The Real And Complex Number Systems

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exists an n ball Bx, r x such that Bx, r x x a, b . For the point a, it is easy to know that Ba, r a a, b . That is, in this case, there is only one accumulation point a of a, b. So, from 1 and 2, we know that the set of the accumulation points of a, b is a, b. Since a a, b, we know that a, b cannot contain its all accumulation points. So, a, b is not closed. Since an n ball Bb, r is not contained in a, b for any r 0, we know that the point b is not interior to a, b. So,a, b is not open. (c) All numbers of the form 1/n, (n 1,2,3,.... Solution: Write the set 1/n : n 1,2,... 1, 1/2, 1/3, . . . , 1/n,... : S. Obviously, 0 is the only one accumulation point of S. So,S is not closed since S does not contain the accumulation point 0. Since 1 S, andB1, r is not contained in S for any r 0, S is not open. Remark: Every point of 1/n : n 1,2,3,... is isolated. (d) All rational numbers. Solutions: Denote all rational numbers by Q. It is trivially seen that the set of accumulation points is R 1 . So, Q is not closed. Consider x Q, anyn ball Bx is not contained in Q. Thatis,x is not an interior point of Q. In fact, every point of Q is not an interior point of Q. So, Q is not open. (e) All numbers of the form 2 n 5 m , (m, n 1,2,.... Solution: Write the set 2 n 5 m : m, n 1, 2, . . m m1 1 2 5m , 1 4 5m ,..., 1 2 n 5m ,... : S 1 2 1 5 , 1 2 1 5 ,..., 1 2 2 5 1 m ,... 1 4 1 5 , 1 4 1 5 ,..., 1 2 4 5 1 m ,... ........................................................ 2 1 n 1 5 , 1 2 n 1 5 ,..., 1 2 2 n 5 1 m ... ..................................................... So,wefindthatS 1 : n 1, 2, . . . 1 : m 1, 2, . . . 2 n 5 0. So,S is not m closed since it does not contain 0. Since 1 S, andB 1 , r is not contained in S for any 2 2 r 0, S is not open. Remark: By (1)-(3), we can regard them as three sequences 1 2 m 5m , 1 5 m m and 1 m1 4 m1 2 n m 5m , respectively. m1 And it means that for (1), the sequence 5 m m m1 moves 1 . Similarly for others. So, it is 2 easy to see why 1 is an accumulation point of 1 2 2 5m m m1 . And thus get the set of all accumulation points of 2 n 5 m : m, n 1, 2, . . . (f) All numbers of the form 1 n 1/m, (m, n 1,2,.... Solution: Write the set of all numbers 1 n 1/m, (m, n 1, 2, . . . as 1 m 1 m m 1 1 m1 m : S. m1 1 2 3
And thus by the remark in (e), it is easy to know that S 1, 1. So,S is not closed since S S. Since2 S, andB2, r is not contained in S for any r 0, S is not open. (g) All numbers of the form 1/n 1/m, (m, n 1,2,.... Solution: Write the set of all numbers 1/n 1/m, (m, n 1, 2, . . . as 1 1/m m m1 1/2 1/m m m1 ...1/n 1/m m m1 ...: S. We find that S 1/n : n N 1/m : m N 0 1/n : n N 0. So,S is not closed since S S. Since1 S, andB1, r is not contained in S for any r 0, S is not open. (h) All numbers of the form 1 n /1 1/n, (n 1,2,.... Soluton: Write the set of all numbers 1 n /1 1/n, (n 1, 2, . . . as k 1 1 1 2k k1 1 1 1 2k1 k k1 : S. We find that S 1, 1. So,S is not closed since S S. Since 1 S, andB 1 2 2 not contained in S for any r 0, S is not open. , r is 3.3 The same as Exercise 3.2 for the following sets in R 2 . (a) All complex z such that |z| 1. Solution: Denote z C :|z| 1 by S. It is easy to know that S z C :|z| 1. So, S is not closed since S S. Letz S, then |z| 1. Consider Bz, |z|1 S, soevery 2 point of S is interior. That is, S is open. (b) All complex z such that |z| 1. Solution: Denote z C :|z| 1 by S. It is easy to know that S z C :|z| 1. So, S is closed since S S. Since1 S, andB1, r is not contained in S for any r 0, S is not open. (c) All complex numbers of the form 1/n i/m, (m, n 1, 2, . . . . Solution: Write the set of all complex numbers of the form 1/n i/m, (m, n 1, 2, . . . as 1 m i m 1 i m m1 2 m ... 1 m1 n m i m ...: S. m1 We know that S 1/n : n 1,2,... i/m : m 1, 2, . . . 0. So,S is not closed since S S. Since1 i S, andB1 i, r is not contained in S for any r 0, S is not open. (d) All points x, y such that x 2 y 2 1. Solution: Denote x, y : x 2 y 2 1 by S. We know that S x, y : x 2 y 2 1. So,S is not closed since S S. Letp x, y S, then x 2 y 2 1. Itiseasytofindthatr 0 such that Bp, r S. So,S is open. (e) All points x, y such that x 0. Solution: Write all points x, y such that x 0asx, y : x 0 : S. It is easy to know that S x, y : x 0. So,S is not closed since S S. Letx S, then it is easy to find r x 0 such that Bx, r x S. So,S is open. (f) All points x, y such that x 0. Solution: Write all points x, y such that x 0asx, y : x 0 : S. It is easy to
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The Real And Complex Number Systems
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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 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: Charpter 3 Elements of Point set To
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 and 106: 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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