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It is a well-known theorem that if $\kappa$ is measurable, then there is a generic extension in which $\kappa$ is no longer weakly compact, but we can force its weak compactness back and recover the full measurability.

Is it possible to refine this result?

Question. Suppose that $\kappa$ is measurable, is there a forcing $\Bbb{P*Q}$ which preserves the measurability of $\kappa$, but $\Bbb P$ only preserves its weak compactness/Ramsey/other large cardinal properties while violating measurability?

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  • $\begingroup$ Apter and Shelah presented a way of doing this different from the well-know Kunen method, but I don't know if it gives what you want. See An interesting way to kill and resurrect measurability. So maybe Miha can say more on this. $\endgroup$
    – Mohammad Golshani
    Commented Dec 23, 2020 at 14:06
  • $\begingroup$ I see. So if the measurable is at least indestructible under Cohen forcing (as a WC cardinal, at the very least), then the first step should "probably" preserve WC, right? $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 23, 2020 at 16:28
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    $\begingroup$ So are you asking for something like "killing them softly" and then resurrecting? $\endgroup$
    – Miha Habič
    Commented Dec 23, 2020 at 23:58
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    $\begingroup$ for weak compactness, how about adding a weakly compact set that has no weakly compact initial segment, followed by shooting a 1-club through its complement (with suitable preparation from below)? $\endgroup$
    – Otto
    Commented Dec 27, 2020 at 10:52
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    $\begingroup$ @AsafKaragila What about the following idea: start with supercompact $\kappa$ and force with reverse Easton iteration to add $\alpha^{++}$-many Cohen subsets to each inaccessible $\alpha \leq \kappa.$ $\kappa$ remains supercompact in the extension. For intermediate submodel consider the iteration below $\kappa$ the same, but at $\kappa$ just add one $\kappa$-Cohen set! $\endgroup$
    – Mohammad Golshani
    Commented Jan 2, 2021 at 12:09

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