The Real And Complex Number Systems

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for every real number y, either there is no x in 0, 1 for which fx y or there are exactly two values of x in 0, 1 for which fx y Assume that there exist a finite many numbers of discontinuities of f, say these points x 1 ,...,x n . By property A, there exists a unique y i such that fx i fy i for 1 i n. Note that the number of the set S : x 1 ,...,x n y 1 ,...,y n x : fx f0, andfx f1 is even, say 2m We remove these points from S, and thus we have 2m 1 subintervals, say I j , 1 j 2m 1. Consider the local extremum in every I j ,1 j 2m 1 and note that every subinterval I j ,1 j 2m 1, has at most finite many numbers of local extremum, say # t I j : fx is the local extremum t j j 1 ,..,t pj p j . And by property A, there exists a unique s j j j k such that f t k f s k for 1 k p j . We again remove these points, and thus we have removed even number of points. And odd number of open intervals is left, call the odd number 2q 1. Note that since the function f is monotonic in every open interval left, R l ,1 l 2q 1, the image of f on these open interval is also an open interval. If R a R b , sayR a a 1 , a 1 and R b b 1 , b 2 with (without loss of generality) a 1 b 1 a 2 b 2 , then R a R b by property A. (Otherwise, b 1 is only point such that fx fb 1 , which contradicts property A.) Note that given any R a , there must has one and only one R b such that R a R b . However, we have 2q 1 open intervals is left, it is impossible. Hence, we know that f has infinite many discontinuities on 0, 1. 4.28 In each case, give an example of a real-valued function f, continuous on S and such that fS T, or else explain why there can be no such f : (a) S 0, 1, T 0, 1. Solution: Let fx 2x if x 0, 1/2, 1ifx 1/2, 1. (b) S 0, 1, T 0, 1 1, 2. Solution: NO! Since a continuous functions sends a connected set to a connected set. However, in this case, S is connected and T is not connected. (c) S R 1 , T the set of rational numbers. Solution: NO! Since a continuous functions sends a connected set to a connected set. However, in this case, S is connected and T is not connected. (d) S 0, 1 2, 3, T 0, 1. Solution: Let (e) S 0, 1 0, 1, T R 2 . fx 0ifx 0, 1, 1ifx 2, 3. Solution: NO! Since a continuous functions sends a compact set to a compact set. However, in this case, S is compact and T is not compact.
(f) S 0, 1 0, 1, T 0, 1 0, 1. Solution: NO! Since a continuous functions sends a compact set to a compact set. However, in this case, S is compact and T is not compact. (g) S 0, 1 0, 1, T R 2 . Solution: Let fx, y cot x,coty. Remark: 1. There is some important theorems. We write them as follows. (Theorem A) Letf : S, d s T, d T be continuous. If X is a compact subset of S, then fX is a compact subset of T. (Theorem B) Letf : S, d s T, d T be continuous. If X is a connected subset of S, then fX is a connected subset of T. 2. In (g), the key to the example is to find a continuous function f : 0, 1 R which is onto. Supplement on Continuity of real valued functions Exercise Suppose that fx : 0, R, is continuous with a fx b for all x 0, , and for any real y, either there is no x in 0, for which fx y or there are finitely many x in 0, for which fx y. Prove that lim x fx exists. Proof: For convenience, we say property A, it means that for any real y, either there is no x in 0, for which fx y or there are finitely many x in 0, for which fx y. We partition a, b into n subintervals. Then, by continuity and property A, asx is large enough, fx is lying in one and only one subinterval. Given 0, there exists N such that 2/N . For this N, we partition a, b into N subintervals, then there is a M 0 such that as x, y M |fx fy| 2/N . So, lim x fx exists. Exercise Suppose that fx : 0, 1 R is continunous with f0 f1 0. Prove that (a) there exist two points x 1 and x 2 such that as |x 1 x 2 | 1/n, wehave fx 1 fx 2 0 for all n. In this case, we call 1/n the length of horizontal strings. Proof: Define a new function gx fx 1 n fx : 0, 1 1 n . Claim that there exists p 0, 1 1 n such that gp 0. Suppose NOT, byIntermediate Value Theorem, without loss of generality, let gx 0, then g0 g 1 n ...g 1 1 n f1 0 which is absurb. Hence, we know that there exists p 0, 1 1 n such that gp 0. That is, f p 1 n fp. So,wehave1/n as the length of horizontal strings. (b) Could you show that there exists 2/3 as the length of horizontal strings Proof: The horizontal strings does not exist, for example,
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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) ∪
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(e) A ∩ (B − C) = (A ∩ B) −
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Proof: Given x ∈ X, then f (x)
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By hyppothesis, we get T ⊆ f (S)
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Proof: Since S is an infinite set,
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Proof: Assume S˜R and let f be a o
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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 and 94: this is because a can be an isolate
Page 95 and 96: irrational. Prove that: (a) ffx x
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: Proof: Since the hypothesis says th
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
Page 145 and 146: f k x f k 0 x 0 fk x x by induct
Page 147 and 148: h k1 h k k jk f j g kj x j0
Page 149 and 150: g x g x 3 2 a d f x a d f x 3 2
Page 151 and 152: which implies that F 0, where
Page 153 and 154: g fx gfx. Assume that there is a
Page 155 and 156: 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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