Given a non-empty set $X$,consider the binary operation $^*: P(X) \times P(X) \rightarrow P(X)$ defined by $A \,^*\, B = A \cap B$ for all $A, B \in P(X)$,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.

(A) It is given that the binary operation $^*: P(X) \times P(X) \rightarrow P(X)$ is defined by $A \,^*\, B = A \cap B$ for all $A, B \in P(X)$.
We know that for any set $A \in P(X)$,$A \cap X = A$ and $X \cap A = A$.
This implies $A \,^*\, X = A$ and $X \,^*\, A = A$ for all $A \in P(X)$.
Thus,$X$ is the identity element for the given binary operation $^*$.
Now,an element $A \in P(X)$ is invertible if there exists an element $B \in P(X)$ such that $A \,^*\, B = X$ and $B \,^*\, A = X$ (since $X$ is the identity element).
This means $A \cap B = X$ and $B \cap A = X$.
Since $A \subseteq X$ and $B \subseteq X$,the intersection $A \cap B$ can only be equal to $X$ if $A = X$ and $B = X$.
Therefore,$X$ is the only invertible element in $P(X)$ with respect to the given operation $^*$.
Hence,the result is proved.