How elementary can we go? - MathOverflow

How elementary can we go?

2011-07-28T22:50:33Z

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?

Answer by Andreas Blass for How elementary can we go?

2011-07-28T23:15:00Z

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.

Answer by Ali Enayat for How elementary can we go?

2011-07-28T23:25:05Z

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.

Answer by Joel David Hamkins for How elementary can we go?

2011-07-28T23:30:09Z

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).