# fake and weak cardinals

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.

• Jesus, Agent Orange is really getting to you, isn't he? SAD. – Asaf Karagila Apr 5 '18 at 14:02
• I'll leave it to you to define sad cardinal. – Monroe Eskew Apr 5 '18 at 14:12
• A sad cardinal is one that won the popular vote, but not the college of cardinals? – Asaf Karagila Apr 5 '18 at 14:25
• 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). – Yair Hayut Apr 5 '18 at 14:57
• No, there is still $H(\kappa^+)^V$. – Monroe Eskew Apr 5 '18 at 15:19

## 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.

• 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. – Monroe Eskew Apr 6 '18 at 10:34
• 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. – Monroe Eskew Apr 6 '18 at 11:22
• 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 $\kappa$. – Monroe Eskew Apr 6 '18 at 11:34