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 ) is defined as A∗B=( A−B )∪( B−A ).

Let x be the identity of ∗, then,

a∗x=x∗a=a

Let A∈P( X ), then,

A∗ϕ=( A−ϕ )∪( ϕ−A ) =A∪ϕ =A

And,

ϕ∗A=( ϕ−A )∪( A−ϕ ) =ϕ∪A =A

Since, A∗ϕ=ϕ∗A=A, so ϕ is the identity of operation ∗.

For invertible element A∈P( X ) of the set, there exist B∈P( X ) such that,

A∗B=ϕ=B∗A

It is observed that,

A∗A=( A−A )∪( A−A ) =ϕ∪ϕ =ϕ

Thus, the elements Aof P( X )are invertible with inverse of A −1 =A.