11
$\begingroup$

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?

$\endgroup$
5
  • 7
    $\begingroup$ The issue is not with diamond holding but rather with its failure. $\endgroup$ Feb 12, 2018 at 23:01
  • $\begingroup$ But that's basically the same question, no? $\endgroup$ Feb 13, 2018 at 8:39
  • 2
    $\begingroup$ 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. $\endgroup$ Feb 13, 2018 at 8:43
  • 1
    $\begingroup$ See Failure of diamond at large cardinals $\endgroup$ Feb 13, 2018 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. $\endgroup$ Feb 13, 2018 at 8:57

2 Answers 2

13
$\begingroup$

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 $A_\alpha = A_\kappa \cap \alpha$.

$\endgroup$
3
  • 4
    $\begingroup$ It's a subtle claim... $\endgroup$
    – Asaf Karagila
    Feb 14, 2018 at 9:59
  • 1
    $\begingroup$ 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 $C$. $\endgroup$ Feb 14, 2018 at 12:27
  • $\begingroup$ Yes. It was a typo. $\endgroup$
    – Yair Hayut
    Feb 14, 2018 at 15:01
2
+50
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Thank you for the answer. I'm afraid that by 'inaccessible' I mean 'strongly inaccessible' though. $\endgroup$ Mar 19, 2018 at 13:11
  • $\begingroup$ @DanSaattrupNielsen I thought it seemed to easy.. thanks. $\endgroup$
    – Not Mike
    Mar 19, 2018 at 13:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.