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

$(Hint:\; (A- \emptyset )\; \cup \; (\emptyset - A)=A\; and ( A-A) \cup \; (A-A)= A \ast A = \emptyset)$

Toolbox:
• An element $e \in X$ is an identify element if $e * A=A=A*e$ for $A \in X$
• An element A will be invertible if there exists B such that $A*B=e=B*A$
Given in a non-empty set $X$, $*:P(X) \times P(X) \to P(X)$ defined by $A*B=(A-B) \cup (B-A) \qquad A,B \in P(X)$
$\textbf {Step 1: Checking if the empty set}\; \phi \; \textbf{is the identity}$:
An element $e \in X$ is an identify element if $e * A=A=A*e$ for $A \in X$
Let $A \in P(X) \rightarrow$ $A * \phi=(A - \phi) \cup (\phi-A)=A \cup \phi = A$
Similarly, $\phi *A=(\phi -A) \cup (A - \phi)= \phi \cup A=A$
$\Rightarrow A * \phi=\phi *A=A \rightarrow$ $\phi$ the empty set is the identify element for given operation *
$\textbf {Step 2: Checking if the elements are invertible with}\; A^{-1} = A$:
An element A will be invertible if there exists B such that $A*B=e=B*A$
$\Rightarrow A *A=(A -A) \cup (A-A) =\phi \cup \phi = \phi$
Hence all the elements of are invertible and $A^{-1}=A$
edited Mar 20, 2013