The Real And Complex Number Systems

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that as n n 0 ,wehave a n M. (3) lim n inf a n a if, and only if, (a) given any 0, there are infinite many numbers n such that a a n and (b) given any 0, there is a positive integer n 0 such that as n n 0 ,wehave a n a . From Theorem 2 an Theorem 3, the sequence is divergent, we give the following definitios. Definition Let a n be a real sequence, then we have (1) If lim n sup a n , then we call the sequence a n diverges to , denoted by lim n a n . Theorem Theorem Theorem (2) If lim n inf a n , then we call the sequence a n diverges to , denoted by lim n a n . Let a n be a real sequence. If a is a limit point of a n , then we have lim n inf a n a lim n sup a n . Some useful results Let a n be a real sequence, then (1) lim n inf a n lim n sup a n . (2) lim n infa n lim n sup a n and lim n supa n lim n inf a n (3) If every a n 0, and 0 lim n inf a n lim n sup a n , then we have lim n sup a 1 n 1 lim n inf a n Let a n and b n be two real sequences. (1) If there is a positive integer n 0 such that a n b n , then we have lim n inf a n lim n inf b n and lim n sup a n lim n sup b n . (2) Suppose that lim n inf a n , lim n inf b n , lim n sup a n , lim n sup b n , then lim n inf a n lim n inf b n and lim n inf 1 a n 1 lim n sup a n . lim n infa n b n lim n inf a n lim n sup b n (or lim n sup a n 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 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.
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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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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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Page 247 and 248: which implies that which implies th
Page 249 and 250: So, by above sayings, we have prove
Page 251 and 252: (a) If fn is multiplicative and if
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Page 255 and 256: (b) Prove that h n (x) does not con
Page 257 and 258: Proof: Since g is continuous on a c
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Page 261 and 262: y continuity of f k(x0 ) (x) − f
Page 263 and 264: In addition, as c ≥ 1/2, if f n
Page 265 and 266: (Lemma) If {a n } and {b n } are tw
Page 267 and 268: 9.16 Let {f n } be a sequence of re
Page 269 and 270: which implies that ∣ lim sup 1 k
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Page 273 and 274: [ ] π And as x ∈ , π , then n+p
Page 275 and 276: 0.1 Supplement on some results on W
Page 277 and 278: which implies that h (x + t) − h
Page 279 and 280: we have ∫ d c |f n − g| 2 dx
Page 281 and 282: we complete it. 9.31 Given that two
Page 283 and 284: if x = λ (≠ 0) is a root of 1 +
Page 285 and 286: So, the series diverges. (ii) As
Page 287 and 288: 9.38 For each real t, define f t (x
Page 289 and 290: So, B 0 = P 0 (0) = C 0 = 1, B 1 =
Page 291 and 292: Remark: (1) The reader can see the
Page 293: That is, lim n inf a n sup c n . n
Page 297 and 298: q! kq1 1 k! kq1 q! k! 1 q 1
Page 299: 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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