State whether the following statements are true or false:
(i) A ∪ A = A
(ii) A ∩ A = φ
(iii) A ∪ A' = ξ
(iv) A ∩ A' = φ
(v) A ⊂ B ⇒ A ∪ B = B
(vi) A ⊂ B ⇒ A ∩ B = A
(vii) A ∩ φ = A
(viii) A ∪ φ = A.
Answer
(i) A ∪ A = A
A ∪ A = {x : x ∈ A or x ∈ A} = {x : x ∈ A} = A. This is the idempotent law for union.
Hence, the statement A ∪ A = A is true.
(ii) A ∩ A = φ
A ∩ A = {x : x ∈ A and x ∈ A} = {x : x ∈ A} = A, and not φ. By the idempotent law, A ∩ A = A.
Hence, the statement A ∩ A = φ is false.
(iii) A ∪ A' = ξ
A ∪ A' = {x : x ∈ A or x ∉ A}. Every element of the universal set either belongs to A or does not belong to A, so this set is ξ.
Hence, the statement A ∪ A' = ξ is true.
(iv) A ∩ A' = φ
For all x ∈ A ∩ A' we must have x ∈ A and x ∉ A, which is impossible. So A ∩ A' contains no element of ξ.
Hence, the statement A ∩ A' = φ is true.
(v) A ⊂ B ⇒ A ∪ B = B
If A ⊂ B, then every element of A is also an element of B, so A contributes no new element to the union.
∴ A ∪ B = B
Hence, the statement A ⊂ B ⇒ A ∪ B = B is true.
(vi) A ⊂ B ⇒ A ∩ B = A
If A ⊂ B, then every element of A is also an element of B, so all the elements of A are common to A and B.
∴ A ∩ B = A
Hence, the statement A ⊂ B ⇒ A ∩ B = A is true.
(vii) A ∩ φ = A
A ∩ φ = {x : x ∈ A and x ∈ φ}. Since φ has no element, no x can satisfy x ∈ φ, so A ∩ φ = φ and not A. This is the boundedness law.
Hence, the statement A ∩ φ = A is false.
(viii) A ∪ φ = A
A ∪ φ = {x : x ∈ A or x ∈ φ} = {x : x ∈ A} = A, since φ contributes no element. This is the identity law.
Hence, the statement A ∪ φ = A is true.
Fill in the blanks to make each of the following a true statement:
(i) φ' ∩ A = ...
(ii) ξ' ∩ A = ...
Answer
(i) φ' ∩ A = ...
By the law of complementation,
φ' = ξ.
∴ φ' ∩ A = ξ ∩ A = A (by the identity law)
Hence, φ' ∩ A = A.
(ii) ξ' ∩ A = ...
By the law of complementation,
ξ' = φ.
∴ ξ' ∩ A = φ ∩ A = φ (by the boundedness law)
Hence, ξ' ∩ A = φ.
For any two sets A and B, is it true that P (A) ∪ P (B) = P (A ∪ B)? Justify your answer.
Answer
No, the given statement is not true in general.
To justify it, we give an example. Let A = {1, 2} and B = {2, 3}, then
P(A) = {φ, {1}, {2}, {1, 2}} and P(B) = {φ, {2}, {3}, {2, 3}}
∴ P(A) ∪ P(B) = {φ, {1}, {2}, {1, 2}, {3}, {2, 3}}
Now A ∪ B = {1, 2, 3}
∴ P(A ∪ B) = {φ, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}
Here {1, 3} ∈ P(A ∪ B) but {1, 3} ∉ P(A) ∪ P(B), because {1, 3} is neither a subset of A nor a subset of B.
∴ P(A) ∪ P(B) ≠ P(A ∪ B)
Hence, it is not true that P(A) ∪ P(B) = P(A ∪ B).
Let A, B be two sets. Prove that (A − B) ∪ B = A if and only if B ⊂ A.
Answer
First we simplify (A − B) ∪ B.
(A − B) ∪ B = (A ∩ B') ∪ B (∵ X − Y = X ∩ Y')
= (A ∪ B) ∩ (B' ∪ B) (Distributive law)
= (A ∪ B) ∩ ξ (∵ B ∪ B' = ξ)
= A ∪ B (∵ X ∩ ξ = X)
∴ (A − B) ∪ B = A ∪ B ...........(1)
First, let (A − B) ∪ B = A, we shall prove that B ⊂ A.
From (1), A ∪ B = A.
But we know that A ∪ B = A if and only if B ⊂ A.
∴ B ⊂ A
Conversely, let B ⊂ A, we shall prove that (A − B) ∪ B = A.
As B ⊂ A, every element of B is also an element of A, so A ∪ B = A.
∴ From (1), (A − B) ∪ B = A ∪ B = A
Hence, proved that (A − B) ∪ B = A if and only if B ⊂ A.
For any sets A and B, prove that P (A) ∪ P (B) ⊂ P (A ∪ B).
Answer
Let X be any arbitrary element of P(A) ∪ P(B), then
X ∈ P(A) ∪ P(B)
⇒ X ∈ P(A) or X ∈ P(B)
⇒ X ⊂ A or X ⊂ B (by the definition of a power set)
Now, we know that A ⊂ A ∪ B and B ⊂ A ∪ B.
If X ⊂ A, then X ⊂ A ⊂ A ∪ B, so X ⊂ A ∪ B.
If X ⊂ B, then X ⊂ B ⊂ A ∪ B, so X ⊂ A ∪ B.
Thus in either case, X ⊂ A ∪ B
⇒ X ∈ P(A ∪ B) (by the definition of a power set)
So every element of P(A) ∪ P(B) is also an element of P(A ∪ B).
Hence, proved that P(A) ∪ P(B) ⊂ P(A ∪ B).
Show that the following four conditions are equivalent :
(i) A ⊂ B
(ii) A − B = φ
(iii) A ∩ B = A
(iv) A ∪ B = B.
Answer
To show that the four conditions are equivalent, it is enough to prove the cycle of implications
(i) ⇒ (ii) ⇒ (iii) ⇒ (iv) ⇒ (i)
(i) ⇒ (ii)
Let A ⊂ B. Then every element of A is also an element of B, so there is no element which belongs to A but does not belong to B.
∴ A − B = φ
(ii) ⇒ (iii)
Let A − B = φ i.e. A ∩ B' = φ.
We know that,
A = A ∩ ξ (Identity law)
⇒ A = A ∩ (B ∪ B') (∵ B ∪ B' = ξ)
⇒ A = (A ∩ B) ∪ (A ∩ B') (Distributive law)
⇒ A = (A ∩ B) ∪ φ (∵ A ∩ B' = φ, given)
⇒ A = A ∩ B
∴ A ∩ B = A
(iii) ⇒ (iv)
Let A ∩ B = A.
Then A ∪ B = (A ∩ B) ∪ B (putting A = A ∩ B)
= B (Absorption law, since B ∪ (B ∩ A) = B)
∴ A ∪ B = B
(iv) ⇒ (i)
Let A ∪ B = B.
We know that,
A ⊂ A ∪ B (Boundedness law)
⇒ A ⊂ B (∵ A ∪ B = B, given)
∴ A ⊂ B
Hence, proved that the four conditions A ⊂ B, A − B = φ, A ∩ B = A and A ∪ B = B are equivalent.
For any two sets A and B, prove that
(i) (A − B) ∪ B = A ∪ B
(ii) (A − B) ∩ B = φ.
Answer
(i) (A − B) ∪ B = A ∪ B
(A − B) ∪ B = (A ∩ B') ∪ B (∵ X − Y = X ∩ Y')
= (A ∪ B) ∩ (B' ∪ B) (Distributive law)
= (A ∪ B) ∩ ξ (∵ B ∪ B' = ξ)
= A ∪ B (∵ X ∩ ξ = X)
Hence, proved that (A − B) ∪ B = A ∪ B.
(ii) (A − B) ∩ B = φ
(A − B) ∩ B = (A ∩ B') ∩ B (∵ X − Y = X ∩ Y')
= A ∩ (B' ∩ B) (Associative law)
= A ∩ φ (∵ B ∩ B' = φ)
= φ (∵ X ∩ φ = φ)
Hence, proved that (A − B) ∩ B = φ.
For any two sets A and B, show that
(i) (A ∪ B) − B = A − B
(ii) A − (A ∩ B) = A − B.
Answer
(i) (A ∪ B) − B = A − B
(A ∪ B) − B = (A ∪ B) ∩ B' (∵ X − Y = X ∩ Y')
= (A ∩ B') ∪ (B ∩ B') (Distributive law)
= (A ∩ B') ∪ φ (∵ B ∩ B' = φ)
= A ∩ B' (∵ X ∪ φ = X)
= A − B (∵ X ∩ Y' = X − Y)
Hence, proved that (A ∪ B) − B = A − B.
(ii) A − (A ∩ B) = A − B
A − (A ∩ B) = A ∩ (A ∩ B)' (∵ X − Y = X ∩ Y')
= A ∩ (A' ∪ B') (De Morgan law)
= (A ∩ A') ∪ (A ∩ B') (Distributive law)
= φ ∪ (A ∩ B') (∵ A ∩ A' = φ)
= A ∩ B' (∵ φ ∪ X = X)
= A − B (∵ X ∩ Y' = X − Y)
Hence, proved that A − (A ∩ B) = A − B.
For any sets A, B and C, using properties of sets, prove that
(i) A − (A − B) = A ∩ B
(ii) (A − B) ∪ (A − C) = A − (B ∩ C)
(iii) (A − B) ∩ (C − B) = (A ∩ C) − B
(iv) A − (B − C) = (A − B) ∪ (A ∩ C).
Answer
(i) A − (A − B) = A ∩ B
Solving L.H.S.,
A − (A − B) = A − (A ∩ B') (∵ X − Y = X ∩ Y')
= A ∩ (A ∩ B')' (∵ X − Y = X ∩ Y')
= A ∩ (A' ∪ (B')') (De Morgan law)
= A ∩ (A' ∪ B) (∵ (B')' = B)
= (A ∩ A') ∪ (A ∩ B) (Distributive law)
= φ ∪ (A ∩ B) (∵ A ∩ A' = φ)
= A ∩ B (∵ φ ∪ X = X)
Hence, proved that A − (A − B) = A ∩ B.
(ii) (A − B) ∪ (A − C) = A − (B ∩ C)
Solving L.H.S.,
(A − B) ∪ (A − C) = (A ∩ B') ∪ (A ∩ C') (∵ X − Y = X ∩ Y')
= A ∩ (B' ∪ C') (Distributive law)
= A ∩ (B ∩ C)' (De Morgan law)
= A − (B ∩ C) (∵ X ∩ Y' = X − Y)
Hence, proved that (A − B) ∪ (A − C) = A − (B ∩ C).
(iii) (A − B) ∩ (C − B) = (A ∩ C) − B
Solving L.H.S.,
(A − B) ∩ (C − B) = (A ∩ B') ∩ (C ∩ B') (∵ X − Y = X ∩ Y')
= (A ∩ C) ∩ (B' ∩ B') (Commutative and Associative laws)
= (A ∩ C) ∩ B' (∵ B' ∩ B' = B', Idempotent law)
= (A ∩ C) − B (∵ X ∩ Y' = X − Y)
Hence, proved that (A − B) ∩ (C − B) = (A ∩ C) − B.
(iv) A − (B − C) = (A − B) ∪ (A ∩ C)
Solving L.H.S.,
A − (B − C) = A − (B ∩ C') (∵ X − Y = X ∩ Y')
= A ∩ (B ∩ C')' (∵ X − Y = X ∩ Y')
= A ∩ (B' ∪ (C')') (De Morgan law)
= A ∩ (B' ∪ C) (∵ (C')' = C)
= (A ∩ B') ∪ (A ∩ C) (Distributive law)
= (A − B) ∪ (A ∩ C) (∵ X ∩ Y' = X − Y)
Hence, proved that A − (B − C) = (A − B) ∪ (A ∩ C).
For any two sets A and B, prove that
(A − B) ∪ (B − A) = (A ∪ B) − (A ∩ B)
Answer
Taking the left hand side,
L.H.S. = (A − B) ∪ (B − A)
= (A ∩ B') ∪ (B ∩ A') (∵ X − Y = X ∩ Y')
= [(A ∩ B') ∪ B] ∩ [(A ∩ B') ∪ A'] (Distributive law)
= [(A ∪ B) ∩ (B' ∪ B)] ∩ [(A ∪ A') ∩ (B' ∪ A')] (Distributive law)
= [(A ∪ B) ∩ ξ] ∩ [ξ ∩ (A' ∪ B')] (∵ B ∪ B' = ξ and A ∪ A' = ξ)
= (A ∪ B) ∩ (A' ∪ B') (∵ X ∩ ξ = X)
= (A ∪ B) ∩ (A ∩ B)' (De Morgan law)
= (A ∪ B) − (A ∩ B) (∵ X ∩ Y' = X − Y)
= R.H.S.
Hence, proved that (A − B) ∪ (B − A) = (A ∪ B) − (A ∩ B).