Suppose a first order set theory $S$ has a countable model. Does it follow that there is a countable ordinal $\alpha$ so that $L_\alpha$ in the constructible hierarchy is a model of $S$?
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asked Feb 24, 2022 at 0:21
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1 Answer
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For the question in the text, the answer is no. Let $S$ consist of (1) a sufficient finite number of ZF axioms to provide an absolute (i.e., $\Delta_1$ in the Lévy hierarchy) definition of the constructible hierarchy and (2) the sentence formalizing $V\neq L$.
For the question in the title, asking only for a constructible model and not necessarily one of the form $L_\alpha$, the answer is yes if the theory itself is constructible, but not in general.
answered Feb 24, 2022 at 1:22
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$\begingroup$ What about any first order S which does not extend ZFC? Will such an S have a model $L_\alpha$ for countable $\alpha$ if it has a countable model? $\endgroup$ Commented Feb 24, 2022 at 14:54
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1$\begingroup$ @FAB The $S$ in the first paragraph of my answer does not extend ZFC. $\endgroup$ Commented Feb 24, 2022 at 15:02
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$\begingroup$ I don't understand. Doesn't (2) invoke an extra axiom? – $\endgroup$ Commented Feb 24, 2022 at 15:37
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1$\begingroup$ Yes, $V\neq L$ is an axiom not in ZFC. But (1) is only part of ZFC. So the combination is not an extension of ZFC. $\endgroup$ Commented Feb 24, 2022 at 16:32
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$\begingroup$ Ah, yes of course! :) What if (as is my real interest) S is ZFC minus the power set axiom, plus, for a countable ordinal $\gamma$, less than $\gamma$ applications of the power set operation? $\endgroup$ Commented Feb 25, 2022 at 16:49