Dual covering theorem - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T22:26:39Z http://mathoverflow.net/feeds/question/71672 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/71672/dual-covering-theorem Dual covering theorem Haim 2011-07-30T18:28:46Z 2011-07-31T09:39:18Z <p>Jensen's covering theorem states that if $0^\sharp$ doesn't exist, then every uncountable set of ordinals can be covered by a constructible set of the same cardinality.</p> <p>Now consider the following (somewhat) dual statements:</p> <blockquote> <ol> <li>Every uncountable set of ordinals covers a constructible uncountable set of ordinals.</li> <li>Every uncountable set of ordinals covers a constructible set of ordinals of the same cardinality.</li> </ol> </blockquote> <p>I have two questions:</p> <blockquote> <ol> <li>Does any of the above statements follow from the non-existence of $0^\sharp$?</li> <li>If the answer is "no", are they still known to be consistent (in order to avoid trivialities, we may assume that V=L doesn't hold)? </li> </ol> </blockquote> http://mathoverflow.net/questions/71672/dual-covering-theorem/71685#71685 Answer by Andres Caicedo for Dual covering theorem Andres Caicedo 2011-07-30T20:57:31Z 2011-07-30T20:57:31Z <p>The answer to the first question is no: You can add a club subset of a stationary subset of $\omega_1$ by forcing. The closure of any uncountable subset of the generic club is a club contained in the stationary set, so if we begin in $L$ with a stationary-costationary subset of $\omega_1$, and add a club through it, the extension contradicts both statements.</p> <p>As for your second question, add a Cohen real to $L$. Suppose $A\subseteq{\rm ORD}$ and $|A|$ has uncountable cofinality in the extension. Since Cohen's forcing is countable, there is a condition $p$ that is in the generic and such that <code>$A_p=\{\alpha\mid p$</code> forces <code>$\alpha\in\dot A\}$</code> has the same size as $A$. Note that this is a constructible subset of $A$ in the ground model. </p> <p>In fact, the same holds for uncountable sets of cofinality $\omega$: Suppose now that $A$ has countable cofinality, let $\alpha$ be its supremum, let $\gamma$ be least such that $A\cap\gamma$ is uncountable. For each $\beta$ between $\gamma$ and $\alpha$ there is a $p_\beta$ in the generic that decides a subset of $A\cap\beta$ that is in $L$ and has size $|A\cap\beta|$. There must be a $p$ that appears as $p_\beta$ unboundedly often, and $A_p$ is as wanted.</p> <p>In summary: The extension of $L$ obtained by adding a Cohen real is a model of your two statements.</p> http://mathoverflow.net/questions/71672/dual-covering-theorem/71719#71719 Answer by Péter Komjáth for Dual covering theorem Péter Komjáth 2011-07-31T09:39:18Z 2011-07-31T09:39:18Z <p>Perhaps Magidor's covering lemma may be mentioned: if $0^\sharp$ does not exist and $A$ is a set of ordinals which is closed under the primitive recursive set functions, then $A$ is the union of countably many constructible sets. M. Magidor: Representing sets of ordinals..., Transactions of the AMS, 317(1990), 91-126.</p>