本文概述
- 二元性
- 延伸原理
- 两套笛卡尔积
表：集合代数定律
| 等幂律 | （a）A∪A = A | （b）A∩A = A |
| 关联法 | （a）（A∪B）∪C = A∪（B∪C） | （b）（A∩B）∩C = A∩（B∩C） |
| 交换律 | （a）A∪B = B∪A | （b）A∩B = B∩A |
| Distributive Laws | （a）A∪（B∩C）=（A∩B）∩（A∪C） | （b）A∩（B∪C）=（A∪B）∪（A∩C） |
| De Morgan’ s Laws | （a）（A∪B）c =Ac∩Bc | （b）（A∩B）c =Ac∪Bc |
| Identity Laws | （a）A = A（b）A = U | （c）U = A（d）A = = |
| 补充法 | （a）A∪Ac = U（b）A∩Ac =? | （c）Uc =?（d）?c= U |
| Involution Law | （a）（Ac）c = A |
示例1：证明幂等定律：
(a) A ∪ A = A解：
Since, B ? A ∪ B, therefore A ? A ∪ ALetx ∈ A ∪ A ? x ∈ Aorx ∈ A ?x ∈ A∴ A ∪ A ? AAsA ∪ A ? A andA ? A ∪ A ? A =A ∪ A. Hence Proved.(b) A ∩ A = A解：
Since, A ∩ B ? B, therefore A ∩ A ? ALet x ∈ A ? x ∈ Aand x ∈ A? x ∈ A ∩ A∴ A ? A ∩ AAs A ∩ A ? A and A ? A ∩ A ? A = A ∩ A. Hence Proved.示例2：证明的关联法则：
(a) (A ∪ B) ∪ C = A ∪ (B ∪ C)解：
Let some x ∈ (A'∪ B) ∪ C?(x ∈ Aorx ∈ B)orx ∈ C?x ∈ Aorx ∈ Borx ∈ C?x ∈ Aor(x ∈ Borx ∈ C)?x ∈ Aorx ∈ B ∪ C ?x ∈ A ∪ (B ∪ C).Similarly, if somex ∈ A ∪ (B ∪ C), thenx ∈ (A ∪ B) ∪ C.Thus, anyx ∈ A ∪ (B ∪ C) ?x ∈ (A ∪ B) ∪ C. Hence Proved.(b) (A ∩ B) ∩ C = A ∩ (B ∩ C)解：
Let some x ∈ A ∩ (B ∩ C) ?x ∈ A and x ∈ B ∩ C ?x ∈ Aand (x ∈ B and x ∈ C)?x ∈ Aand x ∈ B and x ∈ C?(x ∈ Aand x ∈ B) and x ∈ C)?x ∈ A ∩ B and x ∈ C?x ∈ (A ∩ B) ∩ C.Similarly, if somex ∈ A ∩ (B ∩ C), then x ∈ (A ∩ B) ∩ CThus, anyx ∈ (A ∩ B) ∩ C?x ∈ A ∩ (B ∩ C). Hence Proved.例3：证明交换律
(a)A ∪ B = B ∪ A解：
To Prove A ∪ B = B ∪ AA ∪ B = {x: x ∈ A or x ∈ B}= {x: x ∈ B or x ∈ A}(∵ Order is not preserved in case of sets)A ∪ B = B ∪ A. Hence Proved.(b) A ∩ B = B ∩ A解：
To Prove A ∩ B = B ∩ AA ∩ B = {x: x ∈ A and x ∈ B}= {x: x ∈ B and x ∈ A}(∵ Order is not preserved in case of sets)A ∩ B = B ∩ A. Hence Proved.示例4：证明分配定律
(a) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)解：
To Prove Let x ∈ A ∪ (B ∩ C)? x ∈ A orx ∈ B ∩ C ?(x ∈ Aor x ∈ A) or (x ∈ B andx ∈ C)?(x ∈ Aor x ∈ B) and (x ∈ Aor x ∈ C)?x ∈ A ∪ B andx ∈ A ∪ C?x ∈ (A ∪ B) ∩ (A ∪ C)Therefore, A ∪ (B ∩ C) ? (A ∪ B) ∩ (A ∪ C)............(i)Again, Let y ∈ (A ∪ B)∩ (A ∪ C) ?y ∈ A ∪ B and y ∈ A ∪ C?(y ∈ A or y ∈ B) and (y ∈ A or y ∈ C)?(y ∈ A and y ∈ A) or (y ∈ B and y ∈ C)?y ∈ Aory ∈ B ∩ C?y ∈ A∪ (B ∩ C)Therefore, (A ∪ B) ∩ (A ∪ C) ? A ∪ (B ∩ C)............(ii)Combining (i) and (ii), we get A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C). Hence Proved(b) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)解：
To Prove Let x ∈ A ∩ (B ∪ C)?x ∈ A and x ∈ B ∪ C ?(x ∈ A and x ∈ A) and (x ∈ Bor x ∈ C)?(x ∈ A and x ∈ B) or(x ∈ A and x ∈ C)?x ∈ A ∩ B orx ∈ A ∩ C?x ∈ (A ∩ B) ∪ (A ∪ C)Therefore, A ∩ (B ∪ C) ? (A ∩ B) ∪ (A ∪ C)............ (i)Again, Lety ∈ (A ∩ B) ∪ (A ∪ C) ? y ∈ A ∩ B or y ∈ A ∩ C?(y ∈ A and y ∈ B) or (y ∈ A and y ∈ C)?(y ∈ A or y ∈ A) and (y ∈ B or y ∈ C)? y ∈ A andy ∈ B ∪ C? y ∈ A ∩ (B ∪ C)Therefore, (A ∩ B) ∪ (A ∪ C) ? A ∩ (B ∪ C)............ (ii)Combining (i) and (ii), we get A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∪ C). Hence Proved示例5：证明德摩根定律
(a) (A ∪B)c=Ac∩ Bc【集合代数】解：
To Prove (A ∪B)c=Ac∩ BcLet x ∈ (A ∪B)c?x ?A ∪ B(∵ a ∈ A ? a ? Ac)?x ?A and x ? B?x ?Ac and x ? Bc?x ?Ac∩ BcTherefore, (A ∪B)c ? Ac∩ Bc............. (i)Again, let x ∈ Ac∩ Bc ? x ∈ Ac and x ∈ Bc? x ?A and x ? B?x ?A ∪ B? x ∈ (A ∪B)cTherefore, Ac∩ Bc? (A ∪B)c............. (ii)Combining (i) and (ii), we get Ac∩ Bc =(A ∪B)c. Hence Proved.(b) (A ∩B)c = Ac∪ Bc解：
Let x ∈ (A ∩B)c ? x ?A ∩ B(∵ a ∈ A ? a ? Ac)? x ?A or x ? B? x ∈ Ac and x ∈ Bc? x ∈ Ac∪ Bc∴ (A ∩B)c? (A ∪B)c.................. (i)Again, Let x ∈ Ac∪ Bc? x ∈ Ac or x ∈ Bc? x ?A or x ? B? x ?A ∩ B? x ∈ (A ∩B)c∴ Ac∪ Bc? (A ∩B)c.................... (ii)Combining (i) and (ii), we get(A ∩B)c=Ac∪ Bc. Hence Proved.示例6：证明身份法则。
(a) A ∪ ? = A解：
To Prove A ∪ ? = ALetx ∈ A ∪ ? ? x ∈ Aorx ∈ ?? x ∈ A(∵x ∈ ?, as ? is the null set )Therefore, x ∈ A ∪ ? ? x ∈ AHence, A ∪ ? ? A.We know that A ? A ∪ B for any set B. But for B = ?, we have A ? A ∪ ? From above, A ? A ∪ ? , A ∪ ? ? A ? A = A ∪ ?. Hence Proved.(b) A ∩ ? = ?解：
To Prove A ∩ ? = ?Ifx ∈ A, then x ??(∵? is a null set)Therefore, x ∈ A, x ?? ? A ∩ ? = ?. Hence Proved.(c) A ∪ U = U解：
To Prove A ∪ U = UEvery set is a subset of a universal set.∴A ∪ U ? UAlso, U ? A ∪ UTherefore, A ∪ U = U. Hence Proved.(d) A ∩ U = A解：
To Prove A ∩ U = AWe knowA ∩ U ? A................. (i)So we have to show that A ? A ∩ ULetx ∈ A ? x ∈ A and x ∈ U(∵ A ? U so x ∈ A ? x ∈ U )∴x ∈ A ? x ∈ A ∩ U∴A ? A ∩ U................. (ii)From (i) and (ii), we get A ∩ U = A. Hence Proved.例7：证明补充法则
(a) A ∪ Ac= U解：
To Prove A ∪ Ac= UEvery set is a subset of U∴A ∪ Ac ? U.................. (i)We have to show that U ? A ∪ AcLet x ∈ U?x ∈ Aorx ?A?x ∈ Aorx ∈ Ac? x ∈ A ∪ Ac∴ U ? A ∪ Ac................... (ii)From (i) and (ii), we get A ∪ Ac= U. Hence Proved.(b) A ∩ Ac=?解：
As ? is the subset of every set∴? ? A ∩ Ac..................... (i)We have to show that A ∩ Ac ? ?Let x ∈ A ∩ Ac? x ∈ A and x ∈Ac? x ∈ Aand x ?A? x ∈ ?∴A ∩ Ac ??..................... (ii)From (i) and (ii), we get A∩ Ac=?. Hence Proved.(c) Uc= ?解：
Let x ∈ Uc? x ? U ? x ∈ ?∴ Uc= ?. Hence Proved.(As U is the Universal Set).(d) ?c = U解：
Let x ∈ ?c ? x ? ?? x ∈ U(As ? is an empty set) ∴ ?c = U.Hence Proved.示例8：证明对合律
(a) (Ac )c A.解：
Let x ∈ (Ac )c ? x ? Ac?x ∈ a∴ (Ac )c =A. Hence Proved.二元性 E的对偶E ?是通过分别用∩, ∪, ?和U代替E中的每次出现∪, ∩, U和obtained而获得的方程。例如, 双重
(U ∩ A) ∪ (B ∩ A) = A is (? ∪ A) ∩ (B ∪ A) = A作为对偶原理, 要指出的是, 如果任何方程E是一个恒等式, 那么它的对偶E *也是一个恒等式。
延伸原理 根据扩展原理, 只有当A和B具有相同的成员时, 它们才是相同的。我们用A = B表示相等的集合。
If A= {1, 3, 5} and B= {3, 1, 5}, then A=B i.e., A and B are equal sets. If A= {1, 4, 7} and B= {5, 4, 8}, then A≠ B i.e.., A and B are unequal sets.两套笛卡尔积 按此顺序的两个集合P和Q的笛卡尔积是所有有序对的集合, 其第一个成员属于集合P, 第二个成员属于集合Q并用P x Q表示, 即
P x Q = {(x, y): x ∈ P, y ∈ Q}.示例：令P = {a, b, c}和Q = {k, l, m, n}。确定P和Q的笛卡尔积。
解决方案：P和Q的笛卡尔积为
