Timeline for Reinhardt cardinals and iterability
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 27, 2020 at 7:21 | vote | accept | Mohammad Golshani | ||
| Jun 29, 2020 at 1:27 | answer | added | Farmer S | timeline score: 5 | |
| Nov 27, 2014 at 3:49 | comment | added | Mohammad Golshani | @YairHayut Definition is essentially the same as the one given in Woodin's paper "Suitable extender models". | |
| Nov 26, 2014 at 7:46 | comment | added | Yair Hayut | What is your definition of supercompact cardinal? I think that in the absence of choice the first order definition (existence of normal measure on $P_\kappa \lambda$) and the second order one (existence of elementary embedding to a model that is closed under $\lambda$ sequences) are not equivalent. | |
| Nov 25, 2014 at 6:12 | history | edited | Mohammad Golshani | CC BY-SA 3.0 |
added 160 characters in body; edited tags
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| Oct 24, 2014 at 15:37 | comment | added | Andrés E. Caicedo | @JoelDavidHamkins Ah, sure. The situation here is easier than with embeddings $j:L(V_\lambda)\to L(V_\lambda)$. | |
| Oct 24, 2014 at 13:44 | comment | added | Joel David Hamkins | Andres, why doesn't the usual argument show that all iterates are well-founded? Let $\xi$ be least such that $j_{0,\lambda}(\xi)$ is in the ill-founded part of some $M_\lambda$, with $\lambda$ a limit. But in some $M_\alpha$ with $0<\alpha<\lambda$, there will be born a smaller $\xi'<\xi$ with $j_{\alpha,\lambda}(\xi')$ also in the ill-founded part. This argument seems to use only ZF in the language with j, in order that we may refer in any model to the corresponding iterates of $j$. | |
| Oct 24, 2014 at 13:37 | comment | added | Andrés E. Caicedo | @YairHayut However, without additional assumptions, it does not need to be the case that the direct limits taken at limit stages are well-founded. | |
| Oct 24, 2014 at 10:57 | comment | added | Joel David Hamkins | Yair, we may apply $j$ to any $j\upharpoonright V_\alpha$, and take the union to form $j_{1,2}=j(j)=\bigcup_\alpha j(j\upharpoonright V_\alpha)$. This is presumably what Mohammad means by iterating. | |
| Oct 24, 2014 at 10:20 | comment | added | Yair Hayut | I don't understand. Why do we have an elementary embedding $j_1$ with critical point $j(\kappa_0) = \kappa_1$ ($\kappa_0 = \text{crit }j$)? After all, $j$ can't be first order definable in $V$. | |
| Oct 24, 2014 at 6:02 | comment | added | Asaf Karagila♦ | @Andres: Ah, yes. That's what I was missing. I had a hunch it might be in the limit steps. Those can be sneaky bastards sometimes. Thanks! (In fact, it's even easy to see why $M_\omega\neq V$, because $\kappa_\omega$ must be regular in $M_\omega$ but singular in $V$...) | |
| Oct 24, 2014 at 6:01 | comment | added | Andrés E. Caicedo | @AsafKaragila Why is $M_\omega=V$? $M_\omega$ is defined as a direct limit, and the maps are not the identity. | |
| Oct 24, 2014 at 5:47 | history | edited | Asaf Karagila♦ | CC BY-SA 3.0 |
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| Oct 24, 2014 at 5:46 | comment | added | Asaf Karagila♦ | Is $j\colon V\to V$? If so, $M=V$ since $M_\alpha$ is $V$ for all $\alpha$. No? Am I missing something? | |
| Oct 24, 2014 at 5:03 | history | asked | Mohammad Golshani | CC BY-SA 3.0 |