併集

A和B的併集

在集合論和數學的其他分支中，一群集合的併集（Union），是以這群集合的所有元素來構成的集合。

有限聯集

聯集是由公理化集合論的分類公理來確保其唯一存在的特定集合 $A\cup B$ ：

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

也就是直觀上：

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

舉例：

集合$\{1,2,3\}$和$\{2,3,4\}$的併集是$\{1,2,3,4\}$。數$9$不屬於素數集合$\{2,3,5,7,11,\ldots \}$和偶數集合$\{2,4,6,8,10,\ldots \}$的併集，因為$9$既不是素數，也不是偶數。

更通常的，多個集合的併集可以這樣定義： 例如，$A,B$和$C$的併集含有所有$A$的元素，所有$B$的元素和所有$C$的元素，而沒有其他元素。形式上：

$x$是$A\cup B\cup C$的元素，當且僅當$x$屬於$A$或$x$屬於$B$或$x$屬於$C$。

代數性質

二元併集（兩個集合的併集）是一種結合運算，即

$A\cup (B\cup C)=(A\cup B)\cup C$。事實上，$A\cup B\cup C$也等於這兩個集合，因此圓括號在僅進行併集運算的時候可以省略。

相似的，併集運算滿足交換律，即集合的順序任意。

空集是併集運算的單位元。即$\varnothing \cup A=A$，對任意集合$A$。可以將空集當作零個集合的併集。

結合交集和補集運算，併集運算使任意冪集成為布爾代數。例如，併集和交集相互滿足分配律，而且這三種運算滿足德·摩根律。若將併集運算換成對稱差運算，可以獲得相應的布爾環。

無限併集

由公理化集合論的併集公理，有唯一的集合 $\bigcup {\mathcal {M}}$ 滿足：

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

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

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

例如：

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

無限併集有多種表示方法：

可模仿求和符號記為

$\bigcup _{A\in {\mathcal {M}}}A$。

但大多數人會假設指標集 $I$ 的存在，換句話說

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

在指標集 $I$ 是自然數系 $\mathbb {N}$ 的情況下，更可以仿無窮級數來表示，也就是說：

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

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

無限併集的性質

定理(0) — 
$\vdash \bigcup \varnothing =\varnothing$

證明

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

比較性質

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

證明

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

覆蓋性質

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

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

證明

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

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

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

證明

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

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

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

證明

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

運算性質

定理(5) — 
若

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

則

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

證明

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

直觀上這個定理說，交集在「無限併集滿足分配律」，一般會不正式的寫為

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

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

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

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

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

那有

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

這個定理一般會被不正式的寫為

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

參考

• 樸素集合論
• 交集
• 補集
• 對稱差
• 不交並
• 布爾邏輯