第11类RD Sharma解决方案–第1集–练习1.6 |套装2

问题8.找到集合A，B和C，使得A∩B和B∩C和A∩C是非空集，并且A∩B∩C = ϕ

解决方案：

Let us consider the sets, 

A = {5, 6, 10}

B = {6, 8, 9}

C = {9, 10, 11}

Now, we have, 

A ∩ B = 6 ≠ ϕ

B ∩ C = 9 ≠ ϕ

A ∩ C = 10 ≠ ϕ

And, A ∩ B ∩ C = ϕ

Now, we have A ∩ B and B ∩ C and A ∩ C as non-empty sets, but A ∩ B ∩ C is empty set. 

问题9.对于任何两个集合A和B，证明A∩B = ϕ => A⊂B’

解决方案：

Let, a ∈ A => a ∉ B

Thus, 

A ∩ B  = ϕ

=> a ∈ B’

Thus, a ∈ A and a ∈ B’ => A ⊂ B’

问题10.证明以下内容：

(i)A – B和A∩B是不交集

(ii)B – A和A∩B是不交集

(iii)A – B和B – A是不交集

解决方案：

(i) A – B and A ∩ B 

Let a  ∈ A – B => a ∈ A and a ∉ B => a ∉ A ∩ B

Therefore, A – B and A ∩ B are disjoint sets.

(ii) Let a ∈ B – A => a ∈ B and a ∉ A => a ∉ A ∩ B

Hence, B – A and A ∩ B are disjoint sets.

(iii) A – B and B – A,

A – B = x, x : x ∈ A and x ∉ B

A – B and B – A are disjoint sets. 

问题11.使用集合的属性，表明对于任意两个集合A和B，

(A∪B)∩(A∩B’)= A

解决方案：

We have, 

LHS = A ∪ B ∩ A ∩ B’

Solving this, we get, 

= A ∪ B ∩ A ∪ A ∪ B ∩ B’

= A ∪ A ∪ B ∩ B’

Since, B ∩ B’ = ∅

= A ∪ A ∩ B’

= A 

Therefore, LHS = RHS.

问题12。

(i)证明对于任何两个A和B集，

A’UB = U => A⊂B

(ii)证明对于任何两个A和B集，

B’⊂A’= U => A⊂B

解决方案：

(i) Let a ∈ A

= a ∈ U

= a ∈ A’ ∪ B, because, U = A’ ∪ B

= a ∈ B, because a ∉ A’

Hence, A ⊂ B

(ii) Let a ∈ A

= a ∉ A’

= a ∉ B’, because, B’ is a subset of A’

= a ∈ B

Hence, A ⊂ B

问题13：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：

Result is False.

Proof:

Let X ∈ P(A) ∪ P(B)

= X ∈ P(A) or X ∈ P(B) 

= X ⊂ A or X ⊂ B

= X ⊂ A U B

= X ∈ P(A ∩ B)

Thus, P(A) ∪ P(B) ⊂ P(A ∪ B)

Also, Let us assume, 

X ∈ P(A ∪ B). But, X ∉ P(A) or X ∉ P(B) 

For instance, we have X = 1, 2, 3, 4 and A = 2, 5 and B = 1, 3, 4.

So, X ∉ P(A) ∪ P(B) 

Therefore, P(A ∪ B) doesn’t necessarily have to be a subset of P(A) ∪ P(B).

问题14。

(i)证明对于任何集合A和B，

A =(A∩B)∩(A – B)

(ii)证明对于任何集合A和B，

A∪(B – A)= A∪B

解决方案：

(i) We have,

RHS = (A ∩ B) ∪ (A – B)

= (A ∩ B) ∪ (A ∩ B)’

= (A ∩ B) ∪ (A ∩ A) ∩ (B ∪ B)’

= A ∩ (A ∪ B)’ ∩ (B ∪ B)’

= A ∩ (A ∪ B)’ ∩ U

= A ∩ (A ∪ B)’

= A

Therefore, RHS = LHS

(ii) We have,$\bigcup\limits_{r=1}^{20} X_{r} = S$

LHS = A ∪ (B – A)

= A ∪ (B ∩ A)’

= (A ∪ B) ∩ (A ∪ A)’

= (A ∪ B) ∩ U

LHS = A ∪ B = RHS

问题15：每个集合X包含5个元素，每个集合Y包含2个元素，并且$\bigcup\limits_{r=1}^{20} X_{r} = S = \bigcup\limits_{r=1}^{n} Y_{r}$ S的每个元素恰好属于Xr的10个和Yr的4个，然后找到n的值。

解决方案：

We have, Each set X contains 5 elements, and $\bigcup\limits_{r=1}^{20} X_{r} = S$

Therefore, n(S) = 20 x 5 = 100

But, we know, that each of the element of S belong to exactly 10 of the X_r‘s.

Therefore, n(S) = 100/10 = 10        -(1)

Also, Y contains 2 elements and

$\bigcup\limits_{r=1}^{n} Y_{r} = S$

Therefore, n(S) = n x 2 = 2n 

Each of the element of S belong to exactly 4 of the Yr’s.

n(S) = 2n/4 = n/2         -(2)

From equation (1) and (2)

10 = n/2

n = 20