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).

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.

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’.

The number of countable models

1970 ◽  
Vol 35 (1) ◽  
pp. 14-18 ◽  
Author(s):  
Michael Morley
Keyword(s):  
Continuum Hypothesis ◽  
Predicate Calculus ◽  
The Continuum ◽  
Countable Models

A theory formulated in a countable predicate calculus can have at most nonisomorphic countable models. It has been conjected (e.g., in [4]) that if it has an uncountable number of such models then it has exactly such. Of course, this would follow immediately if one assumed the continuum hypothesis.In this paper we show that if a theory has more than ℵ1 (i.e., at least ℵ2) isomorphism types of countable models then it has exactly . Our results generalize immediately to theories in Lω1ω and even to pseudo-axiomatic classes in Lω1ω. In this last case, a result of H. Friedman shows that it is the best possible result.

Flow and Solute Transport in Saturated Porous Media: 1. The Continuum Hypothesis

2008 ◽  
Vol 11 (4) ◽  
pp. 403-413 ◽  
Author(s):  
Amgad Salama ◽  
Paul J. Van Geel
Keyword(s):  
Porous Media ◽  
Solute Transport ◽  
Continuum Hypothesis ◽  
Saturated Porous Media ◽  
The Continuum

scholarly journals Several results on compact metrizable spaces in $$\mathbf {ZF}$$

2021 ◽  
Author(s):  
Kyriakos Keremedis ◽  
Eleftherios Tachtsis ◽  
Eliza Wajch
Keyword(s):  
Continuum Hypothesis ◽  
Compact Spaces ◽  
Power Set ◽  
Compact Metrizable Space ◽  
Permutation Models ◽  
Independence Results ◽  
The Axiom Of Choice ◽  
Metrizable Spaces ◽  
Transfer Theorems ◽  
The Continuum

AbstractIn the absence of the axiom of choice, the set-theoretic status of many natural statements about metrizable compact spaces is investigated. Some of the statements are provable in $$\mathbf {ZF}$$ ZF , some are shown to be independent of $$\mathbf {ZF}$$ ZF . For independence results, distinct models of $$\mathbf {ZF}$$ ZF and permutation models of $$\mathbf {ZFA}$$ ZFA with transfer theorems of Pincus are applied. New symmetric models of $$\mathbf {ZF}$$ ZF are constructed in each of which the power set of $$\mathbb {R}$$ R is well-orderable, the Continuum Hypothesis is satisfied but a denumerable family of non-empty finite sets can fail to have a choice function, and a compact metrizable space need not be embeddable into the Tychonoff cube $$[0, 1]^{\mathbb {R}}$$ [ 0 , 1 ] R .

Kreisel, the continuum hypothesis and second order set theory

1976 ◽  
Vol 5 (2) ◽  
Author(s):  
Thomas Weston
Keyword(s):  
Set Theory ◽  
Continuum Hypothesis ◽  
Second Order ◽  
The Continuum ◽  
Order Set

The Operational Significance of the Continuum Hypothesis in the Theory of Water Movement Through Soils and Aquifers

1984 ◽  
Vol 20 (5) ◽  
pp. 521-530 ◽  
Author(s):  
Philippe Baveye ◽  
Garrison Sposito
Keyword(s):  
Water Movement ◽  
Continuum Hypothesis ◽  
The Continuum

Regular Conditional Expectations and the Continuum Hypothesis

1994 ◽  
pp. 85-93
Author(s):  
I. V. Evstigneev
Keyword(s):  
Continuum Hypothesis ◽  
Conditional Expectations ◽  
The Continuum
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