并集

File:Set union.png

A和B的并集

在集合论和数学的其他分支中，一群集合的并集（Union），是以这群集合的所有元素来构成的集合。

有限并集

并集是由公理化集合论的分类公理来确保其唯一存在的特定集合 <math>A \cup B</math> ：

<math>(\forall A)(\forall B)(\forall x)\left\{

   (x \in A \cup B)
   \Leftrightarrow
   \left[
       (x \in A)
       \vee
       (x \in B)
   \right]

\right\} </math>

也就是直观上：

“对所有 <math>x</math> ， <math>x \in A \cup B</math> 等价于 <math>x \in A</math> 或 <math>x \in B</math>”

举例：

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

<math>x</math>是<math>A \cup B \cup C</math>的元素，当且仅当<math>x</math>属于<math>A</math>或<math>x</math>属于<math>B</math>或<math>x</math>属于<math>C</math>。

代数性质

二元并集（两个集合的并集）是一种结合运算，即

<math>A \cup (B \cup C) =(A \cup B)\cup C</math>。事实上，<math>A \cup B \cup C</math>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<math>\varnothing \cup A = A</math>，对任意集合<math>A</math>。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第摩根定律。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 <math>\bigcup\mathcal{M}</math> 满足：

<math>(\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\}</math>

也就是直观上“对所有 <math>\mathcal{M}</math> 和所有 <math>x</math> ， <math>x \in \bigcup\mathcal{M}</math> 等价于有某个 <math>\mathcal{M}</math> 的下属集合 <math>A</math> ，使得<math>x \in A</math>”。以上的 <math>\mathcal{M}</math> 可以直观的视为一个集合族，而把 <math>\bigcup\mathcal{M}</math> 看成对 <math>\mathcal{M}</math> 内的集合取并集，例如：

取 <math>\mathcal{M} = \{A,B,C\}</math> 则 <math>\bigcup\mathcal{M} = A \cup B \cup C</math> 。

但这个公理并没有对 <math>\mathcal{M}</math> 下属集合的数量做出任何限制，所以这个 <math>\bigcup\mathcal{M}</math> 被俗称为任意并集或无限并集。

若 <math>X \subseteq \bigcup\mathcal{M}</math> ，会称 <math>X</math> 被 <math>\mathcal{M}</math> 覆盖（cover），也就是直观上可以用 <math>\mathcal{M}</math> 里的所有集合叠起来盖住 <math>X</math>。

无限并集有多种表示方法：

可模仿求和符号记为

<math>\bigcup_{A\in \mathcal{M}} A</math>。

但大多数人会假设指标集 <math>I</math> 的存在，换句话说

若 <math>I \,\overset{A}{\cong}\, \mathcal{M} </math> 则 <math>\bigcup_{i\in I} A(i) := \bigcup \mathcal{M}</math>

在指标集 <math>I</math> 是自然数系 <math>\N</math> 的情况下，更可以仿无穷级数来表示，也就是说：

若 <math>\N \,\overset{A}{\cong}\, \mathcal{M}</math> 则 <math>\bigcup^{\infty}_{i = 0} A(i) := \bigcup \mathcal{M}</math>

也可以更粗略直观的将 <math>\bigcup^{\infty}_{i = 0} A(i)</math> 写作<math>A_{0} \cup A_{1} \cup A_{2} \cup \ldots</math>。

无限并集的性质

定理(0) — 
<math> \vdash \bigcup\varnothing = \varnothing </math>

证明

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

比较性质

定理(1) — 
<math> (\mathcal{M} \subseteq \mathcal{N}) \vdash \left(\bigcup\mathcal{M} \subseteq \bigcup\mathcal{N}\right)</math>

证明

注意到可以从(AND)得到 <math> (\mathcal{P} \Rightarrow \mathcal{Q}),\,(\mathcal{P} \Rightarrow \mathcal{R}),\,\mathcal{P} \vdash \mathcal{Q} \wedge \mathcal{R} </math> 换句话说，从演绎元定理有 (u) <math> (\mathcal{P} \Rightarrow \mathcal{Q}),\,(\mathcal{P} \Rightarrow \mathcal{R}) \vdash \mathcal{P} \Rightarrow (\mathcal{Q} \wedge \mathcal{R}) </math> (1) <math>(\forall A)\left[ (A \in \mathcal{M}) \Rightarrow (A \in \mathcal{N}) \right]</math> (Hyp) (2) <math>(A \in \mathcal{M}) \Rightarrow (A \in \mathcal{N})</math>(MP with 1, A4) (3) <math>[(a \in A) \wedge (A \in \mathcal{M})] \Rightarrow (A \in \mathcal{M})</math>(AND) (4)<math>[(a \in A) \wedge (A \in \mathcal{M})] \Rightarrow (a \in A)</math>(AND) (5)<math>[(a \in A) \wedge (A \in \mathcal{M})] \Rightarrow (A \in \mathcal{N})</math>(D1 with 2, 3) (6)<math>[(a \in A) \wedge (A \in \mathcal{M})] \Rightarrow [(a \in A) \wedge (A \in \mathcal{N})] </math>(u with 4, 5) (7)<math>(\exists A \in \mathcal{M})(a \in A) \Rightarrow (\exists A \in \mathcal{N})(a \in A) </math>(GENe with <math>A</math>, 6) (8) <math>(\forall x)\left\{ \left(x \in \bigcup\mathcal{M}\right) \Leftrightarrow (\exists A \in \mathcal{M})(x \in A) \right\}</math>(MP with 并集公理, A4) (9) <math>(\forall x)\left\{ \left(x \in \bigcup\mathcal{N}\right) \Leftrightarrow (\exists A \in \mathcal{N})(x \in A) \right\}</math>(MP with 并集公理, A4) (10) <math>\left(x \in \bigcup\mathcal{M}\right) \Leftrightarrow (\exists A \in \mathcal{M})(x \in A)</math> (MP with 8, A4) (11) <math>\left(x \in \bigcup\mathcal{N}\right) \Leftrightarrow (\exists A \in \mathcal{N})(x \in A)</math> (MP with 9, A4) (12) <math>\left(x \in \bigcup\mathcal{M}\right) \Rightarrow (\exists A \in \mathcal{N})(x \in A)</math>(D1 with 7, 10) (13) <math>\left(x \in \bigcup\mathcal{M}\right) \Rightarrow \left(x \in \bigcup\mathcal{N}\right)</math>(D1 with 11, 12) (14) <math>(\forall x)\left[ \left(x \in \bigcup\mathcal{M}\right) \Rightarrow \left(x \in \bigcup\mathcal{N}\right) \right] </math>(GEN with <math>a</math> , 13)

覆盖性质

定理(2) — 
<math> \vdash A = \bigcup\mathcal{P}(A) </math>

“<math>A</math> 正好就是其幂集的并集”，这个定理直观上可理解成，因为幂集 <math>\mathcal{P}(A)</math> 是以 <math>A</math> 和 <math>A</math> 的子集为元素，所以 <math>\mathcal{P}(A)</math> 的并集理当是 <math>A</math> 。

证明

注意到可以从(AND)得到 <math> (\mathcal{P} \Rightarrow \mathcal{Q}),\,(\mathcal{P} \Rightarrow \mathcal{R}),\,\mathcal{P} \vdash \mathcal{Q} \wedge \mathcal{R} </math> 换句话说，从演绎元定理有 (u) <math> (\mathcal{P} \Rightarrow \mathcal{Q}),\,(\mathcal{P} \Rightarrow \mathcal{R}) \vdash \mathcal{P} \Rightarrow (\mathcal{Q} \wedge \mathcal{R}) </math> (1)<math>(\forall x)\left\{ \left[x \in \bigcup\mathcal{P}(A)\right] \Leftrightarrow (\exists S)\{ [S \in \mathcal{P}(A)] \wedge (x \in S) \} \right\} </math>(MP with 并集公理, A4) (2) <math>(\forall S)\{ [S \in \mathcal{P}(A)] \Leftrightarrow (S \subseteq A) \} </math>(幂集公理) (3) <math>[S \in \mathcal{P}(A)] \Leftrightarrow (S \subseteq A) </math>(MP with A4 ,2) (4) <math>(\forall x)\left\{ \left[x \in \bigcup\mathcal{P}(A)\right] \Leftrightarrow (\exists S)[ (S \subseteq A) \wedge (x \in S) ] \right\} </math> (Equv with 1, 3) (5) <math>[(S \subseteq A) \wedge (x \in S)] \Rightarrow (S \subseteq A) </math>(AND) (6) <math>(\forall x)[ (x \in S) \Rightarrow (x \in A) ] \Rightarrow [(x \in S) \Rightarrow (x \in A)] </math>(A4) (7) <math>[(S \subseteq A) \wedge (x \in S)] \Rightarrow [(x \in S) \Rightarrow (x \in A)] </math>(D1 with 5, 6) (8) <math>[(S \subseteq A) \wedge (x \in S)] \Rightarrow (x \in S) </math>(AND) (9) <math>[(S \subseteq A) \wedge (x \in S)] \Rightarrow \{ (x \in S) \wedge [(x \in S) \Rightarrow (x \in A)] \} </math>(u with 7, 8) 注意到 <math> (x \in S),\, [(x \in S) \Rightarrow (x \in A)] \vdash (x \in A) </math> 再对上式套用(AND)就有 <math> \{ (x \in S) \wedge [(x \in S) \Rightarrow (x \in A)] \} \vdash (x \in A) </math>(a) (10') <math>[(S \subseteq A) \wedge (x \in S)] \Rightarrow (x \in A) </math>(D1 with a, 9) (11') <math>(\exists S)[ (S \subseteq A) \wedge (x \in S) ] \Rightarrow (x \in A) </math>(GENe with <math>S</math>, 10') (12') <math>(\forall S)\{\neg[ (S \subseteq A) \wedge (x \in S) ]\} \Rightarrow \{\neg[ (A \subseteq A) \wedge (x \in A) ]\} </math> (A4) (13') <math>[ (A \subseteq A) \wedge (x \in A) ] \Rightarrow (\exists S)[ (S \subseteq A) \wedge (x \in S) ] </math> (MP with T, 12') (14') <math>(x \in A) \Rightarrow (x \in A)</math> (I) (15') <math>A \subseteq A</math> (GEN with <math>x</math> , 14') 注意到(AND)依据演绎定理可改写为 <math> (A \subseteq A) \vdash (x \in A) \Rightarrow [(A \subseteq A) \wedge (x \in A)] </math>(b) (16'') <math> (x \in A) \Rightarrow [(A \subseteq A) \wedge (x \in A)] </math> (b with 15') (17'') <math>(x \in A) \Rightarrow (\exists S)[ (S \subseteq A) \wedge (x \in S) ] </math> (D1 with 13', 16'') (18'') <math>(x \in A) \Leftrightarrow (\exists S)[ (S \subseteq A) \wedge (x \in S) ] </math> (AND with 11', 17'') (19'') <math>(\forall x)\left\{ \left[x \in \bigcup\mathcal{P}(A)\right] \Leftrightarrow (x \in A) \right\} </math>(Equv with 4, 18''')

定理(3) — 
<math>\left(\bigcup\mathcal{M} \subseteq A \right) \vdash (\forall M \in \mathcal{M})(M \subseteq A)</math>

直观上，这个定理说“一群集合的并集包含于 <math>A</math> ，则它们个个都包含于 <math>A</math> ”

证明

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

定理(4) — 
<math> \vdash \left(A = \bigcup\mathcal{M}\right) \Rightarrow \left\{A \subseteq \bigcup\mathcal[\mathcal{M} \cap \mathcal{P}(A)]\right\} </math>

直观上，这个定理说“集族 <math>\mathcal{M}</math> 的并集为 <math>A</math> ，则对 <math>A</math> 的每点 <math>a</math> ，都可从 <math>\mathcal{M}</math> 里找到一个 <math>a</math> 的邻域 <math>M</math> ，且这个邻域不会比 <math>A</math> 大 ”

证明

注意到可以从(AND)得到 <math> (\mathcal{P} \Rightarrow \mathcal{Q}),\,(\mathcal{P} \Rightarrow \mathcal{R}),\,\mathcal{P} \vdash \mathcal{Q} \wedge \mathcal{R} </math> 换句话说，从演绎元定理有 (u) <math> (\mathcal{P} \Rightarrow \mathcal{Q}),\,(\mathcal{P} \Rightarrow \mathcal{R}) \vdash \mathcal{P} \Rightarrow (\mathcal{Q} \wedge \mathcal{R}) </math> (1) <math>(\forall a)\left[ (a \in A) \Leftrightarrow (\exists M \in \mathcal{M})(a \in M) \right]</math> (Hyp) (2) <math>(\forall M \in \mathcal{M})(M \subseteq A)</math>(MP with 1, 定理3) (3) <math>(M \in \mathcal{M}) \Rightarrow (M \subseteq A)</math>(MP with A4, 2) (4) <math>[(a \in M) \wedge (M \in \mathcal{M})] \Rightarrow (M \in \mathcal{M})</math>(AND) (5) <math>[(a \in M) \wedge (M \in \mathcal{M})] \Rightarrow (a \in M)</math>(AND) (6) <math>[(a \in M) \wedge (M \in \mathcal{M})] \Rightarrow (M \in \mathcal{M})</math>(AND) (7) <math>[(a \in M) \wedge (M \in \mathcal{M})] \Rightarrow (M \subseteq A)</math> (D1 with 3, 4) (8) <math>[(a \in M) \wedge (M \in \mathcal{M})] \Rightarrow [(a \in M) \wedge (M \in \mathcal{M})]</math>(a with 5, 6) (9) <math>[(a \in M) \wedge (M \in \mathcal{M})] \Rightarrow [(a \in M) \wedge (M \in \mathcal{M}) \wedge (M \subseteq A)]</math>(a with 7, 8) (10) <math>(\exists M \in \mathcal{M})(a \in M) \Rightarrow (\exists M \in \mathcal{M})[(a \in M) \wedge (M \subseteq A)]</math>(GENe with <math>M</math>, 9) (11) <math>(a \in A) \Leftrightarrow (\exists M \in \mathcal{M})(a \in M) </math>(MP with A4, 1) (12) <math>(a \in A) \Rightarrow (\exists M \in \mathcal{M})(a \in M) </math>(AND with 11) (13) <math>(a \in A) \Rightarrow (\exists M \in \mathcal{M})[(a \in M) \wedge (M \subseteq A)] </math>(D1 with 10, 12) (14) <math>(\forall a \in A)(\exists M \in \mathcal{M})[(a \in M) \wedge (M \subseteq A)] </math>(GEN with <math>a</math>, 13) (15)<math>(\forall S)\{ [S \in \mathcal{P}(A)] \Leftrightarrow (S \subseteq A) \} </math>(幂集公理) (16)<math>[M \in \mathcal{P}(A)] \Leftrightarrow (M \subseteq A) </math>(MP with A4, 15) (17)<math>(\forall a \in A)(\exists M \in \mathcal{M})\{(a \in M) \wedge [M \in \mathcal{P}(A)]\} </math>(Equv with 14, 16) (18) <math>(\forall A)(\forall B)(\forall x)\left\{ (x \in A \cap B) \Leftrightarrow \left[ (x \in A) \wedge (x \in B) \right] \right\} </math>(有限交集) (19)<math>(\forall B)(\forall x)\left\{ (x \in \mathcal{M} \cap B) \Leftrightarrow \left[ (x \in \mathcal{M}) \wedge (x \in B) \right] \right\} </math>(MP with A4, 18) (20)<math>(\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\} </math>(MP with A4, 19) (21)<math>[M \in \mathcal{M} \cap \mathcal{P}(A)] \Leftrightarrow \left\{ (M \in \mathcal{M}) \wedge [M \in \mathcal{P}(A)] \right\} </math>(MP with A4, 20) (22)<math>(\forall a \in A)(\exists M)\{ (a \in M) \wedge [M \in \mathcal{M} \cap \mathcal{P}(A)] \} </math>(Equv with 17, 21) (23)<math>\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)] \} </math>(MP with 并集公理, A4) (24)<math>(\forall a)\left\{ (a \in A) \Rightarrow \left\{a \in \bigcup[\mathcal{M} \cap \mathcal{P}(A)]\right\} \right\} </math>(Equv with 22, 23)

运算性质

定理(5) — 
若

<math>\mathcal{M}_A := \left\{

   B \,

证明

(1)<math>(\forall B)[ (B \in \mathcal{M}_A) \Leftrightarrow (\exists M \in \mathcal{M})(B = M \cap A) ]</math> (<math>\mathcal{M}_A </math>的定义) (2) <math>(\forall x)\left\{ \left(x \in \bigcup\mathcal{M}\right) \Leftrightarrow (\exists B \in \mathcal{M})(x \in B) \right\} </math>(MP with 并集公理, A4) (3) <math>(\forall A)(\forall B)(\forall x)\left\{ (x \in A \cap B) \Leftrightarrow \left[ (x \in A) \wedge (x \in B) \right] \right\} </math>(有限交集) (4)<math>\left(x \in \bigcup\mathcal{M}_A\right) \Leftrightarrow (\exists B)[ (B \in \mathcal{M}_A) \wedge (x \in B)]</math>(MP with A4, 2) (5) <math>(B \in \mathcal{M}_A) \Leftrightarrow (\exists M \in \mathcal{M})(B = M \cap A) </math>(MP with A4, 1) (6) <math>\left(x \in \bigcup\mathcal{M}_A\right) \Leftrightarrow (\exists B)[ (x \in B) \wedge (\exists M \in \mathcal{M})(B = M \cap A) ] </math>(Equv with 4, 5) (7)<math>\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) ] </math>(Equv with Ce, 6) (8)<math>\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) ] </math>(Equv with 量词可交换性 ,7) (9) <math>(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)] \} </math>(E2) (10)<math>[(x \in B) \wedge ( M \in \mathcal{M}) \wedge (B = M \cap A)] \Rightarrow (B = M \cap A) </math>(AND) (11)<math>[(x \in B) \wedge ( M \in \mathcal{M}) \wedge (B = M \cap A)] </math><math>\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)] \} </math>(D1 with 9,10) (12)<math>\{ [(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)] \} </math> <math>\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)] \} </math>(MP with A2, 11) (13)<math>[(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)] </math>(I) (14)<math>[(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)] </math>(MP with 12, 13) (15)<math>[(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})] </math>(AND) (16)<math>[(x \in B) \wedge ( M \in \mathcal{M}) \wedge (B = M \cap A)] \Rightarrow [(x \in M \cap A) \wedge ( M \in \mathcal{M})] </math>(D1 with 14,15) (17)<math>(\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})] </math>(GENe with <math>B</math> then <math>M</math>) (18)<math>M \cap A = M \cap A </math> (E1) 注意到配合(AND)和演绎定理有 <math>\mathcal{P} \vdash \mathcal{R} \Rightarrow (\mathcal{R} \wedge \mathcal{P}) </math>(a) (19)<math>[(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)] </math>(a with 18) (20)<math>(\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)]\} </math>(A4) (21)<math>[(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)] </math>(MP with T, 20) (22)<math>[(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)] </math>(D1 with 19, 21) (23)<math>(\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)] </math>(GENe with <math>M</math>) (24)<math>(\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})] </math>(AND with 17, 23) (25)<math>\left(x \in \bigcup\mathcal{M}_A\right) \Leftrightarrow (\exists M)[(x \in M \cap A) \wedge ( M \in \mathcal{M})] </math>(Equv with 8, 24) (26) <math>(x \in M \cap A) \Leftrightarrow [(x \in M) \wedge (x \in A)]</math>(MP with A4, 3) (27)<math>\left(x \in \bigcup\mathcal{M}_A\right) \Leftrightarrow (\exists M)[ (x \in M) \wedge (x \in A) \wedge (M \in \mathcal{M}) ] </math>(Equv with 25, 26) (28)<math>\left(x \in \bigcup\mathcal{M}_A\right) \Leftrightarrow \{ (\exists M)[ (x \in M) \wedge (M \in \mathcal{M}) ] \wedge (x \in A) \} </math>(Equv with Ce, 27) (30) <math>\left(x \in \bigcup\mathcal{M}\right) \Leftrightarrow (\exists M \in \mathcal{M})(x \in M) </math>(MP with A4, 2) (31)<math>\left(x \in \bigcup\mathcal{M}_A\right) \Leftrightarrow \left[ \left(x \in \bigcup\mathcal{M}\right) \wedge (x \in A) \right] </math>(Equv with 28, 30) (32)<math>\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] </math>(MP with A4, 3) (33)<math>\left[x \in A \cap \left(\bigcup\mathcal{M}\right) \right] \Leftrightarrow \left(x \in \bigcup\mathcal{M}_A\right) </math>(Equv with 31, 32) (34)<math>(\forall x)\left\{ \left[x \in A \cap \left(\bigcup\mathcal{M}\right) \right] \Leftrightarrow \left(x \in \bigcup\mathcal{M}_A\right) \right\} </math>(GEN with <math>x</math>, 33)

直观上这个定理说，交集在“无限并集满足分配律”，一般会不正式的写为

<math>\bigcup_{i\in I}\left(A \cap B_{i}\right) = A \cap \bigcup_{i\in I} B_{i}</math>

定理(6) — 
<math>\N \,\overset{A}{\cong}\, \mathcal{A} </math>，若对自然数 <math>m \in \N </math> 做以下的符号定义：

<math>\mathcal{A}_m

=

\left\{ S \in \mathcal{A} \,

这个定理一般会被不正式的写为

<math>\bigcup^{\infty}_{i=0}\left(\bigcap^{\infty}_{j=i} A_j \right)

\subseteq \bigcap^{\infty}_{i=0}\left(\bigcup^{\infty}_{j=i} A_j \right)</math>。

参考

• 朴素集合论
• 交集
• 补集
• 对称差
• 不交并
• 布尔逻辑