Given a non-empty set X, let$*$ : P(X) × 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) $=$ A and (A$-$A )$\cup$(A $-$A) $=$ A $*$ A $=$ $\phi$).
Given a non-empty set X, let$*$ : P(X) × 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) $=$ A and (A$-$A )$\cup$(A $-$A) $=$ A $*$ A $=$ $\phi$).
Official Solution
To show : $\phi$ is the identity
For every A $\in P(X)$, we have
$\phi * A = (\phi - A) \cup (A - \phi ) = \phi \cup A = A$
and $A * \phi = (A - \phi ) \cup (\phi - A) = A \cup \phi = A$
$\Rightarrow$ $\phi$ is the identity element for the operation $*$ on P(X).
Also, A$*$A $=$ (A$-$A) $\cup$ ( A$-$A ) $=$ $\phi \cup \phi = \phi$
$\Rightarrow$ Every element A of P(X) is invertible with ${A^{ - 1}} = A$
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