Timeline for Are Berkeley cardinals easier to refute in ZFC than Reinhardt cardinals?
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Nov 22, 2023 at 2:33 | comment | added | Andrés E. Caicedo | @Gabe "(roughly)" 🙂 | |
| Nov 21, 2023 at 19:19 | vote | accept | Tim Campion | ||
| Nov 21, 2023 at 18:54 | comment | added | Gabe Goldberg | (6) Well it may not be definable, but $\vec \kappa$ is in $V_{\gamma+1}$ and $i(\vec \kappa) = \vec \kappa\restriction \mathbb N^+$ (roughly), so $i(\sup \vec \kappa) = \sup (\vec \kappa\restriction \mathbb N^+) = \sup \vec \kappa$. | |
| Nov 21, 2023 at 18:53 | comment | added | Gabe Goldberg | (2) Well, you can make $\gamma$ as big as you want. Depending on how you code the sequence of $M_\alpha$'s, the $\gamma$ I chose will work. But you could go up $\omega$ ranks and never worry about a pairing function again. (3) I'm not sure what you mean about definably Berkeley, $M_\alpha$ isn't definable. (4) If $i(M_\alpha) = M_\alpha$ then $i\restriction M_\alpha$ is elementary from $M_\alpha$ to $M_\alpha$ since the $M_\alpha\vDash \varphi(a)$ if and only if $V_{\gamma+1}$ satisfies the relativization $\varphi^{M_\alpha}(a)$ iff it satisfies $\varphi^{M_\alpha}(i(a))$. | |
| Nov 21, 2023 at 18:48 | history | edited | Gabe Goldberg | CC BY-SA 4.0 |
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| Nov 21, 2023 at 18:22 | comment | added | James E Hanson | @TimCampion Choice is being used in the step where we choose the sequence of $M_\alpha$'s uniformly. | |
| Nov 21, 2023 at 18:09 | history | edited | Gabe Goldberg | CC BY-SA 4.0 |
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| Nov 21, 2023 at 18:00 | comment | added | Tim Campion | (8) Where did we use choice? | |
| Nov 21, 2023 at 17:07 | comment | added | Tim Campion | (5) In the first sentence of the second paragraph, I think $j$ is $i$? (6) In the first sentence of the second paragraph, the point is that $n \mapsto \kappa_n$ is definable in $V_{\gamma+1}$ by the recursion theorem there, and then the sup is also definable, so $i$ preserves the sup, and hence the sup is a fixed point of $i$. (7) Similarly to before, I don't understand why $k$ restricts to an elementary embedding $N_n \to N_n$. | |
| Nov 21, 2023 at 17:07 | comment | added | Tim Campion | (3) In the second sentence, we used that there is no definably-Berkeley below $\delta$, but then in the fourth sentence, used that $\delta$ is in fact Berkeley. (4) In the last sentence of the first paragraph, I don't see why $i$ would restrict to an elementary embedding $M_\alpha \to M_\alpha$. I know this would hold if $M_\alpha$ were an elementary substructure of $V_{\gamma+1}$, but I don't think it is. | |
| Nov 21, 2023 at 17:07 | comment | added | Tim Campion | Thanks! A few notes and questions. (1) In the second sentence, the condition is that there is no elementary embedding $j : M_\alpha \to M_\alpha$ with critical point below $\alpha$. (2) In order to have $\vec M \in V_{\gamma+1}$, we need $\gamma$ to be a limit ordinal -- otherwise we only have $\vec M \in V_{\gamma+3}$ or something like that. I see that the $M_\alpha$ can't be constant in $\alpha$ because $\delta$ is Berkeley, which feels close to implying $\gamma$ is limit, but I don't quite see it. | |
| Nov 21, 2023 at 15:52 | comment | added | Gabe Goldberg | Yes, $\vec M$ is what you guessed. But $i(M_\alpha)$ really means $i(M_\alpha)$!!! Of course since $i(\vec M) = \vec M$, we have $i(M_\alpha)= M_{i(\alpha)}$. | |
| Nov 21, 2023 at 15:05 | comment | added | Tim Campion | Thanks, this is great! Does $\vec M$ denote (the graph of) the function $\delta \to V$, $\alpha \mapsto M_\alpha$? Also, does $i(M_\alpha)$ really mean $i(M_\alpha)$, or does it mean the image of $i$ restricted to $M_\alpha$? | |
| Nov 21, 2023 at 12:51 | history | edited | Gabe Goldberg | CC BY-SA 4.0 |
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| Nov 20, 2023 at 20:39 | history | answered | Gabe Goldberg | CC BY-SA 4.0 |