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  • $\begingroup$ But the definition of the power set is not quantifier free; and we only want a quantifier free preservation. Moreover $j''\omega=\omega$, since $k\in\omega$ is an element definable by a quantifier free formula. $\endgroup$ Commented Nov 6, 2013 at 21:21
  • $\begingroup$ I answered the second question: is there a embedding from $P(\omega)$ into $L$. The embedding $j$ didn't intend to preserve the power set, but it forced many different sets to be in some $P(B)$. I didn't claim that $j(X)\in P^L(B)$ for some $X \in P^V(\omega)$. I don't agree that $j(k)=k$. For example, in the embedding that Hamkins gave in the question you have $\forall k\,j(k)\neq k$. $\endgroup$ Commented Nov 6, 2013 at 21:40
  • $\begingroup$ Right. I didn't re-read the entire question again. My bad... $\endgroup$ Commented Nov 6, 2013 at 21:52
  • $\begingroup$ Thank you very much for this answer! Your answer got me thinking about the question again, and I posted some new (and old) observations in another answer. It seems that we may be converging to a ZFC resolution of the question. $\endgroup$ Commented Nov 9, 2013 at 12:39