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).

Question

Philoid · Accepted Answer

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

A * B = (A – B) ∪ (B – A), A, B ∈ P(X).

Now, let A ϵ P(X). Then, we get,

A * ф = (A – ф) ∪ (ф – A) = A ∪ ф = A

ф * A = (ф - A) ∪ (A - ф) = ф ∪ A = A

⇒ A * ф = A = ф * A, A ϵ P(X)

Therefore, ф is the identity element for the given operation *.

Now, an element A ϵ P(X) will be invertible if there exists B ϵ P(X) such that

A * B = ф = B * A. (as ф is an identity element.)

Now, we can see that A * A = (A –A) ∪ (A – A) = ф ∪ ф = ф A ϵ P(X).

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