Given a non-empty set X, consider the binary operation $*$: P(X) × P(X) $\rightarrow$ P(X) given by A$*$ B $=$ A$\cap$B $\forall$ A, B in P(X), where P(X) is the power set of X. Show that X is the identity element for this operation and X is the only invertible element in $P(X)$ with respect to the operation $*$.
(i) Let E $\in$ P(X) be the identity element.
Then, A $*$ E $=$ E $*$ A $=$ A $\forall$A $\in$ P( X)
$\Rightarrow$ A $\cap$ E $=$ E $\cap$ A $=$ A $\forall$A $\in$ P( X)
$\Rightarrow$ X $\cap$ E $=$ X because X $\in$P(X) $\Rightarrow$ X $\subset$E.
Also,
Thus, E $=$ X. Hence, X is the identity element.
(ii) Let A $\in$ P(X ) be invertible. Then, there exists B $\in$ P(X such that A $*$ B $=$ B $*$ A $=$ X, where X is the identity element.
$\Rightarrow$ A $\cap$ B $=$ B $\cap$ A $=$ X $\Rightarrow$ X $\subset$ A, X $\subset$ B
Also, A, B
$\therefore$ A $=$ X $=$ B.
Hence, X is the only invertible element and A$^{ - 1}$ $=$ B $=$ X.
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