I am trying to establish whether it is consistent that some property holds at the least weakly compact cardinal. I know that the property holds at measureables.

Hence (hoping everything else goes well), a natural approach would be to find some forcing extension in which the measurable cardinal becomes the least weakly compact.

Can this be done? Since a measurable cardinal has a cofinal sequence of weakly compact cardinals, if such a extension is possible, it would necessarily kill the measurability of the cardinal. A weaker question is if there there is a forcing extension where a measurable cardinal $\kappa$ in the ground model is not measurable but still weakly compact in the generic extension?

Are there forcing extensions that kills weakly compact cardinals? Given a positive solution to the weaker question above, perhaps applying the weakly compact killing extension to that extension, one may be able to make the cardinal the least weakly compact. Although there may be some trouble since in the first extension $\kappa$ may still be a limit of weakly compacts.

Thanks for any information that can be given.


1 Answer 1


The answer is yes. The following is theorem 5 in our paper B. Cody, M. Gitik, J. D. Hamkins, J. Schanker The least weakly compact cardinal can be unfoldable, weakly measurable and nearly $\theta$-supercompact, under review.

Theorem. (Cody, Gitik, Hamkins, Schanker) If $\kappa$ is measurable, then there is a forcing extension in which it is the least weakly compact cardinal, and still weakly measurable.

The basic idea in the proof is to perform an iteration where one kills off all the smaller weakly compact cardinals $\delta<\kappa$ by adding $\delta$-Suslin trees. One forces with $\text{Add}(\delta,\delta^+)$ followed by the $\delta$-Suslin tree forcing $\mathbb{S}_\delta$. At stage $\kappa$, one uses $\text{Add}(\kappa,\kappa^{++})$, without any $\kappa$-Suslin tree forcing. The result is that all the smaller weakly compacts are killed, but $\kappa$ itself survives by a weak compactness lifting argument. If you only care about preserving weak compactness, then you can force with $\text{Add}(\delta,1)*\mathbb{S}_\delta$ at each stage and with $\text{Add}(\kappa,\kappa^+)$ at the top.

We have similar results showing that the least weakly compact cardinal can be unfoldable (which answered a long-standing open question about whether this was impossible) as well as nearly $\theta$-supercompact (see the paper for details).

Theorem. If $\kappa$ is measurable and strongly unfoldable, then there is a forcing extension in which it is the least weakly compact cardinal, unfoldable and weakly measurable.

Theorem. If $\kappa$ is $\theta$-supercompact and strongly unfoldable, then there is a forcing extension, preserving all cardinals in the interval $[\kappa,\theta]$, in which $\kappa$ is the least weakly compact cardinal, unfoldable, weakly measurable and nearly $\theta$-supercompact.

Perhaps the property you have in mind for the least weakly compact cardinal will follow from some of these large cardinal properties.

  • $\begingroup$ Thanks for your response. After some time, I started thinking about this problem again and had some questions: I am interested in turning a measurable into the least weakly compact. I understand that $\mathbb{S}_\delta$ is used to add Suslin trees to kill the weak compactness. What is the role of $Add(\delta, 1)$ and the last $Add(\kappa, \kappa^+)$? From the paper, it seems the last one may be used to preserve some property of $\kappa$ in the extension. However, I am not sure how the intermediate $Add(\delta,1)$ are used. $\endgroup$
    – William
    Jan 24, 2015 at 4:38
  • $\begingroup$ Is it possible that in a further extension, the measurability of $\kappa$ can be restored. For instance by forcing with some of the Suslin trees that were added? $\endgroup$
    – William
    Jan 24, 2015 at 4:40

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