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Vopenka's principle implies the existence of weakly compact cardinals (a proper class of them, I believe). My question is whether Vopenka's principle is consistent with the assertion that the universe itself is weakly compact. Alternatively, can a Vopenka cardinal be weakly compact?

There are several versions of this question, depending on how classes are treated in the formulation of the two statements. I'm hopeful that the answer is not too different for the different formulations.

I suspect that VP does not imply that ORD is weakly compact. After all, VP implies that ORD is Woodin, and I've read that the first Woodin cardinal is not weakly compact. So if consistent, the conjunction VP + ORD is weakly compact might have higher consistency strength than either of its conjuncts.

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    $\begingroup$ What do you mean by "ORD is weakly compact"? At least one version of that is outright inconsistent with $\mathsf{ZFC}$. $\endgroup$
    – Noah Schweber
    Commented Jul 31, 2021 at 16:35
  • $\begingroup$ @NoahSchweber Wow thanks -- that's a kind of mind-blowing inconsistency, I was not aware! For definiteness, maybe I should just stick with the large cardinal formulation, i.e. "Can a Vopenka cardinal be weakly compact?". I think that's unambiguous and not manifestly inconsistent. $\endgroup$
    – Tim Campion
    Commented Jul 31, 2021 at 16:51
  • $\begingroup$ @JosephVanName I'm aware that Cantor's attic has been down for some time. Somebody made a mirror on github at some point -- are you saying that the mirror is problematic? Or are you referring to Joel's blog? $\endgroup$
    – Tim Campion
    Commented Aug 1, 2021 at 1:15
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    $\begingroup$ @JosephVanName In case you're unwilling to circumvent your browser's restrictions to visit Noah's link for security reasons, I'll mention that it's a blog post discussing this paper of Joel and Ali Enayat, where they show that ZFC refutes the statement that Ord satisfies a "definable tree property". The journal link appears not to be open-access, but I think this is the arxiv version. $\endgroup$
    – Tim Campion
    Commented Aug 1, 2021 at 1:28

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Yes, a Vopenka cardinal can be weakly compact, at least assuming the consistency of a huge cardinal (though this is certainly a bit of an overkill). A huge cardinal is a weakly compact (in fact, measurable) Vopenka cardinal.

EDIT: Actually almost huge cardinals suffice to get measurable Vopenka cardinals. Theorem 24.18 of Kanamori's "Higher Infinite" says that if $\kappa$ is almost huge, then there is a normal ultrafilter $U$ on $\kappa$ such that there are $U$-many $\alpha < \kappa$ such that $\alpha$ is a Vopenka cardinal. The argument actually shows that there are $U$-many $\alpha$ such that $\alpha$ is Vopenka AND measurable. To see why, note that part (b) of that theorem follows from the fact that if $j: V \to M$ witnesses the almost-hugeness of $\kappa$, then $M \models$ "$\kappa$ is Vopenka". But $M$ also models "$\kappa$ is measurable" (in fact $U$ itself is in $M$), because

  1. $\kappa$ is measurable in $V$
  2. $j(\kappa)$ is inaccessible in $V$; so in particular all $\kappa$-complete ultrafilters on $\kappa$ are in $V_{j(\kappa)}$; and
  3. $V_{j(\kappa)} \subset M$ because $M$ is closed under $<j(\kappa)$ sequences.

EDIT 2 (regarding Tim's comment): actually $\kappa$ itself is a Vopenka cardinal, because ``$\kappa$ is a Vopenka cardinal" is absolute between any transitive models that have the same powerset of $\kappa$ (and $V$ and $M$ in the above argument have the same powerset of $\kappa$, and $M$ believes that $\kappa$ is a Vopenka cardinal).

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  • $\begingroup$ Thanks, this is a relief to know! I had attempted to piece this together, but all I could find along these lines after a brief search was the fact that the first huge cardinal is not supercompact, so I feared it might not work. $\endgroup$
    – Tim Campion
    Commented Jul 31, 2021 at 17:14
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    $\begingroup$ @TimCampion I just edited to add a reference to the relevant part of Kanamori's book. $\endgroup$
    – Sean Cox
    Commented Jul 31, 2021 at 17:19
  • $\begingroup$ Thanks! If I'm reading things right, then huge = 1-huge and measurable = 0-huge, and (n+1)-huge implies n-huge -- so clearly huge implies measurable. I understand 24.18 to be saying that if $\kappa$ is almost huge, then there are a lot of Vopenka cardinals below $\kappa$, but I don't understand why this implies that $\kappa$ itself is Vopenka. Cantor's attic also asserts that huge implies Vopenka, but the reference is simply to Kanamori without so much as a chapter number. $\endgroup$
    – Tim Campion
    Commented Jul 31, 2021 at 17:34
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    $\begingroup$ @TimCampion Yeah I can see how that would be frustrating. I added an edit with the explanation of why the argument in Kanamori's book gets lots of measurable Vopenka cardinals from an almost huge. $\endgroup$
    – Sean Cox
    Commented Jul 31, 2021 at 18:05
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    $\begingroup$ @TimCampion Actually $\kappa$ itself is Vopenka, because that's just about subsets of $\kappa$, and the almost huge target $M$ has access to that stuff. $\endgroup$
    – Sean Cox
    Commented Jul 31, 2021 at 18:26

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