Find a nonempty set $A$ such that $A\cap P(A)=\emptyset$ 
*

*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?

You need to find a set $A$, none of whose members are subsets of $A$. There's lots of those! The simplest possible example would be a set of one element; can you find one which works?
Another hint: the cardinality of a subset of a finite set is less than or equal to the cardinality of the set. So if $A$ is finite, all its subsets are of cardinality less than or equal to $|A|$. Can you make finite $A$ so that all its members have infinite cardinality, but all its subsets have finite cardinality?
