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I know that if there was $V_\kappa$ that would be a model of ZFC (for example assuming that $\kappa$ is an inaccessible cardinal, so we are working in a theory ZFC+"there is an inaccessible cardinal"), it would have a countable elementary submodel due to the Löwenheim–Skolem theorem.

I think that due to the construction, $V_\alpha$ for a countable $\alpha$ would not be closed under power set or replacement, but I'm not sure how to prove this. Is my intuition leading me in the right direction?

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$V_\alpha$ for a countable $\alpha$ is never a model of ZFC. If $\alpha\leq\omega$, then $V_\alpha$ violates the axiom of infinity. If $\alpha$ is a successor ordinal, then $V_\alpha$ violates the axiom of power set. If $\alpha$ is a countable limit ordinal greater than $\omega$, then $V_\alpha$ contains an uncountable element, namely $V_{\omega+1}$, but no uncountable ordinal, so it violates the theorem of ZFC that says every set is in one-to-one correspondence with an ordinal, so it must violate at least one of the axioms on which that theorem is based. (In fact, in the situation of the last sentence, some instances of replacement will fail in $V_\alpha$.)

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  • $\begingroup$ I think he is asking if one takes $(V_{\kappa}, \epsilon)$ with $\kappa$ inaccessible and construct a countable elementary submodel $M (\subseteq V_{\beta})$ then iterate this and take the union of a chain to get a $V_{\alpha}$ countable elementary submodel. Why does this not work ? $\endgroup$ – Rene Schipperus Sep 13 '16 at 18:12
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    $\begingroup$ @ReneSchipperus I assume that by "iterate this" you mean to get a countable elementary submodel $M'$ properly including $M$, then a countable elementary submodel properly including $M'$, etc., taking unions at limit stages. (The word "iterate" isn't in the question.) The answer here is that, if this iteration ever yielded a model of the form $V_\alpha$, then that model would have $V_{\omega+1}$ as an element, and it would have a bijection from that element to an ordinal, so $\alpha$ would not be countable. $\endgroup$ – Andreas Blass Sep 13 '16 at 18:17
  • $\begingroup$ In contrast it is possible for $L_\alpha$ to be a model of ZFC if we work in L with $\kappa$ inaccessible in L. Then the transitive collapse of any countable elementary submodel of $L_\kappa$ will be $L_\alpha$ by condensation and models ZFC. $\endgroup$ – Jing Zhang Sep 16 '16 at 19:01
  • $\begingroup$ @JingZhang Yes, if there is an inaccessible cardinal, then there is a countable $\alpha$ such that $L_\alpha$ is a model of ZFC; in fact, there are uncountably many such $\alpha$'s. If you just want one such $\alpha$, you don't need to assume an inaccessible; just assume that ZFC has a transitive model. And even to get uncountably many such $\lpaha$, you need a lot less than an inaccessible. $\endgroup$ – Andreas Blass Sep 17 '16 at 1:16

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