问题1.找到最小集合A，使A∪{1，2} = {1,2,3,5,9}。
解决方案：
A ∪ {1, 2} = {1, 2, 3, 5, 9}
The union indicates that the summation of elements of both sets should form RHS.
Elements of A and {1, 2} together give us the resultant set.
Therefore, the smallest set will be,
A = {1, 2, 3, 5, 9} – {1, 2}
Hence, A = {3, 5, 9}
问题2。令A = {1,2,4,5} B = {2,3,5,6} C = {4,5,6,7}。验证以下身份：
(i)A∪(B∩C)=(A∩B)∩(A∪C)
(ii)A∩(B∪C)=(A∪B)∪(A∩C)
(iii)A∩(B – C)=(A∩B)-(A∩C)
(iv)A –(B∪C)=(A – B)∩(A – C)
(v)A –(B∩C)=(A – B)∪(A – C)
(vi)A∩(B△C)=(A∩B)△(A∩C)
解决方案：
(i) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
Considering LHS,
(B ∩ C) = {x : x ∈ B and x ∈ C}
Therefore, we get,
= {5, 6}
A ∪ (B ∩ C) = {x : x ∈ A or x ∈ (B ∩ C)}
= {1, 2, 4, 5, 6}
Considering RHS,
(A ∪ B) = {x : x ∈ A or x ∈ B}
= {1, 2, 4, 5, 6}.
(A ∪ C) = {x : x ∈ A or x ∈ C}
= {1, 2, 4, 5, 6, 7}
(A ∪ B) ∩ (A ∪ C) = {x : x ∈ (A ∪ B) and x ∈ (A ∪ C)}
= {1, 2, 4, 5, 6}
Hence, LHS = RHS
(ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
Considering LHS,
(B ∪ C) = {x: x ∈ B or x ∈ C}
= {2, 3, 4, 5, 6, 7}
(A ∩ (B ∪ C)) = {x : x ∈ A and x ∈ (B ∪ C)}
= {2, 4, 5}
Considering RHS,
(A ∩ B) = {x : x ∈ A and x ∈ B}
= {2, 5}
(A ∩ C) = {x : x ∈ A and x ∈ C}
= {4, 5}
(A ∩ B) ∪ (A ∩ C) = {x : x ∈ (A∩B) and x ∈ (A ∩ C)}
= {2, 4, 5}
Hence, LHS = RHS
(iii) A ∩ (B – C) = (A ∩ B) – (A ∩ C)
We know,
B – C is defined as {x ∈ B : x ∉ C}
Given,
B = {2, 3, 5, 6}
C = {4, 5, 6, 7}
B – C = {2, 3}
Considering LHS,
(A ∩ (B – C)) = {x : x ∈ A and x ∈ (B – C)}
= {2}
Considering RHS,
(A ∩ B) = {x: x ∈ A and x ∈ B}
= {2, 5}
(A ∩ C) = {x : x ∈ A and x ∈ C}
= {4, 5}
(A ∩ B) – (A ∩ C) is defined as {x ∈ (A ∩ B) : x ∉ (A ∩ C)}
= {2}
∴ LHS = RHS
(iv) A – (B ∪ C) = (A – B) ∩ (A – C)
Considering LHS,
(B ∪ C) = {x : x ∈ B or x ∈ C}
= {2, 3, 4, 5, 6, 7}
A – (B ∪ C) is defined as {x ∈ A : x ∉ (B ∪ C)}
A = {1, 2, 4, 5}
(B ∪ C) = {2, 3, 4, 5, 6, 7}
A – (B ∪ C) = {1}
Considering RHS,
(A – B) = A – B is defined as {x ∈ A : x ∉ B}
Given,
A = {1, 2, 4, 5}
B = {2, 3, 5, 6}
A – B = {1, 4}
(A – C)
A – C is defined as {x ∈ A : x ∉ C}
A = {1, 2, 4, 5}
C = {4, 5, 6, 7}
A – C = {1, 2}
(A – B) ∩ (A – C) = {x : x ∈ (A – B) and x ∈ (A – C)}.
= {1}
Hence, LHS = RHS
(v) A – (B ∩ C) = (A – B) ∪ (A – C)
Considering LHS,
(B ∩ C) = {x : x ∈ B and x ∈ C}
= {5, 6}
A – (B ∩ C) is defined as {x ∈ A : x ∉ (B ∩ C)}
A = {1, 2, 4, 5}
(B ∩ C) = {5, 6}
(A – (B ∩ C)) = {1, 2, 4}
Considering RHS,
(A – B) = A – B is defined as {x ∈ A : x ∉ B}
A = {1, 2, 4, 5}
B = {2, 3, 5, 6}
A – B = {1, 4}
(A – C) = A – C is defined as {x ∈ A : x ∉ C}
A = {1, 2, 4, 5}
C = {4, 5, 6, 7}
A – C = {1, 2}
(A – B) ∪ (A – C) = {x : x ∈ (A – B) OR x ∈ (A – C)}.
= {1, 2, 4}
Therefore, LHS = RHS
(vi) A ∩ (B △ C) = (A ∩ B) △ (A ∩ C)
A = {1, 2, 4, 5} B = {2, 3, 5, 6} C = {4, 5, 6, 7}.
Considering LHS,
A ∩ (B △ C)
B △ C = (B – C) ∪ (C – B) = {2, 3} ∪ {4, 7} = {2, 3, 4, 7}
A ∩ (B △ C) = {2, 4}
Considering RHS,
A ∩ B = {2, 5}
A ∩ C = {4, 5}
(A ∩ B) △ (A ∩ C) = [(A ∩ B) – (A ∩ C)] ∪ [(A ∩ C) – (A ∩ B)]
= {2} ∪ {4}
= {2, 4}
Therefore, LHS = RHS
问题3.如果U = {2，3，5，7，9}是通用集，而A = {3，7}，B = {2，5，7，9}，则证明：
(i)(A∪B)’= A’∩B’
(ii)(A∩B)’= A’∪B’
解决方案：
(i) (A ∪ B)’ = A’ ∩ B’
Considering LHS,
A ∪ B = {x : x ∈ A or x ∈ B}
= {2, 3, 5, 7, 9}
(A ∪ B)’ means Complement of (A ∪ B) with respect to universal set U.
So, (A ∪ B)’ = U – (A ∪ B)’
Now,
U – (A ∪ B)’ is defined as {x ∈ U: x ∉ (A ∪ B)’}
U = {2, 3, 5, 7, 9}
(A ∪ B)’ = {2, 3, 5, 7, 9}
U – (A ∪ B)’ = ϕ
Considering RHS,
We have, A’ = U – A
(U – A) is defined as {x ∈ U : x ∉ A}
U = {2, 3, 5, 7, 9}
A = {3, 7}
A’ = U – A = {2, 5, 9}
Now, B’ = U – B
(U – B) is defined as {x ∈ U : x ∉ B}
U = {2, 3, 5, 7, 9}
B = {2, 5, 7, 9}
B’ = U – B = {3}
A’ ∩ B’ = {x : x ∈ A’ and x ∈ C’}.
= ϕ
∴ LHS = RHS
(ii) (A ∩ B)’ = A’ ∪ B’
Considering LHS,
(A ∩ B)’
(A ∩ B) = {x: x ∈ A and x ∈ B}.
= {7}
Also,
(A ∩ B)’ = U – (A ∩ B)
U – (A ∩ B) is defined as {x ∈ U : x ∉ (A ∩ B)’}
U = {2, 3, 5, 7, 9}
(A ∩ B) = {7}
U – (A ∩ B) = {2, 3, 5, 9}
(A ∩ B)’ = {2, 3, 5, 9}
Considering RHS,
So, A’ = U – A
(U – A) is defined as {x ∈ U : x ∉ A}
U = {2, 3, 5, 7, 9}
A = {3, 7}
A’ = U – A = {2, 5, 9}
So, B’ = U – B
(U – B) is defined as {x ∈ U : x ∉ B}
U = {2, 3, 5, 7, 9}
B = {2, 5, 7, 9}
B’ = U – B = {3}
A’ ∪ B’ = {x : x ∈ A or x ∈ B}
= {2, 3, 5, 9}
∴ LHS = RHS
问题4.对于任何两个集合A和B，证明
(i)B⊂A∪B
(ii)A∩B⊂A
(iii)A⊂B⇒A∩B = A
解决方案：
(i) B ⊂ A ∪ B
Considering an element ‘p’ belonging to B = p ∈ B
p ∈ B ∪ A
B ⊂ A ∪ B
(ii) A ∩ B ⊂ A
Considering an element ‘p’ belonging to B = p ∈ B
p ∈ A ∩ B
p ∈ A and p ∈ B
A ∩ B ⊂ A
(iii) A ⊂ B ⇒ A ∩ B = A
Considering an element ‘p’ belonging to B = p ∈ B
p ∈ A ⊂ B
Now, x ∈ B
Let and p ∈ A ∩ B
x ∈ A and x ∈ B
x ∈ A and x ∈ A (since, A ⊂ B)
Therefore, (A ∩ B) = A
问题5.对于任何两个集合A和B，证明以下陈述是等效的：
(i)A⊂B
(ii)A – B = ϕ
(iii)A∪B = B
(iv)A∩B = A
解决方案：
(i) A ⊂ B
We need to prove (i) = (ii), (ii) = (iii), (iii) = (iv), (iv) = (v)
Let us prove, (i) = (ii)
We know, A – B = {x ∈ A : x ∉ B} as A ⊂ B,
Now, Each element of A is an element of B,
∴ A – B = ϕ
Therefore, (i) = (ii)
(ii) A – B = ϕ
We need to prove that (ii) = (iii)
Let us assume, A – B = ϕ
∴ Every element of A is an element of B
So, A ⊂ B and so A ∪ B = B
Hence, (ii) = (iii)
(iii) A ∪ B = B
We need to show that (iii) = (iv)
By assuming A ∪ B = B
∴ A⊂ B and so A ∩ B = A
Hence, (iii) = (iv)
(iv) A ∩ B = A
Finally, now we need to show (iv) = (i)
By assuming A ∩ B = A
Since, A ∩ B = A, so A ⊂ B
Hence, (iv) = (i)
问题6.对于三个集合A，B和C，证明
(i)A = B = A = C不一定意味着B =C。
(ii)A⊂B⇒C – B⊂C – A
解决方案：
(i) A ∩ B = A ∩ C need not imply B = C.
Let us assume, A = {1, 2}
B = {2, 3}
C = {2, 4}
Then,
A ∩ B = {2}
A ∩ C = {2}
Therefore, A ∩ B = A ∩ C, where, B is not equal to C
(ii) A ⊂ B ⇒ C – B ⊂ C – A
Given: A ⊂ B
Let us assume x ∈ C – B
⇒ x ∈ C and x ∉ B [by definition C – B]
⇒ x ∈ C and x ∉ A
⇒ x ∈ C – A
Therefore, x ∈ C – B ⇒ x ∈ C – A for all x ∈ C – B.
∴ A ⊂ B ⇒ C – B ⊂ C – A
问题7.对于任何两组，请证明：
(i)A∪(A∩B)= A
(ii)A∩(A∪B)= A
解决方案：
(i) A ∪ (A ∩ B) = A
Since, union is distributive over intersection, we have, A ∪ (A ∩ B)
(A ∪ A) ∩ (A ∪ B) [Since, A ∪ A = A]
A ∩ (A ∪ B)
= A
(ii) A ∩ (A ∪ B) = A
Since, union is distributive over intersection, we have, (A ∩ A) ∪ (A ∩ B)
A ∪ (A ∩ B) [Since, A ∩ A = A]
= A