Question

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

Answer 1

Solve for Identity.
Given: * : P(X) × P(X) $\to$ P(x) be defined as
$A\ast B=(A-B)\cup (B-A),\forall A,B\in P\left(X\right)$.
e is the identity of * if
a * e = e * a = a
Here,
$A\ast \varphi =(A-\varphi )\cup (\varphi -A)$

$=A\cup \varphi =A$

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

$=\varphi \cup A=A$

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

$\varphi$ is the identity of operation *.

solve for Invertible.
An element a in set is invertible if, there is an element b in set such that

a * b = e = b * a

Here, $e=\varphi$
Now,

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

Since, $\mathrm{Sin}ce,A\ast A=\varphi =A\ast A,$
All the elements A of P(X) are invertible.
And the inverse of A = A.