Question

Given a non-empty set X, let *: P(X) × P(X) → P(X) be defined as A * B = (A − B) ∪ (B − A), &mnForE; 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

It is given that *: P(X) × P(X) → P(X) is defined as

A * B = (A − B) ∪ (B − A) &mnForE; A, B ∈ P(X).

Let A ∈ P(X). Then, we have:

A * Φ = (A − Φ) ∪ (Φ − A) = A ∪ Φ = A

Φ * A = (Φ − A) ∪ (A − Φ) = Φ ∪ A = A

∴A * Φ = A = Φ * A. &mnForE; A ∈ P(X)

Thus, Φ 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 the identity element)

Now, we observed that.

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