Vaught's theorem recursively revisited

1981 ◽  
Vol 46 (2) ◽  
pp. 397-411 ◽  
Author(s):  
Terrence Millar
Keyword(s):  
Isomorphism Type ◽  
Complete Theory ◽  
Decidable Model ◽  
Recursive Isomorphism ◽  
The Relationship ◽  
Countable Models

In this paper we investigate the relationship between the number of countable and decidable models of a complete theory. The number of decidable models will be determined in two ways, in §1 with respect to abstract isomorphism type, and in §2 with respect to recursive isomorphism type.Definition 1. A complete theory is (α, β) if the number of countable models of T, up to abstract isomorphism, is β, and similarly the number of decidable models of T is α.Definition 2. A model is ω-decidable if ∣∣= ω and for an effective listing {θn∣n < ω} of all sentences in the language of Th() augmented by new constant symbols i*, i < ω, {n ∣〈, i〉i<ω ⊨ θn} is recursive, where i interprets i* (in these terms, is decidable if is abstractly isomorphic to an ω-decidable model).Definition 3. A complete theory is (α, β)r if it is (γ, β) for some γ and it has exactly αω-decidable models up to recursive isomorphism.Specifically we will show in §1 that there is a (2, ω) theory, and in §2 that although there is a (2, 2ω) theory, there is no (2, β)r theory for any β, β < 2ω.

A complete, decidable theory with two decidable models

1979 ◽  
Vol 44 (3) ◽  
pp. 307-312 ◽  
Author(s):  
Terrence S. Millar
Keyword(s):  
Completeness Theorem ◽  
Countable Model ◽  
Complete Theory ◽  
Consistent Theory ◽  
Decidable Theory ◽  
Decidable Model ◽  
Recursive Set ◽  
Countable Models

A well-known result of Vaught's is that no complete theory has exactly two nonisomorphic countable models. The main result of this paper is that there is a complete decidable theory with exactly two nonisomorphic decidable models.A model is decidable if it has a decidable satisfaction predicate. To be more precise, let T be a decidable theory, let {θn∣n < ω} be an effective enumeration of all formulas in L(T), and let be a countable model of T. For any indexing E = {ai∣ i < ω} of ∣∣, and any formula ϕ ∈ L(T), let ‘ϕE’ denote the result of substituting ‘ai’ for every free occurrence of ‘xi’ in ϕ, i < ω. Then is decidable just in case, for some indexing E of ∣∣, {n ∣ ⊨ θnE} is a recursive set of integers. It is easy to show that the decidability of a model does not depend on the choice of the effective enumeration of the formulas in L(T); we omit details. By a simple ‘effectivization’ of Henkin's proof of the completeness theorem (see Chang [1]) we haveFact 1. Every decidable consistent theory has a decidable model.Assume next that T is a complete decidable theory and {θn ∣ n < ω} is an effective enumeration of all formulas of L(T).

Why some people are excited by Vaught's conjecture

1985 ◽  
Vol 50 (4) ◽  
pp. 973-982 ◽  
Author(s):  
Daniel Lascar
Keyword(s):  
Model Theory ◽  
Continuum Hypothesis ◽  
Complete Theory ◽  
Scott Sentences ◽  
Countable Theory ◽  
The Continuum ◽  
Countable Models

§I. In 1961, R. L. Vaught ([V]) asked if one could prove, without the continuum hypothesis, that there exists a countable complete theory with exactly ℵ1 isomorphism types of countable models. The following statement is known as Vaught conjecture:Let T be a countable theory. If T has uncountably many countable models, then T hascountable models.More than twenty years later, this question is still open. Many papers have been written on the question: see for example [HM], [M1], [M2] and [St]. In the opinion of many people, it is a major problem in model theory.Of course, I cannot say what Vaught had in mind when he asked the question. I just want to explain here what meaning I personally see to this problem. In particular, I will not speak about the topological Vaught conjecture, which is quite another issue.I suppose that the first question I shall have to face is the following: “Why on earth are you interested in the number of countable models—particularly since the whole question disappears if we assume the continuum hypothesis?” The answer is simply that I am not interested in the number of countable models, nor in the number of models in any cardinality, as a matter of fact. An explanation is due here; it will be a little technical and it will rest upon two names: Scott (sentences) and Morley (theorem).

Recursive isomorphism types of recursive Boolean algebras

1981 ◽  
Vol 46 (3) ◽  
pp. 572-594 ◽  
Author(s):  
J. B. Remmel
Keyword(s):  
Boolean Algebra ◽  
Recursive Function ◽  
Boolean Algebras ◽  
Isomorphism Type ◽  
Turing Degrees ◽  
Main Question ◽  
Partial Recursive Function ◽  
Recursive Isomorphism ◽  
Infinite Set ◽  
The Ideal

A Boolean algebra is recursive if B is a recursive subset of the natural numbers N and the operations ∧ (meet), ∨ (join), and ¬ (complement) are partial recursive. Given two Boolean algebras and , we write if is isomorphic to and if is recursively isomorphic to , that is, if there is a partial recursive function f: B1 → B2 which is an isomorphism from to . will denote the set of atoms of and () will denote the ideal generated by the atoms of .One of the main questions which motivated this paper is “To what extent does the classical isomorphism type of a recursive Boolean algebra restrict the possible recursion theoretic properties of ?” For example, it is easy to see that must be co-r.e. (i.e., N − is an r.e. set), but can be immune, not immune, cohesive, etc? It follows from a result of Goncharov [4] that there exist classical isomorphism types which contain recursive Boolean algebras but do not contain any recursive Boolean algebras such that is recursive. Thus the classical isomorphism can restrict the possible Turing degrees of , but what is the extent of this restriction? Another main question is “What is the recursion theoretic relationship between and () in a recursive Boolean algebra?” In our attempt to answer these questions, we were led to a wide variety of recursive isomorphism types which are contained in the classical isomorphism type of any recursive Boolean algebra with an infinite set of atoms.

Number of countable models of a countable complete theory

1981 ◽  
Vol 20 (1) ◽  
pp. 48-65
Author(s):  
T. G. Mustafin
Keyword(s):  
Complete Theory ◽  
Countable Models

A note on a theorem of Vaught

1971 ◽  
Vol 36 (3) ◽  
pp. 439-440 ◽  
Author(s):  
Joseph G. Rosenstein
Keyword(s):  
Predicate Calculus ◽  
Complete Theory ◽  
First Order ◽  
Free Variables ◽  
Order Predicate Calculus ◽  
Countable Models

In [2] Vaught showed that if T is a complete theory formalized in the first-order predicate calculus, then it is not possible for T to have exactly (up to isomorphism) two countable models. In this note we extend his methods to obtain a theorem which implies the above.First some definitions. We define Fn(T) to be the set of well-formed formulas (wffs) in the language of T whose free variables are among x1 x2, …, xn. An n-type of T is a maximal consistent set of wffs of Fn(T); equivalently, a subset P of Fn(T) is an n-type of T if there is a model M of T and elements a1, a2, …, an of M such that M ⊧ ϕ(a1, a2, …, an) for every ϕ ∈ P. In the latter case we say that 〈a1, a2, …, an〉 ony realizes P in M. Every set of wffs of Fn(T) which is consistent with T can be extended to an n-type of T.

Recursive Boolean algebras with recursive atoms

1981 ◽  
Vol 46 (3) ◽  
pp. 595-616 ◽  
Author(s):  
Jeffrey B. Remmel
Keyword(s):  
Boolean Algebra ◽  
Boolean Algebras ◽  
Isomorphism Type ◽  
Natural Numbers ◽  
Recursive Isomorphism ◽  
The Ideal

A Boolean algebra (henceforth abbreviated B.A.) is said to be recursive if B is a recursive subset of the natural numbers N and the operations ∧ (meet), ∨ (join), and ¬ (complement) are partial recursive. Let denote the set of atoms of and denote the ideal generated by the atoms of . Given recursive B.A.s and , we write ≈ if is isomorphic to and ≈r if is recursively isomorphic to , i.e., if there is a partial recursive isomorphism from onto .Recursive B.A.s have been studied by several authors including Ershov [2], Fiener [3], [4], Goncharov [5], [6], [7], LaRoche [8], Nurtazin [7], and the author [10], [11]. This paper continues a study of the recursion theoretic relationships among , , and the recursive isomorphism type of a recursive B.A. we started in [11]. We refer the reader to [11] for any unexplained notation and definitions. In [11], we were mainly concerned with the possible recursion theoretic properties of the set of atoms in recursive B.A.s. We found that even if we insist that be recursive, there is considerable freedom for the properties of . For example, we showed that if is a recursive B.A. such that is recursive and is infinite, then (i) there exists a recursive B.A. such that and both and are recursive and (ii) for any nonzero r.e. degree δ, there exist recursive B.A.s , , … such that for each i, is of degree δ, is recursive, is immune if i is even and is not immune if i is odd, and no two B.A.s in the sequence are recursively isomorphic.

Theories with exactly three countable models and theories with algebraic prime models

1980 ◽  
Vol 45 (2) ◽  
pp. 302-310 ◽  
Author(s):  
Anand Pillay
Keyword(s):  
Complete Theory ◽  
Uniform Bound ◽  
Bounded Degree ◽  
Definable Set ◽  
Countable Theory ◽  
Uniformly Bounded ◽  
Pathological Case ◽  
Countable Models

We prove first that if T is a countable complete theory with n(T), the number of countable models of T, equal to three, then T is similar to the Ehrenfeucht example of such a theory. Woodrow [4] showed that if T is in the same language as the Ehrenfeucht example, T has elimination of quantifiers, and n(T) = 3 then T is very much like this example. All known examples of theories T with n(T) finite and greater than one are based on the Ehrenfeucht example. We feel that such theories are a pathological case. Our second theorem strengthens the main result of [2]. The theorem in the present paper says that if T is a countable theory which has a model in which all the elements of some infinite definable set are algebraic of uniformly bounded degree, then n(T) ≥ 4. It is known [3] that if n(T) > 1, then n(T) > 3, so our result is the first nontrivial step towards proving that n(T) ≥ ℵ0. We would also like, of course, to prove the result without the uniform bound on the finite degrees of the elements in the subset.Theorem 2.1 is included in the author's Ph. D. thesis, as is a weaker version of Theorem 3.7. Thanks are due to Harry Simmons for his suggestions concerning the presentation of the material, and to Wilfrid Hodges for his advice while I was a Ph. D. student.

On computable automorphisms of the rational numbers

2001 ◽  
Vol 66 (3) ◽  
pp. 1458-1470 ◽  
Author(s):  
A. S. Morozov ◽  
J. K. Truss
Keyword(s):  
Rational Numbers ◽  
Isomorphism Type ◽  
Turing Degrees ◽  
First Order ◽  
Principal Ideals ◽  
General Correspondence ◽  
The Relationship ◽  
The Ideal

AbstractThe relationship between ideals I of Turing degrees and groups of I-recursive automorphisms of the ordering on rationals is studied. We discuss the differences between such groups and the group of all automorphisms, prove that the isomorphism type of such a group completely defines the ideal I, and outline a general correspondence between principal ideals of Turing degrees and the first-order properties of such groups.

Number of countable models

1978 ◽  
Vol 43 (3) ◽  
pp. 492-496 ◽  
Author(s):  
Anand Pillay
Keyword(s):  
Finite Number ◽  
Complete Theory ◽  
Prime Model ◽  
Categorical Theory ◽  
Definable Subset ◽  
Algebraic Elements ◽  
Minimal Prime ◽  
Countable Theory ◽  
The Universe ◽  
Countable Models

We prove that a countable complete theory whose prime model has an infinite definable subset, all of whose elements are named, has at least four countable models up to isomorphism. The motivation for this is the conjecture that a countable theory with a minimal model has infinitely many countable models. In this connection we first prove that a minimal prime model A has an expansion by a finite number of constants A′ such that the set of algebraic elements of A′ contains an infinite definable subset.We note that our main conjecture strengthens the Baldwin–Lachlan theorem. We also note that due to Vaught's result that a countable theory cannot have exactly two countable models, the weakest possible nontrivial result for a non-ℵ0-categorical theory is that it has at least four countable models.§1. Notation and preliminaries. Our notation follows Chang and Keisler [1], except that we denote models by A, B, etc. We use the same symbol to refer to the universe of a model. Models we refer to are always in a countable language. For T a countable complete theory we let n(T) be the number of countable models of T up to isomorphism. ∃n means ‘there are exactly n’.

Can you read my mind? Age as a moderator in the relationship between theory of mind and relational aggression

2015 ◽  
Vol 39 (6) ◽  
pp. 552-559 ◽  
Author(s):  
Carlos Gomez-Garibello ◽  
Victoria Talwar
Keyword(s):  
Theory Of Mind ◽  
Relational Aggression ◽  
Cognitive Skills ◽  
Social Behaviors ◽  
Cognitive Factors ◽  
Complete Theory ◽  
Chronological Age ◽  
Older Children ◽  
The Relationship ◽  
Attribution Of Intentions

The present study examined whether age moderates the relationship between cognitive factors (theory of mind and attribution of intentions) and relational aggression. Participants ( N = 426; 216 boys) between 6 and 9 years of age were asked to complete theory of mind tasks and answer an attribution of intentions questionnaire. Teachers evaluated their students’ social behaviors including relational aggressive acts. Results suggest that theory of mind did affect relational aggression, when this association was moderated by chronological age. Specifically, it was found that the association between theory of mind and relational aggression was only significant and positive for younger participants; for older children the direction of this association was inverse. Taken together, findings from this study partially support the assertion that sophisticated cognitive skills are a prerequisite for indirect ways of aggression.

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