并集, 在集合论和数学的其他分支中, 一组集合的, 是这些集合的所有元素构成的集合, 而不包含其他元素, a和b的, 目录, 有限聯集, 代数性质, 无限, 无限的性質, 比較性質, 覆蓋性質, 運算性質, 参考, 参考文献有限聯集, 编辑聯集是由公理化集合论的分類公理來確保其唯一存在的特定集合, displaystyle, nbsp, displaystyle, forall, forall, forall, left, leftrightarrow, left, right, right, nbsp, 也就是直. 在集合论和数学的其他分支中 一组集合的并集 1 是这些集合的所有元素构成的集合 而不包含其他元素 A和B的并集 目录 1 有限聯集 1 1 代数性质 2 无限并集 2 1 无限并集的性質 2 1 1 比較性質 2 1 2 覆蓋性質 2 1 3 運算性質 3 参考 4 参考文献有限聯集 编辑聯集是由公理化集合论的分類公理來確保其唯一存在的特定集合 A B displaystyle A cup B nbsp A B x x A B x A x B displaystyle forall A forall B forall x left x in A cup B Leftrightarrow left x in A vee x in B right right nbsp 也就是直觀上 對所有 x displaystyle x nbsp x A B displaystyle x in A cup B nbsp 等價於 x A displaystyle x in A nbsp 或 x B displaystyle x in B nbsp 举例 集合 1 2 3 displaystyle 1 2 3 nbsp 和 2 3 4 displaystyle 2 3 4 nbsp 的并集是 1 2 3 4 displaystyle 1 2 3 4 nbsp 数9 displaystyle 9 nbsp 不属于素数集合 2 3 5 7 11 displaystyle 2 3 5 7 11 ldots nbsp 和偶数集合 2 4 6 8 10 displaystyle 2 4 6 8 10 ldots nbsp 的并集 因为9 displaystyle 9 nbsp 既不是素数 也不是偶数 更通常的 多个集合的并集可以这样定义 例如 A B displaystyle A B nbsp 和C displaystyle C nbsp 的并集含有所有A displaystyle A nbsp 的元素 所有B displaystyle B nbsp 的元素和所有C displaystyle C nbsp 的元素 而没有其他元素 形式上 x displaystyle x nbsp 是A B C displaystyle A cup B cup C nbsp 的元素 当且仅当x displaystyle x nbsp 属于A displaystyle A nbsp 或x displaystyle x nbsp 属于B displaystyle B nbsp 或x displaystyle x nbsp 属于C displaystyle C nbsp 代数性质 编辑 二元并集 两个集合的并集 是一种结合运算 即 A B C A B C displaystyle A cup B cup C A cup B cup C nbsp 事实上 A B C displaystyle A cup B cup C nbsp 也等于这两个集合 因此圆括号在仅进行并集运算的时候可以省略 相似的 并集运算满足交换律 即集合的顺序任意 空集是并集运算的单位元 即 A A displaystyle varnothing cup A A nbsp 对任意集合A displaystyle A nbsp 可以将空集当作零个集合的并集 结合交集和补集运算 并集运算使任意幂集成为布尔代数 例如 并集和交集相互满足分配律 而且这三种运算满足德 摩根律 若将并集运算换成对称差运算 可以获得相应的布尔环 无限并集 编辑由公理化集合论的并集公理 有唯一的集合 M displaystyle bigcup mathcal M nbsp 滿足 M x x M A A M x A displaystyle forall mathcal M forall x left left x in bigcup mathcal M right Leftrightarrow exists A left left A in mathcal M right wedge x in A right right nbsp 也就是直觀上 對所有 M displaystyle mathcal M nbsp 和所有 x displaystyle x nbsp x M displaystyle x in bigcup mathcal M nbsp 等價於有某個 M displaystyle mathcal M nbsp 的下屬集合 A displaystyle A nbsp 使得x A displaystyle x in A nbsp 以上的 M displaystyle mathcal M nbsp 可以直觀的視為一個集合族 而把 M displaystyle bigcup mathcal M nbsp 看成對 M displaystyle mathcal M nbsp 內的集合取并集 但這個公理並沒有對 M displaystyle mathcal M nbsp 下屬集合的數量做出任何限制 所以這個 M displaystyle bigcup mathcal M nbsp 被俗稱為任意并集或无限并集 若 X M displaystyle X subseteq bigcup mathcal M nbsp 會稱 X displaystyle X nbsp 被 M displaystyle mathcal M nbsp 覆蓋 cover 也就是直觀上可以用 M displaystyle mathcal M nbsp 裡的所有集合疊起來蓋住 X displaystyle X nbsp 例如 對 M A B C displaystyle mathcal M A B C nbsp M A B C displaystyle bigcup mathcal M A cup B cup C nbsp 若 M displaystyle M nbsp 是空集 M displaystyle bigcup mathcal M nbsp 也是空集 无限并集有多种表示方法 可模仿求和符号記為 A M A displaystyle bigcup A in mathcal M A nbsp 但大多數人會假設指标集 I displaystyle I nbsp 的存在 換句話說 若 I A M displaystyle I overset A cong mathcal M nbsp 則 i I A i M displaystyle bigcup i in I A i bigcup mathcal M nbsp 在指标集 I displaystyle I nbsp 是自然数系 N displaystyle mathbb N nbsp 的情况下 更可以仿无穷级数來表示 也就是說 若 N A M displaystyle mathbb N overset A cong mathcal M nbsp 則 i 0 A i M displaystyle bigcup i 0 infty A i bigcup mathcal M nbsp 也可以更粗略直觀的將 i 0 A i displaystyle bigcup i 0 infty A i nbsp 写作A 0 A 1 A 2 displaystyle A 0 cup A 1 cup A 2 cup ldots nbsp 无限并集的性質 编辑 定理 0 displaystyle vdash bigcup varnothing varnothing nbsp 證明 1 x x displaystyle forall x neg x in varnothing nbsp 空集公理 2 S displaystyle neg S in varnothing nbsp MP with A4 1 3 S x S displaystyle S in varnothing Rightarrow neg x in S nbsp M0 with 2 4 S x S displaystyle neg neg S in varnothing Rightarrow neg x in S nbsp Equv with DN 3 5 S x S displaystyle neg neg S in varnothing wedge neg x in S nbsp Equv with De Morgan 4 6 S S x S displaystyle forall S big neg neg S in varnothing wedge neg x in S big nbsp GEN with S displaystyle S nbsp 5 7 S S x S displaystyle neg exists S neg S in varnothing wedge neg x in S nbsp Equv with DN 6 8 x x S S x S displaystyle forall x left left x in bigcup varnothing right Leftrightarrow exists S S in varnothing wedge x in S right nbsp MP with 并集公理 A4 9 x S S x S displaystyle left x in bigcup varnothing right Leftrightarrow exists S S in varnothing wedge x in S nbsp MP with A4 8 10 x S S x S displaystyle left x in bigcup varnothing right Rightarrow exists S S in varnothing wedge x in S nbsp MP with AND 9 11 S S x S x displaystyle neg exists S S in varnothing wedge x in S Rightarrow neg left x in bigcup varnothing right nbsp MP with T 10 12 x displaystyle neg left x in bigcup varnothing right nbsp MP with 7 11 13 x x displaystyle forall x left x not in bigcup varnothing right nbsp GEN with x displaystyle x nbsp 12 14 y x x y displaystyle y varnothing Leftrightarrow forall x neg x in y nbsp E 15 y y x x y displaystyle forall y y varnothing Leftrightarrow forall x neg x in y nbsp GEN with y displaystyle y nbsp 14 16 x x displaystyle left bigcup varnothing varnothing right Leftrightarrow forall x left neg left x in bigcup varnothing right right nbsp MP with A4 15 17 displaystyle bigcup varnothing varnothing nbsp Equv with 13 16 比較性質 编辑 定理 1 M N M N displaystyle mathcal M subseteq mathcal N vdash left bigcup mathcal M subseteq bigcup mathcal N right nbsp 證明注意到可以從 AND 得到 P Q P R P Q R displaystyle mathcal P Rightarrow mathcal Q mathcal P Rightarrow mathcal R mathcal P vdash mathcal Q wedge mathcal R nbsp 換句話說 從演繹元定理有 u P Q P R P Q R displaystyle mathcal P Rightarrow mathcal Q mathcal P Rightarrow mathcal R vdash mathcal P Rightarrow mathcal Q wedge mathcal R nbsp 1 A A M A N displaystyle forall A left A in mathcal M Rightarrow A in mathcal N right nbsp Hyp 2 A M A N displaystyle A in mathcal M Rightarrow A in mathcal N nbsp MP with 1 A4 3 a A A M A M displaystyle a in A wedge A in mathcal M Rightarrow A in mathcal M nbsp AND 4 a A A M a A displaystyle a in A wedge A in mathcal M Rightarrow a in A nbsp AND 5 a A A M A N displaystyle a in A wedge A in mathcal M Rightarrow A in mathcal N nbsp D1 with 2 3 6 a A A M a A A N displaystyle a in A wedge A in mathcal M Rightarrow a in A wedge A in mathcal N nbsp u with 4 5 7 A M a A A N a A displaystyle exists A in mathcal M a in A Rightarrow exists A in mathcal N a in A nbsp GENe with A displaystyle A nbsp 6 8 x x M A M x A displaystyle forall x left left x in bigcup mathcal M right Leftrightarrow exists A in mathcal M x in A right nbsp MP with 并集公理 A4 9 x x N A N x A displaystyle forall x left left x in bigcup mathcal N right Leftrightarrow exists A in mathcal N x in A right nbsp MP with 并集公理 A4 10 x M A M x A displaystyle left x in bigcup mathcal M right Leftrightarrow exists A in mathcal M x in A nbsp MP with 8 A4 11 x N A N x A displaystyle left x in bigcup mathcal N right Leftrightarrow exists A in mathcal N x in A nbsp MP with 9 A4 12 x M A N x A displaystyle left x in bigcup mathcal M right Rightarrow exists A in mathcal N x in A nbsp D1 with 7 10 13 x M x N displaystyle left x in bigcup mathcal M right Rightarrow left x in bigcup mathcal N right nbsp D1 with 11 12 14 x x M x N displaystyle forall x left left x in bigcup mathcal M right Rightarrow left x in bigcup mathcal N right right nbsp GEN with a displaystyle a nbsp 13 覆蓋性質 编辑 定理 2 A P A displaystyle vdash A bigcup mathcal P A nbsp A displaystyle A nbsp 正好就是其冪集的聯集 這個定理直觀上可理解成 因為冪集 P A displaystyle mathcal P A nbsp 是以 A displaystyle A nbsp 和 A displaystyle A nbsp 的子集為元素 所以 P A displaystyle mathcal P A nbsp 的聯集理當是 A displaystyle A nbsp 證明注意到可以從 AND 得到 P Q P R P Q R displaystyle mathcal P Rightarrow mathcal Q mathcal P Rightarrow mathcal R mathcal P vdash mathcal Q wedge mathcal R nbsp 換句話說 從演繹元定理有 u P Q P R P Q R displaystyle mathcal P Rightarrow mathcal Q mathcal P Rightarrow mathcal R vdash mathcal P Rightarrow mathcal Q wedge mathcal R nbsp 1 x x P A S S P A x S displaystyle forall x left left x in bigcup mathcal P A right Leftrightarrow exists S S in mathcal P A wedge x in S right nbsp MP with 并集公理 A4 2 S S P A S A displaystyle forall S S in mathcal P A Leftrightarrow S subseteq A nbsp 幂集公理 3 S P A S A displaystyle S in mathcal P A Leftrightarrow S subseteq A nbsp MP with A4 2 4 x x P A S S A x S displaystyle forall x left left x in bigcup mathcal P A right Leftrightarrow exists S S subseteq A wedge x in S right nbsp Equv with 1 3 5 S A x S S A displaystyle S subseteq A wedge x in S Rightarrow S subseteq A nbsp AND 6 x x S x A x S x A displaystyle forall x x in S Rightarrow x in A Rightarrow x in S Rightarrow x in A nbsp A4 7 S A x S x S x A displaystyle S subseteq A wedge x in S Rightarrow x in S Rightarrow x in A nbsp D1 with 5 6 8 S A x S x S displaystyle S subseteq A wedge x in S Rightarrow x in S nbsp AND 9 S A x S x S x S x A displaystyle S subseteq A wedge x in S Rightarrow x in S wedge x in S Rightarrow x in A nbsp u with 7 8 注意到 x S x S x A x A displaystyle x in S x in S Rightarrow x in A vdash x in A nbsp 再對上式套用 AND 就有 x S x S x A x A displaystyle x in S wedge x in S Rightarrow x in A vdash x in A nbsp a 10 S A x S x A displaystyle S subseteq A wedge x in S Rightarrow x in A nbsp D1 with a 9 11 S S A x S x A displaystyle exists S S subseteq A wedge x in S Rightarrow x in A nbsp GENe with S displaystyle S nbsp 10 12 S S A x S A A x A displaystyle forall S neg S subseteq A wedge x in S Rightarrow neg A subseteq A wedge x in A nbsp A4 13 A A x A S S A x S displaystyle A subseteq A wedge x in A Rightarrow exists S S subseteq A wedge x in S nbsp MP with T 12 14 x A x A displaystyle x in A Rightarrow x in A nbsp I 15 A A displaystyle A subseteq A nbsp GEN with x displaystyle x nbsp 14 注意到 AND 依據演繹定理可改寫為 A A x A A A x A displaystyle A subseteq A vdash x in A Rightarrow A subseteq A wedge x in A nbsp b 16 x A A A x A displaystyle x in A Rightarrow A subseteq A wedge x in A nbsp b with 15 17 x A S S A x S displaystyle x in A Rightarrow exists S S subseteq A wedge x in S nbsp D1 with 13 16 18 x A S S A x S displaystyle x in A Leftrightarrow exists S S subseteq A wedge x in S nbsp AND with 11 17 19 x x P A x A displaystyle forall x left left x in bigcup mathcal P A right Leftrightarrow x in A right nbsp Equv with 4 18 定理 3 M A M M M A displaystyle left bigcup mathcal M subseteq A right vdash forall M in mathcal M M subseteq A nbsp 直觀上 這個定理說 一群集合的聯集包含於 A displaystyle A nbsp 則它們個個都包含於 A displaystyle A nbsp 證明 1 a M M a M a A displaystyle forall a left exists M in mathcal M a in M Rightarrow a in A right nbsp Hyp 2 M M a M M M a M displaystyle M in mathcal M wedge a in M Rightarrow exists M in mathcal M a in M nbsp A4 and T 3 M M a M a A displaystyle exists M in mathcal M a in M Rightarrow a in A nbsp MP with 1 A4 4 M M a M a A displaystyle M in mathcal M wedge a in M Rightarrow a in A nbsp D1 with 2 3 5 M M a M a A displaystyle M in mathcal M Rightarrow a in M Rightarrow a in A nbsp MP with abb 4 6 a M M a M a A displaystyle forall a M in mathcal M Rightarrow a in M Rightarrow a in A nbsp GEN with a displaystyle a nbsp 5 7 M M a a M a A displaystyle M in mathcal M Rightarrow forall a a in M Rightarrow a in A nbsp MP with A5 6 8 M M M a a M a A displaystyle forall M M in mathcal M Rightarrow forall a a in M Rightarrow a in A nbsp GEN with M displaystyle M nbsp 7 定理 4 A M A M P A displaystyle vdash left A bigcup mathcal M right Rightarrow left A subseteq bigcup mathcal mathcal M cup mathcal P A right nbsp 直觀上 這個定理說 集族 M displaystyle mathcal M nbsp 的聯集為 A displaystyle A nbsp 則對 A displaystyle A nbsp 的每點 a displaystyle a nbsp 都可從 M displaystyle mathcal M nbsp 裡找到一個 a displaystyle a nbsp 的鄰域 M displaystyle M nbsp 且這個鄰域不會比 A displaystyle A nbsp 大 證明注意到可以從 AND 得到 P Q P R P Q R displaystyle mathcal P Rightarrow mathcal Q mathcal P Rightarrow mathcal R mathcal P vdash mathcal Q wedge mathcal R nbsp 換句話說 從演繹元定理有 u P Q P R P Q R displaystyle mathcal P Rightarrow mathcal Q mathcal P Rightarrow mathcal R vdash mathcal P Rightarrow mathcal Q wedge mathcal R nbsp 1 a a A M M a M displaystyle forall a left a in A Leftrightarrow exists M in mathcal M a in M right nbsp Hyp 2 M M M A displaystyle forall M in mathcal M M subseteq A nbsp MP with 1 定理3 3 M M M A displaystyle M in mathcal M Rightarrow M subseteq A nbsp MP with A4 2 4 a M M M M M displaystyle a in M wedge M in mathcal M Rightarrow M in mathcal M nbsp AND 5 a M M M a M displaystyle a in M wedge M in mathcal M Rightarrow a in M nbsp AND 6 a M M M M M displaystyle a in M wedge M in mathcal M Rightarrow M in mathcal M nbsp AND 7 a M M M M A displaystyle a in M wedge M in mathcal M Rightarrow M subseteq A nbsp D1 with 3 4 8 a M M M a M M M displaystyle a in M wedge M in mathcal M Rightarrow a in M wedge M in mathcal M nbsp a with 5 6 9 a M M M a M M M M A displaystyle a in M wedge M in mathcal M Rightarrow a in M wedge M in mathcal M wedge M subseteq A nbsp a with 7 8 10 M M a M M M a M M A displaystyle exists M in mathcal M a in M Rightarrow exists M in mathcal M a in M wedge M subseteq A nbsp GENe with M displaystyle M nbsp 9 11 a A M M a M displaystyle a in A Leftrightarrow exists M in mathcal M a in M nbsp MP with A4 1 12 a A M M a M displaystyle a in A Rightarrow exists M in mathcal M a in M nbsp AND with 11 13 a A M M a M M A displaystyle a in A Rightarrow exists M in mathcal M a in M wedge M subseteq A nbsp D1 with 10 12 14 a A M M a M M A displaystyle forall a in A exists M in mathcal M a in M wedge M subseteq A nbsp GEN with a displaystyle a nbsp 13 15 S S P A S A displaystyle forall S S in mathcal P A Leftrightarrow S subseteq A nbsp 幂集公理 16 M P A M A displaystyle M in mathcal P A Leftrightarrow M subseteq A nbsp MP with A4 15 17 a A M M a M M P A displaystyle forall a in A exists M in mathcal M a in M wedge M in mathcal P A nbsp Equv with 14 16 18 A B x x A B x A x B displaystyle forall A forall B forall x left x in A cap B Leftrightarrow left x in A wedge x in B right right nbsp 有限交集 19 B x x M B x M x B displaystyle forall B forall x left x in mathcal M cap B Leftrightarrow left x in mathcal M wedge x in B right right nbsp MP with A4 18 20 x x M P A x M x P A displaystyle forall x big x in mathcal M cap mathcal P A Leftrightarrow left x in mathcal M wedge x in mathcal P A right big nbsp MP with A4 19 21 M M P A M M M P A displaystyle M in mathcal M cap mathcal P A Leftrightarrow left M in mathcal M wedge M in mathcal P A right nbsp MP with A4 20 22 a A M a M M M P A displaystyle forall a in A exists M a in M wedge M in mathcal M cap mathcal P A nbsp Equv with 17 21 23 a M P A M a M M M P A displaystyle left a in bigcup mathcal M cap mathcal P A right Leftrightarrow exists M a in M wedge M in mathcal M cap mathcal P A nbsp MP with 并集公理 A4 24 a a A a M P A displaystyle forall a left a in A Rightarrow left a in bigcup mathcal M cap mathcal P A right right nbsp Equv with 22 23 運算性質 编辑 定理 5 若 M A B M M B M A displaystyle mathcal M A left B exists M in mathcal M B M cap A right nbsp 則 M A A M displaystyle vdash bigcup mathcal M A A cap left bigcup mathcal M right nbsp 證明 1 B B M A M M B M A displaystyle forall B B in mathcal M A Leftrightarrow exists M in mathcal M B M cap A nbsp M A displaystyle mathcal M A nbsp 的定義 2 x x M B M x B displaystyle forall x left left x in bigcup mathcal M right Leftrightarrow exists B in mathcal M x in B right nbsp MP with 并集公理 A4 3 A B x x A B x A x B displaystyle forall A forall B forall x left x in A cap B Leftrightarrow left x in A wedge x in B right right nbsp 有限交集 4 x M A B B M A x B displaystyle left x in bigcup mathcal M A right Leftrightarrow exists B B in mathcal M A wedge x in B nbsp MP with A4 2 5 B M A M M B M A displaystyle B in mathcal M A Leftrightarrow exists M in mathcal M B M cap A nbsp MP with A4 1 6 x M A B x B M M B M A displaystyle left x in bigcup mathcal M A right Leftrightarrow exists B x in B wedge exists M in mathcal M B M cap A nbsp Equv with 4 5 7 x M A B M x B M M B M A displaystyle left x in bigcup mathcal M A right Leftrightarrow exists B exists M x in B wedge M in mathcal M wedge B M cap A nbsp Equv with Ce 6 8 x M A M B x B M M B M A displaystyle left x in bigcup mathcal M A right Leftrightarrow exists M exists B x in B wedge M in mathcal M wedge B M cap A nbsp Equv with 量詞可交換性 7 9 B M A x B M M B M A x M A M M M A M A displaystyle B M cap A Rightarrow x in B wedge M in mathcal M wedge B M cap A Rightarrow x in M cap A wedge M in mathcal M wedge M cap A M cap A nbsp E2 10 x B M M B M A B M A displaystyle x in B wedge M in mathcal M wedge B M cap A Rightarrow B M cap A nbsp AND 11 x B M M B M A displaystyle x in B wedge M in mathcal M wedge B M cap A nbsp x B M M B M A x M A M M M A M A displaystyle Rightarrow x in B wedge M in mathcal M wedge B M cap A Rightarrow x in M cap A wedge M in mathcal M wedge M cap A M cap A nbsp D1 with 9 10 12 x B M M B M A x B M M B M A displaystyle x in B wedge M in mathcal M wedge B M cap A Rightarrow x in B wedge M in mathcal M wedge B M cap A nbsp x B M M B M A x M A M M M A M A displaystyle Rightarrow x in B wedge M in mathcal M wedge B M cap A Rightarrow x in M cap A wedge M in mathcal M wedge M cap A M cap A nbsp MP with A2 11 13 x B M M B M A x B M M B M A displaystyle x in B wedge M in mathcal M wedge B M cap A Rightarrow x in B wedge M in mathcal M wedge B M cap A nbsp I 14 x B M M B M A x M A M M M A M A displaystyle x in B wedge M in mathcal M wedge B M cap A Rightarrow x in M cap A wedge M in mathcal M wedge M cap A M cap A nbsp MP with 12 13 15 x M A M M M A M A x M A M M displaystyle x in M cap A wedge M in mathcal M wedge M cap A M cap A Rightarrow x in M cap A wedge M in mathcal M nbsp AND 16 x B M M B M A x M A M M displaystyle x in B wedge M in mathcal M wedge B M cap A Rightarrow x in M cap A wedge M in mathcal M nbsp D1 with 14 15 17 M B x B M M B M A M x M A M M displaystyle exists M exists B x in B wedge M in mathcal M wedge B M cap A Rightarrow exists M x in M cap A wedge M in mathcal M nbsp GENe with B displaystyle B nbsp then M displaystyle M nbsp 18 M A M A displaystyle M cap A M cap A nbsp E1 注意到配合 AND 和演繹定理有P R R P displaystyle mathcal P vdash mathcal R Rightarrow mathcal R wedge mathcal P nbsp a 19 x M A M M x M A M M M A M A displaystyle x in M cap A wedge M in mathcal M Rightarrow x in M cap A wedge M in mathcal M wedge M cap A M cap A nbsp a with 18 20 B x B M M B M A x M A M M M A M A displaystyle forall B neg x in B wedge M in mathcal M wedge B M cap A Rightarrow neg x in M cap A wedge M in mathcal M wedge M cap A M cap A nbsp A4 21 x M A M M M A M A B x B M M B M A displaystyle x in M cap A wedge M in mathcal M wedge M cap A M cap A Rightarrow exists B x in B wedge M in mathcal M wedge B M cap A nbsp MP with T 20 22 x M A M M B x B M M B M A displaystyle x in M cap A wedge M in mathcal M Rightarrow exists B x in B wedge M in mathcal M wedge B M cap A nbsp D1 with 19 21 23 M x M A M M M B x B M M B M A displaystyle exists M x in M cap A wedge M in mathcal M Rightarrow exists M exists B x in B wedge M in mathcal M wedge B M cap A nbsp GENe with M displaystyle M nbsp 24 M B x B M M B M A M x M A M M displaystyle exists M exists B x in B wedge M in mathcal M wedge B M cap A Leftrightarrow exists M x in M cap A wedge M in mathcal M nbsp AND with 17 23 25 x M A M x M A M M displaystyle left x in bigcup mathcal M A right Leftrightarrow exists M x in M cap A wedge M in mathcal M nbsp Equv with 8 24 26 x M A x M x A displaystyle x in M cap A Leftrightarrow x in M wedge x in A nbsp MP with A4 3 27 x M A M x M x A M M displaystyle left x in bigcup mathcal M A right Leftrightarrow exists M x in M wedge x in A wedge M in mathcal M nbsp Equv with 25 26 28 x M A M x M M M x A displaystyle left x in bigcup mathcal M A right Leftrightarrow exists M x in M wedge M in mathcal M wedge x in A nbsp Equv with Ce 27 30 x M M M x M displaystyle left x in bigcup mathcal M right Leftrightarrow exists M in mathcal M x in M nbsp MP with A4 2 31 x M A x M x A displaystyle left x in bigcup mathcal M A right Leftrightarrow left left x in bigcup mathcal M right wedge x in A right nbsp Equv with 28 30 32 x A M x A x M displaystyle left x in A cap left bigcup mathcal M right right Leftrightarrow left x in A wedge left x in bigcup mathcal M right right nbsp MP with A4 3 33 x A M x M A displaystyle left x in A cap left bigcup mathcal M right right Leftrightarrow left x in bigcup mathcal M A right nbsp Equv with 31 32 34 x x A M x M A displaystyle forall x left left x in A cap left bigcup mathcal M right right Leftrightarrow left x in bigcup mathcal M A right right nbsp GEN with x displaystyle x nbsp 33 直觀上這個定理說 交集在 无限并集满足分配律 一般會不正式的寫為 i I A B i A i I B i displaystyle bigcup i in I left A cap B i right A cap bigcup i in I B i nbsp 定理 6 N A A displaystyle mathbb N overset A cong mathcal A nbsp 若對自然数 m N displaystyle m in mathbb N nbsp 做以下的符號定義 A m S A A 1 S m displaystyle mathcal A m left S in mathcal A A 1 S geq m right nbsp I S m N S A m displaystyle mathcal I left S bigg exists m in mathbb N left S bigcap mathcal A m right right nbsp S S m N S A m displaystyle mathcal S left S bigg exists m in mathbb N left S bigcup mathcal A m right right nbsp 那有 I S displaystyle vdash bigcup mathcal I subseteq bigcap mathcal S nbsp 這個定理一般會被不正式的寫為 i 0 j i A j i 0 j i A j displaystyle bigcup i 0 infty left bigcap j i infty A j right subseteq bigcap i 0 infty left bigcup j i infty A j right nbsp 参考 编辑朴素集合论 交集 补集 对称差 不交并 布尔逻辑参考文献 编辑 程极泰 集合论 应用数学丛书 第一版 国防工业出版社 1985 14 15034 2766 取自 https zh wikipedia org w index php title 并集 amp oldid 80025088, 维基百科,wiki,书籍,书籍,图书馆,