The Real And Complex Number Systems

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continuous at x, then 0 lim n gx n g lim n x n by continuity of f at x gx x. So, the function g may be continuous at 0. In fact, g is continuous at 0 which prove as follows. Given 0, choose , as|x| , wehave |gx g0| |gx| |x| . So,g is continuous at 0. Hence, from the preceding, we know that g is continuous only at x 0. 3. Write 1ifx 0, hx 0ifx R Q 0, 1, 1/n if x m/n, g. c. d. m, n 1. Consider a 0, 1 and given 0, there exists the largest positive integer N such that N 1/. LetT x : hx , then 0, 1 x : hx 1 x : hx 1/2...x : hx 1/N if 1, T if 1. Note that T is at most a finite set, and then we can choose a 0 such that a , a acontains no points of T and a , a 0, 1. So,if x a , a a, wehavehx . It menas that lim xa hx 0. Hence, we know that h is continuous at x 0, 1 R Q. For two points x 1, and y 0, it is clear that h is not continuous at x 1, and not continuous at y 1bythe method mentioned in the exercise of part 1 and part 2. Hence, we have proved that h is continuous only at the irrational points in 0, 1. Remark: 1. Sometimes we call f Dirichlet function. 2. Here is another proof about g, we write it down to make the reader get more. Proof: Write 0ifx R Q 0, 1, gx x if x Q 0, 1. Given a 0, 1, andifg is continuous at a, then given 0 a, there exists a 0 such that as x a , a 0, 1, wehave |gx ga| . If a R Q, choose 0 so that a Q. Then a a , a which implies |ga ga| |ga | a a. But it is impossible. If a Q, choose 0 so that a R Q. a a , a which implies |ga ga| |a| a a. But it is impossible. If a 0, given 0 and choose , thenas0 x , wehave |gx g0| |gx| |x| x . It means that g is continuous at 0. 4.17 For each x 0, 1, letfx x if x is rational, and let fx 1 x if x is
irrational. Prove that: (a) ffx x for all x in 0, 1. Proof: If x is rational, then ffx fx x. Andifx is irrarional, so is 1 x 0, 1. Then ffx f1 x 1 1 x x. Hence, ffx x for all x in 0, 1. (b) fx f1 x 1 for all x in 0, 1. Proof: If x is rational, so is 1 x 0, 1. Then fx f1 x x 1 x 1. And if x is irrarional, so is 1 x 0, 1. Then fx f1 x 1 x 1 1 x 1. Hence, fx f1 x 1 for all x in 0, 1. (c) f is continuous only at the point x 1 . 2 Proof: If f is continuous at x, then choose x n Q and y n Q c such that x n x, and y n x. Then we have, by continuity of f at x, fx f lim n x n lim n fx n lim n x n x and fx f lim n y n lim n fy n lim n 1 y n 1 x. So, x 1/2 is the only possibility for f. Given 0, we want to find a 0 such that as x 1/2 ,1/2 0, 1, wehave |fx f1/2| |fx 1/2| . Choose 0 so that 1/2 ,1/2 0, 1, then as x 1/2 ,1/2 0, 1, wehave |fx 1/2| |x 1/2| if x Q, |fx 1/2| |1 x 1/2| |1/2 x| if x Q c . Hence, we have proved that f is continuous at x 1/2. (d) f assumes every value between 0 and 1. Proof: Given a 0, 1, wewanttofindx 0, 1 such that fx a. Ifa Q, then choose x a, wehavefx a a. Ifa R Q, then choose x 1 a R Q, we have fx 1 a 1 1 a a. Remark: The range of f on 0, 1 is 0, 1. In addition, f is an one-to-one mapping since if fx fy, then x y. (The proof is easy, just by definition of 1-1, so we omit it.) (e) fx y fx fy is rational for all x and y in 0, 1. Proof: We prove it by four steps. 1. If x Q and y Q, then x y Q. So, fx y fx fy x y x y 0 Q. 2. If x Q and y Q c , then x y Q c .So, fx y fx fy 1 x y x 1 y 2x Q. 3. If x Q c and y Q, then x y Q c .So, fx y fx fy 1 x y 1 x y 2y Q. 4. If x Q c and y Q c , then x y Q c or x y Q. So,
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The Real And Complex Number Systems
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Proof: Consider 2n−2 ∑ (x + 1)
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Then S = N. Proof: (A ⇒ B): If S
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Note: There are many and many metho
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In addition, (√ a ) 2 − − b (
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Proof: Say {ar + b : a ∈ Z, b ∈
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(⇐)It is clear since every a n is
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Proof: Choose a 0 = [x], and thus c
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Proof: Use Cauchy-Schwarz inequalit
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In addition, ∑ a k b k + a j b j
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So, z 1 z 2 = ¯z 1¯z 2 (c) z¯z =
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so, |a − b| 2 ≤ ( 1 + |a| 2) (
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Proof: Note that in this text book,
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1.39 State and prove a theorem anal
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1.43 (a) Prove that Log (z w ) = wL
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For complex θ, we show that it hol
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So, ix−y = i (x + iy) = iz = log
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and the sum of their squares is giv
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Some Basic Notations Of Set Theory
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(i) If x ∈ R, it is clear that (x
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(a) A ∪ (B ∪ C) = (A ∪ B) ∪
Page 43 and 44: (e) A ∩ (B − C) = (A ∩ B) −
Page 45 and 46: Proof: Given x ∈ X, then f (x)
Page 47 and 48: By hyppothesis, we get T ⊆ f (S)
Page 49 and 50: Proof: Since S is an infinite set,
Page 51 and 52: Proof: Assume S˜R and let f be a o
Page 53 and 54: 2.22 Let S denote the collection of
Page 55 and 56: Charpter 3 Elements of Point set To
Page 57 and 58: And thus by the remark in (e), it i
Page 59 and 60: 3.7 Prove that a nonempty, bounded
Page 61 and 62: Bx, r x S Bx, r x T . So, at
Page 63 and 64: Claim that Bp, r S as follows. Let
Page 65 and 66: Covering theorems in R n 3.17 If S
Page 67 and 68: uncountable. Hence, by exercise 3.2
Page 69 and 70: Then we have (a) (b) (c) (d) x y
Page 71 and 72: (b) Give an example of a metric spa
Page 73 and 74: Hence from (1)-(4), we know that i
Page 75 and 76: 3.44 intM A M A . Proof: Let B
Page 77 and 78: That is, intB which is absurb. He
Page 79 and 80: Limits And Continuity Limits of seq
Page 81 and 82: x n1 x n 1 1 x n x n 1 2 . Rema
Page 83 and 84: |a m a n| |a m a m1 a m1 a m2
Page 85 and 86: and use the fact if a n converges
Page 87 and 88: lim y0 fx, y 0ifx 0, 1ifx 0. and
Page 89 and 90: fx, y 1 r cos sinr2 cossin if x
Page 91 and 92: then we have lim fx, y L if x 0an
Page 93: this is because a can be an isolate
Page 97 and 98: In addition, since f1 f1 by f0 0,
Page 99 and 100: lim r n0 fx f lim n x n lim n fx
Page 101 and 102: if x Q, andgx 0ifx Q c . Remark:
Page 103 and 104: number f T supfx fy : x, y T is
Page 105 and 106: Proof: Since the hypothesis says th
Page 107 and 108: (f) S 0, 1 0, 1, T 0, 1 0, 1. S
Page 109 and 110: In Exercises 4.29 through 4.33, we
Page 111 and 112: function. Since f0, 1 0, 1 0, 1,
Page 113 and 114: 4.40 If x is a point in a metric sp
Page 115 and 116: from b to A) are disjoint. So, if e
Page 117 and 118: x F k1 F k A B U V which im
Page 119 and 120: Proof: Assume that B C 1 is discon
Page 121 and 122: Proof: By Mean Value Theorem, wehav
Page 123 and 124: x r y r rz r1 x y ry r1 /2. So,
Page 125 and 126: hx gfx if x S. Iff is uniformly c
Page 127 and 128: dx, y , x, y A, wehave dfx, fy
Page 129 and 130: In addition, part (b) comes from in
Page 131 and 132: continuous at a, a. (2) (x y) Sinc
Page 133 and 134: Proof: Let D denote the set of dico
Page 135 and 136: (a) Provet that the formula df, g
Page 137 and 138: so, by induction, we find dp n1 , p
Page 139 and 140: Proof: If p and p are fixed points
Page 141 and 142: (b) Assume there is a subsequence p
Page 143 and 144: Hence, f is increasing on , a/3 a
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f k x f k 0 x 0 fk x x by induct
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h k1 h k k jk f j g kj x j0
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g x g x 3 2 a d f x a d f x 3 2
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which implies that F 0, where
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g fx gfx. Assume that there is a
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Remark: For x 0, we can show that
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Proof: Look at the Generalized Mean
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Proof: By Generalized Mean Value Th
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there is a point p such that fp 0.
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Remark: If we can make sure that fx
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which implies that Hence, g 1 h g
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fx fy x y f x /2 *’ Combi
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lim xa fx gx L. Remark: 1. The pr
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every g k is never zero in c , c
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f x 1 ...x n n f x 1 ...x n1 n x
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f x 1 ...x n n fx 1 ...fx n n . C
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lx fc f cx c fx. (Exercise 2)
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So, we have Partial derivatives fb
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ux, y ey e y cosx 2 and vx, y e
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ux, y (1) arctany/x, ifx 0, y R
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Functions of Bounded Variation and
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For the part fx x fy 1 |fx fy| 1
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Claim that 1 sup A. Suppose NOT, t
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we know that V f is an increasing
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in Section 6.12. Proof: Sinceft : t
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2n h P 2 hx i hx i1 i1 n 2n
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n k1 n ||fb k | |fa k || |fb k
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Supplement on lim sup and lim inf I
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(1) lim n→ inf a n − a n has
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Remark: The exercise is useful in t
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which implies that c ≤ a b since
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If lim sup n→ a n1 a n , then it
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We first prove lim n→ sup n ≤
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0 ≤ d 1 * by Theorem 8.23 (i). N
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Consider which implies that which i
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So, L 1 5 since L ≥ 1. 2 Remark:
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For case (i), since 1 n log nlog lo
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()Assume that ∑ n1 That is, p!
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n S n ∑−1 k1 1 3k − 2 − 1
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(b) Prove that ∑ n1 −1 n−1 /
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then b k sin kx, a k1 − a k 1
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diverges, then ∑ where By (2) and
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Suppose NOT, it means that lim n→
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then since n lim n→ ∑ k1 n S n
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|sin k|r ∑ k So, ∑ |sink|r dive
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n For the part 1 2 0 that x sint
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lim n→ sin ... sin n 1 ... 1 n
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Also, lim q→ fp, q lim q→ p si
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n c n ∑ a k b n−k k0 n ∑ k0
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So, n s n ∑ cos2k − 1x j1 sin
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n u n ∑−1 k a k k1 n ∑−1
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which implies that which implies th
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So, by above sayings, we have prove
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(a) If fn is multiplicative and if
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Sequences of Functions Uniform conv
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(b) Prove that h n (x) does not con
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Proof: Since g is continuous on a c
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Hence, {g (x) x n } converges unifo
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y continuity of f k(x0 ) (x) − f
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In addition, as c ≥ 1/2, if f n
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(Lemma) If {a n } and {b n } are tw
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9.16 Let {f n } be a sequence of re
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which implies that ∣ lim sup 1 k
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have ∫ x 0 ∞∑ t 2n − 1 2 n=
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[ ] π And as x ∈ , π , then n+p
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0.1 Supplement on some results on W
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which implies that h (x + t) − h
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we have ∫ d c |f n − g| 2 dx
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we complete it. 9.31 Given that two
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if x = λ (≠ 0) is a root of 1 +
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So, the series diverges. (ii) As
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9.38 For each real t, define f t (x
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So, B 0 = P 0 (0) = C 0 = 1, B 1 =
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Remark: (1) The reader can see the
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That is, lim n inf a n sup c n . n
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and lim n infa n b n lim n a n
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q! kq1 1 k! kq1 q! k! 1 q 1
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g x log 1 1 x x a x 2 x log
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The Real And Complex Number Systems

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