Bolzano-Weierstrass theorem.

MATH 409 Advanced Calculus I Lecture 7: Monotone sequences. The Bolzano-Weierstrass theorem. MATH 409 Advanced Calculus I Lecture 7: Monotone sequences. The Bolzano-Weierstrass theorem. Limit of a sequence Definition. Sequence {x n } of real numbers is said to converge to a real number a if for

More information

A lower bound for X is an element z F such that

A lower bound for X is an element z F such that Math 316, Intro to Analysis Completeness. Definition 1 (Upper bounds). Let F be an ordered field. For a subset X F an upper bound for X is an element y F such that A lower bound for X is an element z F

More information

Mathematics 324 Riemann Zeta Function August 5, 2005

Mathematics 324 Riemann Zeta Function August 5, 2005 Mathematics 324 Riemann Zeta Function August 5, 25 In this note we give an introduction to the Riemann zeta function, which connects the ideas of real analysis with the arithmetic of the integers. Define

More information

MA131 - Analysis 1. Workbook 6 Completeness II

MA131 - Analysis 1. Workbook 6 Completeness II MA3 - Analysis Workbook 6 Completeness II Autumn 2004 Contents 3.7 An Interesting Sequence....................... 3.8 Consequences of Completeness - General Bounded Sequences.. 3.9 Cauchy Sequences..........................

More information

Math 324 Summer 2012 Elementary Number Theory Notes on Mathematical Induction

Math 324 Summer 2012 Elementary Number Theory Notes on Mathematical Induction Math 4 Summer 01 Elementary Number Theory Notes on Mathematical Induction Principle of Mathematical Induction Recall the following axiom for the set of integers. Well-Ordering Axiom for the Integers If

More information

Appendix A. Sequences and series. A.1 Sequences. Definition A.1 A sequence is a function N R.

Appendix A. Sequences and series. A.1 Sequences. Definition A.1 A sequence is a function N R. Appendix A Sequences and series This course has for prerequisite a course (or two) of calculus. The purpose of this appendix is to review basic definitions and facts concerning sequences and series, which

More information

Introduction to Series and Sequences Math 121 Calculus II Spring 2015

Introduction to Series and Sequences Math 121 Calculus II Spring 2015 Introduction to Series and Sequences Math Calculus II Spring 05 The goal. The main purpose of our study of series and sequences is to understand power series. A power series is like a polynomial of infinite

More information

Consequences of the Completeness Property

Consequences of the Completeness Property Consequences of the Completeness Property Philippe B. Laval KSU Today Philippe B. Laval (KSU) Consequences of the Completeness Property Today 1 / 10 Introduction In this section, we use the fact that R

More information

POL502: Foundations. Kosuke Imai Department of Politics, Princeton University. October 10, 2005

POL502: Foundations. Kosuke Imai Department of Politics, Princeton University. October 10, 2005 POL502: Foundations Kosuke Imai Department of Politics, Princeton University October 10, 2005 Our first task is to develop the foundations that are necessary for the materials covered in this course. 1

More information

Metric Space Topology (Spring 2016) Selected Homework Solutions. HW1 Q1.2. Suppose that d is a metric on a set X. Prove that the inequality d(x, y)

Metric Space Topology (Spring 2016) Selected Homework Solutions. HW1 Q1.2. Suppose that d is a metric on a set X. Prove that the inequality d(x, y) Metric Space Topology (Spring 2016) Selected Homework Solutions HW1 Q1.2. Suppose that d is a metric on a set X. Prove that the inequality d(x, y) d(z, w) d(x, z) + d(y, w) holds for all w, x, y, z X.

More information

MATH NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS

MATH NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS MATH. 4433. NEW HOMEWORK AND SOLUTIONS TO PREVIOUS HOMEWORKS AND EXAMS TOMASZ PRZEBINDA. Final project, due 0:00 am, /0/208 via e-mail.. State the Fundamental Theorem of Algebra. Recall that a subset K

More information

REVIEW OF ESSENTIAL MATH 346 TOPICS

REVIEW OF ESSENTIAL MATH 346 TOPICS REVIEW OF ESSENTIAL MATH 346 TOPICS 1. AXIOMATIC STRUCTURE OF R Doğan Çömez The real number system is a complete ordered field, i.e., it is a set R which is endowed with addition and multiplication operations

More information

Fundamentals of Pure Mathematics - Problem Sheet

Fundamentals of Pure Mathematics - Problem Sheet Fundamentals of Pure Mathematics - Problem Sheet ( ) = Straightforward but illustrates a basic idea (*) = Harder Note: R, Z denote the real numbers, integers, etc. assumed to be real numbers. In questions

More information

Notes for Math 290 using Introduction to Mathematical Proofs by Charles E. Roberts, Jr.

Notes for Math 290 using Introduction to Mathematical Proofs by Charles E. Roberts, Jr. Notes for Math 290 using Introduction to Mathematical Proofs by Charles E. Roberts, Jr. Chapter : Logic Topics:. Statements, Negation, and Compound Statements.2 Truth Tables and Logical Equivalences.3

More information

CONSTRUCTION OF THE REAL NUMBERS.

CONSTRUCTION OF THE REAL NUMBERS. CONSTRUCTION OF THE REAL NUMBERS. IAN KIMING 1. Motivation. It will not come as a big surprise to anyone when I say that we need the real numbers in mathematics. More to the point, we need to be able to

More information

A LITTLE REAL ANALYSIS AND TOPOLOGY

A LITTLE REAL ANALYSIS AND TOPOLOGY A LITTLE REAL ANALYSIS AND TOPOLOGY 1. NOTATION Before we begin some notational definitions are useful. (1) Z = {, 3, 2, 1, 0, 1, 2, 3, }is the set of integers. (2) Q = { a b : aεz, bεz {0}} is the set

More information

Math 320: Real Analysis MWF 1pm, Campion Hall 302 Homework 4 Solutions Please write neatly, and in complete sentences when possible.

Math 320: Real Analysis MWF 1pm, Campion Hall 302 Homework 4 Solutions Please write neatly, and in complete sentences when possible. Math 320: Real Analysis MWF pm, Campion Hall 302 Homework 4 Solutions Please write neatly, and in complete sentences when possible. Do the following problems from the book: 2.6.3, 2.7.4, 2.7.5, 2.7.2,

More information

CSE 1400 Applied Discrete Mathematics Proofs

CSE 1400 Applied Discrete Mathematics Proofs CSE 1400 Applied Discrete Mathematics Proofs Department of Computer Sciences College of Engineering Florida Tech Fall 2011 Axioms 1 Logical Axioms 2 Models 2 Number Theory 3 Graph Theory 4 Set Theory 4

More information

5.5 Deeper Properties of Continuous Functions

5.5 Deeper Properties of Continuous Functions 5.5. DEEPER PROPERTIES OF CONTINUOUS FUNCTIONS 195 5.5 Deeper Properties of Continuous Functions 5.5.1 Intermediate Value Theorem and Consequences When one studies a function, one is usually interested

More information

Logical Connectives and Quantifiers

Logical Connectives and Quantifiers Chapter 1 Logical Connectives and Quantifiers 1.1 Logical Connectives 1.2 Quantifiers 1.3 Techniques of Proof: I 1.4 Techniques of Proof: II Theorem 1. Let f be a continuous function. If 1 f(x)dx 0, then

More information

Math 410 Homework 6 Due Monday, October 26

Math 410 Homework 6 Due Monday, October 26 Math 40 Homework 6 Due Monday, October 26. Let c be any constant and assume that lim s n = s and lim t n = t. Prove that: a) lim c s n = c s We talked about these in class: We want to show that for all

More information

Math 117: Topology of the Real Numbers

Math 117: Topology of the Real Numbers Math 117: Topology of the Real Numbers John Douglas Moore November 10, 2008 The goal of these notes is to highlight the most important topics presented in Chapter 3 of the text [1] and to provide a few

More information