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A theorem of A. Levy says that, if $\kappa$ is an inaccessible cardinal, then $V_\kappa\prec_{\Sigma_1}V$ namely $V_\kappa$ is an elementary submodel when considering only $\Sigma_1$ formulas.

Where can I find a proof of this theorem ?

Is this property true also for some other (non inaccessible) cardinals ?

Asaf Karagila
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Etienne Paul
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  • Related: http://mathoverflow.net/questions/71524/how-elementary-can-we-go/ – Asaf Karagila Mar 14 '15 at 22:22
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    If you know that $V_\kappa = H_\kappa$, for $\kappa$ inaccessible, then the proof it straightforward. Clearly, $\Sigma_1$ formulas are upward absolute for $V_\kappa$. Now suppose $\exists y\phi(y, \vec{x})$. Let $M$ be an elementary substructure of $V$ for $\exists y \phi$ with $\vec{tc({x})}\in M$ and $|M|<\kappa$. Then the Mostowski collapse of $M$, call it $M'$, is a transitive set of size less than $\kappa$ and $M'\vDash \exists y\phi(y,\vec{x})$. So $M'\in H_\kappa = V_\kappa$ and thus $V_\kappa\vDash \exists y\phi(y,\vec{x})$, again by upward absoluteness of $\Sigma_1$ formulas. – Sam Roberts Mar 14 '15 at 22:39
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    Kanamori's The Higher Infinite is worth looking at. He cites Levy for the result about inaccessibles (p. 299), shows that $V_\kappa\prec_2 V$ when $\kappa$ is supercompact (p. 299), and that $V_\kappa\prec_3 V$ when $\kappa$ is extendible (p. 318). – Sam Roberts Mar 14 '15 at 22:49
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    @SamRoberts I would suggest for you to post your comment as an answer. – Joel David Hamkins Mar 15 '15 at 01:52

1 Answers1

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Here's one way to prove it. Let $H_\kappa = \{x: |tc(\{x\})|<\kappa\}$. Then we have:

Theorem 1 If $\kappa$ is an uncountable cardinal, then $H_\kappa\prec_1 V$.

Proof. Let $\phi$ be $\Delta_0$ with free variables among $y,x$. Since $\Sigma_1$ formulas are upward absolute for transitive models, if $H_\kappa\vDash \exists y\phi$, then $\exists y\phi$. So suppose $\exists y\phi$ for $x\in H_\kappa$. Now let $M\prec_1 V$ with $tc(\{x\})\subseteq M$ and $|M|<\kappa$ (such an $M$ exists because $\kappa$ is uncountable and $|tc(\{x\})|<\kappa$ by definition of $H_\kappa$). By the Mostowski collapse lemma, there is an isomorphism $j:M \to M'$ for some transitive $M'$. Since $tc(\{x\})\subseteq M$, $j(x) = x$ and thus $M'\vDash \exists y\phi$. Finally, we note that because $M'$ is transitive and has cardinality less than $\kappa$, $M'\in H_\kappa$ and so $H_\kappa\vDash \exists y\phi$ (again by upward absoluteness). $\Box$

Our result then follows from the fact that $H_\kappa = V_\kappa$, for $\kappa$ inaccessible. It is furthermore optimal for inaccessibles since ``$\kappa$ is inaccessible" is $\Pi_1$. But stronger results hold for other large cardinals. For example:

Theorem 2 If $\kappa$ is supercompact (or even strong), then $V_\kappa\prec_2 V$.

Proof. See Kanamori The Higher Infinite (2003) p. 299 and p. 359. $\Box$

Theorem 3 If $\kappa$ is extendible, then $V_\kappa\prec_3 V$.

Proof. See Kanamori The Higher Infinite (2003) p. 318. $\Box$

This result is also optimal for extendibles since ``$\kappa$ is an extendible" is $\Pi_3$.

Sam Roberts
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  • Hence the property $V_\kappa\prec_{\Sigma_1}V$ is true for any cardinal $\kappa$ such that $V_\kappa = H_\kappa$. Is this equality only valid for inaccessible cardinals ? – Etienne Paul Mar 17 '15 at 21:54
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    Actually, it is true for any uncountable cardinal (see my last edit). On the other question, it is not in general true that $H_\kappa = V_\kappa$ implies that $\kappa$ is inaccessible. To see this, let $\kappa = \beth_\kappa$ have countable co-finality. Then $H_\kappa = V_\kappa$, even though $\kappa$ is not inaccessible. It does hold for regular cardinals, though. That is, if $\kappa$ is regular and $H_\kappa = V_\kappa$, then $\kappa$ is inaccessible. – Sam Roberts Mar 18 '15 at 23:32
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    To see this, note that for $\alpha<\kappa$, $\mathcal P(\alpha)\in V_\kappa$. So if $H_\kappa = V_\kappa$, then $|\mathcal P(\alpha)|\leq |tc(\mathcal P(\alpha))|<\kappa$. – Sam Roberts Mar 18 '15 at 23:34
  • Sam, to your last comment, $\alpha$ is an ordinal and therefore transitive, so its power set is also transitive. No need to take the closure there. – Asaf Karagila Mar 19 '15 at 06:31
  • So, at that point in the argument we have $|tc({\mathcal P(\alpha)})|<\kappa$ (because $\mathcal P(\alpha)\in V_\kappa = H_\kappa$) and we want to show that $|\mathcal P(\alpha)|<\kappa$. To do that we could either note that since $\mathcal P(\alpha)$ is transitive, $|\mathcal P(\alpha)| = |tc({\mathcal P(\alpha)})|$ as you suggest or we could note that $|\mathcal P(\alpha)|\leq |tc({\mathcal P(\alpha)})|$ without appealing to transitivity as I did. I'm not sure why the former would be better here. Am I missing something? – Sam Roberts Mar 19 '15 at 12:16
  • @AsafKaragila Of course, assuming we want the proof to be that explicit. – Sam Roberts Mar 21 '15 at 13:23
  • Sorry, I wasn't pinged in your original reply, so I only now see that you wrote it. The point is that since $\alpha$ is transitive, $\mathcal P(\alpha)=\operatorname{tc}(\mathcal P(\alpha))$. I don't understand why you're insisting that I am appealing to "additional" information, where your argument also appeals to that, because you need to pause and consider what in the name of Zeus is $\operatorname{tc}(\mathcal P(\alpha))$. Oh, right, it's just $\mathcal P(\alpha)$. :-) – Asaf Karagila Mar 21 '15 at 13:29
  • @AsafKaragila Here's perhaps a better way to put my question: you said there's "no need to take the transitive closure" of $\mathcal P(\alpha)$ in the argument I gave. How would you have avoided it? – Sam Roberts Mar 21 '15 at 13:58
  • I wouldn't avoid it. I'd just skip it, since $\mathcal P(\alpha)$ is its own transitive closure. – Asaf Karagila Mar 21 '15 at 14:00
  • @AsafKaragila Ah, I see; I thought that might be what you were getting at. It struck me that what I gave was a suitable level of detail, given the question. – Sam Roberts Mar 21 '15 at 14:12
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    Thanks again Sam. I agree that $H_\kappa\prec_1 V$ for any uncountable cardinal $\kappa$. However, if we are interested in $V_\kappa$ rather than $H_\kappa$, I understand that we can only say that $V_\kappa\prec_1 V$ if $\kappa$ is a fixed point of the $\beth$ function. This is already an interesting extension of the initial formulation ``$V_\kappa\prec_1 V$ if $\kappa$ is an inaccessible cardinal'', since there are much more fixed points of the $\beth$ function than inaccessible cardinals. – Etienne Paul Mar 22 '15 at 08:56
  • @EtiennePaul You're right; silly mistake. I've fixed it now. Thanks! – Sam Roberts Mar 22 '15 at 09:38