Question

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 $x$.

Answer 1

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 $A*B = E = B*A$.