Timeline for Countable $L_\alpha$ model for $S$ if $S$ has a countable well founded model?
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9 events
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| Mar 1, 2022 at 10:12 | comment | added | Fedor Pakhomov | @FarmerS At least for me the natural interpretation of the the term "fragment of $T$" is that it is a theory $U$ in the same language as $T$ such that $U$ as a set of theorems is contained in $T$ as a set of theorems. | |
| Mar 1, 2022 at 10:09 | comment | added | Fedor Pakhomov | It isn't provable in $\mathsf{ZFC}$. Consider $T=\mathsf{ZFC}^-+(V=L)\to \mathsf{PowerSet}$. In $\mathsf{ZFC}$ we could construct a countable transitive model of $T$ by adding a Cohen real to a countable transitive model of $\mathsf{ZFC}^-$. However, any constructive model of $T$ is also a model of $\mathsf{ZFC}$ thus by 2-nd incompleteness theorem $\mathsf{ZFC}$ couldn't prove that there are constructive models of $T$. Therefore $\mathsf{ZFC}$ doesn't prove the implication "there is a countable transitive model of $T$" $\Rightarrow$ "there is a countable transitive constructive model of $T$". | |
| Feb 27, 2022 at 22:19 | comment | added | Ali Enayat | @EmilJeřábek you are right Emil. In the mean time Farmer has pointed out that the current formulation is defective. | |
| Feb 27, 2022 at 18:01 | comment | added | Farmer S | There are no proper fragments of ZFC which properly extend the ZFC$^-$, since there is only one axiom difference between ZFC and ZFC$^-$. | |
| Feb 27, 2022 at 6:32 | comment | added | Emil Jeřábek | @Ali Does it? The answer involves $V\ne L$, hence it is not a subtheory of ZFC. | |
| Feb 27, 2022 at 4:04 | comment | added | Ali Enayat | The negative nswer to your other question (mathoverflow.net/questions/416884/…) by Andreas Blass also answers this one in the negative. | |
| Feb 26, 2022 at 21:39 | history | edited | Frode Alfson Bjørdal | CC BY-SA 4.0 |
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| Feb 26, 2022 at 20:54 | history | edited | LSpice | CC BY-SA 4.0 |
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| Feb 26, 2022 at 18:34 | history | asked | Frode Alfson Bjørdal | CC BY-SA 4.0 |