北大02-03集合论与图论__期末试卷及答案.pdf

2002-2003 2003 6 1-5 3 6-10 3 10 60 40 100 1⊕ A B P(A)⊕P(B)=P(A)⊕P(C)⇔B=C10 2 ⊕P(A)⊕P(B)=P(A)⊕P(C)⇔P(B)=P(C) 3 P(B)=P(C)⇔B=C 3 2 2R A B C⊆A D⊆BR=C×D A x ,x B y ,y (x Ry ∧x Ry )→x Ry . 1 2 1 2 1 1 2 2 1 2 2 ⇒x 1∈Cy 2 ∈D 3 - 0 - ⇒C=dom(R)D=ran(R) 3 2 3A={1,2,…,10}A R={x ,y |x ,y ∈A∧x+y =10}R 5 2 1,1A 5,5 ∈A 3,7,7,3 ∈A 7 3 3,7,7,3 ∈A 3,3A 4A ×A P(A)2→A A→2. |A ×A | = |2→A | = |A |2 2 |P(A)| = |A →2| = 2|A | 2 (1) A A →2={} |A ×A | = |2→A | = 0 |P(A)| = |A→2| = 11 (2) A |A |=1 |A ×A | = |2→A | = |A |2 = 1 2 = 2|A | = |P(A)| = |A →2| 1 (3) A |A |=2 2 |A | |A ×A | = |2→A | = |A | = 4 = 2 = |P(A)| = |A→2| 1 (4) A |A |=3 2 |A | |A ×A | = |2→A | = |A | =9 8 = 2 = |P(A)| = | A →2| 1 (5) A |A |4 |A ×A | = |2→A | = |A |2 2|A | = |P(A)| = |A →2| 1 (6) A |A ×A | = |2→A | = |A |2 = |A | 2|A | = |P(A)| = | A →2| 1 (1)(2)(5)(6) 5 - 1 - D={d ,d ,…,d }K={k ,k ,…,k