Find a nonempty set $A$ such that $A\cap P(A)=\emptyset$ 
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*Find a nonempty set $A$ such that $A\cap P(A)=\emptyset$. (as $P(A)$ is power set of$A$).


My solition. Let $A=${$\emptyset, 1$} be set. Then, $P(A)=${$\emptyset$, {$\emptyset$},{$1$},$A$}. Hence, $A\cap P(A)=\emptyset$. Can you check my solition?

Actually, $\{\emptyset,1\} \cap \{\emptyset,\{\emptyset\},\{1\},A\} = \{\emptyset\}$
Hint: $\emptyset \in p(A)$ for any set $A$. So if $\emptyset \in A$, we can never have $A \cap P(A) = \emptyset$. It's important to distinguish $\emptyset=\{\}$, and $\{\emptyset\}$.
