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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 ?

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  • $\begingroup$ Related: mathoverflow.net/questions/71524/how-elementary-can-we-go $\endgroup$
    – Asaf Karagila ♦
    Commented Mar 14, 2015 at 22:22
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    $\begingroup$ 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. $\endgroup$
    – Sam Roberts
    Commented Mar 14, 2015 at 22:39
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    $\begingroup$ 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). $\endgroup$
    – Sam Roberts
    Commented Mar 14, 2015 at 22:49
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    $\begingroup$ @SamRoberts I would suggest for you to post your comment as an answer. $\endgroup$
    – Joel David Hamkins
    Commented Mar 15, 2015 at 1:52

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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 $\Pi_3$.

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    $\begingroup$ 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 ? $\endgroup$
    – Etienne Paul
    Commented Mar 17, 2015 at 21:54
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    $\begingroup$ 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. $\endgroup$
    – Sam Roberts
    Commented Mar 18, 2015 at 23:32
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    $\begingroup$ To see this, note that for $\alpha<\kappa$, $\mathcal P(\alpha)\in V_\kappa$. So if $H_\kappa = V_\kappa$, then $|\mathcal P(\alpha)|\leq |tc(\mathcal P(\alpha))|<\kappa$. $\endgroup$
    – Sam Roberts
    Commented Mar 18, 2015 at 23:34
  • $\begingroup$ 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. $\endgroup$
    – Asaf Karagila ♦
    Commented Mar 19, 2015 at 6:31
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    $\begingroup$ 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. $\endgroup$
    – Etienne Paul
    Commented Mar 22, 2015 at 8:56

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