scholarly journals The Complexity of Classification Problems for Models of Arithmetic

2010 ◽  
Vol 16 (3) ◽  
pp. 345-358 ◽  
Author(s):  
Samuel Coskey ◽  
Roman Kossak
Keyword(s):  
Classification Problem ◽  
The Other ◽  
Saturated Model ◽  
Classification Problems ◽  
Finitely Generated ◽  
Saturated Models ◽  
Other Hand ◽  
Recursively Saturated ◽  
Recursively Saturated Model ◽  
Models Of Arithmetic

AbstractWe observe that the classification problem for countable models of arithmetic is Borel complete. On the other hand, the classification problems for finitely generated models of arithmetic and for recursively saturated models of arithmetic are Borel; we investigate the precise complexity of each of these. Finally, we show that the classification problem for pairs of recursively saturated models and for automorphisms of a fixed recursively saturated model are Borel complete.

ω1-like recursively saturated models of Presburger's arithmetic

1986 ◽  
Vol 51 (2) ◽  
pp. 421-429
Author(s):  
Victor Harnik
Keyword(s):  
The Other ◽  
Isomorphism Type ◽  
Complete Theory ◽  
Saturated Model ◽  
Technical Result ◽  
Saturated Models ◽  
Starting Point ◽  
Introductory Discussion ◽  
Recursively Saturated ◽  
Recursively Saturated Model

Let Pr be Presburger's arithmetic, i.e., the complete theory of the structure (ω, +). Lipshitz and Nadel showed in [4] that a countable model of Pr is recursively saturated iff it can be expanded to a model of Peano arithmetic, PA. As a starting point for an introductory discussion, let us mention one more fact about countable recursively saturated models of Pr. If is such a model then the following is readily seen (as explained also in §1):Any two countable recursively saturated elementary endextensions of are isomorphic.If we drop “countable” from the assumption of this statement, we can still say that the two models are ∞ω-equivalent. Must they be isomorphic if both have cardinality ℵ1? Certainly not, since one of the models can be ω1-like while the other is not. Once we realized this much, let us concentrate on ω1-like structures. We prove in §3:Theorem. Any countable recursively saturated modelof Pr hasisomorphism types of ω1-like recursively saturated elementary endextensions. Only one of these is the isomorphism type of a structure that can be expanded to a model of PA.The key technical result is proven in §2. It says that an as above has precisely two countable recursively saturated elementary endextensions which are nonisomorphic over .

Models with the ω-property

1989 ◽  
Vol 54 (1) ◽  
pp. 177-189 ◽  
Author(s):  
Roman Kossak
Keyword(s):  
Structural Properties ◽  
Interesting Property ◽  
Order Type ◽  
Countable Model ◽  
The Other ◽  
Recursive Saturation ◽  
Saturated Models ◽  
Other Hand ◽  
To Come ◽  
Recursively Saturated

In [KP] we have studied the problem of determining when a subset of a (countable) model M of PA can be coded in an elementary end extension of M. Sets with this property are called elementary extensional. In particular we can ask whether there are elementary extensional subsets of a model which have order type ω. It turns out that having elementary extensional subsets of order type ω is an interesting property connected with other structural properties of models of PA. We will call this property the ω-property. In [KP] the problem of characterizing models with the ω-property was left open. It is still open, and the aim of this paper is to present a collection of results pertaining to it. It should be mentioned that the same notion was studied by Kaufmann and Schmerl in [KS2] in connection with some weak notions of saturation which they discuss there. Our notion of a model with the ω-property corresponds to the notion of an upward monotonically ω-lofty cut.It is fairly easy to see that countable recursively saturated models (or in fact all recursively saturated models with cofinality ω) and all short recursively saturated models have the ω-property (Proposition 1.2 below). On the other hand, if we had asked the question about the existence of models with the ω-property before 1975 (when recursively saturated models were introduced) the answer would probably not have been that easy and we would have to come to notions close to recursive saturation.

scholarly journals Interpreting the compositional truth predicate in models of arithmetic

2021 ◽  
Author(s):  
Cezary Cieśliński
Keyword(s):  
Proof Theory ◽  
Formal Proof ◽  
Truth Predicate ◽  
Saturated Model ◽  
First Order ◽  
Order Arithmetic ◽  
Recursively Saturated ◽  
Recursively Saturated Model ◽  
Models Of Arithmetic

AbstractWe present a construction of a truth class (an interpretation of a compositional truth predicate) in an arbitrary countable recursively saturated model of first-order arithmetic. The construction is fully classical in that it employs nothing more than the classical techniques of formal proof theory.

Large resplendent models generated by indiscernibles

1989 ◽  
Vol 54 (4) ◽  
pp. 1382-1388 ◽  
Author(s):  
James H. Schmerl
Keyword(s):  
Peano Arithmetic ◽  
Saturated Model ◽  
Infinite Cardinal ◽  
Common Generalization ◽  
Saturated Models ◽  
Uncountable Cardinal ◽  
Finite Language ◽  
Recursively Saturated ◽  
Recursively Saturated Model

The motivation for the results presented here comes from the following two known theorems which concern countable, recursively saturated models of Peano arithmetic.(1) if is a countable, recursively saturated model of PA, then for each infinite cardinal κ there is a resplendent which has cardinality κ. (See Theorem 10 of [1].)(2) if is a countable, recursively saturated model of PA, then is generated by a set of indiscernibles. (See [4].)It will be shown here that (1) and (2) can be amalgamated into a common generalization.(3) if is a countable, recursively saturated model of PA, then for each infinite cardinal κ there is a resplendent which has cardinality κ and which is generated by a set of indiscernibles.By way of contrast we will also get recursively saturated models of PA which fail to be resplendent and yet are generated by indiscernibles.(4) if is a countable, recursively saturated model of PA, then for each uncountable cardinal κ there is a κ-like recursively saturated generated by a set of indiscernibles.None of (1), (2) or (3) is stated in its most general form. We will make some comments concerning their generalizations. From now on let us fix a finite language L; all structures considered are infinite L-structures unless otherwise indicated.

Models of arithmetic and closed ideals

1982 ◽  
Vol 47 (4) ◽  
pp. 833-840 ◽  
Author(s):  
Julia Knight ◽  
Mark Nadel
Keyword(s):  
Closed Ideal ◽  
Turing Degrees ◽  
Saturated Model ◽  
Recursive Saturation ◽  
New Information ◽  
Recursively Saturated ◽  
Recursively Saturated Model ◽  
Saturated Structure ◽  
Models Of Arithmetic ◽  
Natural Way

A set J of Turing degrees is called an ideal if (1) J ≠ ∅, (2) for any pair of degrees ã, , if ã, ϵ J, then ã ⋃ ϵJ, and (3) for any ⋃ ϵ J and any , if < ⋃, then ϵ J. A set J of degrees is said to be closed if for any theory T with a set of axioms of degree in J, T has a completion of degree in J.Closed ideals of degrees arise naturally in the following way. If is a recursively saturated structure, let I() = { for some ā ϵ }. Let D() = {: is recursive in d-saturated}. (Recursive in d-saturation is defined like recursive saturation except that the sets of formulas considered are recursive in d.) These two sets of degrees were investigated in [2]. It was shown that if is a recursively saturated model of P, Pr = Th(ω, +), or Pr′ = Th(Z, +, 1), then I() = D(), and this set is a closed ideal. Any closed ideal J can be represented as I() = D() for some recursively saturated model of Pr′. For sets J of power at most ℵ1, Pr′ can be replaced by P.Assuming CH, all closed ideals have power at most ℵ1, but if CH fails, there are closed ideals of power greater than ℵ1, and it is not known whether these can be represented as I() = D() for a recursively saturated model of P.In the present paper, it will first be shown that information about representation of closed ideals provides new information about an old problem of MacDowell and Specker [6] and extends an old result of Scott [8] in a natural way. It will also be shown that the representation results from [2] answer a problem of Friedman [1]. This part of the paper is aimed at convincing the reader that representation problems are worth investigating.

scholarly journals Generating Adjoint Groups

2019 ◽  
Vol 62 (3) ◽  
pp. 733-738 ◽  
Author(s):  
Be'eri Greenfeld
Keyword(s):  
Open Problem ◽  
Jacobson Radical ◽  
The Other ◽  
Adjoint Group ◽  
Finitely Generated ◽  
Infinite Dimensional ◽  
Other Hand ◽  
Nil Ring

AbstractWe prove two approximations of the open problem of whether the adjoint group of a non-nilpotent nil ring can be finitely generated. We show that the adjoint group of a non-nilpotent Jacobson radical cannot be boundedly generated and, on the other hand, construct a finitely generated, infinite-dimensional nil algebra whose adjoint group is generated by elements of bounded torsion.

Saturation of homogeneous resplendent models

1986 ◽  
Vol 51 (1) ◽  
pp. 222-224 ◽  
Author(s):  
Julia F. Knight
Keyword(s):  
Continuum Hypothesis ◽  
Homogeneous Model ◽  
Peano Arithmetic ◽  
Saturated Model ◽  
Relation Symbol ◽  
First Order ◽  
Recursively Saturated ◽  
Recursively Saturated Model ◽  
Saturated Structure ◽  
The Continuum

The complete diagram of a structure , denoted by Dc(), is the set of all sentences true in the structure (, a)a∈. A structure is said to be resplendent if for every sentence θ involving a new relation symbol R in addition to symbols occurring in Dc(), if θ is consistent with Dc(), then there is a relation P on such that (see[1]).Baldwin asked whether a homogeneous recursively saturated structure is necessarily resplendent. Here it is shown that this need not be the case. It is shown that if is an uncountable homogeneous resplendent model of an unstable theory, then must be saturated. The proof is related to the proof in [5] that an uncountable homogeneous recursively saturated model of first order Peano arithmetic must be saturated. The example for Baldwin's question is an uncountable homogeneous model for a particular unstable theory, such that is recursively saturated and omits some type. (The continuum hypothesis is needed to show the existence of such a model in power ℵ1.)The proof of the main result requires two lemmas.

scholarly journals Condensable models of set theory

2021 ◽  
Author(s):  
Ali Enayat
Keyword(s):  
Set Theory ◽  
Countable Model ◽  
Saturated Model ◽  
Infinitary Logic ◽  
First Order ◽  
Recursively Saturated ◽  
Recursively Saturated Model ◽  
Models Of Set Theory

AbstractA model $${\mathcal {M}}$$ M of ZF is said to be condensable if $$ {\mathcal {M}}\cong {\mathcal {M}}(\alpha )\prec _{\mathbb {L}_{{\mathcal {M}}}} {\mathcal {M}}$$ M ≅ M ( α ) ≺ L M M for some “ordinal” $$\alpha \in \mathrm {Ord}^{{\mathcal {M}}}$$ α ∈ Ord M , where $$\mathcal {M}(\alpha ):=(\mathrm {V}(\alpha ),\in )^{{\mathcal {M}}}$$ M ( α ) : = ( V ( α ) , ∈ ) M and $$\mathbb {L}_{{\mathcal {M}}}$$ L M is the set of formulae of the infinitary logic $$\mathbb {L}_{\infty ,\omega }$$ L ∞ , ω that appear in the well-founded part of $${\mathcal {M}}$$ M . The work of Barwise and Schlipf in the 1970s revealed the fact that every countable recursively saturated model of ZF is cofinally condensable (i.e., $${\mathcal {M}}\cong {\mathcal {M}}(\alpha ) \prec _{\mathbb {L}_{{\mathcal {M}}}}{\mathcal {M}}$$ M ≅ M ( α ) ≺ L M M for an unbounded collection of $$\alpha \in \mathrm {Ord}^{{\mathcal {M}}}$$ α ∈ Ord M ). Moreover, it can be readily shown that any $$\omega $$ ω -nonstandard condensable model of $$\mathrm {ZF}$$ ZF is recursively saturated. These considerations provide the context for the following result that answers a question posed to the author by Paul Kindvall Gorbow.Theorem A.Assuming a modest set-theoretic hypothesis, there is a countable model $${\mathcal {M}}$$ M of ZFC that is bothdefinably well-founded (i.e., every first order definable element of $${\mathcal {M}}$$ M is in the well-founded part of $$\mathcal {M)}$$ M ) andcofinally condensable. We also provide various equivalents of the notion of condensability, including the result below.Theorem B.The following are equivalent for a countable model$${\mathcal {M}}$$ M of $$\mathrm {ZF}$$ ZF : (a) $${\mathcal {M}}$$ M is condensable. (b) $${\mathcal {M}}$$ M is cofinally condensable. (c) $${\mathcal {M}}$$ M is nonstandard and $$\mathcal {M}(\alpha )\prec _{\mathbb {L}_{{\mathcal {M}}}}{\mathcal {M}}$$ M ( α ) ≺ L M M for an unbounded collection of $$ \alpha \in \mathrm {Ord}^{{\mathcal {M}}}$$ α ∈ Ord M .

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