Question

Given a none-empty set $X,let\ast :P\left(X\right)\times P\left(X\right)\to P\left(X\right)$ be defined as $A\times B=(A-B)\cup (B-A),\forall A,B\in P\left(X\right)$. Show that the empty set $\varphi$ is the identity for the operation $\ast$ and all the elements A of P(X) are invertible with $A^{-1}=A.$

Answer 1

Given that $\ast :P\left(X\right)\times P\left(X\right)\to P\left(X\right)$ is defined as
$A\ast B=(A-B)\cup (B-A)$ for $A,B\in P\left(X\right)$
Let $A\in P\left(X\right)$, Then, we have
$A\ast \varphi =(A-\varphi )\cup (\varphi -A)=A\cup \varphi =A\varphi \ast A=(\varphi -A)\cup (A-\varphi )=\varphi \cup A=A$
Therefore, $A\ast \varphi =A=\varphi \ast A$ for all $A\in P\left(X\right)$

Thus, $\varphi$ is the identity element for the given operation $\ast .$
Now, an element $A\in P\left(X\right)$ wil be invertible, if there exists $B\in P\left(X\right)$ such that $A\ast B=\varphi =B\ast A$. As $\varphi$ is the identity element.
Now, we observed that $A\ast A=(A-A)\cup (A-A)=\varphi \cup \varphi =\varphi \forall A\in P\left(X\right).$
Hence, all the elements A of P(X) are invertible with $A^{-1}=A.$