Let \[X\] be a non-empty set and \[P\left( X \right)\] be its power set. Let \[*\] be an operation defined on elements of \[P\left( X \right)\] by \[A*B = A \cap B{\rm{ }}\forall A,B \in P\left( X \right)\], then:
(a) Prove that \[*\] is a binary operation on \[P\left( X \right)\].
(b) Is \[*\] commutative?
(c) Is \[*\] associative?
(d) Find the identity element and inverse element in \[P\left( X \right)\] with respect to \[x\].

Hint: Here, we need to use the binary operations to prove the given conditions and find the identity and inverse element. We will prove that \[*\] is a binary operation on \[P\left( X \right)\] . Then, we will use the binary operations to prove that it is commutative and associative. Finally, we will use the definitions of identity element and inverse element, and find them

Complete step-by-step answer:
(a)
It is given that \[A*B = A \cap B{\rm{ }}\forall A,B \in P\left( X \right)\].
We know that the intersection of two sets belongs to those two sets, that is \[\left( {A \cap B} \right) \in A,B\].
If \[A,B \in P\left( X \right)\], then \[\left( {A \cap B} \right) \in P\left( X \right)\].
Thus, we get
\[*:P\left( X \right) \times P\left( X \right) \to P\left( X \right)\]
Therefore, \[*\] is a binary operation on \[P\left( X \right)\].
(b)
The binary operation \[A*B = A \cap B\] is commutative if \[A*B = B*A{\rm{ }}\forall A,B \in P\left( X \right)\].
Using the binary operation, we get
\[A*B = A \cap B\] and \[B*A = B \cap A\]
The intersection of set A and B, is the same as the intersection of set B and A.
Therefore, we get
\[A \cap B = B \cap A\]
Substituting \[A*B = A \cap B\] and \[B*A = B \cap A\] in the equation, we get
\[A*B = B*A\]
Therefore, we have proved that the binary operation \[*\] is commutative.
(c)
The binary operation \[A*B = A \cap B\] is associative if \[\left( {A*B} \right)*C = A*\left( {B*C} \right){\rm{ }}\forall A,B,C \in P\left( X \right)\].
Using the binary operation, we get
\[A*B = A \cap B\] and \[B*C = B \cap C\]
The intersection of \[A \cap B\] and \[C\] is \[A \cap B \cap C\].
Therefore, we get
\[\left( {A*B} \right)*C = \left( {A \cap B} \right)*C = A \cap B \cap C\]
Now, the intersection of \[A\] and \[B \cap C\] is \[A \cap B \cap C\].
Therefore, we get
\[A*\left( {B*C} \right) = A*\left( {B \cap C} \right) = A \cap B \cap C\]
Therefore, from the equations \[\left( {A*B} \right)*C = \left( {A \cap B} \right)*C = A \cap B \cap C\] and \[A*\left( {B*C} \right) = A*\left( {B \cap C} \right) = A \cap B \cap C\], we get
\[\left( {A*B} \right)*C = A*\left( {B*C} \right)\]
Therefore, we have proved that the binary operation \[*\] is associative.
(d)
Let the set \[E\] be the identity element in \[P\left( X \right)\].
We know that if \[E\] is the identity element in \[P\left( X \right)\], then
\[A*E = A = E*A{\rm{ }}\forall A \in P\left( X \right)\]
Using the binary operation, we get
\[A \cap E = A = E \cap A{\rm{ }}\forall A \in P\left( X \right)\]
Since \[A \in P\left( X \right)\], therefore \[A \subset X\].
Therefore, we get
\[A \cap X = A = X \cap A\]
Hence, \[E = X\].
Therefore, \[X\] is the identity element in \[P\left( X \right)\].
Let the set \[B\] be the inverse element in \[P\left( X \right)\].
We know that if \[B\] is the identity element in \[P\left( X \right)\], then
\[A*B = E = B*A{\rm{ }}\forall A,B \in P\left( X \right)\]
Using the binary operation, we get
\[A \cap B = E = B \cap A{\rm{ }}\forall A,B \in P\left( X \right)\]
Substituting \[E = X\], we get
\[A \cap B = X = B \cap A{\rm{ }}\forall A,B \in P\left( X \right)\]
Since \[A \subset X\], therefore \[B \subset X\].
Therefore, we get
\[X \cap X = X = X \cap X\]
Hence, \[A = B = X\].
Therefore, \[X\] is the inverse element in \[P\left( X \right)\].

Note: In the equation \[A \cap E = A = E \cap A\], we do not need to prove that \[A \cap E = E \cap A\], because the binary operation is commutative. Similarly, in the equation \[A \cap B = E = B \cap A\], we do not need to prove that \[A \cap B = B \cap A\]. Also we need to know if \[E\] is the identity element, then \[A*E = A = E*A\]. If \[B\] is the identity element, then \[A*B = E = B*A\].