6
$\begingroup$

Suppose $\lambda$ is a successor of a singular cardinal. We will say $\lambda$ fake if there is a transitive set $M$ such that $\lambda \subseteq M$ satisfying $\mathrm{ZFC}^-$ (ZFC without powerset) in which there is a largest $M$-cardinal $\kappa < \lambda$ which is regular in $M$. We will say $\lambda$ is weak if we can find such $M$ and $\kappa$ such that $M \models \kappa^{<\kappa} = \kappa$.

Question: If $\lambda$ is a fake successor of a singular, is it also weak?

Some motivation: To obtain some properties around singular cardinals of high consistency strength, one often creates weak successors of singulars using Prikry-type forcing. But to obtain other such properties, one needs to use successors of singulars that are not weak. These two methods are in tension. In practice, the examples of fake successors of singulars are also weak, since the witnesses may be taken from inner models satisfying GCH. But I am wondering if there is a deeper explanation.

Remark: If $\kappa$ is supercompact and $\mathrm{cf}(\mu)<\kappa<\mu$, then $\mu^+$ is not weak. Using Radin forcing, we can produce a model with many measurable cardinals in which every successor of a singular is weak.

Improve this question
$\endgroup$
17
  • 2
    $\begingroup$ Jesus, Agent Orange is really getting to you, isn't he? SAD. $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 5, 2018 at 14:02
  • 1
    $\begingroup$ I'll leave it to you to define sad cardinal. $\endgroup$
    – Monroe Eskew
    Commented Apr 5, 2018 at 14:12
  • 3
    $\begingroup$ A sad cardinal is one that won the popular vote, but not the college of cardinals? $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 5, 2018 at 14:25
  • 1
    $\begingroup$ Will you get an example of fake non-weak cardinal by taking a measurable cardinal $\kappa$, add a Prikry sequence and $\kappa^+$ many Cohen reals? Take $M$ to be a model of height $\lambda=\kappa^+$ with all the Cohen reals but without the Prikry sequence, ($M = H(\kappa^+)$ of the generic extension by the Cohen reals). $\endgroup$
    – Yair Hayut
    Commented Apr 5, 2018 at 14:57
  • 1
    $\begingroup$ No, there is still $H(\kappa^+)^V$. $\endgroup$
    – Monroe Eskew
    Commented Apr 5, 2018 at 15:19

1 Answer 1

Reset to default
1
$\begingroup$

This doesn't work. I'm leaving this as it might helpful to someone later on.

I think that the answer is yes, but my idea has a gap.

This happens for silly reasons: $V=L$ satisfies $\sf GCH$, and fake cardinals imply the existence of sharps.

Note that if $\lambda$ is a successor of singular cardinal which is not its successor in $L$, then $0^\#$ exists. In particular, if $\kappa$ is the predecessor of $\lambda$, then $\kappa$ is strongly inaccessible in $L$.

So you'd like to look at $L_\lambda$, but remember, $0^\#$ exists. So that's actually a model of $\sf ZFC$, and in particular, has no last cardinal. But now force with $\operatorname{Col}(\kappa,<\lambda)^L$. Now I think that th existence of $0^\#$ implies the we can find an $L$-generic filter, $G$, for this forcing, which in particular, does not add any bounded sets to $\kappa$. But now look at $L_\lambda[G]$ and it is a witness for the weakness of $\lambda$. Sad.

Improve this answer
$\endgroup$
8
  • 1
    $\begingroup$ I will ask Sy about this. $\endgroup$
    – Monroe Eskew
    Commented Apr 6, 2018 at 9:28
  • 1
    $\begingroup$ He says $\kappa$-Cohen generics exist in $L[0^\sharp]$ iff $\kappa$ has countable cofinality in $L[0^\sharp]$. So your argument doesn't work in general. $\endgroup$
    – Monroe Eskew
    Commented Apr 6, 2018 at 10:34
  • 2
    $\begingroup$ Believe me, I have the best sharps. But I don't know if we can derive anything but $0^\sharp$ from the existence of a fake successor of singular. $\endgroup$
    – Monroe Eskew
    Commented Apr 6, 2018 at 11:22
  • 2
    $\begingroup$ I misunderstood Sy. Actually he can show that no matter what $V$ is, if $0^\sharp$ exists and $\kappa$ is regular in $L$ and of uncountable cofinality in $V$, then there cannot exist a $g \in V$ which is $\kappa$-Cohen generic over $L$. So this means if we force with $\mathrm{Add}(\kappa)^L$ over $V$, then somehow this collapses $\kappa$. $\endgroup$
    – Monroe Eskew
    Commented Apr 6, 2018 at 11:34
  • 2
    $\begingroup$ This question is relevant to the remark Sy gave, in fact sharps are way too strong than what's necessary there. $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 10, 2018 at 15:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .