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併集公理（配上配對公理）來確保其唯一存在的特定集合 $A \cup B$：

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

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，以及所有$B$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，若且唯若$x$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第摩根定律。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：

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

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，例如：

取 $\mathcal{M} = \{A,B,C\}$ 則 $\bigcup\mathcal{M} = A \cup B \cup C$。

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 $X$。

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 $\N$ 的情況下，更可以仿無窮級數來表示，也就是說：

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

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

### 無限併集的性質

$$\vdash \bigcup\varnothing = \varnothing$$

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

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

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

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

「$A$ 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`'''`)

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

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)

$$\vdash \left(A = \bigcup\mathcal{M}\right) \Rightarrow \left\{A \subseteq \bigcup\mathcal[\mathcal{M} \cap \mathcal{P}(A)]\right\}$$

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

若

$$\mathcal{M}_A := \left\{ <pre> B \,|\, (\exists M \in \mathcal{M})(B = M \cap A) </pre> \right\}$$

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

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

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

$$\mathcal{A}_m := \left\{ S \in \mathcal{A} \,|\, <pre> A^{-1}(S) \geq m </pre> \right\}$$

$$\mathcal{I} := \left\{ S \,\bigg|\, <pre> (\exists m \in \N)\left( S = \bigcap \mathcal{A}_m \right) </pre> \right\}$$

$$\mathcal{S} := \left\{ S \,\bigg|\, <pre> (\exists m \in \N)\left( S = \bigcup \mathcal{A}_m \right) </pre> \right\}$$

那有

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

這個定理一般會被不正式的寫為

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