并集

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

有限聯集

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 ${\displaystyle A\cup B}$  ：

${\displaystyle (\forall A)(\forall B)(\forall x)\left\{(x\in A\cup B)\Leftrightarrow \left[(x\in A)\vee (x\in B)\right]\right\}}$ 

也就是直觀上：

「對所有 ${\displaystyle x}$  ， ${\displaystyle x\in A\cup B}$  等價於 ${\displaystyle x\in A}$  或 ${\displaystyle x\in B}$ 」

举例：

集合${\displaystyle \{1,2,3\}}$ 和${\displaystyle \{2,3,4\}}$ 的并集是${\displaystyle \{1,2,3,4\}}$ 。数${\displaystyle 9}$ 不属于素数集合${\displaystyle \{2,3,5,7,11,\ldots \}}$ 和偶数集合${\displaystyle \{2,4,6,8,10,\ldots \}}$ 的并集，因为${\displaystyle 9}$ 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，${\displaystyle A,B}$ 和${\displaystyle C}$ 的并集含有所有${\displaystyle A}$ 的元素，所有${\displaystyle B}$ 的元素和所有${\displaystyle C}$ 的元素，而没有其他元素。形式上：

${\displaystyle x}$ 是${\displaystyle A\cup B\cup C}$ 的元素，当且仅当${\displaystyle x}$ 属于${\displaystyle A}$ 或${\displaystyle x}$ 属于${\displaystyle B}$ 或${\displaystyle x}$ 属于${\displaystyle C}$ 。

代数性质

二元并集（两个集合的并集）是一种结合运算，即

${\displaystyle A\cup (B\cup C)=(A\cup B)\cup C}$ 。事实上，${\displaystyle A\cup B\cup C}$ 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${\displaystyle \varnothing \cup A=A}$ ，对任意集合${\displaystyle A}$ 。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

由公理化集合论的并集公理，有唯一的集合 ${\displaystyle \bigcup {\mathcal {M}}}$  滿足：

${\displaystyle (\forall {\mathcal {M}})(\forall x)\left\{\left(x\in \bigcup {\mathcal {M}}\right)\Leftrightarrow (\exists A)\left[\left(A\in {\mathcal {M}}\right)\wedge (x\in A)\right]\right\}}$ 

也就是直觀上「對所有 ${\displaystyle {\mathcal {M}}}$  和所有 ${\displaystyle x}$  ， ${\displaystyle x\in \bigcup {\mathcal {M}}}$  等價於有某個 ${\displaystyle {\mathcal {M}}}$  的下屬集合 ${\displaystyle A}$  ，使得${\displaystyle x\in A}$ 」。以上的 ${\displaystyle {\mathcal {M}}}$  可以直觀的視為一個集合族，而把 ${\displaystyle \bigcup {\mathcal {M}}}$  看成對 ${\displaystyle {\mathcal {M}}}$  內的集合取并集，但這個公理並沒有對 ${\displaystyle {\mathcal {M}}}$  下屬集合的數量做出任何限制，所以這個 ${\displaystyle \bigcup {\mathcal {M}}}$  被俗稱為任意并集或无限并集。

若 ${\displaystyle X\subseteq \bigcup {\mathcal {M}}}$  ，會稱 ${\displaystyle X}$  被 ${\displaystyle {\mathcal {M}}}$  覆蓋（cover），也就是直觀上可以用 ${\displaystyle {\mathcal {M}}}$  裡的所有集合疊起來蓋住 ${\displaystyle X}$ 。

例如：

對 ${\displaystyle {\mathcal {M}}=\{A,B,C\}}$ ，${\displaystyle \bigcup {\mathcal {M}}=A\cup B\cup C}$  ，若 ${\displaystyle M}$ 是空集， ${\displaystyle \bigcup {\mathcal {M}}}$  也是空集。

无限并集有多种表示方法：

可模仿求和符号記為

${\displaystyle \bigcup _{A\in {\mathcal {M}}}A}$ 。

但大多數人會假設指标集 ${\displaystyle I}$  的存在，換句話說

若 ${\displaystyle I\,{\overset {A}{\cong }}\,{\mathcal {M}}}$  則 ${\displaystyle \bigcup _{i\in I}A(i):=\bigcup {\mathcal {M}}}$ 

在指标集 ${\displaystyle I}$  是自然数系 ${\displaystyle \mathbb {N} }$  的情况下，更可以仿无穷级数來表示，也就是說：

若 ${\displaystyle \mathbb {N} \,{\overset {A}{\cong }}\,{\mathcal {M}}}$  則 ${\displaystyle \bigcup _{i=0}^{\infty }A(i):=\bigcup {\mathcal {M}}}$ 

也可以更粗略直觀的將 ${\displaystyle \bigcup _{i=0}^{\infty }A(i)}$  写作${\displaystyle A_{0}\cup A_{1}\cup A_{2}\cup \ldots }$ 。

无限并集的性質

定理(0) — 
${\displaystyle \vdash \bigcup \varnothing =\varnothing }$ 

證明

(1) ${\displaystyle (\forall x)[\neg (x\in \varnothing )]}$  (空集公理) (2) ${\displaystyle \neg (S\in \varnothing )}$ (MP with A4, 1) (3)${\displaystyle (S\in \varnothing )\Rightarrow [\neg (x\in S)]}$ (M0 with 2) (4)${\displaystyle \neg \neg (S\in \varnothing )\Rightarrow [\neg (x\in S)]}$ (Equv with DN, 3) (5)${\displaystyle \neg \{[\neg (S\in \varnothing )]\wedge [\neg (x\in S)]\}}$ (Equv with De Morgan, 4) (6)${\displaystyle (\forall S){\big \{}\neg \{[\neg (S\in \varnothing )]\wedge [\neg (x\in S)]\}{\big \}}}$ (GEN with ${\displaystyle S}$  , 5) (7)${\displaystyle \neg (\exists S)\{[\neg (S\in \varnothing )]\wedge [\neg (x\in S)]\}}$ (Equv with DN, 6) (8)${\displaystyle (\forall x)\left\{\left(x\in \bigcup \varnothing \right)\Leftrightarrow (\exists S)\{(S\in \varnothing )\wedge (x\in S)\}\right\}}$ (MP with 并集公理, A4) (9)${\displaystyle \left(x\in \bigcup \varnothing \right)\Leftrightarrow (\exists S)\{(S\in \varnothing )\wedge (x\in S)\}}$ (MP with A4, 8) (10)${\displaystyle \left(x\in \bigcup \varnothing \right)\Rightarrow (\exists S)\{(S\in \varnothing )\wedge (x\in S)\}}$ (MP with AND ,9) (11)${\displaystyle \neg (\exists S)\{(S\in \varnothing )\wedge (x\in S)\}\Rightarrow \neg \left(x\in \bigcup \varnothing \right)}$ (MP with T, 10) (12)${\displaystyle \neg \left(x\in \bigcup \varnothing \right)}$ (MP with 7, 11) (13)${\displaystyle (\forall x)\left(x\not \in \bigcup \varnothing \right)}$ (GEN with ${\displaystyle x}$  , 12) (14)${\displaystyle (y=\varnothing )\Leftrightarrow (\forall x)[\neg (x\in y)]}$  (E) (15)${\displaystyle (\forall y)\{(y=\varnothing )\Leftrightarrow (\forall x)[\neg (x\in y)]\}}$  (GEN with ${\displaystyle y}$  , 14) (16)${\displaystyle \left(\bigcup \varnothing =\varnothing \right)\Leftrightarrow (\forall x)\left[\neg \left(x\in \bigcup \varnothing \right)\right]}$ (MP with A4, 15) (17) ${\displaystyle \bigcup \varnothing =\varnothing }$  (Equv with 13, 16)

比較性質

定理(1) — 
${\displaystyle ({\mathcal {M}}\subseteq {\mathcal {N}})\vdash \left(\bigcup {\mathcal {M}}\subseteq \bigcup {\mathcal {N}}\right)}$ 

證明

注意到可以從(AND)得到 ${\displaystyle ({\mathcal {P}}\Rightarrow {\mathcal {Q}}),\,({\mathcal {P}}\Rightarrow {\mathcal {R}}),\,{\mathcal {P}}\vdash {\mathcal {Q}}\wedge {\mathcal {R}}}$  換句話說，從演繹元定理有 (u) ${\displaystyle ({\mathcal {P}}\Rightarrow {\mathcal {Q}}),\,({\mathcal {P}}\Rightarrow {\mathcal {R}})\vdash {\mathcal {P}}\Rightarrow ({\mathcal {Q}}\wedge {\mathcal {R}})}$  (1) ${\displaystyle (\forall A)\left[(A\in {\mathcal {M}})\Rightarrow (A\in {\mathcal {N}})\right]}$  (Hyp) (2) ${\displaystyle (A\in {\mathcal {M}})\Rightarrow (A\in {\mathcal {N}})}$ (MP with 1, A4) (3) ${\displaystyle [(a\in A)\wedge (A\in {\mathcal {M}})]\Rightarrow (A\in {\mathcal {M}})}$ (AND) (4)${\displaystyle [(a\in A)\wedge (A\in {\mathcal {M}})]\Rightarrow (a\in A)}$ (AND) (5)${\displaystyle [(a\in A)\wedge (A\in {\mathcal {M}})]\Rightarrow (A\in {\mathcal {N}})}$ (D1 with 2, 3) (6)${\displaystyle [(a\in A)\wedge (A\in {\mathcal {M}})]\Rightarrow [(a\in A)\wedge (A\in {\mathcal {N}})]}$ (u with 4, 5) (7)${\displaystyle (\exists A\in {\mathcal {M}})(a\in A)\Rightarrow (\exists A\in {\mathcal {N}})(a\in A)}$ (GENe with ${\displaystyle A}$ , 6) (8) ${\displaystyle (\forall x)\left\{\left(x\in \bigcup {\mathcal {M}}\right)\Leftrightarrow (\exists A\in {\mathcal {M}})(x\in A)\right\}}$ (MP with 并集公理, A4) (9) ${\displaystyle (\forall x)\left\{\left(x\in \bigcup {\mathcal {N}}\right)\Leftrightarrow (\exists A\in {\mathcal {N}})(x\in A)\right\}}$ (MP with 并集公理, A4) (10) ${\displaystyle \left(x\in \bigcup {\mathcal {M}}\right)\Leftrightarrow (\exists A\in {\mathcal {M}})(x\in A)}$  (MP with 8, A4) (11) ${\displaystyle \left(x\in \bigcup {\mathcal {N}}\right)\Leftrightarrow (\exists A\in {\mathcal {N}})(x\in A)}$  (MP with 9, A4) (12) ${\displaystyle \left(x\in \bigcup {\mathcal {M}}\right)\Rightarrow (\exists A\in {\mathcal {N}})(x\in A)}$ (D1 with 7, 10) (13) ${\displaystyle \left(x\in \bigcup {\mathcal {M}}\right)\Rightarrow \left(x\in \bigcup {\mathcal {N}}\right)}$ (D1 with 11, 12) (14) ${\displaystyle (\forall x)\left[\left(x\in \bigcup {\mathcal {M}}\right)\Rightarrow \left(x\in \bigcup {\mathcal {N}}\right)\right]}$ (GEN with ${\displaystyle a}$  , 13)

覆蓋性質

定理(2) — 
${\displaystyle \vdash A=\bigcup {\mathcal {P}}(A)}$ 

「${\displaystyle A}$  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 ${\displaystyle {\mathcal {P}}(A)}$  是以 ${\displaystyle A}$  和 ${\displaystyle A}$  的子集為元素，所以 ${\displaystyle {\mathcal {P}}(A)}$  的聯集理當是 ${\displaystyle A}$  。

證明

注意到可以從(AND)得到 ${\displaystyle ({\mathcal {P}}\Rightarrow {\mathcal {Q}}),\,({\mathcal {P}}\Rightarrow {\mathcal {R}}),\,{\mathcal {P}}\vdash {\mathcal {Q}}\wedge {\mathcal {R}}}$  換句話說，從演繹元定理有 (u) ${\displaystyle ({\mathcal {P}}\Rightarrow {\mathcal {Q}}),\,({\mathcal {P}}\Rightarrow {\mathcal {R}})\vdash {\mathcal {P}}\Rightarrow ({\mathcal {Q}}\wedge {\mathcal {R}})}$  (1)${\displaystyle (\forall x)\left\{\left[x\in \bigcup {\mathcal {P}}(A)\right]\Leftrightarrow (\exists S)\{[S\in {\mathcal {P}}(A)]\wedge (x\in S)\}\right\}}$ (MP with 并集公理, A4) (2) ${\displaystyle (\forall S)\{[S\in {\mathcal {P}}(A)]\Leftrightarrow (S\subseteq A)\}}$ (幂集公理) (3) ${\displaystyle [S\in {\mathcal {P}}(A)]\Leftrightarrow (S\subseteq A)}$ (MP with A4 ,2) (4) ${\displaystyle (\forall x)\left\{\left[x\in \bigcup {\mathcal {P}}(A)\right]\Leftrightarrow (\exists S)[(S\subseteq A)\wedge (x\in S)]\right\}}$  (Equv with 1, 3) (5) ${\displaystyle [(S\subseteq A)\wedge (x\in S)]\Rightarrow (S\subseteq A)}$ (AND) (6) ${\displaystyle (\forall x)[(x\in S)\Rightarrow (x\in A)]\Rightarrow [(x\in S)\Rightarrow (x\in A)]}$ (A4) (7) ${\displaystyle [(S\subseteq A)\wedge (x\in S)]\Rightarrow [(x\in S)\Rightarrow (x\in A)]}$ (D1 with 5, 6) (8) ${\displaystyle [(S\subseteq A)\wedge (x\in S)]\Rightarrow (x\in S)}$ (AND) (9) ${\displaystyle [(S\subseteq A)\wedge (x\in S)]\Rightarrow \{(x\in S)\wedge [(x\in S)\Rightarrow (x\in A)]\}}$ (u with 7, 8) 注意到 ${\displaystyle (x\in S),\,[(x\in S)\Rightarrow (x\in A)]\vdash (x\in A)}$  再對上式套用(AND)就有 ${\displaystyle \{(x\in S)\wedge [(x\in S)\Rightarrow (x\in A)]\}\vdash (x\in A)}$ (a) (10') ${\displaystyle [(S\subseteq A)\wedge (x\in S)]\Rightarrow (x\in A)}$ (D1 with a, 9) (11') ${\displaystyle (\exists S)[(S\subseteq A)\wedge (x\in S)]\Rightarrow (x\in A)}$ (GENe with ${\displaystyle S}$ , 10') (12') ${\displaystyle (\forall S)\{\neg [(S\subseteq A)\wedge (x\in S)]\}\Rightarrow \{\neg [(A\subseteq A)\wedge (x\in A)]\}}$  (A4) (13') ${\displaystyle [(A\subseteq A)\wedge (x\in A)]\Rightarrow (\exists S)[(S\subseteq A)\wedge (x\in S)]}$  (MP with T, 12') (14') ${\displaystyle (x\in A)\Rightarrow (x\in A)}$  (I) (15') ${\displaystyle A\subseteq A}$  (GEN with ${\displaystyle x}$  , 14') 注意到(AND)依據演繹定理可改寫為 ${\displaystyle (A\subseteq A)\vdash (x\in A)\Rightarrow [(A\subseteq A)\wedge (x\in A)]}$ (b) (16'') ${\displaystyle (x\in A)\Rightarrow [(A\subseteq A)\wedge (x\in A)]}$  (b with 15') (17'') ${\displaystyle (x\in A)\Rightarrow (\exists S)[(S\subseteq A)\wedge (x\in S)]}$  (D1 with 13', 16'') (18'') ${\displaystyle (x\in A)\Leftrightarrow (\exists S)[(S\subseteq A)\wedge (x\in S)]}$  (AND with 11', 17'') (19'') ${\displaystyle (\forall x)\left\{\left[x\in \bigcup {\mathcal {P}}(A)\right]\Leftrightarrow (x\in A)\right\}}$ (Equv with 4, 18''')

定理(3) — 
${\displaystyle \left(\bigcup {\mathcal {M}}\subseteq A\right)\vdash (\forall M\in {\mathcal {M}})(M\subseteq A)}$ 

直觀上，這個定理說「一群集合的聯集包含於 ${\displaystyle A}$  ，則它們個個都包含於 ${\displaystyle A}$  」

證明

(1) ${\displaystyle (\forall a)\left[(\exists M\in {\mathcal {M}})(a\in M)\Rightarrow (a\in A)\right]}$  (Hyp) (2) ${\displaystyle [(M\in {\mathcal {M}})\wedge (a\in M)]\Rightarrow (\exists M\in {\mathcal {M}})(a\in M)}$  (A4 and T) (3) ${\displaystyle (\exists M\in {\mathcal {M}})(a\in M)\Rightarrow (a\in A)}$  (MP with 1, A4) (4) ${\displaystyle [(M\in {\mathcal {M}})\wedge (a\in M)]\Rightarrow (a\in A)}$  (D1 with 2, 3) (5) ${\displaystyle (M\in {\mathcal {M}})\Rightarrow [(a\in M)\Rightarrow (a\in A)]}$  (MP with abb, 4) (6) ${\displaystyle (\forall a)\{(M\in {\mathcal {M}})\Rightarrow [(a\in M)\Rightarrow (a\in A)]\}}$  (GEN with ${\displaystyle a}$  , 5) (7) ${\displaystyle (M\in {\mathcal {M}})\Rightarrow (\forall a)[(a\in M)\Rightarrow (a\in A)]}$  (MP with A5 , 6) (8) ${\displaystyle (\forall M)\{(M\in {\mathcal {M}})\Rightarrow (\forall a)[(a\in M)\Rightarrow (a\in A)]\}}$  (GEN with ${\displaystyle M}$  , 7)

定理(4) — 
${\displaystyle \vdash \left(A=\bigcup {\mathcal {M}}\right)\Rightarrow \left\{A\subseteq \bigcup {\mathcal {[}}{\mathcal {M}}\cup {\mathcal {P}}(A)]\right\}}$ 

直觀上，這個定理說「集族 ${\displaystyle {\mathcal {M}}}$  的聯集為 ${\displaystyle A}$  ，則對 ${\displaystyle A}$  的每點 ${\displaystyle a}$  ，都可從 ${\displaystyle {\mathcal {M}}}$  裡找到一個 ${\displaystyle a}$  的鄰域 ${\displaystyle M}$  ，且這個鄰域不會比 ${\displaystyle A}$  大 」

證明

注意到可以從(AND)得到 ${\displaystyle ({\mathcal {P}}\Rightarrow {\mathcal {Q}}),\,({\mathcal {P}}\Rightarrow {\mathcal {R}}),\,{\mathcal {P}}\vdash {\mathcal {Q}}\wedge {\mathcal {R}}}$  換句話說，從演繹元定理有 (u) ${\displaystyle ({\mathcal {P}}\Rightarrow {\mathcal {Q}}),\,({\mathcal {P}}\Rightarrow {\mathcal {R}})\vdash {\mathcal {P}}\Rightarrow ({\mathcal {Q}}\wedge {\mathcal {R}})}$  (1) ${\displaystyle (\forall a)\left[(a\in A)\Leftrightarrow (\exists M\in {\mathcal {M}})(a\in M)\right]}$  (Hyp) (2) ${\displaystyle (\forall M\in {\mathcal {M}})(M\subseteq A)}$ (MP with 1, 定理3) (3) ${\displaystyle (M\in {\mathcal {M}})\Rightarrow (M\subseteq A)}$ (MP with A4, 2) (4) ${\displaystyle [(a\in M)\wedge (M\in {\mathcal {M}})]\Rightarrow (M\in {\mathcal {M}})}$ (AND) (5) ${\displaystyle [(a\in M)\wedge (M\in {\mathcal {M}})]\Rightarrow (a\in M)}$ (AND) (6) ${\displaystyle [(a\in M)\wedge (M\in {\mathcal {M}})]\Rightarrow (M\in {\mathcal {M}})}$ (AND) (7) ${\displaystyle [(a\in M)\wedge (M\in {\mathcal {M}})]\Rightarrow (M\subseteq A)}$  (D1 with 3, 4) (8) ${\displaystyle [(a\in M)\wedge (M\in {\mathcal {M}})]\Rightarrow [(a\in M)\wedge (M\in {\mathcal {M}})]}$ (a with 5, 6) (9) ${\displaystyle [(a\in M)\wedge (M\in {\mathcal {M}})]\Rightarrow [(a\in M)\wedge (M\in {\mathcal {M}})\wedge (M\subseteq A)]}$ (a with 7, 8) (10) ${\displaystyle (\exists M\in {\mathcal {M}})(a\in M)\Rightarrow (\exists M\in {\mathcal {M}})[(a\in M)\wedge (M\subseteq A)]}$ (GENe with ${\displaystyle M}$ , 9) (11) ${\displaystyle (a\in A)\Leftrightarrow (\exists M\in {\mathcal {M}})(a\in M)}$ (MP with A4, 1) (12) ${\displaystyle (a\in A)\Rightarrow (\exists M\in {\mathcal {M}})(a\in M)}$ (AND with 11) (13) ${\displaystyle (a\in A)\Rightarrow (\exists M\in {\mathcal {M}})[(a\in M)\wedge (M\subseteq A)]}$ (D1 with 10, 12) (14) ${\displaystyle (\forall a\in A)(\exists M\in {\mathcal {M}})[(a\in M)\wedge (M\subseteq A)]}$ (GEN with ${\displaystyle a}$ , 13) (15)${\displaystyle (\forall S)\{[S\in {\mathcal {P}}(A)]\Leftrightarrow (S\subseteq A)\}}$ (幂集公理) (16)${\displaystyle [M\in {\mathcal {P}}(A)]\Leftrightarrow (M\subseteq A)}$ (MP with A4, 15) (17)${\displaystyle (\forall a\in A)(\exists M\in {\mathcal {M}})\{(a\in M)\wedge [M\in {\mathcal {P}}(A)]\}}$ (Equv with 14, 16) (18) ${\displaystyle (\forall A)(\forall B)(\forall x)\left\{(x\in A\cap B)\Leftrightarrow \left[(x\in A)\wedge (x\in B)\right]\right\}}$ (有限交集) (19)${\displaystyle (\forall B)(\forall x)\left\{(x\in {\mathcal {M}}\cap B)\Leftrightarrow \left[(x\in {\mathcal {M}})\wedge (x\in B)\right]\right\}}$ (MP with A4, 18) (20)${\displaystyle (\forall x){\big \{}[x\in {\mathcal {M}}\cap {\mathcal {P}}(A)]\Leftrightarrow \left\{(x\in {\mathcal {M}})\wedge [x\in {\mathcal {P}}(A)]\right\}{\big \}}}$ (MP with A4, 19) (21)${\displaystyle [M\in {\mathcal {M}}\cap {\mathcal {P}}(A)]\Leftrightarrow \left\{(M\in {\mathcal {M}})\wedge [M\in {\mathcal {P}}(A)]\right\}}$ (MP with A4, 20) (22)${\displaystyle (\forall a\in A)(\exists M)\{(a\in M)\wedge [M\in {\mathcal {M}}\cap {\mathcal {P}}(A)]\}}$ (Equv with 17, 21) (23)${\displaystyle \left\{a\in \bigcup [{\mathcal {M}}\cap {\mathcal {P}}(A)]\right\}\Leftrightarrow (\exists M)\{(a\in M)\wedge [M\in {\mathcal {M}}\cap {\mathcal {P}}(A)]\}}$ (MP with 并集公理, A4) (24)${\displaystyle (\forall a)\left\{(a\in A)\Rightarrow \left\{a\in \bigcup [{\mathcal {M}}\cap {\mathcal {P}}(A)]\right\}\right\}}$ (Equv with 22, 23)

運算性質

定理(5) — 
若

${\displaystyle {\mathcal {M}}_{A}:=\left\{B\,|\,(\exists M\in {\mathcal {M}})(B=M\cap A)\right\}}$ 

則

${\displaystyle \vdash \bigcup {\mathcal {M}}_{A}=A\cap \left(\bigcup {\mathcal {M}}\right)}$ 

證明

(1)${\displaystyle (\forall B)[(B\in {\mathcal {M}}_{A})\Leftrightarrow (\exists M\in {\mathcal {M}})(B=M\cap A)]}$  (${\displaystyle {\mathcal {M}}_{A}}$ 的定義) (2) ${\displaystyle (\forall x)\left\{\left(x\in \bigcup {\mathcal {M}}\right)\Leftrightarrow (\exists B\in {\mathcal {M}})(x\in B)\right\}}$ (MP with 并集公理, A4) (3) ${\displaystyle (\forall A)(\forall B)(\forall x)\left\{(x\in A\cap B)\Leftrightarrow \left[(x\in A)\wedge (x\in B)\right]\right\}}$ (有限交集) (4)${\displaystyle \left(x\in \bigcup {\mathcal {M}}_{A}\right)\Leftrightarrow (\exists B)[(B\in {\mathcal {M}}_{A})\wedge (x\in B)]}$ (MP with A4, 2) (5) ${\displaystyle (B\in {\mathcal {M}}_{A})\Leftrightarrow (\exists M\in {\mathcal {M}})(B=M\cap A)}$ (MP with A4, 1) (6) ${\displaystyle \left(x\in \bigcup {\mathcal {M}}_{A}\right)\Leftrightarrow (\exists B)[(x\in B)\wedge (\exists M\in {\mathcal {M}})(B=M\cap A)]}$ (Equv with 4, 5) (7)${\displaystyle \left(x\in \bigcup {\mathcal {M}}_{A}\right)\Leftrightarrow (\exists B)(\exists M)[(x\in B)\wedge (M\in {\mathcal {M}})\wedge (B=M\cap A)]}$ (Equv with Ce, 6) (8)${\displaystyle \left(x\in \bigcup {\mathcal {M}}_{A}\right)\Leftrightarrow (\exists M)(\exists B)[(x\in B)\wedge (M\in {\mathcal {M}})\wedge (B=M\cap A)]}$ (Equv with 量詞可交換性 ,7) (9) ${\displaystyle (B=M\cap A)\Rightarrow \{[(x\in B)\wedge (M\in {\mathcal {M}})\wedge (B=M\cap A)]\Rightarrow [(x\in M\cap A)\wedge (M\in {\mathcal {M}})\wedge (M\cap A=M\cap A)]\}}$ (E2) (10)${\displaystyle [(x\in B)\wedge (M\in {\mathcal {M}})\wedge (B=M\cap A)]\Rightarrow (B=M\cap A)}$ (AND) (11)${\displaystyle [(x\in B)\wedge (M\in {\mathcal {M}})\wedge (B=M\cap A)]}$ ${\displaystyle \Rightarrow \{[(x\in B)\wedge (M\in {\mathcal {M}})\wedge (B=M\cap A)]\Rightarrow [(x\in M\cap A)\wedge (M\in {\mathcal {M}})\wedge (M\cap A=M\cap A)]\}}$ (D1 with 9,10) (12)${\displaystyle \{[(x\in B)\wedge (M\in {\mathcal {M}})\wedge (B=M\cap A)]\Rightarrow [(x\in B)\wedge (M\in {\mathcal {M}})\wedge (B=M\cap A)]\}}$  ${\displaystyle \Rightarrow \{[(x\in B)\wedge (M\in {\mathcal {M}})\wedge (B=M\cap A)]\Rightarrow [(x\in M\cap A)\wedge (M\in {\mathcal {M}})\wedge (M\cap A=M\cap A)]\}}$ (MP with A2, 11) (13)${\displaystyle [(x\in B)\wedge (M\in {\mathcal {M}})\wedge (B=M\cap A)]\Rightarrow [(x\in B)\wedge (M\in {\mathcal {M}})\wedge (B=M\cap A)]}$ (I) (14)${\displaystyle [(x\in B)\wedge (M\in {\mathcal {M}})\wedge (B=M\cap A)]\Rightarrow [(x\in M\cap A)\wedge (M\in {\mathcal {M}})\wedge (M\cap A=M\cap A)]}$ (MP with 12, 13) (15)${\displaystyle [(x\in M\cap A)\wedge (M\in {\mathcal {M}})\wedge (M\cap A=M\cap A)]\Rightarrow [(x\in M\cap A)\wedge (M\in {\mathcal {M}})]}$ (AND) (16)${\displaystyle [(x\in B)\wedge (M\in {\mathcal {M}})\wedge (B=M\cap A)]\Rightarrow [(x\in M\cap A)\wedge (M\in {\mathcal {M}})]}$ (D1 with 14,15) (17)${\displaystyle (\exists M)(\exists B)[(x\in B)\wedge (M\in {\mathcal {M}})\wedge (B=M\cap A)]\Rightarrow (\exists M)[(x\in M\cap A)\wedge (M\in {\mathcal {M}})]}$ (GENe with ${\displaystyle B}$  then ${\displaystyle M}$ ) (18)${\displaystyle M\cap A=M\cap A}$  (E1) 注意到配合(AND)和演繹定理有 ${\displaystyle {\mathcal {P}}\vdash {\mathcal {R}}\Rightarrow ({\mathcal {R}}\wedge {\mathcal {P}})}$ (a) (19)${\displaystyle [(x\in M\cap A)\wedge (M\in {\mathcal {M}})]\Rightarrow [(x\in M\cap A)\wedge (M\in {\mathcal {M}})\wedge (M\cap A=M\cap A)]}$ (a with 18) (20)${\displaystyle (\forall B)\{\neg [(x\in B)\wedge (M\in {\mathcal {M}})\wedge (B=M\cap A)]\}\Rightarrow \{\neg [(x\in M\cap A)\wedge (M\in {\mathcal {M}})\wedge (M\cap A=M\cap A)]\}}$ (A4) (21)${\displaystyle [(x\in M\cap A)\wedge (M\in {\mathcal {M}})\wedge (M\cap A=M\cap A)]\Rightarrow (\exists B)[(x\in B)\wedge (M\in {\mathcal {M}})\wedge (B=M\cap A)]}$ (MP with T, 20) (22)${\displaystyle [(x\in M\cap A)\wedge (M\in {\mathcal {M}})]\Rightarrow (\exists B)[(x\in B)\wedge (M\in {\mathcal {M}})\wedge (B=M\cap A)]}$ (D1 with 19, 21) (23)${\displaystyle (\exists M)[(x\in M\cap A)\wedge (M\in {\mathcal {M}})]\Rightarrow (\exists M)(\exists B)[(x\in B)\wedge (M\in {\mathcal {M}})\wedge (B=M\cap A)]}$ (GENe with ${\displaystyle M}$ ) (24)${\displaystyle (\exists M)(\exists B)[(x\in B)\wedge (M\in {\mathcal {M}})\wedge (B=M\cap A)]\Leftrightarrow (\exists M)[(x\in M\cap A)\wedge (M\in {\mathcal {M}})]}$ (AND with 17, 23) (25)${\displaystyle \left(x\in \bigcup {\mathcal {M}}_{A}\right)\Leftrightarrow (\exists M)[(x\in M\cap A)\wedge (M\in {\mathcal {M}})]}$ (Equv with 8, 24) (26) ${\displaystyle (x\in M\cap A)\Leftrightarrow [(x\in M)\wedge (x\in A)]}$ (MP with A4, 3) (27)${\displaystyle \left(x\in \bigcup {\mathcal {M}}_{A}\right)\Leftrightarrow (\exists M)[(x\in M)\wedge (x\in A)\wedge (M\in {\mathcal {M}})]}$ (Equv with 25, 26) (28)${\displaystyle \left(x\in \bigcup {\mathcal {M}}_{A}\right)\Leftrightarrow \{(\exists M)[(x\in M)\wedge (M\in {\mathcal {M}})]\wedge (x\in A)\}}$ (Equv with Ce, 27) (30) ${\displaystyle \left(x\in \bigcup {\mathcal {M}}\right)\Leftrightarrow (\exists M\in {\mathcal {M}})(x\in M)}$ (MP with A4, 2) (31)${\displaystyle \left(x\in \bigcup {\mathcal {M}}_{A}\right)\Leftrightarrow \left[\left(x\in \bigcup {\mathcal {M}}\right)\wedge (x\in A)\right]}$ (Equv with 28, 30) (32)${\displaystyle \left[x\in A\cap \left(\bigcup {\mathcal {M}}\right)\right]\Leftrightarrow \left[(x\in A)\wedge \left(x\in \bigcup {\mathcal {M}}\right)\right]}$ (MP with A4, 3) (33)${\displaystyle \left[x\in A\cap \left(\bigcup {\mathcal {M}}\right)\right]\Leftrightarrow \left(x\in \bigcup {\mathcal {M}}_{A}\right)}$ (Equv with 31, 32) (34)${\displaystyle (\forall x)\left\{\left[x\in A\cap \left(\bigcup {\mathcal {M}}\right)\right]\Leftrightarrow \left(x\in \bigcup {\mathcal {M}}_{A}\right)\right\}}$ (GEN with ${\displaystyle x}$ , 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

${\displaystyle \bigcup _{i\in I}\left(A\cap B_{i}\right)=A\cap \bigcup _{i\in I}B_{i}}$ 

定理(6) — 
${\displaystyle \mathbb {N} \,{\overset {A}{\cong }}\,{\mathcal {A}}}$ ，若對自然数 ${\displaystyle m\in \mathbb {N} }$  做以下的符號定義：

${\displaystyle {\mathcal {A}}_{m}:=\left\{S\in {\mathcal {A}}\,|\,A^{-1}(S)\geq m\right\}}$ 

${\displaystyle {\mathcal {I}}:=\left\{S\,{\bigg |}\,(\exists m\in \mathbb {N} )\left(S=\bigcap {\mathcal {A}}_{m}\right)\right\}}$ 

${\displaystyle {\mathcal {S}}:=\left\{S\,{\bigg |}\,(\exists m\in \mathbb {N} )\left(S=\bigcup {\mathcal {A}}_{m}\right)\right\}}$ 

那有

${\displaystyle \vdash \bigcup {\mathcal {I}}\subseteq \bigcap {\mathcal {S}}}$ 。

這個定理一般會被不正式的寫為

${\displaystyle \bigcup _{i=0}^{\infty }\left(\bigcap _{j=i}^{\infty }A_{j}\right)\subseteq \bigcap _{i=0}^{\infty }\left(\bigcup _{j=i}^{\infty }A_{j}\right)}$ 。

参考

• 朴素集合论
• 交集
• 补集
• 对称差
• 不交并
• 布尔逻辑

参考文献

1. 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.