1事件之间的关系 (1)从A-B=C推不出A=BUC 当A-B=C时,只能推出 AC BUC.事实上,令 A={1,2,3,4,5},B={1,3,5,7 于是 C=A-B={2,4} 而 B∪C={1,2,3,4,5,7 从而B∪C2A,但是A≠BUC 仅当A→B时,方能得出A=BUC (2)从A=B∪C推不出A-B=C 当A=B∪C成立时,只能推出A-BcC.当BcA,CCA且B∩C≠0 时,可以得出A-B=C.例如,令 A={1,2,3,4,5,6},B={1,2,3},C={2,4,5,6} 则有 {1,2,3,4,5,6}=A. A-B={4,5,6} 从而A-B≠C (3)Uk Ak-Uk Bk#Uk (Ak- Bk) 对于一般情况的A,B,C,有 U4k-∪Bkc∪U 例如,令k=2,取 A2={1,2,3,4,5,6},B1={1,2},B2={5,6} 则有 (A1∪A2)-(B1UB2)={3,4},(A1-B1)U(A2-B2)={1,2,3,4,5,6} 从而(A1UA2)-(B1∪B2)≠(A1-B1)U(A2-B2)1 ¯‡ƒm'X (1) lA − B = CíØÑA = B ∪ C A − B = Cž§UíÑA ⊂ B ∪ C. ¯¢þ§- A = {1, 2, 3, 4, 5}, B = {1, 3, 5, 7}, u´ C = A − B = {2, 4}. B ∪ C = {1, 2, 3, 4, 5, 7}, l B ∪ C ⊃ A§´A 6= B ∪ C. =A ⊃ Bž§UÑA = B ∪ C. (2) lA = B ∪ CíØÑA − B = C A = B ∪ C¤áž§UíÑA − B ⊂ C. B ⊂ A§C ⊂ A … B ∩ C 6= ∅ ž§Œ±ÑA − B = C. ~X§- A = {1, 2, 3, 4, 5, 6}, B = {1, 2, 3}, C = {2, 4, 5, 6}. Kk B ∪ C = {1, 2, 3, 4, 5, 6} = A. A − B = {4, 5, 6}. l A − B 6= C. (3) S k Ak − S k Bk 6= S k (Ak − Bk) éu„œ¹A§B§C§k [ k Ak − [ k Bk ⊂ [ k (Ak − Bk). ~X§-k = 2§ A1 = A2 = {1, 2, 3, 4, 5, 6}, B1 = {1, 2}, B2 = {5, 6}. Kk (A1 ∪ A2) − (B1 ∪ B2) = {3, 4}, (A1 − B1) ∪ (A2 − B2) = {1, 2, 3, 4, 5, 6}. l (A1 ∪ A2) − (B1 ∪ B2) 6= (A1 − B1) ∪ (A2 − B2). 1