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

$\Phi$ is the empty set. the operation * is defined as A * B = (A - B) U ( B - A).
putting, B = $\Phi$
then, A * $\Phi$= (A - Φ) U (Φ - A)
= A U Φ = A
Φ * A = (Φ - A) U (A - Φ) = Φ U A = A
here, A * Φ = Φ * A = A ∈ P(X).
Therefore, ф is the identity element for the given operation *.
Now, an element A ∈  P(X) will be invertible if there exists B ∈  P(X) such that
A * B = Φ = B * A. (as Φ is an identity element.)
Now, we can see that A * A = (A –A) ∪ (A – A) =Φ ∪ Φ = Φ $\forall$ A ∈ P(X).

Therefore, all the element A of P(X) are invertible with A⁻¹ = A.