# 并集

## 有限聯集

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

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

$x$ $A\cup B\cup C$ 的元素，当且仅当$x$ 属于$A$ $x$ 属于$B$ $x$ 属于$C$

### 代数性质

$A\cup (B\cup C)=(A\cup B)\cup C$ 。事实上，$A\cup B\cup C$ 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。

## 无限并集

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

$X\subseteq \bigcup {\mathcal {M}}$  ，會稱 $X$ ${\mathcal {M}}$  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 $X$

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

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

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

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

### 无限并集的性質

$\vdash \bigcup \varnothing =\varnothing$

(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 并集公理, 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

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

$({\mathcal {P}}\Rightarrow {\mathcal {Q}}),\,({\mathcal {P}}\Rightarrow {\mathcal {R}}),\,{\mathcal {P}}\vdash {\mathcal {Q}}\wedge {\mathcal {R}}$

(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 并集公理, A4)

(9) $(\forall x)\left\{\left(x\in \bigcup {\mathcal {N}}\right)\Leftrightarrow (\exists A\in {\mathcal {N}})(x\in A)\right\}$ (MP with 并集公理, 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 $A$ $A$ 子集為元素，所以 ${\mathcal {P}}(A)$  的聯集理當是 $A$

$({\mathcal {P}}\Rightarrow {\mathcal {Q}}),\,({\mathcal {P}}\Rightarrow {\mathcal {R}}),\,{\mathcal {P}}\vdash {\mathcal {Q}}\wedge {\mathcal {R}}$

(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 并集公理, A4)

(2) $(\forall S)\{[S\in {\mathcal {P}}(A)]\Leftrightarrow (S\subseteq A)\}$ (幂集公理)

(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)

$(x\in S),\,[(x\in S)\Rightarrow (x\in A)]\vdash (x\in A)$

$\{(x\in S)\wedge [(x\in S)\Rightarrow (x\in A)]\}\vdash (x\in A)$ (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')

$(A\subseteq A)\vdash (x\in A)\Rightarrow [(A\subseteq A)\wedge (x\in A)]$ (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''')

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

(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)

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

$({\mathcal {P}}\Rightarrow {\mathcal {Q}}),\,({\mathcal {P}}\Rightarrow {\mathcal {R}}),\,{\mathcal {P}}\vdash {\mathcal {Q}}\wedge {\mathcal {R}}$

(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)$(\forall S)\{[S\in {\mathcal {P}}(A)]\Leftrightarrow (S\subseteq A)\}$ (幂集公理)

(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 并集公理, 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

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

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

(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 并集公理, 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)

${\mathcal {P}}\vdash {\mathcal {R}}\Rightarrow ({\mathcal {R}}\wedge {\mathcal {P}})$ (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)

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

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

${\mathcal {A}}_{m}:=\left\{S\in {\mathcal {A}}\,|\,A^{-1}(S)\geq m\right\}$
${\mathcal {I}}:=\left\{S\,{\bigg |}\,(\exists m\in \mathbb {N} )\left(S=\bigcap {\mathcal {A}}_{m}\right)\right\}$
${\mathcal {S}}:=\left\{S\,{\bigg |}\,(\exists m\in \mathbb {N} )\left(S=\bigcup {\mathcal {A}}_{m}\right)\right\}$

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

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

## 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.