i)记 B k = ⋃ i = k n A i , k ≥ 2 B_k =\bigcup_{i=k}^n A_i,k \ge 2Bk=⋃i=knAi,k≥2,
B k C = ( ⋃ i = k n A i ) C = ( A k ∪ B k + 1 ) C = A k C ∩ B k + 1 C B_k^C=(\bigcup_{i=k}^n A_i)^C=(A_k \cup B_{k+1})^C=A_k^C \cap B^C_{k+1}BkC=(i=k⋃nAi)C=(Ak∪Bk+1)C=AkC∩Bk+1C
这样我们就得到了 B k B_kBk的递推关系,下面我们重复使用这个递推关系
( ⋃ i = 1 n A i ) C = ( A 1 ∪ B 2 ) C = A 1 C ∩ B 2 C = A 1 C ∩ A 2 C ∩ B 3 C = ⋯ = A 1 C ∩ A 2 C ∩ ⋯ A k C = ⋂ i = 1 n A i C (\bigcup_{i=1}^n A_i)^C=(A_1\cup B_2)^C=A_1^C \cap B_2^C \\ = A_1^C \cap A_2^C \cap B_3^C = \cdots =A_1^C \cap A_2^C \cap \cdots A_k^C =\bigcap_{i=1}^nA_i^C(i=1⋃nAi)C=(A1∪B2)C=A1C∩B2C=A1C∩A2C∩B3C=⋯=A1C∩A2C∩⋯AkC=i=1⋂nAiC
ii)记 D k = ⋂ i = k n A i , k ≥ 2 D_k=\bigcap_{i=k}^n A_i,k \ge 2Dk=⋂i=knAi,k≥2
D k C = ( ⋂ i = k n A i ) C = ( A k ∩ D k + 1 ) C = A k C ∪ D k + 1 C D_k^C=(\bigcap_{i=k}^n A_i)^C=(A_k \cap D_{k+1})^C=A_k^C \cup D^C_{k+1}DkC=(i=k⋂nAi)C=(Ak∩Dk+1)C=AkC∪Dk+1C
重复使用这个递推关系
( ⋂ i = 1 n A i ) C = ( A 1 ∩ D 2 ) C = A 1 C ∪ D 2 C = A 1 C ∪ A 2 C ∪ D 3 C = ⋯ = A 1 C ∪ A 2 C ∪ ⋯ A k C = ⋃ i = 1 n A i C (\bigcap_{i=1}^n A_i)^C=(A_1\cap D_2)^C=A_1^C \cup D_2^C \\ = A_1^C \cup A_2^C \cup D_3^C = \cdots =A_1^C \cup A_2^C \cup \cdots A_k^C =\bigcup_{i=1}^nA_i^C(i=1⋂nAi)C=(A1∩D2)C=A1C∪D2C=A1C∪A2C∪D3C=⋯=A1C∪A2C∪⋯AkC=i=1⋃nAiC
证毕
这个公式是非常有用的,未来在组合学、分析、概率论等课程中大家会频繁使用到它。
定义0.4 集列类比数列,称一系列集合为集列,记为 { A i } i = 1 n = { A 1 , A 2 , ⋯ , A n } , n ∈ N \{A_i\}_{i=1}^n=\{A_1,A_2,\cdots,A_n\}, n \in \mathbb{N}{Ai}i=1n={A1,A2,⋯,An},n∈N。
定义0.5 集列的单调性与数列类似,我们也可以定义集列的单调性
集列递增记为 A n ↑ A_n \uparrowAn↑,表示 A i ⊆ A i + 1 , 1 ≤ i ≤ n − 1 A_i \subseteq A_{i+1},1 \le i \le n-1Ai⊆Ai+1,1≤i≤n−1集列递减记为 A n ↓ A_n \downarrowAn↓,表示 A i ⊇ A i + 1 , 1 ≤ i ≤ n − 1 A_i \supseteq A_{i+1},1 \le i \le n-1Ai⊇Ai+1,1≤i≤n−1
例0.3在实分析中,我们经常要基于一个递增的集列构造一个不交的集列,假设 { A i } i = 1 n \{A_i\}_{i=1}^n{Ai}i=1n是一个递增的集列,定义
B 1 = A 1 , B k = A k ∖ A k − 1 , ∀ k ≥ 2 B_1=A_1,B_{k} = A_k \setminus A_{k-1},\forall k \ge 2B1=A1,Bk=Ak∖Ak−1,∀k≥2
验证 B k ∩ B l = ϕ , ∀ k ≠ l B_k \cap B_l = \phi,\forall k\ne lBk∩Bl=ϕ,∀k=l以及 ⋃ k = 1 n B k = ⋃ i = 1 n A i \bigcup_{k=1}^n B_k=\bigcup_{i=1}^n A_i⋃k=1nBk=⋃i=1nAi。
证明
i)不妨假设 k