Coherent sequence of ultrafilters in iterated forcing extensions

Question

Remember that if $\kappa$ is strongly compact, then any ${<}\kappa$-complete filter extends to a ${<}\kappa$-complete ultrafilter.

Let $\Bbb P_\delta=\langle\Bbb P_\alpha,\dot{\Bbb Q}_\alpha\mid \alpha\in\delta\rangle$ be a ${<}\kappa$-distributive forcing iteration with ${<}\kappa$-support such that for each $\alpha\leq\delta$ we have$$\Vdash_{\Bbb P_\alpha}\text{"$\kappa$ is strongly compact"}$$ and assume that in the $\Bbb P_\delta$-extension there is a chain $\langle U_\alpha\mid \alpha\in\delta\rangle$ with $U_\beta\subseteq U_\alpha$ for $\beta<\alpha$ and each $U_\alpha\subseteq(\mathcal P(\kappa))^{\mathbf V^{\Bbb P_\alpha}}$ is a ${<}\kappa$-complete ultrafilter on $\kappa$ in the $\Bbb P_\alpha$-extension.

1. Does there always exist an ultrafilter $U\subseteq\mathcal P(\kappa)$ in the $\Bbb P_\delta$-extension such that $U_\alpha\subseteq U$ for each $\alpha\in\delta$?

   For finite $\delta$ this is clear, since any ${<}\kappa$-complete ultrafilter from the ground model remains a generating set for a ${<}\kappa$-complete filter in the extension, and $\kappa$ is still strongly compact. So the question is really about limit $\delta$.

   Moreover, since $\Bbb P_\delta$ forces $\kappa$ to be strongly compact, we may take any ${<}\kappa$-complete ultrafilter $U$ in the $\Bbb P_\delta$-extension that contains the Frechet filter, and since the Frechet filter is the one from the ground model (by ${<}\kappa$-distributivity), it follows that $\langle U\cap\mathbf V^{\Bbb P_\alpha}\mid \alpha\in\delta\rangle$ is a chain of ultrafilters that all extend to $U$.

2. In case the answer to (1) is No, can we get a Yes under additional assumptions?

   E.g.

   • if $\Bbb P_\delta$ does not collapse cardinals, or
   • if $\kappa$ is Laver-indestructibly supercompact and each $\dot{\Bbb Q}_\alpha$ is ${<}\kappa$-directed closed, or
   • if each $U_\alpha$ is a specific flavour of ultrafilter.

Any references to literature on such generic chains of ${<}\kappa$-complete ultrafilters on uncountable $\kappa$ are welcome as well.

Answer 1

I believe that the answer is no, using some methods from my old paper:

• Hamkins, Joel David, Destruction or preservation as you like it, Ann. Pure Appl. Logic 91, No. 2-3, 191-229 (1998). arXiv:1607.00683, ZBL0949.03047.

The paper shows how to have a supercompact cardinal $\kappa$, say, that is indestructible by certain kinds of forcing but not others. For example, the methods show that we can have a supercompact cardinal $\kappa$ that is indestructible by any length strictly less than $\kappa$ iteration of adding Cohen subsets to $\kappa$, but not by the length $\kappa^+$-iteration, which will destroy even the $\kappa^+$-supercompactness of $\kappa$.

Thus, we will have that $\kappa$ is $\kappa^+$-supercompact in all the partial extensions $V[G_\alpha]$, but not in the $\kappa^+$th extension $V[G]$. We can now proceed to pick a $\kappa$-complete fine measure $U_\alpha$ on $P_\kappa\kappa^+$ in $V[G_\alpha]$, extending the previous, if possible. If we can proceed through all stages $\alpha<\kappa^+$, then we get a counterexample at $\kappa^+$ since $\kappa$ is not $\kappa^+$-supercompact there. And if things go awry earlier, then this is simply a counterexample at that earlier stage $\delta$.

I believe that one can make this work with measures on $\kappa$ instead of measures on $P_\kappa\kappa^+$, since I think my As You Like It methods will show that measurability also is destroyed by the $\kappa^+$ iteration but not the earlier ones. For this, the situation would be that one runs into a problem strictly before $\kappa^+$ since otherwise the union of the ultrafilters on $\kappa$ in $V[G_\alpha]$ would be an ultrafilter in $V[G]$. So we would have a model $V[G_\delta]$ in which $\kappa$ is actually fully supercompact, but there is a tower of measures on $\kappa$ whose union cannot be contained in any measure on $\kappa$.