Timeline for Can we have a model of ZFC for any increasing cardinality function over its ranks?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Sep 23, 2022 at 12:14 | comment | added | Zuhair Al-Johar | Nice! Well my naive attempt was for the $\alpha^M$ to be what $M$ sees as $\alpha$, in other words $\alpha^M$ is a definable ordinal in $M$ whose definition (externally) is exactly equal to the definition of $\alpha$ with relativization to $M$, so $V_\alpha^M$ is what $M$ sees as the $\alpha$ iterative stage of $\emptyset$, i.e. the definition of $V_\alpha^M$ is exactly equal to the definition of $V_\alpha$ relativized to $M$. so if $\alpha$ is parameter free definable after $\phi$ then $\alpha^M$ is definable after $\phi^M$, i.e. the formula $\phi$ with all quantifiers bounded by $M$. | |
| Sep 22, 2022 at 9:40 | comment | added | Asaf Karagila♦ | If $M$ is not a well-founded, indeed transitive, in what sense does $\alpha$, an external ordinal corresponds to $V_\alpha^M$? If it is transitive, then $V_\alpha^M\subseteq V_\alpha$. | |
| Sep 21, 2022 at 19:51 | comment | added | Zuhair Al-Johar | @AsafKaragila, is $|V^M_\alpha| \leq |V_\alpha|$? | |
| Sep 21, 2022 at 19:45 | comment | added | Asaf Karagila♦ | Then my first comment holds. | |
| Sep 21, 2022 at 19:27 | comment | added | Zuhair Al-Johar | @AsafKaragila, $f$ is not inside $M$. | |
| Sep 21, 2022 at 15:57 | comment | added | Asaf Karagila♦ | In what sense is $f$ inside $M$, then? | |
| Sep 21, 2022 at 14:31 | comment | added | Zuhair Al-Johar | Why? I thought the upward L-K can pump up the cardinality of the next stage to whatever we wish to | |
| Sep 21, 2022 at 13:07 | comment | added | Asaf Karagila♦ | Clearly not. If $f(\alpha+1)>2^{f(\alpha)}$, then this cannot hold. | |
| Sep 21, 2022 at 11:02 | history | asked | Zuhair Al-Johar | CC BY-SA 4.0 |