On Spectrum of kappa-Resplendent Models
  We prove that some natural "outside" property is equivalent (for a first order class) to being stable.   For a model, being resplendent is a strengthening of being kappa-saturated. Restricting ourselves to the case kappa > |T| for transparency, a model M is kappa-resplendent means: when we expand M by <kappa individual constants (c_i:i < alpha), if (M, c_i)_ {i<alpha} has an elementary extension expandable to be a model of T' where Th((M,c_i)_{i<alpha}) subseteq T', |T'| < kappa then already (M,c_i)_{i<\alpha} can be expanded to a model of T'.   Trivially any saturated model of cardinality lambda is lambda-resplendent. We ask: how may kappa-resplendent models of a (first order complete) theory T of cardinality lambda are there? Naturally we restrict ourselves to cardinals lambda=lambda^ kappa + 2^ {|T|}. Then we get a complete and satisfying answer: this depends just on T being stable or unstable. In this case proving that for stable T we get few, is not hard; in fact every resplendent model of T is saturated hence determined by its cardinality up to isomorphism. The inverse is more problematic because naturally we have to use Skolem functions with any alpha<kappa places. Normally we use relevant partition theorems (Ramsey theorem or Erd\H{o}s-Rado theorem), but in our case the relevant partitions theorems fails so we have to be careful. 
Saharon Shelah
math.LO