Question

For a non-empty set $X$, if operation $\ast :P\left(X\right)\times P\left(X\right)\Rightarrow P\left(X\right)$ is defined as $A\ast B=(A-B)\cup (B-A),\forall A,B\in P\left(X\right)$ then show that empty set $\varphi$ is the identity for $\ast ,$ and all elements $A$ of $P\left(X\right)$ are invertible with $A^{-1}=A$

Answer 1

$\pi \in P\left(X\right)$ is an identity element
To prove it let $E\in P\left(X\right)$ be the identity element such that $A\ast E=E$
$\ast A=A$
$\forall A\in P\left(X\right)$
$(A-E)\cup (E-A)=A\Rightarrow E=\varphi$
i.e. $(A-\varphi )\cup (E-\varphi )=A$
$A\ast \varphi =\varphi \ast A=A$
$\ast A=A$, hence $\varphi$ is the identity element
Let $BeP\left(X\right)$ be the inverse of $A$
$\therefore A\ast B=B\ast A=\varphi \forall AeP\left(X\right)$
$(A-B)\cup (B-A)=0\Rightarrow B=A$
becasue $A-B=\varphi B-A=\varphi \Rightarrow A=B$
$\therefore \forall A\in P\left(X\right),A\ast A=\varphi$
$A$ ie the invertible element of $A$
$\therefore A^{-1}=A$