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This is a part of exercise E4 from Chapter VII of Kunen's Set Theory.

The hint (courtesy of A. Miller) goes like this: let ${\bf P} = Fn(I,2)$, $(I \geq \omega_{1})^M$. Let G be ${\bf P}$-generic over M, N = ${\bf L}(\mathcal P \left({\omega}\right))^{M[G]}$, and assume AC in N such that $(\kappa = |\mathcal P \left({\omega}\right)|)^N$. Then $\mathbb{1} \Vdash (\check{\kappa} = |\mathcal P \left({\omega}\right)|)^{{\bf L}(\mathcal P \left({\omega}\right))}$.

At first glance it seems that this follows from the 'almost homogeneity' of $Fn(I, 2)$. But $\mathcal P \left({\omega}\right)^{M[G]}$ is not in M, and thus does not have a canonical name in M. I cannot determine if $\mathcal P \left({\omega}\right)^{M[G]}$ is independent of G. If not, does the transition from M[G] to ${\bf L}(\mathcal P \left({\omega}\right))^{M[G]}$ somehow manage to preserve the cardinality of $\mathcal P \left({\omega}\right)$? What am I missing here?

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I'm a little confused by your question, since if we add uncountably many Cohen reals, which is what your forcing does, then the inner model $L(\mathbb{R})$, which is the same as your $L(P(\omega))$, does not actually satisfy the axiom of choice, so your assumption that $N\models\text{AC}$ is not warranted. In particular, $P(\omega)$ does not have a (well-ordered) cardinality there, and there is no such cardinal $\kappa$, if you mean cardinal in the sense of initial ordinal. To see why $L(\mathbb{R})$ doesn't satisfy AC, see Andres Caicedo's beautiful answer to mbsq's question on elementary embeddings.

But perhaps you are making that assumption towards contradiction, or perhaps you have another definable model in mind, where $P(\omega)$ does have a cardinality. In this case, let me answer that yes, homogeneity is all you need. You don't need to worry about having a name for $P(\omega)$, since this is a definable set in the extension. The point is that you don't need to use parameters explicitly when they are definable. If you define $\kappa$ as some cardinal obtained in a definable way from the new $P(\omega)$ in the extension, then if some condition forces that a particular $\kappa$ satisfies that definition, then by homogeniety all conditions must force that that cardinal satisfies the definition, since the only parameter in the assertion that $\kappa$ satisfied the definition was $\check\kappa$ itself, which is invariant under the induced automorphisms of the forcing. So in this case, such a $\kappa$ would indeed be independent of $G$.

(I don't have Kunen's text here, so I'm not sure what the exercise was.)

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Thank you for your reply. Yes the exercise is to prove the negation of Axiom of Choice, so the assumption is used to work toward a contradiction. Can you briefly explain why definable sets need not be taken into account as free variables in the formula? Kunen's treatment of forcing only mentions the correlation between forcing and semantic truth in the extension in the context of names. – Z H Lee Apr 14 '12 at 16:18
The point is that you can state the property as a property about $\kappa$ alone, not a property about $\kappa$ as it relates to some other object. Let $\varphi(x)=$ "$x$ is the cardinality of the reals inside the inner model which consists of all sets constructible from the set of reals". This is the property that you are assuming defines $\kappa$ in $M[G]$. – Joel David Hamkins Apr 14 '12 at 16:28
Ah ok this more or less clears it up. Thanks a lot for the explanation! – Z H Lee Apr 14 '12 at 16:38

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