Question

Given a non-empty set $X$, let $\ast :P\left(X\right)\times P\left(X\right)\to P\left(X\right)$ be defined as $A\ast B=(A-B)\cup (B-A),\forall A,BϵP\left(X\right)$. Show that the empty set $\varphi$ is the identity for the operation $\ast$ and all the elements $A$ of $P\left(X\right)$ are invertible with $A^{-1}=A$

Answer 1

Identity

$e$ is the identity of $\ast$ if

$a\ast e=e\ast a=a$

here,$A\ast \varphi =(A-\varphi )\cup (\varphi -A)=A\cup \varphi =A$

and$\varphi \ast A=(\varphi -A)\cup (A-\varphi )=\varphi \cup A=A$

Since $A\ast \varphi =\varphi \ast A=A$

$\varphi$ is the identity operation $\ast$.

Invertible:-

An element $a$ in set is invertible if,

there is an element in set such that,

$a\ast b=e=b\ast a$

Here,$e=\varphi ,b=A$

Now $A\ast A=(A-A)\cup (A-A)=\varphi \cup \varphi =\varphi$

$EgA\ast A=(A-A)\cup (A-A)=\varphi \cup \varphi =\varphi$

Since $A\ast A=\varphi =A\ast A$

Hence,all the elements $A$ of $P\left(X\right)$ are invertible with inverse of $A=A$.