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

One might expect that the "amount" of elementarity will grow quickly as we progress with large cardinal axioms, however for the next step, $V_\kappa\prec_{\Sigma_2}V$ we need to get much higher. In order to assure this level of elementarity a supercompact is enough (is it too strong? judging by the stage this theorem appears in Jech's and Kanamori's textbooks I would say that if it is too strong then it is not strong by that much)

To have $\Sigma_3$ we need to go even further to extendible cardinals (again, this might be too strong. I am not too familiar with this notion yet).

  • Is there a known large cardinal notion to give $\Sigma_4$ elementarity of $V_\kappa$? What about larger $n$?
  • I would expect complete elementarity to fail due to some Kunen inconsistency theorem sort of argument, is this true?
  • Are there results in the reverse direction? Namely if $\kappa$ is such that $V_\kappa\prec_{\Sigma_k}V$ then $\kappa$ has to be inaccessible/supercompact/extendible/etc

If we use all sort of set theoretic notions to measure how far $V$ is from an inner model (forcing axioms, large cardinals, how the cardinals behave in the inner model compared to $V$, sharps and covering theorems, etc etc).

Assuming the answer to the first question is not "It is inconsistent.", is there a useful way to use this approach to measure the difference between $V$ and its inner models?


4 Answers 4


The hypothesis that $V_\kappa$ is $\Sigma_k$ elementary or even fully elementary in $V$ is much weaker than you say.

One can see part of this quite easily by observing that for any inaccessible cardinal $\delta$, then $V_\delta\models\text{ZFC}$ and there are a club of ordinals $\alpha$ with $V_\alpha\prec V_\delta$. In particular, if $\delta$ is Mahlo, then there are a stationary set of inaccessible cardinals $\kappa$ with $V_\kappa$ fully elementary in $V_\delta$.

In particular, if we lived inside $V_\delta$, we would believe that there is a stationary proper class of inaccessible cardinals $\kappa$ with $V_\kappa$ as fully elementary in the universe as desired.

It turns out that although we can express $V_\kappa\prec_{\Sigma_k} V$ as a first-order assertion of $\kappa$ and $k$, it is not possible to express full elementary $V_\kappa\prec V$ as a single first-order assertion of set theory. Instead, we may use a scheme.

Thus, we introduce $\kappa$ as a constant symbol, and consider the scheme, denoted "$V_\kappa\prec V$ ", asserting of every formula $\varphi$ that $$\forall x\in V_\kappa\ (\varphi(x) \iff V_\kappa\models\varphi[x]\ ).$$ If we add the assumption that $\kappa$ is inaccessible, then this is known as the Levy scheme.

Theorem. The following are equiconsistent over ZFC.

  • The Levy scheme. That is, the scheme "$V_\kappa\prec V$ " plus "$\kappa$ is inaccessible."
  • "ORD is Mahlo". That is, the scheme asserting of every definable (with parameters) proper class club, that it contains an inaccessible cardinal.

Proof. The first implies that $V_\kappa$ satisfies ORD is Mahlo, since $\kappa$ will be a limit point and hence an element of any such club as defined in $V$ using parameters below $\kappa$. If the second is consistent, then so is the first by a compactness argument, using the reflection theorem. QED

Meanwhile, if you drop the inaccessibility requirement, then you can attain the following, which many set theorists find surprising.

Theorem. The scheme "$V_\kappa\prec V$ " is equiconsistent merely with ZFC.

Proof. If ZFC is consistent, then so is every finite fragment of the scheme $V_\kappa\prec V$, by the reflection theorem. QED

One can even attain a proper class club $C\subset\text{ORD}$ of cardinals, with each $\kappa\in C$ satisfying the scheme $V_\kappa\prec V$, without going beyond ZFC in consistency strength.

Both versions of the axiom $V_\kappa\prec V$ were important in my paper on the maximality principle, the principle asserting that any statement that is forceable in such a way that it remains true in all further extensions is already true. It turned out that one can force the maximality principle only from a model of $V_\kappa\prec V$ (and you need $\kappa$ inaccessible for the boldface maximality principle).

  • $\begingroup$ Oh, after posting I find now an abundance of expert answers! $\endgroup$ Jul 28, 2011 at 23:32
  • 2
    $\begingroup$ It was recently suggested to me to ask these sort of questions here rather than on math.SE :-) $\endgroup$
    – Asaf Karagila
    Jul 28, 2011 at 23:35
  • $\begingroup$ (Since MO has a crappy comments notification): So the theorem "If $\kappa$ is X then $V_\kappa\prec_{\Sigma_n}V$" is in fact a metatheorem? $\endgroup$
    – Asaf Karagila
    Jul 31, 2011 at 8:34
  • $\begingroup$ I'm not sure I follow your question. For any fixed finite k, then we can assert $V_\kappa\prec_{\Sigma_k} V$ by a single assertion. There are a closed unbounded class of such $\kappa$, provided by the reflection theorem, and so under Ord is Mahlo, there are many inaccessible such $\kappa$. This hypothesis is weaker in consistency strength than a Mahlo cardinal. For fixed k, you don't need full Ord is Mahlo, but only Ord is Mahlo for certain $\Sigma_n$ definable class clubs, which is a single assertion, and you can figure out $n$ from $k$. $\endgroup$ Jul 31, 2011 at 16:51
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    $\begingroup$ No, a correct cardinal will satisfy every (standard) axiom of ZFC, as a separate matter for each axiom. This is not the same as being internally a model of ZFC, which would involve the nonstandard axioms. Worldly cardinals use this latter notion. $\endgroup$ Nov 19, 2017 at 19:15

This following result answers the third bullet item question in the negative.

Proposition. Suppose $(M,\in)$ is a transitive model of $ZF$ of uncountable cofinality. Then there is some ordinal $\alpha$ in $M$ of countable cofinality such that $(V_{\alpha})^M$ is a full elementary submodel of $M$.

Proof: Use the reflection theorem to produce an increasing sequence $\alpha_k$ for each $k \in \omega$ such that $(V_{\alpha_k})^M$ is a $\Sigma_k$-elementary submodel of $M$. The desired $\alpha$ is the union of the $\alpha_k$'s. QED

So it is quite possible to have $\kappa$ such that $V_\kappa$ is a full elementary submodel of $V$, without $\kappa$ being even regular, let alone inacessible.

  • $\begingroup$ I see that Andreas was finishing his answer when I was beginning mine. $\endgroup$
    – Ali Enayat
    Jul 28, 2011 at 23:26

The second and third of the bulleted questions are answered by an old theorem of Montague and Vaught. Suppose $\mu$ is the first inaccessible cardinal. Then there is $\kappa<\mu$ such that $V_\kappa\prec V_\mu$. Thus, from the point of view of $V_\mu$, there is an elementary submodel of the universe of the form $V_\kappa$, even though there is no inaccessible cardinal.

  • $\begingroup$ @Andreas: Many thanks for the very quick reply. In your answer you write about downwards reflection, that is in $V$ we go down from an inaccessible, and then chop off the "end" of the universe (above $\mu$, that is), and we are done. Does $V_\mu$ know internally that $V_\kappa$ is an elementary submodel? $\endgroup$
    – Asaf Karagila
    Jul 28, 2011 at 23:55
  • $\begingroup$ Asaf, the concept of "being an elementary submodel of the universe" is not expressible by a single assertion of set theory (if it were, you could get a contradiction in these models by letting $\kappa$ be least such, but then $V_\kappa$ would have also to have one, contradiction). But for any finite level of elementarity, $V_\mu$ can indeed observe that $V_\kappa$ is $\Sigma_n$-elementarity, since the satisfaction of statements in $V_\kappa$ is absolute between $V_\mu$ and $V$. $\endgroup$ Jul 29, 2011 at 0:04

I would like to mention, on the second question, that superhuge cardinals are in fact $\Sigma_5$-reflecting, that is to say inaccessible and $V_\kappa\prec_{\Sigma_5} V$. The reason is first that every superhuge cardinal is extendbile, and so $\Pi_4$ formulas are downward absolute in $V_\kappa$. Suppose we have $\exists x(\phi(x,x_0,\dots,x_n))$, where $\phi(x,x_0,\dots,x_n)$ is $\Pi_4$. Let $z$ be a witness to that, and let $\lambda$ be $\Pi_4$-reflecting such that $z\in V_\lambda$. Then, we can find some $j\colon V\rightarrow M$ such that $M\vDash(\Pi_4\text{ formulas are downward absolute in }V_{j(\kappa)})$ and $j(\kappa)>\lambda$. Then $z\in V_{j(\kappa)}$, and so $M\vDash(V_{j(\kappa)}\vDash\exists x(\phi(x,x_0,\dots,x_n)))$, and so $V_{j(\kappa)}\vDash\exists x(\phi(x,x_0,\dots,x_n))$. As $V_\kappa\prec V_{j(\kappa)}$, we have $V_\kappa\vDash\exists x(\phi(x,x_0,\dots,x_n))$ and so $V_\kappa\prec_{\Sigma_5} V$.

In fact, the set of such cardinals form a normal measure below $\kappa$. Let $D=\{X\subseteq\kappa\mid\kappa\in j(X)\}$. Then by a similar argument $M\vDash(\text{$\kappa$ is $\Sigma_5$-reflecting})$, then $U\in D$, where $U=\{\lambda<\kappa\mid\text{$\lambda$ is $\Sigma_5$-reflecting}\}$.


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