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

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 simplesimply 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 herenot the earlier ones. For this, the situation would have to 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 $\kappa$.

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

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 simple a counterexample at that earlier stage $\delta$.

I believe that one can make this work with measures on $\kappa$ instead of measures on $P_\kappa\kappa^+$, but here the situation would have to 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 $V[G]$.

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

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 $\kappa$.

Source Link
Joel David Hamkins
Joel David Hamkins
  • 236.3k
  • 44
  • 777
  • 1.4k

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

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 simple a counterexample at that earlier stage $\delta$.

I believe that one can make this work with measures on $\kappa$ instead of measures on $P_\kappa\kappa^+$, but here the situation would have to 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 $V[G]$.