Question

Given a non-empty set X, let * : P(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. 

(Hint: (A – ϕ) ∪ (ϕ – A) = A and (A – A) ∪ (A – A) = A * A = ϕ)

Answer 1

ϕ ∈ P (X) be an identity element. 

To prove it let E ∈ P (X) be the identity element such that 

A * E = E * A = A 

∀ A ∈ P (X) (A – E) ∪ (E – A) = A 

⇒ E = φ i.e. (A – ϕ) ∪ (E – ϕ)) = A 

A * ϕ = ϕ * A = A * A = A, 

hence ϕ is the identity element. 

Let Be P(X) be the inverse of A 

∴ A* B = B * A = φ ∀ A e P(X) 

(A – B) ∪ (B – A) = 0 ⇒ B = A 

because A – B = ϕ B – A = ϕ⇒ A = B 

∴ ∀ A ∈ P(X), A * A = ϕ 

A is the invertible element of A 

∴ A^-1 = A