User scott cramer - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T20:03:57Z http://mathoverflow.net/feeds/user/1178 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/3462/characterizations-of-non-wellfounded-models Characterizations of non-wellfounded models? Scott Cramer 2009-10-30T18:18:35Z 2009-12-29T22:35:13Z <p>My question is whether there are any characterizations of non-wellfounded models of set theory. A wellfounded model is one that does not have any \epsilon-descending infinite sequences. I'm not asking about models that satisfy ZF-Foundation, but rather ones that satisfy ZF, but are not wellfounded in V. For example, taking the ultrapower by an ultrafilter which is not countably complete produces such a model. </p> <p>I would assume that the reason one does not work with non-wellfounded models is the inability to collapse them, and therefore the inability to study their structure in relation to V. And perhaps there is a theorem which says that "anything is possible" when it comes to these models, and therefore it's a hopeless cause to understand them. </p> http://mathoverflow.net/questions/4418/definition-modifications-without-choice Definition modifications without choice Scott Cramer 2009-11-06T19:24:57Z 2009-11-08T18:50:15Z <p>What definitions or equivalencies between definitions for standard set theory objects (such as large cardinals) do not hold or do not carry through in the expected manner to the world without choice? Also, equivalences that DO carry over, but that you might expect wouldn't, would be interesting as well.</p> <p>For instance I recently heard that the equivalence between the diagonal intersection and the regressive function definitions of normal ultrafilter does not hold without AC, though under closer inspection I don't see why it doesn't. Clarification on this would be helpful as well.</p> http://mathoverflow.net/questions/3462/characterizations-of-non-wellfounded-models/3489#3489 Comment by Scott Cramer Scott Cramer 2009-10-31T19:28:56Z 2009-10-31T19:28:56Z Well, I suppose I mean &quot;any information&quot; not characterization. http://mathoverflow.net/questions/2776/cofinality-of-theta-if-sharps-exist/2862#2862 Comment by Scott Cramer Scott Cramer 2009-10-27T21:01:54Z 2009-10-27T21:01:54Z I was a little confused as to why the A_n are Wadge cofinal, but I suppose it's just by contradiction: if not they would all be Wadge reducible to something in L(R), which would imply that R^# is in L(R), which is ridiculous. Thanks. http://mathoverflow.net/questions/2776/cofinality-of-theta-if-sharps-exist Comment by Scott Cramer Scott Cramer 2009-10-27T20:58:36Z 2009-10-27T20:58:36Z You're right, motivation should have been added. While it's not the most interesting question in the world, I've been rather confused attempting to work with sharps and the structure that they impose beyond just covering. I've also been confused about the cofinality of theta and it's impact. This seemed like a simple enough question that might help elucidate both ideas since it gives a connection between them.