User grigor - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T19:26:19Z http://mathoverflow.net/feeds/user/20584 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/81939/why-does-inner-model-theory-needs-so-much-descriptive-set-theory-and-vice-versa/85697#85697 Answer by Grigor for Why does inner model theory needs so much descriptive set theory (and vice versa)? Grigor 2012-01-14T22:55:10Z 2012-01-15T14:28:54Z <p>Philip gave a nice answer but let me add some points to it.</p> <p>The thing is that while IMT and DST were far apart back in the days, over the years it became clear that part of DST and IMT deal with exactly the same problem. </p> <p>The above sentence may not be completely objective, but at any rate, that seems to be the point of view of some people who do inner model theory (including mine). The main goal of IMT is to construct models that are correct about the universe. How you measure this correctness? Well, there is Levy hierarchy and there is a more refined and more useful version of it, namely the Wadge hierarchy. So to make the question more precise, what we do in inner model theory is that we try to construct and analyze canonical models called mice that capture the levels of the Wadge hierarchy. Back in the days, it wasn't so clear that there is such a deep relationship between Wadge hierarchy (or their more refined forms, the Universally Baire sets, or homogeneously Suslin sets) and the large cardinal hierarchy. Now, the connection is much more clear and transparent. </p> <p>DST is really inevitable. The technical problem that people doing IMT are trying to solve is the construction of $\omega_1+1$ iteration strategies. These iteration strategies allow one to build trees of height $\omega_1$ and then it is guaranteed that any such tree must have a branch. But wait a minute, in ZFC alone, there are trees of height $\omega_1$ with no branches. So how are we going to make sure that these strategies will build only trees of height $\omega_1$ for which there are branches. The answer must be that the strategy has various DST like properties, like it is Universally Baire or hom Suslin and etc (it is a nice exercise to show that such strategies are indeed nice).</p> <p>At any rate, when doing IMT, running to DST seems to be inevitable. It is probably not so true if you are doing DST (by DST we mean Moschovakis' book not the directions it went since 90s). But still there are many theorems you can prove under AD using IMT and no proof avoiding IMT is known. For instance, every regular cardinal below Theta is a measurable cardinal, a fact proven using IMT and no proof avoiding IMT is known. </p> <p>The reason that the subjects are so close, though, is just that they study the same object (namely the Wadge hierarchy) from different point of views. DST does it using recursion theoretic methods (coding lemma, pointclass arguments and etc) and IMT does it by translating the Wadge hierarchy into hierarchy of iteration strategies, which is what IMT studies. </p> http://mathoverflow.net/questions/3462/characterizations-of-non-wellfounded-models/3489#3489 Answer by Grigor for Characterizations of non-wellfounded models? Grigor 2009-10-30T23:29:50Z 2009-10-30T23:29:50Z <p>What do you mean by characterization? We don't really have a characterization of well founded models of set theory, do we? </p> <p>Of course, any consistent extension of ZFC has ill founded models. </p> http://mathoverflow.net/questions/3219/controlling-ultrapowers/3486#3486 Answer by Grigor for Controlling Ultrapowers Grigor 2009-10-30T23:26:14Z 2009-10-30T23:26:14Z <p>As much as you wish. Lowenheim Skolem give you such situations and then you can affect it too much. For instance, you can construct situations where the critical point is singular in the ultrapower. You cannot do much if U is amenable to M. Then it is like a real ultrafilter. I don't know what you are asking actually.</p> <p>An interesting question is whether you can have an ultrafilter on kappa such that the powerset of kappa^+ is in the ultrapower and James Cummings solved this by showing that you can. I don't know if it is interesting to look for ultrafilters that can code powerset(kappa^++) into the ultrapower. probably his proof already gives that.</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> http://mathoverflow.net/questions/743/what-do-models-where-the-ch-is-false-look-like/2867#2867 Answer by Grigor for What do models where the CH is false look like? Grigor 2009-10-27T18:46:48Z 2009-10-27T18:46:48Z <p>People have done a lot on CH. I don't know how to answer the comments but at a recent conference in Bedlewo, Hugh Woodin actually claimed that CH is true in the true model of ZFC. A new picture of V has been emerging in which CH is true along with GCH. There are of course many set theorists who don't believe in true V, but Woodin's work, if everything works out, will be a phenomenal piece of set theory which will identify unarguable the true core model of ZFC. Think of the situation like this. If 0^# doesn't exist that there are few who will argue that L isn't the core model of ZFC, core" here just means the largest canonical. In that sense, he basically identified the model without putting any large cardinal restrictions on the universe. Whether the model is itself the true universe of set theory is an entirely different matter.</p> <p>There is an alternative approach to CH via forcing axioms and etc. Justin Moore has obtained some really deep recent results on this. </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/81939/why-does-inner-model-theory-needs-so-much-descriptive-set-theory-and-vice-versa/85697#85697 Comment by Grigor Grigor 2012-01-16T17:21:13Z 2012-01-16T17:21:13Z One can see omega_1-iteration strategies as sets of reals. It is a theorem of AD^+ that every set of reals is Wadge reducible to an omega_1 iteration strategy. Yes, I meant Borel equivalence stuff, which is immune IMT, I am not aware of single application either way. The answer to the last question is probably about the same. There is no proof of the partition property of delte^1_3 using IMT (in fact, this is a very nice open problem). So I would say there are many theorems on the structure of cardinals that have proofs using one set of methods but no proof is known using the other.