Q. 13   Given a non-empty set X, let* : P(X) \times P(X) \rightarrow P(X) be defined as
A * B = (A - B) \cup (B -A), \;\forall A, B \in 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) = Aand (A -A) \cup (A - A) = A * A = \phi).

Answers (1)

Let  * : P(X) \times P(X) \rightarrow P(X) be defined as  A * B = (A - B) \cup (B -A), \;\forall A, B \in P(X).

       Let  A \in P(X). Then 

                       A * \phi = (A - \phi ) \cup (\phi -A) = A\cup \phi =A

                      \phi * A = (\phi - A ) \cup (A -\phi ) = \phi \cup A =A

          \therefore \, \, \, A*\phi =A= \phi * A      for all  A \in P(X)

 Thus, \phi is identity element for operation *.

An element   A \in P(X)  will be invertible if there exists   B\in P(X),

such that A*B=A=B*A.     (here \phi is identity element)

   A * A = (A - A ) \cup (A -A) = \phi \cup \phi =\phi  \forall A \in P(X)

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