Let $S$ be a proper fragment of $\mathit{ZFC}$, and let $S$ properly extend $\mathit{ZFC}^-$, i.e. $\mathit{ZFC}$ minus the power set axiom. Is there a countable ordinal $\alpha$ so that $L_\alpha$ is a model of $S$ if $S$ has a countable well founded model?
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asked Feb 26, 2022 at 18:34
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1$\begingroup$ @Ali Does it? The answer involves $V\ne L$, hence it is not a subtheory of ZFC. $\endgroup$– Emil JeřábekCommented Feb 27, 2022 at 6:32
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1$\begingroup$ There are no proper fragments of ZFC which properly extend the ZFC$^-$, since there is only one axiom difference between ZFC and ZFC$^-$. $\endgroup$– Farmer SCommented Feb 27, 2022 at 18:01
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1$\begingroup$ @EmilJeřábek you are right Emil. In the mean time Farmer has pointed out that the current formulation is defective. $\endgroup$– Ali EnayatCommented Feb 27, 2022 at 22:19
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1$\begingroup$ 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 $T$". $\endgroup$– Fedor PakhomovCommented Mar 1, 2022 at 10:09
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1$\begingroup$ @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. $\endgroup$– Fedor PakhomovCommented Mar 1, 2022 at 10:12
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