If A = {a, b, c, d, e}, B = {a, c, e, g}, verify that:
(i) A ∪ B = B ∪
A
(ii) A ∪ C = C ∪ A
(iii) B ∪ C = C ∪ B
(iv) A ∩ B = B ∩ A
(v) B ∩ C = C ∩ B
(vi) A ∩ C = C ∩ A
(vii) (A ∪ B ∪ C = A ∪ (B ∪ C)
(viii) (A ∩ B) ∩ C = A ∩ (B ∩ C)
(i) LHS = A ∪ B
= {a, b, c, d, e}∪ {a, c, e, g}
= { a, b, c, d, e, g}
= {a, c, e, g}∪{a, b, c, d, e}
= B ∪ A
= RHS
Hence proved.
(ii) To prove: A ∪ C = C ∪
A
Since the element of set C is not provided,
let x be any element of C.
LHS = A ∪ C
= {a, b, c, d, e} ∪ { x |x
C}
= { a, b, c, d, e, x}
= {x, a, b, c, d, e }
= { x |x
C} ∪ { a, b, c, d, e}
= C ∪ A
= RHS
Hence proved.
(iii) To prove: B ∪ C = C ∪
B
Since the element of set C is not provided,
let x be any element of C.
LHS = B ∪ C
= { a, c, e, g } ∪ { x |x
C}
= { a, c, e, g, x}
= {x, a, c, e, g }
= { x |x
C} ∪ { a, c, e, g }
= C ∪ B
= RHS
Hence proved.
(iv) LHS = A ∩ B
= {a, b, c, d, e}∪ {a, c, e, g}
= {a, c, e}
RHS = B ∩ A
= {a, c, e, g}∩{a, b, c, d, e}
= {a, c, e}
A ∩ B = B ∩ A
(v) Let x be an element of B ∩ C
x
B ∩ C
x
B and x
C
x
C and x
B [by definition of intersection]
x
C ∩ B
B ∩ C 
C ∩ B ….(i)
Now let x be an element of C ∩ B
Then, x
C ∩ B
x
C and x
B
x
B and x
C [by definition of intersection]
x
B ∩ C
C ∩ B 
B ∩ C ….(ii)
From (i) and (ii) we have,
B ∩ C = C ∩ B [ every set is a subset of itself]
Hence proved.
(vi) Let x be an element of A ∩ C
x
A ∩ C
x
A and x
C
x
C and x
A [by definition of intersection]
x
C ∩ A
A ∩ C 
C ∩ A ….(i)
Now let x be an element of C ∩ A
Then, x
C ∩ A
x
C and x
A
x
A and x
C [by definition of intersection]
x
A ∩ C
C ∩ A 
A ∩ C ….(ii)
From (i) and (ii) we have,
A ∩ C = C ∩ A [ every set is a subset of itself]
Hence proved.
(vii) Let x be any element of (A ∪ B) ∪ C
x
(A ∪ B) or x 
C
x
A or x
B or x 
C
x
A or x
(B ∪ C)
x
A ∪ (B ∪ C)
(A ∪ B) ∪ C 
A ∪ (B ∪ C) …..(i)
Now, let x be an element of A ∪ (B ∪ C)
Then, x
A or (B ∪ C)
x
A or x
B or x 
C
x
(A ∪ B) or x 
C
x
(A ∪ B) ∪ C
A ∪ (B ∪ C) 
(A ∪ B) ∪ C …..(ii)
From i and ii, (A ∪ B) ∪ C = A ∪ (B ∪ C)
[ every set is a subset of itself]
Hence , proved.
(viii) Let x be any element of (A ∩ B) ∩ C
x
(A ∩ B) and x 
C
x
A and x
B and x 
C
x
A and x
(B ∩ C)
x
A ∩ (B ∩ C)
(A ∩ B) ∩ C 
A ∩ (B ∩ C) …..(i)
Now, let x be an element of A ∩ (B ∩ C)
Then, x
A and (B ∩ C)
x
A and x
B and x 
C
x
(A ∩ B) and x 
C
x
(A ∩ B) ∩ C
A ∩ (B ∩ C) 
(A ∩ B) ∩ C …..(ii)
From i and ii, (A ∩ B) ∩ C = A ∩ (B ∩ C)
[every set is a subset of itself]
Hence, proved.