The Real And Complex Number Systems

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Since S R, wehaveS inf S,supS. (Note that we accept that inf S or sup S .) If S is not an interval, then there exists x inf S,supS such that x S. (Otherwise, inf S,supS S which implies that S is an interval.) Then we have 1. , x S : A is open in S 2. x, S : B is open in S 3. A B S. Claim that both A and B are not empty. Asume that A is empty, then every s S, wehave s x inf S. By the definition of infimum, it is impossible. So, A is not empty. Similarly for B. Hence, we have proved that S is disconnected, a contradiction. That is, S is an interval. Remark: 1. We note that any interval in R is connected. It is immediate from Exercise 4.44. But we give another proof as follows. Suppose there exists an interval S is not connceted, then there exist two subsets A and B such that 1. A, B are open in S, 2.A and B , 3.A B , and 4. A B S. Since A and B , we choose a A and b B, and let a b. Consider c : supA a, b. Note that c clA A implies that c B. Hence, we have a c b. In addition, c B clB, then there exists a B S c; B . Choose d B S c; c , c S so that 1. c d b and 2. d B. Then d A. (Otherwise, it contradicts c supA a, b. Note that d a, b S A B which implies that d A or d B. We reach a contradiction since d A and d B. Hence, we have proved that any interval in R is connected. 2. Here is an application. Is there a continuous function f : R R such that fQ Q c ,andfQ c Q Ans: NO! If such f exists, then both fQ and fQ c are countable. Hence, fR is countable. In addition, fR is connected. Since fR contains rationals and irrationals, we know fR is an interval which implies that fR is uncountable, a cotradiction. Hence, such f does not exist. 4.39 Let X be a connected subset of a metric space S. LetY be a subset of S such that X Y clX, whereclX is the closure of X. Prove that Y is also connected. In particular, this shows that clX is connected. Proof: Given a two valued function f on Y, we know that f is also a two valued function on X. Hence, f is constant on X, (without loss of generality) say f 0onX. Consider p Y X, it ,means that p is an accumulation point of X. Then there exists a dequence x n X such that x n p. Note that fx n 0 for all n. So, we have by continuity of f on Y, fp f lim n x n lim n fx n 0. Hence, we have f is constant 0 on Y. Thatis,Y is conneceted. In particular, clX is connected. Remark: Of course, we can use definition of a connected set to show the exercise. But, it is too tedious to write. However, it is a good practice to use definition to show it. The reader may give it a try as a challenge.
4.40 If x is a point in a metric space S, letUx be the component of S containing x. Prove that Ux is closed in S. Proof: Let p be an accumulation point of Ux. Letf be a two valued function defined on Ux p, then f is a two valud function defined on Ux. SinceUx is a component of S containing x, then Ux is connected. That is, f is constant on Ux, (without loss of generality) say f 0onUx. And since p is an accumulation point of Ux, there exists a sequence x n Ux such that x n p. Note that fx n 0 for all n. So, we have by continuity of f on Ux p, fp f lim n x n lim n fx n 0. So, Ux p is a connected set containing x. SinceUx is a component of S containing x, wehaveUx p Ux which implies that p Ux. Hence, Ux contains its all accumulation point. That is, Ux is closed in S. 4.41 Let S be an open subset of R. By Theorem 3.11, S is the union of a countable disjoint collection of open intervals in R. Prove that each of these open intervals is a component of the metric subspace S. Explain why this does not contradict Exercise 4.40. Proof: By Theorem 3.11, S n1 I n ,whereI i is open in R and I i I j if i j. Assume that there exists a I m such that I m is not a component T of S. Then T I m is not empty. So, there exists x T I m and x I n for some n. Note that the component Ux is the union of all connected subsets containing x, then we have T I n Ux. * In addition, Ux T ** since T is a component containing x. Hence, by (*) and (**), we have I n T. So, I m I n T. SinceT is connected in R 1 , T itself is an interval. So, intT is still an interval which is open and containing I m I n . It contradicts the definition of component interval. Hence, each of these open intervals is a component of the metric subspace S. Since these open intervals is open relative to R, not S, this does not contradict Exercise 4.40. 4.42 Given a compact S in R m with the following property: For every pair of points a and b in S and for every 0 there exists a finite set of points x 0 , x 1 ,...,x n in S with x 0 a and x n b such that x k x k1 for k 1,2,..,n. Prove or disprove: S is connected. Proof: Suppose that S is disconnected, then there exist two subsets A and B such that 1. A, B are open in S, 2.A and B , 3.A B , and 4. A B S. * Since A and B , we choose a A, andb A and thus given 1, then by hypothesis, we can find two points a 1 A, andb 1 B such that a 1 b 1 1. For a 1 , and b 1 ,given 1/2, then by hypothesis, we can find two points a 2 A, andb 2 B such that a 2 b 2 1/2. Continuous the steps, we finally have two sequence a n A and b n B such that a n b n 1/n for all n. Sincea n A, andb n B, we have a n S and b n S by S A B. Hence, there exist two subsequence a nk A and b nk B such that a nk x, andb nk y, wherex, y S since S is compact. In addition, since A is closed in S, andB is closed in S, wehavex A and y B. On the other hand, since a n b n 1/n for all n, wehavex y. Thatis,
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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
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Charpter 3 Elements of Point set To
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And thus by the remark in (e), it i
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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
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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
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Page 79 and 80: Limits And Continuity Limits of seq
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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
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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,
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Page 101 and 102: if x Q, andgx 0ifx Q c . Remark:
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Page 105 and 106: Proof: Since the hypothesis says th
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Page 109 and 110: In Exercises 4.29 through 4.33, we
Page 111: function. Since f0, 1 0, 1 0, 1,
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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,
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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
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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
Page 157 and 158: Proof: Look at the Generalized Mean
Page 159 and 160: Proof: By Generalized Mean Value Th
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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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