Timeline for Are Berkeley cardinals easier to refute in ZFC than Reinhardt cardinals?
Current License: CC BY-SA 4.0
21 events
| when toggle format | what | by | license | comment | |
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| Sep 30 at 16:39 | answer | added | Dmytro Taranovsky | timeline score: 5 | |
| Nov 21, 2023 at 19:19 | vote | accept | Tim Campion | ||
| Nov 21, 2023 at 18:57 | comment | added | Gabe Goldberg | @TimCampion By the way, the two versions of Berkeley are equivalent (since you can take $\alpha = \text{rank}(M)$ and then code $M$ into $A$). | |
| Nov 21, 2023 at 17:10 | comment | added | Tim Campion | @GabeGoldberg Yes, thanks -- that helps! | |
| Nov 21, 2023 at 17:10 | history | edited | Tim Campion | CC BY-SA 4.0 |
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| Nov 21, 2023 at 15:55 | comment | added | Gabe Goldberg | @TimCampion I think you are using a different definition of Reinhardt than the standard one. Typically, a cardinal is Reinhardt if it is the critical point of an elementary embedding from $V$ to $V$. But as far as I can tell from the comments, you seem to be calling a cardinal $\lambda$ Reinhardt if there is an elementary embedding from $V_\lambda$ to $V_\lambda$. Does this clear up the confusion? | |
| Nov 21, 2023 at 15:04 | comment | added | Tim Campion | @AsafKaragila I'm confused -- your argument seems to say that every Berkeley cardinal satisfies the stronger axiom I stated, which in particular means that every Berkeley cardinal is Reinhardt. But Hanul explained that the smallest Berkeley cardinal can't be Reinhardt... | |
| Nov 21, 2023 at 1:17 | history | became hot network question | |||
| Nov 20, 2023 at 20:42 | comment | added | Gabe Goldberg | @KeithMillar I should have said in my slides that this argument is due to Woodin. | |
| Nov 20, 2023 at 20:39 | answer | added | Gabe Goldberg | timeline score: 21 | |
| Nov 20, 2023 at 20:05 | comment | added | Asaf Karagila♦ | Tim, towards the question addressed to @Hanul, if $\kappa\in M$, then this is the definition of Berkeley. If not, then $M\{\kappa\}$ is transitive again, so by the definition of Berkeley we get embeddings. All we need to note that in this case it must be that $j(\kappa)=\kappa$ when $j:M\to M$ is an elementary embedding, since $\kappa$ is the largest ordinal in $M\cup\{\kappa\}$ in that case. | |
| Nov 20, 2023 at 18:58 | comment | added | Hanul Jeon | @TimCampion I have no idea, unfortunately. | |
| Nov 20, 2023 at 18:55 | comment | added | Noah Schweber | @TimCampion Yes, this is done in Rosenstein's book Linear orderings which is excellent but out of print. The theory of an ordinal can be read off of its Cantor normal form. | |
| Nov 20, 2023 at 18:54 | comment | added | Keith Millar | See slide 9 of math.berkeley.edu/~goldberg/Slides/…. It gives a pretty digestible argument (imo) that Berkeley cardinals are incompatible with AC. Although, it does rely on an alternative (equivalent) characterization with ranks instead of transitive sets. | |
| Nov 20, 2023 at 18:53 | comment | added | Tim Campion | @NoahSchweber Thanks, that answers Question 2! From this MSE answer of yours I learned that every ordinal is elementarily equivalent to an ordinal $< \omega^\omega \cdot 2$, and that $Ord$ is elementarily equivalent to $\omega^\omega \cdot 2$. Do you know whether one can explicitly classify the complete first-order theories of ordinals? | |
| Nov 20, 2023 at 18:49 | comment | added | Tim Campion | @HanulJeon Thanks, that's clarifying. Do you know what is the status of the assertion "There is a cardinal $\kappa$ such that for any transitive $M$ with $\kappa \subseteq M$ there are elementary embeddings $M \to M$ with arbitrarily large critical point $\alpha < \kappa$"? This sort of $\kappa$ would be manifestly Reinhardt, and it also implies that $\kappa$ is Berkeley... | |
| Nov 20, 2023 at 18:46 | comment | added | James E Hanson | Another question in this vein one could ask is whether you can write down a large cardinal notion not known to be inconsistent with ZF with an 'immediate' proof of inconsistency with ZFC. (Obviously this is a somewhat vague question.) | |
| Nov 20, 2023 at 18:02 | comment | added | Hanul Jeon | Berkeley cardinals are not Reinhardt cardinals as $\mathsf{ZF}$ proves the least Berkeley cardinal is singular but every Reinhardt cardinal must be regular. However, if $\delta$ is a Berkeley cardinal, then there is $\lambda<\delta$ such that $(V_\lambda, V_{\lambda+1})$ is a model of $\mathsf{NGB}$ with a Reinhardt cardinal, which becomes a model of choice if we work over $\mathsf{ZFC}$. | |
| Nov 20, 2023 at 17:53 | comment | added | Noah Schweber | Is your question 2 really just asking about elementary embeddings between ordinals? If so, the answer is yes, $\mathsf{ZFC}$-provably there are lots of nontrivial elementary embeddings from $\kappa+1$ to itself whenever $\kappa$ is an uncountable cardinal. Basically, ordinals by themselves have very little expressive power (although things get more interesting if we move up from FOL, see this old question of mine). | |
| Nov 20, 2023 at 17:23 | comment | added | Tim Campion | Actually, I just noticed that since $\kappa \not \in V_\kappa$, it's maybe not so clear why Berkeley cardinals are Reinhardt... | |
| Nov 20, 2023 at 17:14 | history | asked | Tim Campion | CC BY-SA 4.0 |