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Usually the question whether the diamond principle $\diamondsuit(\kappa)$ holds for some large cardinal $\kappa$ only concerns large cardinal notions of very low consistency (among the weakly compacts). Partly since it does hold for all subtle cardinals, which are only barely stronger than the weakly compacts, and pretty much every large cardinal notion below a weakly compact has been shown to consistently not satisfy it (see Failure of diamond at large cardinals and Ben Neria ('17)).

That subtle cardinals satisfy diamond of course means that almost all large cardinals do satisfy it as well, but there are some strange ones lying around though, including Woodin cardinals and inaccessible Jónsson cardinals. Is anything known about diamond holding for any of these two?

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    $\begingroup$ The issue is not with diamond holding but rather with its failure. $\endgroup$ – Andrés E. Caicedo Feb 12 '18 at 23:01
  • $\begingroup$ But that's basically the same question, no? $\endgroup$ – Dan Saattrup Nielsen Feb 13 '18 at 8:39
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    $\begingroup$ Diamond hold at them in known core models. The question of whether diamond can fail at a weakly compact cardinal is a major open question. $\endgroup$ – Mohammad Golshani Feb 13 '18 at 8:43
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    $\begingroup$ See Failure of diamond at large cardinals $\endgroup$ – Mohammad Golshani Feb 13 '18 at 8:44
  • $\begingroup$ Yes, but since Woodins aren't weakly compact in general and whether Jonssons are is still open, the 'weakly compact diamond programme' probably wouldn't be able to give an answer to my question. My question is perhaps then if diamond provably holds for Woodins and inaccessible Jonssons. I'll change the title. $\endgroup$ – Dan Saattrup Nielsen Feb 13 '18 at 8:57
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This is a partial answer. I will show that if $\delta$ is Woodin then $\diamondsuit_\delta$ holds.

Claim: Any Woodin cardinal is subtle.

Proof: Let $\delta$ be a Woodin cardinal. Let $\vec{A} = \langle A_\alpha \mid \alpha < \delta\rangle$ be a sequence a sets, $A_\alpha \subseteq \alpha$ and let $C$ be a club in $\delta$. We want to find $\alpha < \beta$ in $C$ such that $A_\alpha = A_\beta \cap \alpha$.

Since $\delta$ is Woodin, there is a cardinal $\kappa < \delta$ which is $\vec{A} \times C$-strong up to $\delta$. Thus, $\kappa \in C$ and there is an elementary emebedding $j\colon V\to M$, such that :

  • $\mathrm{crit}\ j = \kappa$,
  • $j(\vec{A}) \restriction \kappa + 1 = \vec{A} \restriction \kappa + 1$,

In $M$, $j(\vec{A})(j(\kappa)) \cap \kappa = j(\vec{A})(\kappa) = A_\kappa$ and $\kappa, j(\kappa) \in j(C)$. By elementarity, there is $\alpha < \kappa$ in $C$ such that $A_\alpha = A_\kappa \cap \alpha$.

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    $\begingroup$ It's a subtle claim... $\endgroup$ – Asaf Karagila Feb 14 '18 at 9:59
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    $\begingroup$ Ah, wonderful! Thanks Yair. One question: do you mean that $\kappa,j(\kappa)\in j(C)$, where $\kappa\in j(C)$ since $j$ is $C$-strong? I at least can't see why $j(\kappa)$ should be in $C$. $\endgroup$ – Dan Saattrup Nielsen Feb 14 '18 at 12:27
  • $\begingroup$ Yes. It was a typo. $\endgroup$ – Yair Hayut Feb 14 '18 at 15:01
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Assuming that by "inaccessible Jónsson" you meant a regular limit cardinal of uncountable cofinality which is Jónsson; then using the arguments of [1] (Theorem 15 p.115), we have

if $\mathbb{P}$ is c.c.c. and $\kappa$ is Jónsson then for any $V$-generic $G\subset \mathbb{P}$, $V[G]\vDash$ "$\kappa$ is Jónsson".

In particular, if $\kappa$ is Jónsson, and $G\subset \mathbb{P}=\mathsf{Fn}(\kappa^{+}, 2)$ is $V$-generic, then

$V[G] \vDash $ "$\kappa < 2^{\aleph_0}$ and $\kappa$ is Jónsson."

hence $V[G] \vDash \neg \diamondsuit_\kappa$ and $\kappa$ is Jónsson. Moreover, If we started with $\kappa$ which was a regular limit cardinal then the same holds for $\kappa$ in $V[G]$.

[1] Devlin, Keith J., Some weak versions of large cardinal axioms, Ann. Math. Logic 5, 291-325 (1973). ZBL0279.02051.

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  • $\begingroup$ Thank you for the answer. I'm afraid that by 'inaccessible' I mean 'strongly inaccessible' though. $\endgroup$ – Dan Saattrup Nielsen Mar 19 '18 at 13:11
  • $\begingroup$ @DanSaattrupNielsen I thought it seemed to easy.. thanks. $\endgroup$ – Not Mike Mar 19 '18 at 13:42

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