The Vaught Conjecture: Do Uncountable Models Count?
نویسنده
چکیده
We give a model theoretic proof, replacing admisssible set theory by the LopezEscobar theorem, of Makkai’s theorem: Every counterexample to Vaught’s conjecture has an uncountable model which realizes only countably many Lω1,ω-types. The following two results are new. Theorem I. If a first order theory is a counterexample to the Vaught conjecture then it has 2א1 models of cardinality א1. Theorem II. [2א0 < 2א1 ] If a sentence of Lω1,ω is a counterexample to the Vaught conjecture then it has 2 א1 models of cardinality א1. In this paper we prove several properties of counterexamples to the Vaught conjecture. Specifically, results concern the number of models the counterexample has in power א1. One of these results was proved 30 years ago using admissible model theory; we give a more straightforward argument. The following question guides our discussion. Is the Vaught Conjecture model theory ? Here are some possible ways in which this question would have a clear answer. Shelah, Buechler, Newelski have shown using rather difficult techniques from stability theory that the conjecture holds for first order theories that are ‘simple’ from the stability theoretic standpoint: ω-stable or superstable with finite U -rank. If a counterexample were found for a first order theory of slightly greater complexity (e.g. a stable but not superstable first order theory), this would indicate the issue was a model theoretic one. If on the other hand, a uniform proof for sentences of Lω1,ω were given using methods of descriptive set theory, then it would not be a model theoretic problem. The results below give partial answers to the following methodological questions. What specific model theoretic as opposed to descriptive set theoretic techniques can attack the problem? Can one use more direct model theoretic arguments to obtain some result of admissible model theory?
منابع مشابه
5 Counterexamples to Vaught ’ s Conjecture
A sentence σ ∈ L ω1,ω is a counterexample to Vaught's Conjecture, or simply a counterexample, if ℵ 0 < I(σ, ℵ 0) < 2 ℵ0 , i.e., σ has uncountably many countable models but fewer than continuum many models. Fix σ a counterexample. A key fact about counterexamples is that there are few types for any count-able fragment. Let F be a fragment. Let
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عنوان ژورنال:
- Notre Dame Journal of Formal Logic
دوره 48 شماره
صفحات -
تاریخ انتشار 2007