问题1：在以下几组中确定哪一组是一个子集和另一个子集：
A = {x：x∈R并且x满足x 2 – 8x + 12 = 0}，B = {2，4，6}，C = {2，4，6，8，。 。 。 }，D = {6}
解决方案：
At first, simplifying for set A
x2– 8x + 12 = 0
(x-6)(x-2) = 0
x= 6 or 2
Now, A = {2,6}
A set X is said to be a subset of a set Y if every element of X is also an element of Y.
Therefore, we can write : A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B, D ⊂ C
问题2：在下面的每一个中，确定该语句是对还是错。
如果是真的，请证明这一点。如果为假，请举一个例子。
(i)如果x∈A和A∈B，则x∈B
(ii)如果A⊂B且B∈C，则A∈C
(iii)如果A⊂B和B⊂C，则A⊂C
(iv)如果A⊄B和B⊄C，则A⊄C
(v)如果x∈A且A⊄B，则x∈B
(vi)如果A⊂B和x∉B，则x∉A
解决方案：
(i) False
Example : Let x=1, A={1,2,3} and B = {{1,2,3},{4,5,6}}
Here x ∈ A and A ∈ B but x ∉ B
(ii) False
Example : Let A = {1}, B = {1,3} and C = {{1,3},{5,7}}
Here A ⊂ B and B ∈ C but A ∉ C
(iii) True
Proof : A ⊂ B : Set B contains all the elements in set A
B ⊂ C : Set C contains all the elements in set B
Hence, by transitivity property, set C contains all the elements in set A i.e. A ⊂ C
(iv) False
Example : Let A = {1,2}, B = {3,4,5} and C = {1,2,6,7}
Here A ⊄ B and B ⊄ C but A ⊂ C
(v) False
Example : Let x=1, A = {1,2} and B = {3,4,5}
Here x ∈ A and A ⊄ B but x ∉ B
(vi) True
Proof : A ⊂ B : Set B contains all the elements in set A
So if x ∉ B then also x ∉ A
问题3：令A，B和C为集合，使得A∪B = A∪C和A∩B = A∩C。证明B =C。
解决方案：
Given : A ∪ B = A ∪ C …(1)
A ∩ B = A ∩ C …(2)
A ∪ B = A + B – (A ∩ B) // by principle of inclusion-exclusion
A ∪ C = A + C – (A ∩ C) // by principle of inclusion-exclusion
A + B – (A ∩ B) = A + C – (A ∩ C) // from (1)
A + B – (A ∩ B) = A + C – (A ∩ B) // from (2)
B = C
问题4：证明以下四个条件是等效的：
(i)A⊂B(ii)A – B =φ(iii)A∪B = B(iv)A∩B = A
解决方案：
//Showing (i)=(ii)
A ⊂ B means Set B contains all the elements in set A i.e. set A does not have any different element from B
It means A – B = φ
//Showing (i)=(iii)
All elements of set A are there in Set B so A ∪ B = B
//Showing (i)=(iv)
All elements of set A are there in Set B so A ∩ B = A
From above explanation we can say that all above conditions are equivalent.
问题5：证明如果A⊂B，则C – B⊂C – A
解决方案：
//By taking an example
Let A = {1,2} and B = {1,2,3,4,5}
and C = {2,5,6,7,8}
Set B contains all the elements in set A so A ⊂ B
Now C-B = {6,7,8} //elements present in C but not in B
C-A = {5,6,7,8} //elements present in C but not in A
It is clearly seen that, Set C-A contains all the elements in Set C-B hence C – B ⊂ C – A is proved.
问题6：假设P(A)= P(B)。证明A = B
解决方案：
P(X) represents power set of set X
To prove A = B we have to prove that A ⊂ B and B ⊂ A
(eg : if A = {1,2} then P(A) = {φ, {1}, {2}, {1,2}})
Power set of any set contains all the possible subsets of it.
A ∈ P(A)
as P(A)=P(B) so A ∈ P(B)
If A is P(B) then clearly A is subset of B.
A ⊂ B …(1)
Repeating above process for B ∈ P(B) we get
B ⊂ A …(2)
From above equations,
A = B
问题7：对于任何集合A和B，P(A)∪P(B)= P(A∪B)是否成立?证明你的答案
解决方案：
It is False
Let A = {1,2} and B = {2,3}
A ∪ B = {1,2,3}
P(A) = {φ, {1}, {2}, {1,2}}
P(B) = {φ, {2}, {3}, {2,3}}
P(A) ∪ P(B) = {φ, {1}, {2}, {3}, {1,2}, {2,3}} …(1)
P(A ∪ B) = {φ, {1}, {2}, {3}, {1,2}, {2,3}, {1,3}, {1,2,3}} …(2)
P(A) ∪ P(B) ≠ P(A ∪ B) //from (1) and (2)
问题8：证明对于任何集合A和B，A =(A∩B)∪(A – B)和A∪(B – A)=(A∪B)
解决方案：
(A ∩ B) ∪ (A – B)
(A ∩ B) ∪ (A ∩ B’) //(A – B) = (A ∩ B’)
A ∩ (B ∪ B’)
A ∩ U //B ∪ B’ = U where U represents universal set
A //by identity property
Hence, it is proved that A = (A ∩ B) ∪ (A – B)
A ∪ (B – A)
A ∪ (B ∩ A’)
//B – A = B ∩ A’
(A ∪ B) ∩ (A ∪ A’)
//distributive law
(A ∪ B) ∩ U
//A ∪ A’ = U where U represents universal set
(A ∪ B)
Hence, it is proved that A ∪ (B – A) = (A ∪ B)
问题9：使用集合的属性表明
(i)A∪(A∩B)= A(ii)A∩(A∪B)= A
解决方案：
(i) A ∪ (A ∩ B)
(A ∪ A) ∩ (A ∪ B) //distributive law
A ∩ (A ∪ B) //A ∪ A = A
A //by absorption law
(ii) A ∩ (A ∪ B)
(A ∩ A) ∪ (A ∩ B) //distributive law
A ∪ (A ∩ B) //A ∩ A = A
A //by absorption law
问题10：证明A∩B = A∩C不一定意味着B = C
解决方案：
Let us assume that B ≠ C
Take A = {1,2}, B ={2,3} and C = {3,4} // here B ≠ C
A ∩ B = {2}
A ∩ C = φ
Here we can see that A ∩ B ≠ A ∩ C
Therefore, our assumption was wrong
Hence, B = C is must for A ∩ B = A ∩ C