If $P(A)$ is a set, $A$ is a set?

Question

I know from a ZFC axiom that if $A$ is a set then $P(A)$ (the powerset of A) is a set. Is it possible to prove the converse? Or is it an independence property?

Answer 1

The axioms of union and extensionanilty can be used to prove that if $B$ is a set, then $\cup B$ is also a set. Now consider $B=P(A)$.

Answer 2

To get round the objections, rephrase your question as: Suppose B is a set and $\phi(x)$ a formula such that $\forall s(s\in B \iff (\forall x\in s) \phi(x))$. Then is there a set A such that $B = P(A)$? And the answer is yes, $A = \bigcup B = \{x|\phi(x)\}$.