併集

在集合論和數學的其他分支中，一群集合的併集（Union）^[1]，是以這群集合的所有元素來構成的集合。

A和B的併集

有限併集

併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 ${\displaystyle A\cup B}$ ：

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

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 ${\displaystyle x\in B}$」

舉例：

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

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

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

代數性質

二元併集（兩個集合的併集）是一種結合運算，即

${\displaystyle A\cup (B\cup C)=(A\cup B)\cup C}$。事實上，${\displaystyle A\cup B\cup C}$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素。即${\displaystyle \varnothing \cup A=A}$，對任意集合${\displaystyle A}$。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林代數。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，可以獲得相應的布林環。

無限併集

由公理化集合論的併集公理，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族，而把 ${\displaystyle \bigcup {\mathcal {M}}}$ 看成對 ${\displaystyle {\mathcal {M}}}$ 內的集合取併集，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 ${\displaystyle \bigcup {\mathcal {M}}}$ 被俗稱為任意併集或無限併集。

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 ${\displaystyle {\mathcal {M}}}$ 裏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。

無限併集有多種表示方法：

可模仿求和符號記為

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

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

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

在指標集 ${\displaystyle I}$ 是自然數系 ${\displaystyle \mathbb {N} }$ 的情況下，更可以仿無窮級數來表示，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 ${\displaystyle \bigcup _{i=0}^{\infty }A(i)}$ 寫作${\displaystyle A_{0}\cup A_{1}\cup A_{2}\cup \ldots }$。

無限併集的性質

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 併集公理, 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 併集公理, 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 併集公理, 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) — 
${\displaystyle \vdash A=\bigcup {\mathcal {P}}(A)}$

「${\displaystyle A}$ 正好就是其冪集的併集」，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 ${\displaystyle {\mathcal {P}}(A)}$ 的併集理當是 ${\displaystyle A}$ 。

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 併集公理, A4) (2) ${\displaystyle (\forall S)\{[S\in {\mathcal {P}}(A)]\Leftrightarrow (S\subseteq A)\}}$(冪集公理) (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「一群集合的併集包含於 ${\displaystyle A}$ ，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 ${\displaystyle {\mathcal {M}}}$ 的併集為 ${\displaystyle A}$ ，則對 ${\displaystyle A}$ 的每點 ${\displaystyle a}$ ，都可從 ${\displaystyle {\mathcal {M}}}$ 裏找到一個 ${\displaystyle a}$ 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)${\displaystyle (\forall S)\{[S\in {\mathcal {P}}(A)]\Leftrightarrow (S\subseteq A)\}}$(冪集公理) (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 併集公理, 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 併集公理, 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「無限併集滿足分配律」，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) — 
${\displaystyle \mathbb {N} \,{\overset {A}{\cong }}\,{\mathcal {A}}}$，若對自然數 ${\displaystyle m\in \mathbb {N} }$ 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

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

參考

• 樸素集合論
• 交集
• 補集
• 對稱差
• 互斥併集
• 布林運算