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Prikry forcing is easily seen to be cone homogeneous (for any $p, q \in \mathbb{P}$, there are $p' \leq p, q' \leq q$ and an isomorphism $\Phi: \mathbb{P}/p' \simeq \mathbb{P}/q'$); in particular for any generic filter $G$, $HOD^{V[G]} \subseteq V.$

Now consider the following variation of the Prikry forcing. Let $\kappa$ be measurable and suppose $(U_i: i<\kappa)$ are normal measures on $\kappa$ such that there are $A_i \in U_i,$ which are pairwise disjoint. Let $\mathbb{P}_{\bar{U}}$ consists of all pairs $(t, T)$ where $T \subseteq [\kappa]^{<\omega}$ is a tree with trunk $t$, and for all $t \unlhd u\in T, suc_T(v) \in U_i, suc_T(v) \subseteq A_i,$ where $i=max(v).$ The order is defined in the natural way.

It doesn't seem that the forcing has to be cone homogeneous, except we assume some connection between the measures $U_i$ (they are Rudin-Keisler equivalent, or ...). Now my question is the following:

Question. Assume $G$ is $\mathbb{P}_{\bar{U}}$-generic over $V$. Does $HOD^{V[G]} \subseteq V?$

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  • $\begingroup$ A more general version would allow the ultrafilters to be indexed by elements of $[\kappa]^{<\omega}$. $\endgroup$
    – Goldstern
    Sep 30, 2015 at 15:35

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