Suppose $ZFC^-$ is $ZFC$ minus the power set axiom, and that, for $\gamma$ a countable ordinal, $\mathcal{P}^\gamma$ is an axiom that allows less than $\gamma$ applications of the power set operation. Let S be $ZFC^-$ plus $\mathcal{P}^\gamma$. Is there a countable ordinal $\alpha$ so that $L_\alpha$ is a model of S if S has a countable model?
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asked Feb 26, 2022 at 2:07
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$\begingroup$ (Ignoring the issue of precisely phrasing $\mathcal{P}^\gamma$) Yes, as with your earlier question: take the Mostowski collapse of a large enough countable elementary submodel of $L_\kappa$, for $\kappa$ appropriate, and apply Condensation. $\endgroup$– Noah SchweberCommented Feb 26, 2022 at 2:25
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$\begingroup$ @NoahSchweber My understanding was that the appropriate $\kappa$ for your $L_\kappa$ was uncountable, to the level $\beth_\gamma$, and so $L_\kappa$ would be a standard model. But I do not ask whether there is a countable ordinal $\alpha$ so that $L\alpha$ is a model of S if S has a standard model. $\endgroup$– Frode Alfson BjørdalCommented Feb 26, 2022 at 2:43
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$\begingroup$ Even taking $\gamma=1$ (so not changing $\mathsf{ZFC}$ at all), it's consistent with $\mathsf{ZFC}$ that $\mathsf{ZFC}$ has a countable model but no well-founded model, and so in particular no level of $L$ satisfying $\mathsf{ZFC}$. Does this answer your question? $\endgroup$– Noah SchweberCommented Feb 26, 2022 at 4:55
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$\begingroup$ @NoahSchweber I don't understand. Do you say that the fact that there are unfounded models of ZFC rules out the existence of well founded models of ZFC? $\endgroup$– Frode Alfson BjørdalCommented Feb 26, 2022 at 5:53
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$\begingroup$ @NoahSchweber I would interpret $S$ with $\gamma=1$ as $S=S\cup\{\mathcal{P}(A)|A\in ZFC\}$ $\endgroup$– Frode Alfson BjørdalCommented Feb 26, 2022 at 5:58
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