The Real And Complex Number Systems

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Section 4.18.) Proof: Let f be a two valued function defined on S. SinceA, andB are connected in S, then we have fA a, andfB b, wherea, b 0, 1. Given a sequence x n A with x n 0, 0, then we have a lim n fx n f limx n by continuity of f at 0 n0 f0, 0 b. So, we have f is a constant. That is, S is connected. Assume that S is arcwise connected, then there exists a continuous function g : 0, 1 S such that g0 0, 0 and g1 1, sin 1. Given 1/2, there exists a 0 such that as |t| , wehave gt g0 gt 1/2. * 1 Let N be a positive integer so that 2N Define two subsets U and V as follows: U V 1 , thus let ,0 : p and 1 ,0 : q. 2N 2N1 x, y : x p q 2 x, y : x p q 2 gq, p gq, p Then we have (1). U V gq, p, (2). U , sincep U and V , sinceq V, (3). U V by the given set A, and (*) Since x, y : x pq 2 gq, p. So, we have and x, y : x pq 2 are open in R2 , then U and V are open in (4). U is open in gq, p and V is open in gq, p. From (1)-(4), we have gq, p is disconnected which is absurb since a connected subset under a continuous function is connected. So, such g cannot exist. It means that S is not arcwise connected. Remark: This exercise gives us an example to say that connectedness does not imply path-connectedness. And it is important example which is worth keeping in mind. 4.47 Let F F 1 , F 2 ,... be a countable collection of connected compact sets in R n such that F k1 F k for each k 1. Prove that the intersection k1 F k is connected and closed. Proof: Since F k is compact for each k 1, F k is closed for each k 1. Hence, k1 F k : F is closed. Note that by Theorem 3.39, we know that F is compact. Assume that F is not connected. Then there are two subsets A and B with 1.A , B . 2.A B . 3.A B F. 4.A, B are closed in F. Note that A, B are closed and disjoint in R n . By exercise 4.57, there exist U and V which are open and disjoint in R n such that A U, andB V. Claim that there exists F k such that F k U V. Suppose NOT, thenthereexistsx k F k U V. Without loss of generality, we may assume that x k F k1 . So, we have a sequence x k F 1 which implies that there exists a convergent subsequence x kn , say lim kn x kn x. Itis clear that x F k for all k since x is an accumulation point of each F k . So, we have
x F k1 F k A B U V which implies that x is an interior point of U V since U and V are open. So, Bx; U V for some 0, which contradicts to the choice of x k . Hence, we have proved that there exists F k such that F k U V. LetC U F k ,andD V F k , then we have 1. C since A U and A F k ,andD since B V and B F k . 2. C D U F k V F k U V . 3. C D U F k V F k F k . 4. C is open in F k and D is open in F k by C, D are open in R n . Hence, we have F k is disconnected which is absurb. So, we know that F k1 F k is connected. 4.48 Let S be an open connected set in R n .LetT be a component of R n S. Prove that R n T is connected. Proof: If S is empty, there is nothing to proved. Hence, we assume that S is nonempty. Write R n S xR n S Ux, whereUx is a component of R n S. So, we have R n S xR n S Ux. Say T Up, forsomep. Then R n T S xR n ST Ux. Claim that clS Ux for all x R n S T. If we can show the claim, given a, b R n T, and a two valued function on R n T. Note that clS is also connected. We consider three cases. (1) a S, b Ux for some x. (2)a, b S. (3)a Ux, b Ux . For case (1), let c clS Ux, then there are s n S and u n Ux with s n c and u n c, then we have fa lim n fs n f lim n s n fc f lim n u n lim n fu n fb which implies that fa fb. For case (2), it is clear fa fb since S itself is connected. For case (3), we choose s S, and thus use case (1), we know that fa fs fb. By case (1)-(3), we have f is constant on R n T. Thatis,R n T is connected. It remains to show the claim. To show clS Ux for all x R n S T, i.e., to show that for all x R n S T, clS Ux S S Ux S Ux . Suppose NOT, i.e., for some x, S Ux which implies that Ux R n clS which is open. So, there is a component V of R n clS contains Ux, whereV is open by Theorem 4.44. However, R n clS R n S, sowehaveV is contained in Ux. Therefore, we have Ux V. Note that Ux R n S, andR n S is closed. So, clUx R n S. By definition of component, we have clUx Ux, whichis closed. So, we have proved that Ux V is open and closed. It implies that Ux R n or which is absurb. Hence, the claim has proved. 4.49 Let S, d be a connected metric space which is not bounded. Prove that for every
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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
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Bx, r x S Bx, r x T . So, at
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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
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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 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: from b to A) are disjoint. So, if e
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
Page 157 and 158: Proof: Look at the Generalized Mean
Page 159 and 160: Proof: By Generalized Mean Value Th
Page 161 and 162: there is a point p such that fp 0.
Page 163 and 164: Remark: If we can make sure that fx
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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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