Model Theory

Home , Elementary diagram, Elementary equivalence

Pascal Vanier

Saturday 26th March, 2016 09:22 2 Sources

The basis of this course are the model theory courses of Université Claude Bernard Lyon 1 and in particular the lecture notes of Emmanuel Jeandel for the course of Frank Wagner and Bruno Poizat of which this document is mainly an english translation. We also used the lecture notes of the course of Tuna Altinel and Julien Melleray and the proof of the compactness theorem of the course A crèche course on model theory of Domenico Zambella (Università di Torino). Contents

I Model theory 5

1 Languages and structures 7 1.1 Formulae ...... 7 1.2 Structures ...... 8 1.3 Theories and models ...... 10 1.4 Diagrams ...... 11 1.5 Exercises ...... 12

2 Equivalence between models 13 2.1 Elementary equivalence ...... 13 2.2 Substructures ...... 14 2.3 Morphisms ...... 15 2.4 Diagrams ...... 17 2.5 Tarski’s test ...... 17 2.6 Skolemisation ...... 18 2.7 Deﬁnability ...... 20 2.8 Exercises ...... 20

3 Compactness theorem 21 3.1 Compactness of quantiﬁer free sets ...... 21 3.2 Upward Löwenheim-Skolem ...... 23 3.3 Axiomatisation ...... 24 3.4 κ-categoricity ...... 25 3.5 Exercises ...... 25

4 Back and forth 27 4.1 ∞-isomorphisms ...... 27 4.2 Isomorphic structures ...... 29 4.3 Elementary equivalence ...... 30 4.4 ω-saturated models ...... 30 4.5 Applications to completeness ...... 33 4.5.1 Dense orders without extremities ...... 33 4.6 Discrete orders without extremities ...... 34 4.7 Exercises ...... 35

3 4 CONTENTS Part I

Model theory

Chapter 1

Languages and structures

1.1 Formulae

Deﬁnition 1.1.1 Language

A language L is a set of : • constant symbols (ci), • function symbols (fi), • relation symbols (Ri). We will always assume that equality is among the relation symbols.

Example 1.1.2

−1 • The language of groups Lgroups = {1, , ·} in the multiplicative notation and Lgroups = {0, +, −}. −1 • The language of rings Lﬁeld = {0, 1, +, −, ·, }. At this stage, a language does not give any meaning to the symbols, it should only be considered as an alphabet with which sentences may be written. These sentences will be called terms: Deﬁnition 1.1.3 Term

Let L be a language, the set of L-terms is the smallest set T such that: • any ci is in T , • T contains some set of symbols V disjoint from the (ci) called variables.

• T contains all fi(a1, . . . , ani ) with ai ∈ T .

A term is just the composition of function symbols with constants and some variables of the set V . With this we are now able to deﬁne what a formula is. Formulae on L correspond to ﬁrst order logic on the language L:

7 8 CHAPTER 1. LANGUAGES AND STRUCTURES

Deﬁnition 1.1.4 Formula

Let L be a language and T be the set of L-terms. The set of L-formulae F is the smallest set such that: • F contains all atomic formulae: the Ri(t1, . . . , tn) with tj ∈ T and Ri a relation in L. • F is closed by boolean combinations: if φ, ψ are formulae of F , then so are φ ∨ ψ, φ ∧ ψ and ¬φ. • F is closed by existential and universal quantiﬁers : for any variable v ∈ V , and formula φ ∈ F , ∀vφ and ∃vφ are formulae in F .

As the attentive reader may have noted, formulae may contain variable of V not "caught" by any quantiﬁer, such variables are said to be free variables, and this notion may be deﬁned more formally:

Deﬁnition 1.1.5 Free variable

A free variable for a formula is deﬁned inductively in the following way: • The free variables of an atomic formula are the variables v ∈ V appearing in this formula. • If φ = ψ ∨ ρ or φ = ψ ∧ ρ, the free variables of φ are the variables that are free in either ψ or ρ. • If φ = ¬ψ, the free variables of φ are the free variables of ψ. • If φ = ∃vψ or φ = ∀vψ the free variables of phi are the free variables of ψ minus v.

In the rest of the course, we will note φ(x) ∈ F for a formula whose free variables are x = (x1, . . . , xn).A sentence is a formula without free variables.

1.2 Structures

For the moment, we only deﬁned languages and formulae without giving them any meaning or semantics. The aim of model theory is to study the structures in which some set of sentences are true : 1.2. STRUCTURES 9

Deﬁnition 1.2.1 Structure

Let L be a language, an L-structure M is formed of a base set M, also called the universe or domain of M and of: M • a set of constants (ci ) of M associated to each of the symbols ci of L, M ni • a set of functions (fi : M → M) associated to each of the symbols fi of L, M mi • a set of relations (Ri ⊆ M ) associated to each of the symbols Ri of L. This set of relations always associates the equality symbol to the relation {(x, x) | x ∈ M}.

Now that we have associated a meaning to each of the symbols in our language, we can deﬁne the meaning of a term in a structure very naturally: Deﬁnition 1.2.2

Let L be a language, M be an L-structure and m elements of M, the universe of M. The interpretation of a term t(x) with parameter m may be deﬁned in an inductive way: • The interpretation of a constant c is cM. • The interpretation of x ∈ V is the corresponding element in m. M M M M • The interpretation of f(t1, . . . , tn) is f (t1 , . . . , tn ) with ti the interpretation of ti in M.

This in turn allows us to define what the satifiability of a formula is: Definition 1.2.3 Satisfiability of a formula

Let L be a language and M be an L-structure. If m is a vector of elements of M, the satisfaction of a formula φ(x) with parameter m is defined inductively by : • φ(m) is an atomic formula R(t1, . . . , tn), φ(m) is satisfied if and M M M only if (t1 , . . . , tn ) ∈ R where ti is the interpretation of ti with parameter m. • φ(m) = ψ(m) ∧ ρ(m) is satisfied iff ψ(m) and ρ(m) are both satisfied. • φ(m) = ψ(m) ∨ ρ(m) is satisfied iff at least one of ψ(m) or ρ(m) is satisfied. • φ = ¬ρ is satisfied iff ρ is not satisfied. • φ(m) = ∃yψ(m, y) is satisfied with parameter m iff there exists y ∈ M such that ψ is satisfied with the parameter m, y. • φ(m) = ∀yψ(m, y) is satisfied iff for all y ∈ M, ψ(m, y) is satisfied. When φ(x) is satisfied in M with parameter m, we use the notation M |= φ(m), which means "the structure M satisfies φ(m). 10 CHAPTER 1. LANGUAGES AND STRUCTURES

Deﬁnition 1.2.4 Equivalent formulae

Two formulae φ(x) and ψ(x) are equivalent iﬀ for any L-structure M and parameter m ∈ M, M |= φ(m) iﬀ M |= ψ(m).

Exercise 1.2.1. Prove that the following formulae are equivalent: • φ and ¬¬φ • ¬(φ ∧ ψ) and (¬φ ∨ ¬ψ) • ¬(φ ∨ ψ) and (¬φ ∧ ¬ψ) • ¬∃xφ and ∀x¬φ • ¬∀xφ and ∃x¬φ

The usual notation φ ⇒ ψ stands for ¬φ ∨ ψ and φ ⇔ ψ stands for (φ ⇒ ψ) ∧ (ψ ⇒ φ).

1.3 Theories and models

We may now get to the core of this course: what is a theory and a model for this theory:

Deﬁnition 1.3.1 Theory

Let L be a language, T a set of sentences on L. T is a theory if and only if there exists an L-structure M such that

∀φ ∈ T, M |= φ

M is then called a model of theory T .

In order for a theory to have a model, it needs to be coherent, in the sense that there exists a structure satisfying all of its sentences. For instance a theory containing ¬φ and φ at the same time cannot be coherent, as they cannot be simultaneously satisﬁed by deﬁnition.

Deﬁnition 1.3.2 Consistent theory

A theory T is consistent if it has a model. 1.4. DIAGRAMS 11

Example 1.3.3 We may get back to example 1.1.2, where we deﬁned the language of groups, we may now give a meaning to this language with the theory corresponding to groups: • ∀x, x · 1 = x • ∀x, x · x−1 = 1 • ∀x, x · x−1 · x = 1 • ∀x, y, z, (x · y) · z = x · (y · z)

A theory that will be used sometimes as an example in this course is the theory of orders:

Example 1.3.4 The language of the theory of orders contains only one relation and is the following: • ∀x, x < y • ∀x, y, (x < y) ⇒ ¬(y < x) • ∀x, y, z, (x < y) ∧ (y < z) ⇒ (x < z) If we want a theory of total orders, we can add the sentence ∀x, y, (x < y) ∨ (y < x) ∨ (x = y). We will note T |= φ if and only if for any model M of T , M |= φ, that is to say if the truth of φ depends only on the theory and not on the model.

Property 1.3.1. If φ ∈ T , then T |= φ.

Warning! T 6|= φ is not the same as T |= ¬φ: in the first case, there exists a model of T that does not satisfy φ (i.e.it satisfies ¬φ), and in the second case every model of T satisfies ¬φ.

Deﬁnition 1.3.5 Completeness

A theory T is complete if for any φ we have:

T |= φ or T |= ¬φ or equivalently: if T 6|= φ then T |= ¬φ.

1.4 Diagrams

To each L-structure we may associate the set of sentences true in this structure. 12 CHAPTER 1. LANGUAGES AND STRUCTURES

Deﬁnition 1.4.1 L(A)-theory

Let M be an L-structure and A ⊆ M. An L(A)-theory of M, denoted Th (M,A) is the set of formulae with parameters in A satisﬁed in M. If A = M, then Th (M,A) is called an elementary diagram of M. The diagram of M, ∆M is the set of formulae without quantiﬁers of the elementary diagram.

The theory Th (M,A) is not in the language L but in the language L(A): for each a ∈ A we added a symbol ca to L interpreted by a in M. The new structure obtained is noted MA and is a model of Th (M,A). We note Th (M) instead of Th (M, ∅), this is the set of all sentences satisﬁed in M. Proposition 1.4.2

Let M be an L-structure and A ⊆ M. The L(A)-theory Th (M,A) of M is complete.

1.5 Exercises

Exercise 1.5.1. Write formulae equivalent to each of these propositions: • φn : the structure has at least n elements. • ψn : the structure has at most n elements. • En : the structure has exactly n elements. Exercise 1.5.2. Show that a theory that admits two ﬁnite models with diﬀerent cardinalities is not complete. Exercise 1.5.3. Let Φ and Ψ be two theories: 1. Construct a theory Θ such that M is a model of Θ if and only if it is a model of Φ and a model of Φ. 2. Construct a theory Θ such that M is a model of Θ if and only if it is a model of Φ or a model of Ψ.

Exercise 1.5.4. We work in the structure of the integers N with the usual interpretation for the symbols. We look at it as a {×, <}-structure. 1. In the language {×,S}, where S is the unary successor function, show that there exists a formula φ(x, y, z) such that N |= φ(a, b, c) if and only if a + b = c. 2. Same question with the language {×, <}. Chapter 2

Equivalence between models

When you have two models of a same theory, one of the questions that arises often is to know which properties they share. We introduce here deﬁnitions that appear in almost all mathematical formalisms : substructures and isomorphisms.

2.1 Elementary equivalence

Deﬁnition 2.1.1 Elementary equivalence

Two L-structures M and N are elementarily equivalent, if they satisfy the same sentences, that is to say:

∀φ, M |= φ ⇔ N |= φ

This is denoted M ≡ N.

This is equivalent to the fact that Th (M) = Th (N). There are strong links between elementary equivalence and completeness as is shown by the following proposition:

Proposition 2.1.2

Two L-structures are elementarily equivalent if and only if they are both models of a same complete theory.

Proof. If M and N are elementarily equivalent, they are models of the theory Th (M) = Th (N) which is complete by property 1.4.2. Conversely, let M and N be two models of some complete theory T , then M |= ¬¬φ, so M 6|= ¬φ, hence T 6|= ¬φ, and T |= ¬¬φ since T is complete. Since N is a model of T , N |= φ. The same reasoning applies for a sentence satisﬁed in N.

13 14 CHAPTER 2. EQUIVALENCE BETWEEN MODELS

This proposition is quite important: the main technique to prove that two models are elementarily equivalent is to ﬁnd a theory of which they are both models and to prove that this theory is complete.

Example 2.1.3 The language of orders (Q, <) and (Z, <) are not elementarily equivalent as the formula φ : ∀x, y, (x < y) ⇒ ∃z(x < z ∧ z < y) is true in (Q, <) but not in (Z, <). From this we can also deduce that the theory of orders is not complete.

2.2 Substructures

Deﬁnition 2.2.1 Substructure

Let M be an L-structure, a substructure of M is a structure N such that: • N ⊆ M, • The association of the functions and relations in N is the same as in M. We write N ⊆ M. M is also said to be an extension of N.

Deﬁnition 2.2.2 Elementary substructure

N ⊆ M is an elementary substructure if and only if for any formula φ(m) with parameters in N, N |= φ(m) ⇔ M |= φ(m). We note M ≺ N.

An elementary substructure N of M is elementarily equivalent to M by deﬁnition, but the converse is not always true:

Example 2.2.3

(Z × Z,

Deﬁnition 2.3.1 L-morphism

Let M and N be two L-structures. An L-morphism from M to N is a map σ : M → N such that: • ∀cM, σ(cM) = cN, M n M N • ∀f , ∀mi ∈ M , σ f (m1, . . . , mn) = f (σ(m1), . . . , σ(mn)), M n M • ∀R , ∀mi ∈ M , (m1, . . . , mn) ∈ R ⇔ σ(σ(m1), . . . , σ(mn)) ∈ RN.

Example 2.3.2 The morphisms of two L-structures of the theory of orders are the strictly increasing functions.

Proposition 2.3.3

Every L-morphism is injective.

Proof. Let m, m0 ∈ M such that σ(m) = σ(m0), then (σ(m), σ(m0)) ∈=N, hence by deﬁnition of an L-morphism, (m, m0) ∈=M and m = m0.

Deﬁnition 2.3.4 L-isomorphism

An L-isomorphism is a surjective L-morphism.

Morphisms do not necessarily preserve the satisfaction of formulae: for instance there exists a canonical injection from (Z, <) to (Q, <), but (Z, <) is not an elementary substructure of (Q, <). Therefore, we introduce the notion of elementary morphism:

Deﬁnition 2.3.5 Elementary morphism

An L-morphism σ between M and N is elementary if and only if

∀φ, ∀m ∈ M, M |= φ(m) ⇔ N |= φ(σ(m)) 16 CHAPTER 2. EQUIVALENCE BETWEEN MODELS

Proposition 2.3.6

Let M and N be two L-structures such that M ⊆ N, then M is an elementary substructure of N if and only if the canonical injection M,→ N is an elementary L-morphism.

Proposition 2.3.7

Any L-isomorphism is elementary.

Proof. We show by induction on the structure of φ(m) that

M |= φ(m) ⇔ N |= φ(m)

• If φ is atomic, φ = R(m1, . . . , mn), then by deﬁnition of an L-morphism:

M M |= φ(m1, . . . , mn) ⇔ (m1, . . . , mn) ∈ R N ⇔ (σ(m1), . . . , σ(mn)) ∈ R ⇔ N |= φ(σ(m1), . . . , σ(mn))

• If φ = φ1 ∧ φ2, then

M |= φ(m1, . . . , mn) ⇔ M |= φ1 and M |= φ2 ⇔ N |= φ1 and N |= φ2 by induction hypothesis ⇔ N |= φ

• The proof is similar for φ = φ1 ∨ φ2. • If φ = ¬ψ, then

M |= φ ⇔ M 6|= ψ ⇔ N 6|= ψ by induction hypothesis ⇔ N |= φ

Apart from the induction steps for the quantiﬁers, the bijectivity of σ was not used in the proof, therefore:

Proposition 2.3.8

An L-morphism preserves formulae without quantiﬁers. 2.4. DIAGRAMS 17 2.4 Diagrams

An equivalent statement to the preceding one would be:

Proposition 2.4.1

If σ is an L-morphism from M to N, then N is a model of ∆M, the diagram of M.

Or even better:

Lemma 2.4.2 First diagram lemma

There exists an L-morphism from M to N if and only if N may be extended in a model of ∆M, the diagram of M.

Here, "may be extended" should be understood as extending N by giving an interpretation to each constant cm for m ∈ M. Proof. Let σ be an L-morphism from M to N. It is suﬃcient to interpret each constant cm, with m ∈ M by σ(m). Conversely, if N is a model of ∆M, the map σ : m 7→ cm is an L-morphism.

Lemma 2.4.3 Second diagram lemma

There exists an elementary L-morphism from M to N if and only if N may be extended in a model of Th (M,M), the elementary diagram of M.

Proof. Homework!

2.5 Tarski’s test

For the moment, the only way we have to check that a substructure is elementary is to show that all formulae in this substructure with the symbols of its elements are satisﬁed in both or in none. Tarski’s test will allow us to prove elementary inclusion in an easier way:

Theorem 2.5.1 Tarski’s test

Let N ⊆ M. N is an elementary substructure of M if and only if for any formula φ(x) with parameters in N such that M |= ∃xφ(x), there exists n ∈ N such that M |= φ(n). 18 CHAPTER 2. EQUIVALENCE BETWEEN MODELS

This theorem means that a substructure is an elementary substructure if each time we want to prove a formula in M containing an ∃x, we can take x in N. Proof. [⇒] Suppose N ≺ M. If φ(x) has parameters in N and M |= ∃xφ(x), since N is an elementary substructure of M, N |= ∃xφ(x), so there exists n ∈ N such that N |= φ(n) and M |= φ(n). [⇐] Conversely, suppose for any formula φ(x) with parameters in N such that M |= ∃xφ(x), there exists n ∈ N such that M |= φ(m). We prove by induction on the structure of φ(n) a formula with parameters in N that N |= φ(n) ⇔ M |= φ(n) • For the cases without quantifiers, the proof is essentially the same as in proposition 2.3.7. • Now for the induction step when the formula has an existential quantifier: • N |= ∃xφ(x) ⇒ M |= ∃xφ(x): by definition, there exists n ∈ N such that N |= φ(n) and by induction there exists m ∈ M such that M |= φ(m) hence M |= ∃xφ(x). • M |= ∃xφ(x) ⇒ N |= ∃xφ(x): by Tarski’s test hypothesis, there exists n ∈ N such that M |= φ(n) by induction hypothesis N |= φ(n) and hence N |= ∃xφ(x). • The universal quantifier may be treated in the same fashion.

2.6 Skolemisation

For any formula ∃xφ(x) true in some model, there exists an element a of this model such that φ(a) is satisﬁed. Skolemisation is adding the symbol corresponding to this element in the language.

Deﬁnition 2.6.1 Skolemisation of a theory

Let T be an L-theory. For each formula φ(x, y) we deﬁne a new function symbol fφ(y) and we add these new symbols to the language:

Lskolem = L ∪ {fφ | φ a formula}

The skolemisation Tskolem of theory T is the set of sentences:

Tskolem = T ∪ {∃xφ(x, y) → φ(fφ(y), y) | φ a formula}

Proposition 2.6.2

The skolemisation Tskolem of a theory T is a theory and any model of T is also a model of Tskolem. 2.6. SKOLEMISATION 19

Proof. The proof of this statement is exactly what one would expect: if a formula ∃xφ(x, y) is satisfied, then we can choose an x satisfying it, and that depends only on y, so we may define fφ(y) = x. If the formula is not satisfied, then we can choose any fφ(y) = m. Seeing how we chose x, this proof is entirely dependent on the axiom of choice.

Theorem 2.6.3 Downward Löwenheim-Skolem

Let N be an inﬁnite L-structure, A a set of parameters in N and κ an inﬁnite cardinal such that max(|A| , |L|) ≤ κ ≤ |N|. Then there exists an elementary substructure M ≺ N containing A with |M| = κ.

This proof relies on the fact that for any inﬁne cardinals κ1, κ2:

κ1 + κ2 = κ1κ2 = max(κ1, κ2)

Proof. We may suppose that |A| = κ without loss of generality. We construct by induction a chain (M1)i∈ω of substructures of N such that M0 contains A, for all n i ∈ ω, |Mi| = κ and for any formula φ(x, y) and mi ∈ Mi , if N |= ∃xφ(x, mi), then there exists mi+1 ∈ Mi+1 such that N |= φ(mi+1, mi). Let M0 be the substructure of N generated by A. |M0| = κ since |L| ≤ κ = |A|. Suppose Mi has already been constructed, then for any formula φ(x, y) (there are n at most max(|L| , ℵ0)) and any parameter m ∈ Mi such that N |= ∃φ(x, mi), we choose fφ(m) ∈ N such that N |= φ(fφ(m), m). We then define Mi+1 as the substructure generated by Mi and the fφ(m). This substructure is of cardinal κ and satisfies the induction hypothesis. S Now let M = i∈ω Mi, M is a substructure of N of cardinal κ which verifies Tarski’s test, and is hence an elementary substructure of N.

Corollary 2.6.4

A theory on a ﬁnite language with an inﬁnite model has a countable model.

Theorem 2.6.5

A theory with a ﬁnite or countable language for which all countable models are inﬁnite and isomorphic is complete.

Proof. Let N and M be two countable models of the theory. They are inﬁnite so by taking A = ∅ and κ = ℵ0 we may deduce the existence of two substructures M0 ≺ M and N0 ≺ N. Since M0 and N0 are countable, they are isomorphic, thus elementarily equivalent and M and N are elementarily equivalent. 20 CHAPTER 2. EQUIVALENCE BETWEEN MODELS 2.7 Deﬁnability

First order logic allows to deﬁne very easily what a deﬁnable object, it is an object for which there is a formula that describes it:

Deﬁnition 2.7.1 Deﬁnable function

A function f is deﬁnable in structure M if there exists a formula φ(a, x, y) such that

f(x) = y ⇔ M |= φ(a, x, y)

Deﬁnition 2.7.2 Deﬁnable set

A set S ⊆ M n is deﬁnable in structure M if there exists a formula φ(a, x) such that x ∈ M n ⇔ M |= φ(a, x)

2.8 Exercises

Exercise 2.8.1. Show that ≺ is transitive. Exercise 2.8.2. Let M,N and O be three structures such that M ≺ O, N ≺ O and M ⊆ N. Show that M ≺ N. Exercise 2.8.3. Let L be the empty language i.e.containing only equality. 1. Write a sentence φn stating that there exists at least n distinct elements. 2. Show that the theory {φn | n ∈ N} is complete. Chapter 3

Compactness theorem

The aim of this chapter is to prove the very important Compactness theorem of which the Completeness theorem is a consequence. But ﬁrst a deﬁnition:

Deﬁnition 3.0.1 Finite consistence

A set of formulae S is ﬁnitely consistent if every ﬁnite subset of S is consistent

We may now state the theorem:

Theorem 3.0.2 Compactness

A set of L-sentences Φ is consistent iﬀ it is ﬁnitely consistent.

It is quite clear that if Φ has a model, then this model is a model for any ﬁnite subset of Φ, however the converse is much harder to prove. There are mainly two proofs for this theorem, a proof using ultraﬁlters and a proof based on the so-called Henkin constants. We will use Henkin’s proof here, which gives a stronger statement, by ensuring the existence of a model of cardinality at most max(|L| , |Φ| , ℵ0).

3.1 Compactness of quantiﬁer free sets

We start by giving a proof for the quantiﬁer free case:

Proposition 3.1.1 Quantiﬁer free compactness

Let Φ be a ﬁnitely consistent set of quantiﬁer-free formulae. Then Φ is consistent and there is a structure M of cardinality at most max(|L| , |Φ| , ℵ0) such that M |= Φ.

21 22 CHAPTER 3. COMPACTNESS THEOREM

Before getting to the proof of the quantiﬁer-free compactness theorem, we will need the following lemma:

Proposition 3.1.2 Completion of a set of formulae

Let Φ be a set of L-formulae for which any ﬁnite subset is consistent. Then there exists a complete set of L-formulae Φ0 for which every ﬁnite 0 0 subset is consistent and such that Φ ⊆ Φ and |Φ | ≤ max(|Φ| , |L| , ℵ0).

Proof. Let X be the set of free variables and parameters of Φ, of course |X| ≤ |Φ|. Let φα be an enumeration of all formulae whose parameters/variables are in X, such 0 an enumeration has at most length =max(|Φ| , |L| , |ℵ0|). We construct Φ as the limit of sets of formulae Φα deﬁned inductively:

• Let Φ0 = Φ. • Let Φα+1 = Φα ∪ {φα} if every ﬁnite subset of Φα ∪ {φα} is consistent and Φα+1 = Φα ∪ {¬φα} otherwise. S • Φλ = {φλ} ∪ α<λ Φα 0 S The obtained Φ = α Φα is complete and all its ﬁnite subsets are consistent.

Proof of Quantifier-free compactness. Let A be the set of all parameters of Φ. Let’s start with the two following assumptions: (1) Φ is complete (2) For every term t of Φ, there is a variable y ∈ A such that (t = y) ∈ Φ Let ≈ be the equivalence relation defined by t ≈ s iff (t = s) ∈ Φ for s, t two atomic terms. We now construct a model M for Φ: take M ⊆ A containing exactly one variable per equivalence class this is possible by assumption (2). For every function symbol f and every x ∈ M n, we define f M as the unique element y ∈ M in the equivalence class of f(x) (meaning f(x) = y is in Φ. For every relation symbol R and x ∈ M n, x ∈ RM iff R(x) ∈ Φ. It is clear that M |= Φ if we take the obvious interpretation of each constant symbol a by the unique variable of its equivalence class represented in M. The cardinality bound is a direct consequence of the fact that |A| ≤ |Φ|. We now need a procedure to extend a finitely consistent set Φ into one that verifies the assumptions. The second one can be easily satisfied by adding for each term t the formula (t = y) where y is a fresh variable. And the first one can be satisfied by applying the previous lemma. They both need to be satisfied simultaneously however, so what we do is we construct sets (Ψi)i∈ω inductively:

• Ψ0 = Φ • For i odd, we add (t = y) with y a fresh variable for each term t ∈ Ψi−1. • For i even, we obtain Ψi by applying the previous lemma on Ψi−1. S Both assumptions are veriﬁed by taking the limit Ψ = i∈ω Ψi. Furthermore, the cardinality does not grow with each step, so we also have the cardinality bound.

Let’s authorize some quantiﬁers now, but at the same time some quantiﬁer 3.2. UPWARD LÖWENHEIM-SKOLEM 23 free formulae corresponding to them:

Deﬁnition 3.1.3

A set of formulae Φ is full iﬀ Φ is complete and for any formula ∃yφ(x, y) ∈ Φ then there exists a variable z such that φ(x, z) ∈ Φ.

Proposition 3.1.4

Let Φ be a full set of L-formulae, and Φqf be Φ minus the quantiﬁed formulae. If M |= Φqf, then M |= Φ.

Proof. Let us prove the result by induction: • The induction step for the connectors and the initialisation also. • Now suppose that some formula ∃xφ(y, x) ∈ Φ, then by hypothesis, there exists a formula φ(y, z) for some parameter z. By induction hypothesis M |= φ(y, z) and therefore M |= ∃xφ(y, x) since the interpretation of z is an element satisfying the formula.

Now the only thing left to prove is that from any ﬁnitely constistent set of formulae we may construct a full ﬁnitely consistent set of formulae Φ.

Proposition 3.1.5

Let Φ be a ﬁnitely consistent set of formulae. Then there is a full ﬁnitely consistent set of formulae Φ0 containing Φ with |Φ0| ≤ max(|Φ| , |L| , ℵ0).

Proof. In order to fulfill the completeness requirement of the fullness we can use lemma 3.1.2. And in order to fulfill the assumption that for any formula ∃xφ(y, x) ∈ Φ there exists a formula φ(y, z) ∈ Φ, for each of them we can add a φ(y, z) with z a fresh variable. As before, these two steps may contradict each other, and as before, it suffices to to inductively apply them one step out of two and taking the infinite union in order to obtain the result.

The compactness theorem then follows immediately.

3.2 Upward Löwenheim-Skolem

The ﬁrst consequence of the compactness theorem is the upward Löwenheim- Skolem theorem: 24 CHAPTER 3. COMPACTNESS THEOREM

Proposition 3.2.1

Let M be an inﬁnite L-structure, then for any cardinal κ, there exists an L-structure N such that M ≺ N and such that |N| ≥ κ.

Proof. Add some constants cα for α < κ, take the theory Th (M,M) to which we add the sentences cα 6= cβ, for α 6= β. Each ﬁnite subset of this theory only has a ﬁnite number of constants cα that we can interpret in M. And thus by compactness, there is a model for the theory and this model is an elementary extension of M.

Proposition 3.2.2 Upward Löwenheim-Skolem

Let M be an inﬁnite L-structure, then for any κ > max(|M| , |L|), there exists an elementary extension N of M with |N| = κ.

Proof. The previous proposition in conjunction with the downward Löwenheim- Skolem.

3.3 Axiomatisation

Deﬁnition 3.3.1

Let T be a theory, an axiomatisation of T is a set of formulae Φ such that Φ and T have the same models. T is said to be ﬁnitely axiomatisable if it has a ﬁnite axiomatisation.

Proposition 3.3.2

A theory is ﬁnitely axiomatisable if and only if from any axiomatisation we may extract a ﬁnite axiomatisation.

Proof. [⇐] Trivial. [⇒] Let T be finitely axiomatisable and Φ be a finite axiomatisation of T . We may assume without loss of generality that Φ contains only one sentence φ (take V φ = ψ∈Φ ψ). Let Ψ be an axiomatisation of T , then Ψ ∪ {¬φ} does not have a model since any model of Ψ is a model of φ. By the compactness theorem, there is a finite subset Θ of Ψ ∪ {φ} that does not have a model. Θ cannot contain sentences of only Ψ since Ψ admits a model and hence so would this subset. This implies that any model of Θ is a model of φ and hence a model of T . 3.4. κ-CATEGORICITY 25 3.4 κ-categoricity

Deﬁnition 3.4.1 κ-categoricity

Let κ be an inﬁnite cardinal, a theory T is κ-categorical if all its models of cardinal κ are isomorphic.

In theorem 2.6.5 we proved that if a theory is ω-categorical and has an at most countable language, then it is complete. Let us now generalize this result:

Theorem 3.4.2 Loś-Vaught test

Let T be an L-theory, and κ ≥ |L|. If T is κ-categorical and does not admit any ﬁnite model, then it is complete.

Proof. Let N and M be any two models of T , we will show that they are elementarily equivalent: • if |M| ≥ κ then by the downward Löwenheim-Skolem theorem there exists M0 ≺ M with |M| = κ. • if |M| < κ, then since κ > |L| and M is infinite,then by the upward Löwenheim-Skolem theorem, there exists M0 M with |M0| = κ. Therefore, we can always find M0 of cardinal κ that is elementarily equivalent to M. The same is true for N: we can find a structure N0 with cardinality κ elementary equivalent to N. By κ-categoricity, N0 ≡ M0 and hence M ≡ N. Therefore T is complete.

3.5 Exercises

Exercise 3.5.1. 1. Let (Ti)i∈ω be an increasing sequence of theories, such S that there exists a model of Ti that is not a model of Ti+1. Show that Ti is a theory and that it is not finitely axiomatisable. 2. Let T be a finitely axiomatisable theory on a finite language L. Show that the following are equivalent: • There is a finite number of complete theories T 0 ⊇ T • There is a finite number of theories T 0 ⊇ T • Every theory T 0 ⊇ T is finitely axiomatisable. Exercise 3.5.2. Show that a theory that has models of arbitrarily large finite cardinals has an infinite model. Give an example of a theory that has arbitrarily large finite cardinals but for which all infinite models hava at least cardinal ℵ1.

Exercise 3.5.3. Let M, N1, N2 be structures such that M ≺ N1 and M ≺ N2. Show that there exists N such that N1 ≺ N and N2 is isomorphic to an elementary substructure of N. 26 CHAPTER 3. COMPACTNESS THEOREM Chapter 4

Back and forth

Back and forth is THE method to prove elementary equivalence, and in other terms the completeness of a theory.

4.1 ∞-isomorphisms

Deﬁnition 4.1.1 Local isomorphims

Let M and N be two L-structures. Let (ai) and (bi) be n-tuples of M and N respectively. The map σ :(ai) → (bi) is a local isomorphism if it can be extended into an isomorphim of the L-structure generated by the (ai) on the L-structure generated by the (bi).

What this definition means is that the ai and bi satisfy the same atomic M M formulae in their respective structures. For instance f (a1) = f (a2) iff N N f (b1) = f (b2). It is in general quite easy to see that some map is a local isomorphism. Example 4.1.2 • Take the language of orders and consider two models of the theory of orders. Since there is no function, the map which to ai maps bi will be a local isomorphism if it preserves the relations. Since there is only one relation <, it suffices that ai < aj ⇔ bi < bj. The local isomorphisms are the strictly increasing functions defined on a finite subset of the universe. • Take two finite models (Z/2Z, +) and (Z/4Z, +) and look at all its local isomorphisms. Since they are morphisms, 0 must be sent to 0. Since 1 + 1 = 0 in the first structure, the element to which 1 is sent must also verify this property : hence the image of 1 must be 2. The local isomorphims are then all the restrictions of this particular local isomorphism. Quite often, one looks at the local isomorphism which associates ∅ to ∅. To check whether it is an isomorphism, one must recall that the L-structure

27 28 CHAPTER 4. BACK AND FORTH obtained may not be empty, since it contains interpretations for all constants and hence the closure by the functions.

Deﬁnition 4.1.3 Karpian family

A set K of local isomorphisms between M and N is a karpian family if and only if: • K 6= ∅ • K statiﬁes the FORTH property: For any σ ∈ K and any a ∈ M, there exists τ ∈ K such that • τ|Dom σ = σ • a ∈ Dom τ • K satisﬁes the BACK property: For any σ ∈ K and any b ∈ N, there exists τ ∈ K such that • τ|Dom σ = σ • b ∈ Im τ

Essentially, what this deﬁnition says is that for any local isomorphism σ and any element a ∈ M (resp b ∈ N), one may ﬁnd a local isomorphims τ extending σ which also maps a (resp. b). Karpian families are interesting because of the following:

Proposition 4.1.4

• The union of two karpian families is a karpian family. • If K is a karpian family, then set of all functions which are restrictions of functions of K is a karpian family.

From this we can deduce that if there exists a karpian family, then there exists a maximal one, and it is stable by restriction.

Deﬁnition 4.1.5 ∞-isomorphism

If there exists a karpian family K from M to N, we say that M and N are ∞-equivalent. Let K be the biggest karpian family from M to N. Every element of K is then called an ∞-isomorphism and the support of this isomorphism are said ∞-equivalent.

∞-isomorphism is quite simple actually: the map that sends a to b is an ∞-isomorphism if for any a ∈ M, one can ﬁnd a b ∈ N (and conversely) such 4.2. ISOMORPHIC STRUCTURES 29 that the local isomorphism is preserved.

Example 4.1.6 • Consider the language of orders and the structures (Z, <) and (N, < ). The mapping that sends 0 to 0 is not an ∞-isomorphism, even though it is a local isomorphism: take −1 ∈ Z, to satisfy the back property, one must find an element a ∈ N such that there is always an ∞-isomorphism. But an ∞-isomorphism is a local isomorphism, so (0, a) and (0, −1) must satisfy the same properties. However, −1 < 0, but it is impossible that a < 0. • Now take the same language and the structures (Q, <) and (R, <). Let’s show that the family of strictly increasing functions from Q to R with finite support is karpian (and hence that these functions are ∞-isomorphisms). Let σ be a strictly increasing function from a finite subset of Q to R. That is to say a set {m1, . . . , mk} and a set {n1, . . . , nk} such that mi < mi+1 and ni < ni+1 for anyn i. Let us show the FORTH property: take some q ∈ Q and let us now find somr r ∈ R in order to satisfy the same properties. That is to say that the map which associates {m1, . . . , mk, q} to {n1, . . . , nk, r} is also strictly increasing: • If q is already among the mi, we take as r the corresponding ni. • If ∀i, q < mi, then we must find some r ∈ R inferior to any ni. We can take for instance r = min ni − 1. • If ∀i, q > mi, then we can take r = max ni + 1 for instance. • If there exists some i such that mi < q < mi+1 for some i, mi+mi+1 then we can take r = 2 . The BACK property is treated in the same way. We thus see that (Q, <) and (R, <) are ∞-isomorphic.

4.2 Isomorphic structures

Proposition 4.2.1

Two isomorphic structures are ∞-isomorphic.

Proof. Take σ the isomorphism between the structures, the set of restrictions to ﬁnite domains of σ is a karpian family.

Proposition 4.2.2

Two ∞-isomorphic countably inﬁnite structures are isomorphic. 30 CHAPTER 4. BACK AND FORTH

Proof. Let mi and ni be enumerations of the two ∞-isomorphic structures. We construct by induction ∞-isomorphisms σk such that: • ∀i ≤ k, mi ∈ Dom σk • ∀i ≤ k, ni ∈ Im σk The base case is immediate, since σ0 : ∅ → ∅ is an ∞-isomorphism. Suppose σk constructed, by using the FORTH hypothesis with mk+1 and the BACK hypothesis with nk+1, we get σk+1. The limit of the σk is an isomorphism.

4.3 Elementary equivalence

A very important property of ∞-isomorphism is the following: Proposition 4.3.1

Let M and N be two ∞-equivalent structures, then for a and b ∞-equivalent we have:

M |= φ(a) ⇔ N |= φ(b)

This means in particular that M and N are elementary equivalent.

The converse is not true in general. Proof. By induction on φ let’s show that if a and b are ∞-equivalent, then M |= φ(a) ⇔ N |= φ(b)

Let us now do the induction. The atomic and basic connectors are easy. We treat the ∃ case. Suppose M |= ∃xφ(a, x), then there exists m ∈ M such that M |= φ(a, m) and by the FORTH hypothesis, there exists n ∈ N such that a, m and b, n are ∞-isomorphic since M |= φ(a, m), by induction hypothesis N |= φ(b, n) and hence N |= ∃xφ(b, x). The converse, when N |= ∃xφ(b, x) is treated the same way but using the BACK hypothesis. Taking a = ∅ and b = ∅ shows the elementary equivalence part. In this proof, we can see how ∞-isomorphisms are useful, the usually tedious case of the induction proofs was treated in a very straightforward way. Example 4.3.2 A straightforward corollary of our previous example is that (R, <) and (Q, <) are elementarily equivalent.

4.4 ω-saturated models

The previous proposition allowed us to easily show that two ∞-isomorphic structures are elementarily equivalent. The converse is false in general, however it 4.4. ω-SATURATED MODELS 31 is true for ω-saturated models. But before being able to talk about ω-saturated models, we need to introduce the notion of type

Deﬁnition 4.4.1 Type

A type in M is a finitely satisfiable set of formulae φ(a, x), i.e.for any finite subset of a type, one may find x ∈ M such that M |= φ(a, x) for all formulae of this subset. We note Sn(a) the set of types with n variables with parameters in a.

Example 4.4.2

In the structure of non-negative integers, the set of φn(x): x > 1+···+1 (n times) is finitely satisfiable. We can always find an integer bigger than any integer of some finite set. As shown by the previous example, a type may be finitely satisfiable without being satisfiable : there is no integer bigger than all the integers. However this might contradict the intuition one got from the compactness theorem and indeed, one may find a model in which a type is realized:

Proposition 4.4.3

Every type of S1(a) is realized in an elementary extension of M.

Proof. Let Φ(x) ∈ S1(a), and consider the language with an added constant c. Then any ﬁnite fragment of Th (M) ∪ Φ(c) has M as a model where c is interpreted as necessary (it can be done since Φ(x) is a type), hence by compactness Th (M) ∪ Φ(c) has a model. Which by construction is isomorphic to an elementary extension of M.

ω-saturation corresponds to the notion of being realized inside the structure itself:

Deﬁnition 4.4.4 ω-saturation

M is a ω-saturated if for any tuple a ∈ M, every type Φ(x) ∈ S1(a) is realized in M, that is to say there exists m ∈ M such that for any φ(a, x) ∈ Φ(x), M |= φ(a, m).

The exemple above shows that (N, <, +) is not an ω-saturated structure. We saw above that it was possible to ﬁnd, for a given type, and extension realizing 32 CHAPTER 4. BACK AND FORTH this type. Proposition 4.4.5

Any structure has an ω-saturated extension.

Proof. Let M be the initial structure. We introduce for each type Φ(a, x) a new constant cφ. We then take [ [ Th (M,M) ∪ {φ(a, cφ)} Φ φ∈Φ

Since any of its finite subsets admits M as a model, there exists a model M1 such n that any type with a ∈ M is realized. Inductively, we may define Mi+1 which realizes the types on Mi. We thus obtained a sequence M1 ≺ M2 ≺ ... We can S take N = i Mi, now let Φ ∈ S1(a) be a type, a is finite and thus included in Mi for some i and thus realized in Mi+1 and hence in N.

Theorem 4.4.6

Let M and N be two ω-saturated structures. M and N are elementarily equivalent if and only if M and N are ∞-equivalent.

Proof. [⇐] Already done in proposition 4.3.1. [⇒] Let M and N be elementarily equivalent and ω-saturated, it suffices to show that the set F of all morphisms that maps a to b such that a and b satisfy the same formulae is a karpian family. • ∅ ∈ F • We show the FORTH property. Let σ ∈ F which maps a to b. Let α ∈ M. Take A = {φ(a, x) | M |= φ(a, α)}, now let φ1, . . . , φn be a finite subset of A satisfied by α. We then have

M |= ∃xφ1(a, x) ∧ · · · ∧ φn(a, x)

Since σ ∈ F , N |= ∃xφ1(b, x) ∧ · · · ∧ φn(b, x) So B(x) = φ(b, x) φ ∈ A is ﬁnitely realisable, and hence by ω-saturation it is realised for some β. By construction the map that associates (b, β) to (a, α) is in F . • The BACK property is shown the same way.

Proposition 4.4.7

A theory is complete if and only if all its ω-saturated models are ∞-equivalent. 4.5. APPLICATIONS TO COMPLETENESS 33

Proof. [⇒] follows from the preceeding theorem. [⇐] Let M and N be two models of T . They both have ω-saturated elemntary extensions M0 and N0 which are ∞-equivalent and hence elementarily equivalent. Thus M and N are elementarily equivalent. So we now have a method to prove that a theory is complete: • Find a family A containing the ω-saturated models of T . • Show that any two elements of A are ∞-equivalent. It seems quite hard to ﬁnd exactly the ω-saturated models, so we give a family containing them: since the family contains all ω-saturated models, ω- saturated models are equivalent and any model can be elementarily extended in an ω-staturated one, one gets the result. We have even stronger: Proposition 4.4.8

Any structure ∞-equivalent to an ω-saturated structure is ω-saturated.

Therefore A is exactly the set of the ω-saturated models. Proof. Let M be ∞-equivalent to ω-saturated N. Let a and b be ∞-equivalent and take some type A(x) = {φ(a, x)}. We want to show that it is realized in M. By hypothesis A(x) is finitely satisfiable. Take any finite subset {φ1, . . . , φk} of A(x), ^ M |= ∃x φi(a, x) i By ∞-equivalence (prop. 4.3.1) we thus have ^ N |= ∃x φi(b, x) i So B(x) = φ(b, x) φ ∈ A is finitely satisfiable, and by ω-saturation of N satisfiable by some β. Using the BACK property of the karpian family, there exists some α such that b, β and a, α satisfy the same properties and in particular α realizes A(x).

4.5 Applications to completeness 4.5.1 Dense orders without extremities

Theorem 4.5.1

Two dense orders without extremities are ∞-isomorphic.

Corollary 4.5.2

The theory of dense orders without extremities is complete. 34 CHAPTER 4. BACK AND FORTH

Proof. Let M and N be two models of T the theory of dense orders without extremities. We will show that the set of strictly increasing functions from M to N form a karpian family. Note first that these functions are local isomorphisms, hence the set non-empty. • FORTH: Let σ be a strictly increasing function sending a to b and α ∈ M: • If α is among a, then σ satisfies the property. • If α is lower than all the a, then since N is without extremities, there exists a β lower than all the b and it suffices to take the map σ∪{α 7→ β}. • If α is greater than all the a, in the same way we can find a map by taking a β above all b. • If α is between ai and ai+1 then since N is dense there exists an element β between bi and bi+1. in every case the FORTH property is satisfied. • BACK: We treat this case in the same way.

4.6 Discrete orders without extremities

The proof here is more tedious and will use ω-saturated models. In a discrete order, every element has a successor and a predecessor. We introduce the notion of distance between two elements d(x, y): |{z | x < z ≤ y}| x ≤ y d(x, y) = if d(y, x) otherwise

It is easy to build a formula φn(x, y) satisﬁed iﬀ x and y are at distance at least n. Proposition 4.6.1

If a model is ω-saturated then there exists an inﬁnite number of points xi such that : 1. For all i 6= j, the distance from xi to xj is inﬁnite. 2. The set of the xi forms a dense order without extremities.

Proof. Take an ω-saturated model M: 1. It suffices to check that if x1, . . . , xn are at infinite distance then there exists an xn+1 at infinite distance from them. Take the following type:

A(x) = {φm(x, xi) | m ∈ N, i ≤ n} This set is finitely satisfiable: any finite subset can only evoke distances bounded by some constant M, we then need to find some element at distance more than M from all the xi, which can be found since the order is infinite. Since the model is ω-saturated, the type A(x) is satisfied by some xn+1. 2. It suffices to show that for any x there exists a y at infinite distance from x greater (resp. lower) than x. Since

A(y) = {φm(x, y) ∧ x < y | m ∈ N} 4.7. EXERCISES 35

is a type, we can conclude that there are extremities. For the denseness, it suﬃces to show that for all x < x0 there exists y at inﬁnite distance from both x and x0 such that x < y < x0. Since

0 0 A(y) = {φm(x, y) ∧ φ(x , y) ∧ x < y < x | m ∈ N}

is a type we can also conclude.

Hence a discrete order modulo the relation to be at ﬁnite distance gives a dense order without extremities. Proposition 4.6.2

Let M and N be two models satisfying the above property. Then the set of distance preserving ﬁnite mappings from M to N is a karpian family.

Proof. Homework back and forth!

Corollary 4.6.3

The theory of discrete orders without extremities is complete.

4.7 Exercises

Exercise 4.7.1. Show that two structures with the same language L may be elementary equivalent without being ∞-equivalent.