併集

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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）[1]，是以這群集合的所有元素來構成的集合。

有限併集[編輯]

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 <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>

也就是直觀上：

「對所有 <math>x</math> ， <math>x \in A \cup B</math> 等價於 <math>x \in A</math> 或 <math>x \in B</math>」

舉例：

集合<math>\{1, 2, 3\}</math>和<math>\{2, 3, 4\} </math>的併集是<math>\{1, 2, 3, 4\}</math>。數<math>9</math>不屬於質數集合<math>\{2, 3, 5, 7, 11,\ldots\}</math>和偶數集合<math>\{2, 4, 6, 8, 10,\ldots\}</math>的併集，因為<math>9</math>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，<math>A,B</math>和<math>C</math>的併集含有所有<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，而沒有其他元素。形式上：

<math>x</math>是<math>A \cup B \cup C</math>的元素，當且僅當<math>x</math>屬於<math>A</math>或<math>x</math>屬於<math>B</math>或<math>x</math>屬於<math>C</math>。

代數性質[編輯]

二元併集（兩個集合的併集）是一種結合運算，即

<math>A \cup (B \cup C) =(A \cup B)\cup C</math>。事實上，<math>A \cup B \cup C</math>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<math>\varnothing \cup A = A</math>，對任意集合<math>A</math>。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，可以獲得相應的布林環。

無限併集[編輯]

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

<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>

也就是直觀上「對所有 <math>\mathcal{M}</math> 和所有 <math>x</math> ， <math>x \in \bigcup\mathcal{M}</math> 等價於有某個 <math>\mathcal{M}</math> 的下屬集合 <math>A</math> ，使得<math>x \in A</math>」。以上的 <math>\mathcal{M}</math> 可以直觀的視為一個集族，而把 <math>\bigcup\mathcal{M}</math> 看成對 <math>\mathcal{M}</math> 內的集合取併集，例如：

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

但這個公理並沒有對 <math>\mathcal{M}</math> 下屬集合的數量做出任何限制，所以這個 <math>\bigcup\mathcal{M}</math> 被俗稱為任意併集或無限併集。

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

無限併集有多種表示方法：

可模仿求和符號記為

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

但大多數人會假設指標集 <math>I</math> 的存在，換句話說

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

在指標集 <math>I</math> 是自然數系 <math>\N</math> 的情況下，更可以仿無窮級數來表示，也就是說：

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

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

無限併集的性質[編輯]

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

證明
(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)

比較性質[編輯]

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

證明
注意到可以從(AND)得到
<math>

(\mathcal{P} \Rightarrow \mathcal{Q}),\,(\mathcal{P} \Rightarrow \mathcal{R}),\,\mathcal{P} \vdash \mathcal{Q} \wedge \mathcal{R} </math> 換句話說，從演繹元定理有

(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 併集公理, 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 併集公理, 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)

覆蓋性質[編輯]

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

「<math>A</math> 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 <math>\mathcal{P}(A)</math> 是以 <math>A</math> 和 <math>A</math> 的子集為元素，所以 <math>\mathcal{P}(A)</math> 的併集理當是 <math>A</math> 。

證明
注意到可以從(AND)得到
<math>

(\mathcal{P} \Rightarrow \mathcal{Q}),\,(\mathcal{P} \Rightarrow \mathcal{R}),\,\mathcal{P} \vdash \mathcal{Q} \wedge \mathcal{R} </math> 換句話說，從演繹元定理有

(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 併集公理, A4)

(2) <math>(\forall S)\{ 

   [S \in \mathcal{P}(A)]
   \Leftrightarrow
   (S \subseteq A)

\} </math>(冪集公理)

(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> 再對上式套用(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)依據演繹定理可改寫為

<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>

直觀上，這個定理說「一群集合的併集包含於 <math>A</math> ，則它們個個都包含於 <math>A</math> 」

證明
(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>

直觀上，這個定理說「集族 <math>\mathcal{M}</math> 的併集為 <math>A</math> ，則對 <math>A</math> 的每點 <math>a</math> ，都可從 <math>\mathcal{M}</math> 裏找到一個 <math>a</math> 的鄰域 <math>M</math> ，且這個鄰域不會比 <math>A</math> 大 」

證明
注意到可以從(AND)得到
<math>

(\mathcal{P} \Rightarrow \mathcal{Q}),\,(\mathcal{P} \Rightarrow \mathcal{R}),\,\mathcal{P} \vdash \mathcal{Q} \wedge \mathcal{R} </math> 換句話說，從演繹元定理有

(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>(冪集公理)

(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 併集公理, 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)

運算性質[編輯]

定理(5) — 
若

<math>\mathcal{M}_A := \left\{
   B \,
證明
(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>的定義)

(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 併集公理, 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 量詞可交換性 ,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)和演繹定理有

<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

<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>，若對自然數 <math>m \in \N </math> 做以下的符號定義：

<math>\mathcal{A}_m
=

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

這個定理一般會被不正式的寫為

<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>。

參考[編輯]

參考文獻[編輯]

  1. ↑ 程極泰. 集合论. 應用數學叢書 第一版. 國防工業出版社. 1985: 14. 15034.2766. 
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