The Real And Complex Number Systems

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Proof: Write G (x) = ( x + 1 2) 2 + 11 4 , then G [F (x)] = ( F (x) + 1 ) 2 + 11 2 4 = x2 − 3x + 5 which implies that which implies that (2F (x) + 1) 2 = (2x − 3) 2 F (x) = x − 2 or − x + 1. 2.4 Given three functions F, G, H, what restrictions must be placed on their domains so that the following four composite functions can be defined G ◦ F, H ◦ G, H ◦ (G ◦ F ) , (H ◦ G) ◦ F. Proof: It is clear for answers, R (F ) ⊆ D (G) and R (G) ⊆ D (H) . Assuming that H ◦ (G ◦ F ) and (H ◦ G) ◦ F can be defined, prove that associative law: H ◦ (G ◦ F ) = (H ◦ G) ◦ F. Proof: Given any x ∈ D (F ) , then ((H ◦ G) ◦ F ) (x) = (H ◦ G) (F (x)) = H (G (F (x))) = H ((G ◦ F ) (x)) = (H ◦ (G ◦ F )) (x) . So, H ◦ (G ◦ F ) = (H ◦ G) ◦ F. 2.5 Prove the following set-theoretic identities for union and intersection: 4
(a) A ∪ (B ∪ C) = (A ∪ B) ∪ C, A ∩ (B ∩ C) = (A ∩ B) ∩ C. Proof: For the part, A∪(B ∪ C) = (A ∪ B)∪C : Given x ∈ A∪(B ∪ C) , we have x ∈ A or x ∈ B ∪ C. That is, x ∈ A or x ∈ B or x ∈ C. Hence, x ∈ A∪B or x ∈ C. It implies x ∈ (A ∪ B)∪C. Similarly, if x ∈ (A ∪ B)∪C, then x ∈ A ∪ (B ∪ C) . Therefore, A ∪ (B ∪ C) = (A ∪ B) ∪ C. For the part, A∩(B ∩ C) = (A ∩ B)∩C : Given x ∈ A∩(B ∩ C) , we have x ∈ A and x ∈ B ∩C. That is, x ∈ A and x ∈ B and x ∈ C. Hence, x ∈ A∩B and x ∈ C. It implies x ∈ (A ∩ B) ∩ C. Similarly, if x ∈ (A ∩ B) ∩ C, then x ∈ A ∩ (B ∩ C) . Therefore, A ∩ (B ∩ C) = (A ∩ B) ∩ C. (b) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) . Proof: Given x ∈ A ∩ (B ∪ C) , then x ∈ A and x ∈ B ∪ C. We consider two cases as follows. If x ∈ B, then x ∈ A ∩ B. So, x ∈ (A ∩ B) ∪ (A ∩ C) . If x ∈ C, then x ∈ A ∩ C. So, x ∈ (A ∩ B) ∪ (A ∩ C) . So, we have shown that A ∩ (B ∪ C) ⊆ (A ∩ B) ∪ (A ∩ C) . (*) Conversely, given x ∈ (A ∩ B) ∪ (A ∩ C) , then x ∈ A ∩ B or x ∈ A ∩ C. We consider two cases as follows. If x ∈ A ∩ B, then x ∈ A ∩ (B ∪ C) . If x ∈ A ∩ C, then x ∈ A ∩ (B ∪ C) . So, we have shown that (A ∩ B) ∪ (A ∩ C) ⊆ A ∩ (B ∪ C) . (**) By (*) and (**), we have proved it. (c) (A ∪ B) ∩ (A ∪ C) = A ∪ (B ∩ C) Proof: Given x ∈ (A ∪ B) ∩ (A ∪ C) , then x ∈ A ∪ B and x ∈ A ∪ C. We consider two cases as follows. If x ∈ A, then x ∈ A ∪ (B ∩ C) . If x /∈ A, then x ∈ B and x ∈ C. So, x ∈ B ∩ C. It implies that x ∈ A ∪ (B ∩ C) . Therefore, we have shown that (A ∪ B) ∩ (A ∪ C) ⊆ A ∪ (B ∩ C) . (*) 5
Page 1 and 2: The Real And Complex Number Systems
Page 3 and 4: Proof: Consider 2n−2 ∑ (x + 1)
Page 5 and 6: Then S = N. Proof: (A ⇒ B): If S
Page 7 and 8: Note: There are many and many metho
Page 9 and 10: In addition, (√ a ) 2 − − b (
Page 11 and 12: Proof: Say {ar + b : a ∈ Z, b ∈
Page 13 and 14: (⇐)It is clear since every a n is
Page 15 and 16: Proof: Choose a 0 = [x], and thus c
Page 17 and 18: Proof: Use Cauchy-Schwarz inequalit
Page 19 and 20: In addition, ∑ a k b k + a j b j
Page 21 and 22: So, z 1 z 2 = ¯z 1¯z 2 (c) z¯z =
Page 23 and 24: so, |a − b| 2 ≤ ( 1 + |a| 2) (
Page 25 and 26: Proof: Note that in this text book,
Page 27 and 28: 1.39 State and prove a theorem anal
Page 29 and 30: 1.43 (a) Prove that Log (z w ) = wL
Page 31 and 32: For complex θ, we show that it hol
Page 33 and 34: So, ix−y = i (x + iy) = iz = log
Page 35 and 36: and the sum of their squares is giv
Page 37 and 38: Some Basic Notations Of Set Theory
Page 39: (i) If x ∈ R, it is clear that (x
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
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then we have lim fx, y L if x 0an
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this is because a can be an isolate
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irrational. Prove that: (a) ffx x
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In addition, since f1 f1 by f0 0,
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lim r n0 fx f lim n x n lim n fx
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if x Q, andgx 0ifx Q c . Remark:
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number f T supfx fy : x, y T is
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Proof: Since the hypothesis says th
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(f) S 0, 1 0, 1, T 0, 1 0, 1. S
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In Exercises 4.29 through 4.33, we
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function. Since f0, 1 0, 1 0, 1,
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4.40 If x is a point in a metric sp
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from b to A) are disjoint. So, if e
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x F k1 F k A B U V which im
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Proof: Assume that B C 1 is discon
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Proof: By Mean Value Theorem, wehav
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x r y r rz r1 x y ry r1 /2. So,
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hx gfx if x S. Iff is uniformly c
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dx, y , x, y A, wehave dfx, fy
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In addition, part (b) comes from in
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continuous at a, a. (2) (x y) Sinc
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Proof: Let D denote the set of dico
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(a) Provet that the formula df, g
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so, by induction, we find dp n1 , p
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Proof: If p and p are fixed points
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(b) Assume there is a subsequence p
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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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