How elementary can we go? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T15:00:02Z http://mathoverflow.net/feeds/question/71524 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71524/how-elementary-can-we-go How elementary can we go? Asaf Karagila 2011-07-28T22:50:33Z 2011-07-28T23:36:51Z <p>It is a theorem of A. Levy, if $\kappa$ is an <em>inaccessible cardinal</em>, then $V_\kappa\prec_{\Sigma_1} V$ namely $V_\kappa$ is an elementary submodel when considering only $\Sigma_1$ sentences.</p> <p>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 <em>supercompact</em> 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)</p> <p>To have $\Sigma_3$ we need to go even further to <em>extendible</em> cardinals (again, this might be too strong. I am not too familiar with this notion yet).</p> <ul> <li>Is there a known large cardinal notion to give $\Sigma_4$ elementarity of $V_\kappa$? What about larger $n$? </li> <li>I would expect complete elementarity to fail due to some Kunen inconsistency theorem sort of argument, is this true?</li> <li>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</li> </ul> <p>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).</p> <p>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?</p> http://mathoverflow.net/questions/71524/how-elementary-can-we-go/71526#71526 Answer by Andreas Blass for How elementary can we go? Andreas Blass 2011-07-28T23:15:00Z 2011-07-28T23:15:00Z <p>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 $\kappa&lt;\mu$ such that <code>$V_\kappa\prec V_\mu$</code>. Thus, from the point of view of <code>$V_\mu$</code>, there is an elementary submodel of the universe of the form <code>$V_\kappa$</code>, even though there is no inaccessible cardinal.</p> http://mathoverflow.net/questions/71524/how-elementary-can-we-go/71527#71527 Answer by Ali Enayat for How elementary can we go? Ali Enayat 2011-07-28T23:25:05Z 2011-07-28T23:25:05Z <p>This following result answers the <strong>third bullet item question</strong> in the negative.</p> <p><strong>Proposition.</strong> 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 $M$.</p> <p><strong>Proof:</strong> 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. <strong>QED</strong></p> <blockquote> <p>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.</p> </blockquote> http://mathoverflow.net/questions/71524/how-elementary-can-we-go/71528#71528 Answer by Joel David Hamkins for How elementary can we go? Joel David Hamkins 2011-07-28T23:30:09Z 2011-07-28T23:36:51Z <p>The hypothesis that $V_\kappa$ is $\Sigma_k$ elementary or even fully elementary in $V$ is much weaker than you say.</p> <p>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 $V_\delta$.</p> <p>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.</p> <p>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.</p> <p>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.</p> <p><b>Theorem.</b> The following are equiconsistent over ZFC.</p> <ul> <li>The Levy scheme. That is, the scheme "$V_\kappa\prec V$ " plus "$\kappa$ is inaccessible."</li> <li>"ORD is Mahlo". That is, the scheme asserting of every definable (with parameters) proper class club, that it contains an inaccessible cardinal.</li> </ul> <p>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</p> <p>Meanwhile, if you drop the inaccessibility requirement, then you can attain the following, which many set theorists find surprising.</p> <p><b>Theorem.</b> The scheme "$V_\kappa\prec V$ " is equiconsistent merely with ZFC.</p> <p>Proof. If ZFC is consistent, then so is every finite fragment of the scheme $V_\kappa\prec V$, by the reflection theorem. QED</p> <p>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.</p> <p>Both versions of the axiom $V_\kappa\prec V$ were important in <a href="http://de.arxiv.org/abs/math.LO/0009240" rel="nofollow">my paper on the maximality principle</a>, 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).</p>