第五章习题答&

第五章习题答&

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Unformatted text preview: P151∼ P1531 11 Ð8 1 21 {-31 2} 1 41 {51 61 71 81 91 101 111 121 131 141 15} 21 Ð8 1 21 {x| x 1 1 41 {x|ƒ x 8ª* µì 1 61 {x| x=2k+11 k∈I} * ª 31 ì 8 µƒ *ª¹`x †XC~¿ª*} 1 61 * 41 ` x ª ¹ †XC~¿ª*} 1 6Ð8 1 21 1 1 {11 1 }1 ={∅ {1}1 {∅}1 {11 1 }} 1 4 1 1 1 {∅ {a} 1 {∅}} 1 ={∅ {∅} 1 {{a}} 1 {{∅}} 1 {∅ A=G1 B=E1 C=F ª µ ƒ 78 ì 21 31 1Ð ª µ 1ì 8 ƒ 51 ª µ ƒ 88 ì I1 11 41 10< x< 20} } {a}}1 {∅ {∅}}1 {{a}1 {∅}}1 {∅ {a}1 {∅}}} 71 “ 1 ”ø. » { 11 81 Ð A∈B1 B∈C1 A∈C 1 21 31 ƒ 68µ ìª 41 51 A∈C 8µ ƒì ª A1 B1 ƒ C 8µ ì* ª A∈B1 B∈C1 A={a} 1 B={{a}} 1 C={{{a}}} 1 101 4E ¾ ª* » U={11 21 31 41 5}1 A={11 4}1 B={11 21 5}1 C={21 4} 1 81 1 1 A1 1 1 1 1 A1 1 1 C1 1 C1 ={∅ {1}1 {4}1 {11 4}}− ∅ {2}1 {4}1 {21 4}} { ={{1}1 {11 4}} 11E ª 4* » ¾ 1 11 1 21 A− B∩C1 =1 A− 1 1 1 A− 1 B C B 1 A− 1 1 B=∅ 11 A− B∩ C1 = A∩ ∼ = A∩ B∩ C1 B∪ ∼ C1 =1 A∩ ∼ B1 1 1 A∩ ∼ C1 =1 A− 1 1 1 A− 1 B C 1 21 A− 1 1 B B =1 A∩ ∼ B1 1 B = A∩ =∅ 121 1 11 1 21 A1 Bª1¹ C † x ` *XC~¿ª} A− 1 1 1 A− 1 =A B C A− 1 1 1 A− 1 = ∅ B C 11 A− 1 1 1 A− 1 B C B∩ B1 = 1 A∩ ∼ B1 1 1 A∩ ∼ C1 = A∩ = A∩ ∼ B∪ ∼ C1 B∩ C1 = A− B∩ C1 1 1 4 » ¸¾ª 1 21 A− 1 1 1 A− 1 =A 1 B C A∩ 1 4 1¸¾ª » B∩ C1 =∅ A− B∩ C1 =A A− 1 1 1 A− 1 =A− B∩ C1 B C A− B∩C1 =∅ 1 A− 1 1 1 A− 1 =∅ 1 B C 5.11 1 A⊆ B∩ C1 13. 1 A1 B 1 1 11 1 21 A− B=B1 A 1 ¾ 4B ¸ª » A− B=B− 1 A 1 4 ¸ª A, B »¾ 11 A=B=φ A− B=B 1 B B ⇔ A− ⊆ B∧B⊆ A− B) B B) ⇔( A− ∩ B= A− ∧B∩ ( A− = B B ⇔ A ∩ ~ B ∩ B= A− ∧B ∩A ∩ ~ B = B B= ⇔ A− φ∧ B=φ ⇔ A ⊆ B∧B=φ ⇔ A=B=φ A− B=B B B B ⇔ A− ⊆ B∧ ⊆ A− B B= B ⇔A− − φ∧ B− A− 1 =φ B= ⇔ A− φ∧ B∩~A1 ∪ B= A= ⇔ A− φ∧ B− φ∧ B=φ B ⇔ A ⊆ B∧ ⊆ A∧B=φ ⇔ A=B=φ (2) A=B1 A− B=B− A B A BA ⇔A− ⊆ B− ∧ − ⊆ A-B B ⇔A∩ ~B⊆ B∩~A ∧ ∩ ~A⊆ A∩ ~B ⇔ A∩ ~B∩B∩~A=A∩ ~B1 ∧ B∩~A∩ A ∩~B=B∩~A1 ⇔A∩ ~B=φ∧B∩ ~A=φ B-A=φ ⇔A-B=φ ∧ B ⇔A⊆ B∧ ⊆ A ⇔A=B 15. 1 A={{φ }1 {{φ }}}1 1 41 1 1 1 1 A1 1 1 1 1 1 A1 =∪ {φ {{φ }}1 {{{φ }}}1 {{φ }1 {{φ }}} B∩ B1 =φ ={{φ }1 {{φ }}} =A 16. 1 B={{11 2}1 {21 3}1 {11 3}1 ª φ d}Ø *F ¾ 31 ∩ B1 1 41 1 1 ∩ B1 1 31 ∩ B1 =∩ {11 21 3}=1∩2∪ 3 1 41 1 1 ∩ B1 =∪φ =φ 17. ·( ¾ª* (A) x x∈A⇔{x}⊆ A ⇔{x}∈P1 A1 ⇔{x}∈P(B) ⇔{x}⊆ B ⇔x∈B 1 24.1 A=B A={01 1}1 B={11ª 4· ( * 2} ¾ 31 B× A1 2 (B) A B 1 1 (A)1 1 (B)1 1 21 A× B1 1 21 A× B={<0,1>,<0,2>,<1,1>,<1,2>} 1 31 B× A1 2 ={<1,0>,<1,1>,<2,0>,<2,1>}× {<1,0>,<1,1>,<2,0>,<2,1>} ={<<1,0>,<1,0>>,<<1,0>,<1,1>>,<<1,0>,<2,0>>,<<1,0>,<2,1>>,<<1,1 >,<1,0>>,<<1,1>,<1,1>>,<<1,1>,<2,0>>,<<1,1>,<2,1>>,<<2,0>,<1,0>>, <<2,0>,<1,1>>,<<2,0>,<2,0>>,<<2,0>,<2,1>>,<<2,1>,<1,0>>,<<2,1>,< 1,1>>,<<2,1>,<2,0>>,<<2,1>,<2,1>>} 26.1 AE 1J É *Bª 1º Ï C ˜ Ð ¬ 1 1 21 A× 1 31 1 (A∩B)×(C∩D) A× C1 1 (A×C)∩(B×D) B∪ C1 =1 A× B1 ∪ A×(B1 C)1 (A×B)1 (A×C) ª ¾ } 18 ø x, y y 1 x, y ∈ ( A I B ) × ( C I D ) ⇔ x ∈ A IB ∧ y ∈ C ID ⇔ ( x ∈ A ∧ x ∈ B) ∧ ( y ∈C ∧ y ∈ D) ⇔ ( x ∈ A ∧ y ∈C ) ∧ ( x ∈ B ∧ y ∈ D) ⇔ x , y ∈ A × C ∧ x, y ∈ B × D ⇔ x , y ∈ ( A × C ) I( B × D ) (A∩B)×(C∩D)1 (A×C)∩(B×D) 1 21 1 x, y y x, y ∈ A × ( B U C ) ⇔ x ∈ A ∧ y ∈ B UC ⇔ x ∈ A ∧ ( y ∈ B ∨ y ∈C) ⇔ ( x ∈ A ∧ y ∈ B) ∨ ( x ∈ A ∧ y ∈C ) ⇔ ( x, y ∈ A × B ) ∨ ( x, y ∈ A × C ) ⇔ x, y ∈ ( A × B ) U ( A × C ) ∈ A× ( B UC ) = ( A × B ) U( A × C ) ∈ 31 1 x, y ∈ x, y y A × B1 − A × C1 A × C1 A × C1 ⇔ x, y ∈ ⇔ x, y ∈ ⇔ ⇔ A × B1 ∧ x, y ∉ A × B1 ∧ ¬ x, y ∈ x∈ A ∧ y∈ B1 ∧ ¬ x ∈ A ∧ y ∈ B1 ∧ ¬ x ∈ A ∧ y ∈ C1 x∈ A1 ∨ ¬ y∈ C1 ⇔ x∈ A ∧ y∈ B ∧ ¬ x∈ A1 ∨ x∈ A ∧ y∈ B ∧ ¬ y ∈ C1 ⇔ x∈ A ∧ y∈ B ∧ ¬ ⇔ x∈ A ∧ y∈ y ∈ C1 B− 1 C ⇔ x, y ∈ A×(B1 C) 1 27. ªø ¾* È A×(B1 C)1 (A×B)1 (A×C) A⊆C B⊆D x,y x,y ∈ A×B ⇔ x∈ A ∧ y ∈ B ⇒ x ∈C ∧ y ∈ D ⇔ A ×B ⊆ C×D ∈A⊆C B⊆D ( ) x,y ∈ C×D¨ Ó A ×B ⊆ C×D 28.1 ¾* ªH ù A=φ A ×B ⊆ C×D B ={a b} C={1 2} D ={x} A ×B= φ A ×B ⊆ C×DªH ¾* ù A⊆C B⊆D B⊆D ...
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This note was uploaded on 02/06/2011 for the course CS 343 taught by Professor Zhoujunli during the Spring '08 term at BUPT.

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