Timeline for fake and weak cardinals
Current License: CC BY-SA 3.0
22 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 9, 2018 at 14:41 | comment | added | Yair Hayut | ... then the supercompact and thus $I(\lambda)$ is full except a single cofinality ($\mathrm{cf} \kappa$), which is impossible. | |
| Apr 9, 2018 at 14:39 | comment | added | Yair Hayut | @NotMike: note that $\mu_\omega$ is not a cardinal in $V$. Under GCH, it has size $\mu^+$. In fact, strengthening a bit the assumptions on $M$, I think that I can show that there are no fake cardinals above a supercompact cardinal: let's assume that M has a definable global well order. Then, in L(M), the axiom of choice holds and $L(M) \cap P(\kappa) = M \cap P(\kappa)$. In particular, $\lambda$ is successor of regular in $L(M)$ and we can apply Shelah's theorem and obtain that the approachability ideal $I(\lambda)$ contains $\lambda \setminus (S^\lambda_\kappa)^M$. Now, $\kappa$ is larger... | |
| Apr 9, 2018 at 6:55 | comment | added | Not Mike | @Yair I think the following might yield a fake cardinal above the supercompact; fix some measurable cardinal $\mu$ above the supercompact, and consider the model $M$ which is the direct limit of the first $\omega$ many iterates of the ultrapower generated by a normal measure on $\mu$; letting $\mu_\omega = \sup \{ j_k(\mu): k \in \omega\}\in M$, we have that $M$ thinks $\mu_\omega$ is regular, so unless I'm mistaken, $M$ should witness that $\mu_\omega^{+}$ is fake. (Fixed some typos) | |
| Apr 8, 2018 at 18:59 | comment | added | Yair Hayut | Do you know if successor of a singular cardinal of small cofinality above a supercompact cardinal can be fake? | |
| Apr 8, 2018 at 8:06 | comment | added | Monroe Eskew | It also follows from the existence of $0^\sharp$ that every cardinal is inaccessible in L. | |
| Apr 8, 2018 at 7:58 | comment | added | Not Mike | @Monroe Ohhh! I see it now, it might be the case that you've done something like, take a pair of measurable cardinals, force the larger to be the successor of the smaller, then force the smaller to have countable cofinality.. Cute. (I see my mistake now; interesting) | |
| Apr 8, 2018 at 7:23 | comment | added | Monroe Eskew | $\lambda$ might not be a successor in L. | |
| Apr 8, 2018 at 4:29 | comment | added | Not Mike | Assuming $\lambda$ is fake; why wouldn't the model $L_{\lambda^{+}}$ witness that $\lambda$ is both fake and weak? Being regular is a $\Pi_1$-statement and since $\lambda$ being "fake" is equivalent to $\mu = \cup \{ \kappa \in \lambda: \text{cf}(\kappa) = \kappa \} \in \lambda$ being regular; you've automatically got that $L_{\lambda^{+}}$ witnesses that $\lambda$ is fake. Isn't it the case that $L_{\lambda^{+}} \Vdash \mu^{<\mu} = \mu$? | |
| Apr 6, 2018 at 8:28 | answer | added | Asaf Karagila♦ | timeline score: 1 | |
| Apr 5, 2018 at 19:49 | history | edited | Monroe Eskew |
added 26 characters in body
|
|
| Apr 5, 2018 at 19:44 | history | edited | Monroe Eskew | CC BY-SA 3.0 |
added 26 characters in body
|
| Apr 5, 2018 at 19:39 | history | edited | Monroe Eskew | CC BY-SA 3.0 |
added 26 characters in body
|
| Apr 5, 2018 at 16:03 | comment | added | Monroe Eskew | @Asaf, this sounds right, but I’m not sure if inner model theory is developed enough to make this claim in full generality. Maybe the question can be answered with some anti-LC assumption, or some crazy conjecture about ultimate L. | |
| Apr 5, 2018 at 15:53 | comment | added | Asaf Karagila♦ | If $\lambda$ is fake, isn't some sufficiently nice core model will give you the witness for being weak too? | |
| Apr 5, 2018 at 15:37 | comment | added | Monroe Eskew | Yes, now you have it right. Both are supposed to be $\Sigma_1$-ZF properties. | |
| Apr 5, 2018 at 15:35 | comment | added | Yair Hayut | So $\lambda$ is weak if there is a transitive $M$ that witnesses its fakeness while having $\kappa^{<\kappa} = \kappa$? I see that I misunderstood the definition. I thought that you require that any transitive $M$ that witnesses the fakeness also witness the weakness. | |
| Apr 5, 2018 at 15:19 | comment | added | Monroe Eskew | No, there is still $H(\kappa^+)^V$. | |
| Apr 5, 2018 at 14:57 | comment | added | Yair Hayut | 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). | |
| Apr 5, 2018 at 14:25 | comment | added | Asaf Karagila♦ | A sad cardinal is one that won the popular vote, but not the college of cardinals? | |
| Apr 5, 2018 at 14:12 | comment | added | Monroe Eskew | I'll leave it to you to define sad cardinal. | |
| Apr 5, 2018 at 14:02 | comment | added | Asaf Karagila♦ | Jesus, Agent Orange is really getting to you, isn't he? SAD. | |
| Apr 5, 2018 at 13:52 | history | asked | Monroe Eskew | CC BY-SA 3.0 |