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 \(A^{-1}=A\).