Cofinality of Theta if sharps exist - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T07:33:59Z http://mathoverflow.net/feeds/question/2776 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/2776/cofinality-of-theta-if-sharps-exist Cofinality of Theta if sharps exist Scott Cramer 2009-10-27T05:11:07Z 2010-10-17T05:08:24Z <p>If &#8477;<sup>#</sup> exists then why is cof(&theta;<sup>L(&#8477;)</sup>) = &omega;? Also I have the same question for the L(V<sub>&lambda;+1</sub>) generalization (if it's actually a different proof; I presume it isn't), i.e. if &theta; is defined as the sup of the surjections in L(V<sub>&lambda;+1</sub>) of V<sub>&lambda;+1</sub> onto an ordinal, then if V<sub>&lambda;+1</sub><sup>#</sup> exists why is cof(&theta;<sup>L(V<sub>&lambda;+1</sub>)</sup>) = &omega;? </p> http://mathoverflow.net/questions/2776/cofinality-of-theta-if-sharps-exist/2862#2862 Answer by Grigor for Cofinality of Theta if sharps exist Grigor 2009-10-27T18:34:38Z 2009-10-27T18:34:38Z <p>This is because the pieces of the sharp singularize Theta. Let s_n be the sequence of the first n cardinals above continuum and let a_n be the nth cardinal above continuum. Then the theory of reals with a parameter s_n in L_{a_n+1}(R) is a set of reals A_n. They are Wadge cofinal in Theta, another words the sequence is not in L(R) but each A_n is and that is why you get a singularization.</p> http://mathoverflow.net/questions/2776/cofinality-of-theta-if-sharps-exist/3129#3129 Answer by Grigor for Cofinality of Theta if sharps exist Grigor 2009-10-28T20:00:06Z 2009-10-28T20:00:06Z <p>Scott, the best way to think of sharps is via mice. Think of x^# as a mouse over x with one measure which is iterable. R^# is a mouse over R with one measure which is iterable. Things become very easy ones you make the move from sharps as reals or sets of reals or etc to sharps as mice. </p>