The Real And Complex Number Systems

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In particular, if a n converges, we have lim n→ supa n b n lim n→ a n lim n→ sup b n and lim n→ infa n b n lim n→ a n lim n→ inf b n . (3) Suppose that − lim n→ inf a n , lim n→ inf b n , lim n→ sup a n , lim n→ sup b n , anda n 0, b n 0 ∀n, then lim n→ inf a n lim n→ inf b n ≤ lim n→ infa n b n ≤ lim n→ inf a n lim n→ sup b n (or lim n→ inf b n ≤ lim n→ supa n b n ≤ lim n→ sup a n lim n→ sup b n . In particular, if a n converges, we have and lim n→ supa n b n lim n→ infa n b n lim n→ a n lim n→ a n lim n→ sup b n lim n→ inf b n . lim n→ sup a n ) Theorem Let a n be a positive real sequence, then lim n→ inf a n1 a n ≤ lim n→ infa n 1/n ≤ lim n→ supa n 1/n ≤ lim n→ sup a n1 a n . Remark We can use the inequalities to show n! 1/n lim n→ n 1/e. Theorem Let a n be a real sequence, then lim n→ inf a n ≤ lim n→ inf a 1 ...a n n ≤ lim n→ sup a 1 ...a n n ≤ lim n→ sup a n . Exercise Let f : a, d → R be a continuous function, and a n is a real sequence. If f is increasing and for every n, lim n→ inf a n , lim n→ sup a n ∈ a, d, then lim n→ sup fa n f lim n→ sup a n and lim n→ inf fa n f lim n→ inf a n . Remark: (1) The condition that f is increasing cannot be removed. For example, fx |x|, and 1/k if k is even a k −1 − 1/k if k is odd. (2) The proof is easy if we list the definition of limit sup and limit inf. So, we omit it. Exercise Let a n be a real sequence satisfying a np ≤ a n a p for all n, p. Show that an n converges. Hint: Consider its limit inf.
Remark: The exercise is useful in the theory of Topological Entorpy. Infinite Series And Infinite Products Sequences 8.1 (a) Given a real-valed sequence a n bounded above, let u n supa k : k ≥ n. Then u n ↘ and hence U lim n→ u n is either finite or −. Prove that U lim n→ sup a n lim n→ supa k : k ≥ n. Proof: It is clear that u n ↘ and hence U lim n→ u n is either finite or −. If U −, then given any M 0, there exists a positive integer N such that as n ≥ N, we have u n ≤−M which implies that, as n ≥ N, a n ≤−M. So, lim n→ a n −. Thatis,a n is not bounded below. In addition, if a n has a finite limit supreior, say a. Thengiven 0, and given m 0, there exists an integer n m such that a n a − which contradicts to lim n→ a n −. From above results, we obtain U lim n→ sup a n in the case of U −. If U is finite, then given 0, there exists a positive integer N such that as n ≥ N, we have U ≤ u n U . So, as n ≥ N, u n U which implies that, as n ≥ N, a n U . In addition, given ′ 0, and m 0, there exists an integer n m, U − ′ a n by U ≤ u n supa k : k ≥ n if n ≥ N. From above results, we obtain U lim n→ sup a n in the case of U is finite. (b)Similarly, if a n is bounded below, prove that V lim n→ inf a n lim n→ infa k : k ≥ n. Proof: Since the proof is similar to (a), we omit it. If U and V are finite, show that: (c) There exists a subsequence of a n which converges to U and a subsequence which converges to V. Proof: SinceU lim sup n→ a n by (a), then (i) Given 0, there exists a positive integer N such that as n ≥ N, wehave a n U . (ii) Given 0, and m 0, there exists an integer Pm m, U − a Pm . Hence, a Pm is a convergent subsequence of a n with limit U. Similarly for the case of V.
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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
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Covering theorems in R n 3.17 If S
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uncountable. Hence, by exercise 3.2
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Then we have (a) (b) (c) (d) x y
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(b) Give an example of a metric spa
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Hence from (1)-(4), we know that i
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3.44 intM A M A . Proof: Let B
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That is, intB which is absurb. He
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Limits And Continuity Limits of seq
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x n1 x n 1 1 x n x n 1 2 . Rema
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|a m a n| |a m a m1 a m1 a m2
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and use the fact if a n converges
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lim y0 fx, y 0ifx 0, 1ifx 0. and
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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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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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Page 163 and 164: Remark: If we can make sure that fx
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Page 169 and 170: lim xa fx gx L. Remark: 1. The pr
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Page 175 and 176: f x 1 ...x n n fx 1 ...fx n n . C
Page 177 and 178: lx fc f cx c fx. (Exercise 2)
Page 179 and 180: So, we have Partial derivatives fb
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Page 185 and 186: Functions of Bounded Variation and
Page 187 and 188: For the part fx x fy 1 |fx fy| 1
Page 189 and 190: Claim that 1 sup A. Suppose NOT, t
Page 191 and 192: we know that V f is an increasing
Page 193 and 194: in Section 6.12. Proof: Sinceft : t
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Page 197 and 198: n k1 n ||fb k | |fa k || |fb k
Page 199 and 200: Supplement on lim sup and lim inf I
Page 201: (1) lim n→ inf a n − a n has
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Page 207 and 208: If lim sup n→ a n1 a n , then it
Page 209 and 210: We first prove lim n→ sup n ≤
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Page 213 and 214: Consider which implies that which i
Page 215 and 216: So, L 1 5 since L ≥ 1. 2 Remark:
Page 217 and 218: For case (i), since 1 n log nlog lo
Page 219 and 220: ()Assume that ∑ n1 That is, p!
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Page 223 and 224: (b) Prove that ∑ n1 −1 n−1 /
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Page 229 and 230: Suppose NOT, it means that lim n→
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Page 233 and 234: |sin k|r ∑ k So, ∑ |sink|r dive
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Page 239 and 240: Also, lim q→ fp, q lim q→ p si
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Page 243 and 244: So, n s n ∑ cos2k − 1x j1 sin
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Page 247 and 248: which implies that which implies th
Page 249 and 250: So, by above sayings, we have prove
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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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