The Real And Complex Number Systems

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1/|x| if x 0, gx 0ifx 0. The difference of two examples is that the limit of |g0 h g0| does not exist as h tends to 0. 4.11 Let f be defined on R 2 .If lim x,ya,b fx, y L and if the one-dimensional lim xa fx, y and lim yb fx, y both exist, prove that lim xa lim yb fx, y lim yb lim xa fx, y L. Proof: Since lim x,ya,b fx, y L, thengiven 0, there exists a 0 such that as 0 |x, y a, b| , wehave |fx, y L| /2, which implies lim|fx, y L| yb which implies limfx, y L yb /2 if 0 |x, y a, b| lim xa limfx, y L /2 if 0 |x, y a, b| . yb Hence, we have proved lim xa|lim yb x, y L| /2 . Since is arbitrary, we have which implies that lim xa limfx, y L 0 yb lim xa limfx, y L 0. yb So, lim xa lim yb fx, y L. The proof of lim yb lim xa fx, y L is similar. Remark: 1. The exercise is much important since in mathematics, we would encounter many and many similar questions about the interchange of the order of limits. So, we should keep the exercise in mind. 2. In the proof, we use the concept: |lim xa fx| 0if,andonlyiflim xa fx 0. 3. The hypothesis fx, y L as x, y a, b tells us that every approach form these points x, y to the point a, b, fx, y approaches to L. Use this concept, and consider the special approach from points x, y to x, b and thus from x, b to a, b. Note that since lim yb fx, y exists, it means that we can regrad this special approach as one of approaches from these points x, y to the point a, b. So, it is natural to have the statement. 4. The converse of statement is not necessarily true. For example, x y if x 0ory 0 fx, y 1 otherwise. Trivially, we have the limit of fx, y does not exist as x, y 0, 0. However,
lim y0 fx, y 0ifx 0, 1ifx 0. and lim x0 fx, y 0ify 0, 1ify 0. lim x0 lim y0 fx, y lim y0 limfx, y 1. x0 In each of the preceding examples, determine whether the following limits exist and evaluate those limits that do exist: lim x0 limfx, y ; lim y0 y0 limfx, y ; x0 lim fx, y. x,y0,0 Now consider the functions f defined on R 2 as follows: (a) fx, y x2 y 2 if x, y 0, 0, f0, 0 0. x 2 y 2 Proof: 1. Since(x 0) we have 2. Since (y 0) we have x limfx, y lim 2 y 2 x0 x0 x 2 y 2 lim y0 x limfx, y lim 2 y 2 y0 y0 x 2 y 2 lim x0 limfx, y 1. x0 limfx, y 1. y0 y 2 1 ify 0, y 2 1ify 0, x 2 1ifx 0, x 2 1ifx 0, 3. (x, y 0, 0) Letx r cos and y r sin , where 0 2, and note that x, y 0, 0 r 0. Then lim fx, y lim x 2 y 2 x,y0,0 x,y0,0 x 2 y 2 lim r0 r 2 cos 2 sin 2 r 2 cos 2 sin 2 . So, if we choose and /2, we find the limit of fx, y does not exist as x, y 0, 0. Remark: 1. This case shows that 1 lim x0 lim y0 fx, y lim y0 lim x0 fx, y 1 2. Obviously, the limit of fx, y does not exist as x, y 0, 0. Since if it was, then by (*), (**), and the preceding theorem, we know that which is absurb. lim x0 lim y0 fx, y lim y0 lim x0 fx, y **
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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
Page 35 and 36: and the sum of their squares is giv
Page 37 and 38: Some Basic Notations Of Set Theory
Page 39 and 40: (i) If x ∈ R, it is clear that (x
Page 41 and 42: (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 use the fact if a n converges
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 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
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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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