Question

Given a non-empty set X, let *:P (X)×P (X)$\to$ P (X) be defined as A * B= (A−B)∪(B−A),∀ A,B∈ P (X).
Show that the empty set $\Phi$ is the identity for the operation * and all the elements A of P (X) are invertible with $A^{-1}=A.$
(Hint: $(A-\Phi)\cup(\Phi-A)=A\,and\,(A-A)\cup(A-A)=A*A=\Phi)$.

Answer 1

It is given that : P (X) × P (X) $\to$ P (X) is defined as
A * B = (A − B) ∪ (B − A) ∀ A, B ∈ P (X).
Let A ∈ P (X).
Then, we have:
A * $\Phi$ = (A − $\Phi$) ∪ ($\Phi$ − A) = A ∪ $\Phi$ = A
$\Phi$ * A = ($\Phi$ − A) ∪ (A − $\Phi$) = $\Phi$ ∪ A = A
∴A * $\Phi$ = A = $\Phi$ * A.
∀ A ∈ P (X)
Thus, $\Phi$ 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 = $\Phi$ = B * A.
(As $\Phi$ is the identity element) Now, we observed that .

Hence, all the elements A of P (X) are invertible with $A^{-1}=A$.