Given a non-empty set X, let * : P(X) x P(X) → P(X) be defined as A * B = (A – B) ∪ (B – A), ∀ A, B ∈ P(X). Show that the empty set ϕ is the identity for the operation * and all the elements A of P(X) are invertible with A–1 = A.
Let A be any element of P(X).
(i) Now A * ϕ = (A – ϕ) ∪ (ϕ – A)
= (A ∪ ϕ’) ∪ (ϕ ∩ A') [∵ A – B = A ∩ B']
= (A ∩ U) ∩ ϕ
= A u ϕ = A
∴ A * ϕ = A ...(1)
Also ϕ * A = (ϕ – A) ∪ (A – ϕ)
= (ϕ ∩ A') ∪ (A ∩ ϕ')
= ϕ ∪ (A ∩U)
= ϕ ∪ A = A
∴ ϕ * A = A ...(2)
From (1) and (2). we get,
A * ϕ = A = ϕ * A ∀ A ∈ P(X)
∴ ϕ is the identity element of (P(A), *)
(ii) A * A = (A – A) ∪ (A – A)
= ϕ ∪ ϕ = ϕ ∴ A * A = ϕ A is invertible and its inverse is A.
⇒ A is invertible ∀ A ∈ P(X) and A –1 = A.