An introduction to Model Theory (work in progress )

**An** **in**troduction**to** **Model** **Theory**(**work** **in** **progress** )Rafel Farré CireraJuly 20, 2011

Contents1 Prelim**in**aries 21.1 The compactness Theorem . . . . . . . . . . . . . . . . . . . . . . 41.2 Löwenheim-SkolemTheorems . . . . . . . . . . . . . . . . . . . . 51.3 Complete theories: the test of ̷Loš-Vaught . . . . . . . . . . . . . 101.4 Elementary cha**in**s . . . . . . . . . . . . . . . . . . . . . . . . . . 131.5 Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 Types and Saturation 222.1 Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.2 Saturation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.3 Partial elementary maps . . . . . . . . . . . . . . . . . . . . . . . 292.4 More about saturation . . . . . . . . . . . . . . . . . . . . . . . . 313 Omitt**in**g Types 373.1 A little bit of **to**pology . . . . . . . . . . . . . . . . . . . . . . . . 373.2 The spaces of complete types of a theoty . . . . . . . . . . . . . . 403.3 The omitt**in**g types theorem . . . . . . . . . . . . . . . . . . . . . 414 Countable models 444.1 a**to**mic models . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444.2 The Ryll-Nardzewski’s theorem . . . . . . . . . . . . . . . . . . . 454.3 countable a**to**mic and countable saturated models of a completetheory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474.4 On the number of countable models . . . . . . . . . . . . . . . . 494.5 restes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 515 Partial isomorphism 535.1 Partial isomorphism . . . . . . . . . . . . . . . . . . . . . . . . . 535.2 Karp’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 555.3 Partial isomorphism and completeness . . . . . . . . . . . . . . . 575.4 Elim**in**ation of quantifiers . . . . . . . . . . . . . . . . . . . . . . 585.5 Partial isomorphism and Q.E. . . . . . . . . . . . . . . . . . . . . 611

Chapter 1Prelim**in**ariesWe will denote the language by L, or more precisely L will denote the set of nonlogicalsymbols: symbols of constant, function symbols and relation symbols.The relation symbols will be denoted by capital letters (usually P, Q, R . . .), thefunction symbols will be denoted usually by f, g, h . . ., while the constants willbe denoted by c, d . . ..The L-terms are build recursively, start**in**g from constants and variables, byapply**in**g the follow**in**g rule: if f is an n-ary function symbol and t 1 , . . . , t n areterms then f(t 1 , . . . , t n ) is also a term. To denote terms we will use parenthesisf(t 1 , . . . , t n ) although they are not necessary. We assume a countable set ofvariables V ar usually denoted by, x, y, z, x 1 , . . . , x n , . . ..The firs-order formulas are built start**in**g with a**to**mic formulas (formulasof type t 1 = t 2 or Rt 1 , . . . , t n ) us**in**g the connectives ¬, ∧, ∨, →, ↔ and thequantifiers ∀, ∃. We will use the fact that each formula is equivalent **to** oneus**in**g only ¬, ∧, ∃ (i.e., where ∨, →, ↔, ∀ do not occur) **to** simplify proofs by**in**duction.Usually we will use the same letter M, N, . . . **to** **in**dicate the structure andits doma**in**. The card**in**al of a structure M is the card**in**al of its doma**in** and isdenoted by |M|. A tuple a of M is a f**in**ite sequence a 1 , . . . , a n of elements ofthe doma**in** of M and this will be denoted **in**formally by putt**in**g a ∈ M. Thelength of the tuple a 1 , . . . , a n is n.If t is a term, t(x) **in**dicates that the variables of t are among those of x, i.e.,a subset of {x 1 , . . . , x n } where x denotes x 1 , . . . , x n . If M is a structure and ais a tuple of elements of M of the same length as x, t M (a) (or also t(a) if thereis no possible confusion) will denote the **in**terpretation of t **in** M when assign**in**ga i **to** x i . When we use this notation we are assum**in**g that x and a have thesame length and ordered **in** some manner, although not explicitly stated.The **in**terpretation of a term t M (a) is def**in**ed by recursion below. Observethat if t = f(t 1 , . . . , t n ) has its variables among x the same happens with eachterm t i . So we will always assume that if t = f(t 1 , . . . , t n ) = t(x) then alsot i = t i (x).2

CHAPTER 1. PRELIMINARIES 3Def**in**ition 1.0.1. Given a term t(x) a structure M and a tuple a ∈ M, wedef**in**e the **in**terpretation t M (a) as follows:• If t(x) is a constant c we put t M = c M ,• If t(x) is a variable x i we put t M = a i ,• If t(x) is f(t 1 , . . . , t n ), we put t M (a) = f M (t M 1 (a), . . . , t M n (a)).For a formula φ, we write φ(x) **to** **in**dicate that x is a tuple of variablesand the free variables φ are among those of x . If a is a tuple of M we writeM |= φ(a) **to** **in**dicate that that the formula φ holds **in** M when assign**in**g thevariables **in** the tuple x **to** the elements of the tuple a **in** an ordered manner(send**in**g x i by a i for each and i). Aga**in**, we are assum**in**g that x and a havethe same length.Below we def**in**e by recursion the satisfaction relation M |= φ(a). In order **to**simplify, we assume here the formulas are built us**in**g the connectives ¬, ∧ andthe existential quantifier ∃. Observe that if ∃yψ has its free variables among xthen ψ has its free variables among y, x.Def**in**ition 1.0.2. Given a formula φ(x), a structure M and a tuple a ∈ M, wesay that φ holds **in** M with the assignment x i ↦→ a i , **in** symbols M |= φ(a), asfollows:• If φ(x) is t 1 = t 2 we put M |= φ(a) iff t M 1 (a) = t M 2 (a).• If φ(x) is R(t 1 , . . . , t n ) we put M |= φ(a) iff (t M 1 (a), . . . , t M n (a)) ∈ R M .• If φ(x) is ¬ψ we put M |= ¬ψ(a) iff M ̸|= ψ(a).• If φ(x) is ψ 1 ∧ ψ 2 we put M |= φ(a) iff M |= ψ 1 (a) and M |= ψ 2 (a).• If φ(x) is ∃yψ(y, x) we put M |= φ(a) iff M |= ψ(b, a) for some b ∈ M.Given a formula φ(y 1 , . . . , y m ) and terms t 1 (x), . . . , t m (x) we will denote byφ(t 1 , . . . , t m ) the result of substitut**in**g y i by t i simultaneously **in** a lexicographicvariant of φ. A lexicographic variant of a formula is obta**in**ed by replac**in**g boundvariables by other new variables. It is known that replac**in**g bound variables **in**φ by variables not occurr**in**g **in** any term t i (**in** fact it is enough **to** replace onlythose that occur **in** some t i ) the substitution x i /t i turns out **to** be free. Thesubstitution lemma becomes, with this notationM |= φ(t 1 , . . . , t m )(a) iff M |= φ(t M 1 (a), . . . , t M m (a)) (1.0.1)Very often these terms t i will be constants c i . In this case, the substitutionlemma becomes:M |= φ(c 1 , . . . , c m ) iff M |= φ(c M 1 , . . . , c M m ) (1.0.2)To end, a little bit more notation:

CHAPTER 1. PRELIMINARIES 4Notation 1.0.3. When **work****in**g with particular examples we will freely writeassignments **in**side formulas. For **in**stance, let L denote the language that conta**in**sonly a s**in**gle b**in**ary relation < and M be the structure (Z, < Z ), wherewe denote here also by < Z the strict order relation between **in**tegers. Let theformula φ(x, y) = ∃z(x < z ∧ z < y). We will write (Z, < Z ) |= ∃z(1 < z ∧ z < 3)**to** denote M |= φ(1, 3). This is very useful notation **in** order **to** ‘understand theformula’. However one must know that ∃z(1 < z ∧ z < 3) is not a formula, s**in**ce1 and 3 are not constants of the language.Likewise, if φ(x) is ∃yψ(y, x) we put M |= ∃yψ(y, a) **to** denote M |= φ(a).Aga**in**, here ∃yψ(y, a) is not a formula.remark 1.0.4. The notation φ( . . . , ) has many (maybe **to**o many) uses andone has **to** be careful with the notation. If we write φ(x 1 , . . . , x n ) it usuallydenotes that the free variables of φ is a subset of {x 1 , . . . , x n }. Of course,this is because we assume that x 1 , . . . , x n are variables. If y 1 , . . . , y n is a newtuple of variables φ(y 1 , . . . , y n ) denotes the same formula after the substitutionx i → y i . If c 1 , . . . , c n is a tuple of constants φ(c 1 , . . . , c n ) denotes the resultafter the substitution x i → c i . If a 1 , . . . , a n is a tuple of elements **in** some L-structure M, φ(a 1 , . . . , a n ) is not a formula (so may no be written!), but we writeM |= φ(a 1 , . . . , a n ) **to** denote the satisfaction of φ **in** M under the assignmentx i → a i .Inf**in**itely many free variables can occur **in** an **in**f**in**ite set of formulas. Inthis situation one needs **to** assign **in**f**in**itely many variables, say all variables **to**simplify. If V ar denotes the set of all variables, an assignment **in****to** M is a maps : V ar → M. For a given formula φ(x), we denote M |= φ(s) when M |= φ(a),where a i = s(x i ).Def**in**ition 1.0.5. Let Σ be a set of formulas and φ a formula.• The set Σ is called satisfiable iff there is some structure M and someassignment s : V ar → M such that M |= φ(s) for each formula φ ∈ Σ.This fact is denoted by M |= Σ(s).• the formula φ is a logical consequence of Σ (or Σ entails φ), **in** symbols:Σ |= φ, if M |= φ(s) for each M and s such that M |= Σ(s).Usually, the formulas will be closed (without free variables) and called sentences.The card**in**al of L, denoted by |L|, is the number of symbols of L (constant,function and relation). Unless otherwise stated, we **work** with a countableset of variables. In this situation there are |L| + ℵ 0 formulas.1.1 The compactness TheoremOur start**in**g po**in**t are the theorems of compactness (Theorem 1.1.1), statedwithout proof.

CHAPTER 1. PRELIMINARIES 5Theorem 1.1.1 (compactness theorem, first version). Given a set of formulasΣ and a formula φ then:Σ |= φ iff there is a f**in**ite subset Σ 0 of Σ such that Σ 0 |= φEquivalently, the compactness theorem can be stated as follows:Theorem 1.1.2 (compactness theorem, second version). Every set of formulasf**in**itely satisfiable is satisfiable.Corollary 1.1.3. Let Σ be a set of sentences. If every f**in**ite subset of Σ has amodel then Σ has a model.Exercise 1.1.4.other.1. Prove each versions of the compactness theorem us**in**g the2. prove that from corollary 1.1.3 one can recover the compactness theorem1.1.1.Exercise 1.1.5. A class K of L-structures is called axiomatizable (also called∆-elementary class or EC ∆ class) if it is of the form Mod(T ) for some L-theoryT . In case T is f**in**ite we call K f**in**itely axiomatizable (also called an elementaryclass). Use compactness **to** show:1. The class of **to**rsion groups is not axiomatizable.2. The class of connected graphs is not axiomatizable.Exercise 1.1.6. Let M be an L-structures. A subset A ⊆ M n is called def**in**ableif there is an L-formula φ(x) such that A = {a ∈ M n | M |= φ(a)}. Here weimplicitly assume that the length of x is n. Use compactness **to** show:1. The **to**rsion part of a group is not def**in**able (more precisely: there is noformula φ(x) such that **in** any group G φ def**in**es its **to**rsion part).2. The connected component of a graph is not def**in**able (more precisely:there is no formula φ(x, y) such that **in** any graph G M |= φ(a, b) iff aand b are **in** the same connected component).3. The set of all algebraic (over the prime field) elements **in** a given field arenot def**in**able (more precisely: there is no formula φ(x) such that **in** anyfield K, φ def**in**es the algebraic elements of K).1.2 Löwenheim-SkolemTheoremsDef**in**ition 1.2.1 (embedd**in**gs and substructures). **An** embedd**in**g is an**in**jective map h : M → N satisfy**in**g the follow**in**g properties:1. h(c M ) = c N for every constant symbol c.

CHAPTER 1. PRELIMINARIES 62. h(f M (a)) = f N (ha) for every function symbol f and every tuple a ofelements of M of the same length than the arity of f.3. a ∈ R M iff ha ∈ R N for every tuple a of M and every relation symbol R.Here a denotes a 1 , . . . , a n and ha denotes ha 1 , . . . , ha n . When the identity isan embedd**in**g of M **in****to** N it is said that M is a substructure of N and wewill denote it by M ⊆ N. **An** Isomorphism is a surjective embedd**in**g, i.e., abijection preserv**in**g all the symbols of the language.If M is a substructure of N, then obviously the doma**in** of M is a subset ofthe doma**in** of N. Moreover, the follow**in**g holds:1. c M = c N for every constant symbol c.2. f M (a) = f N (a) for every function symbol f and every tuple a of elementsof M of the same length than the arity of f.3. a ∈ R M iff a ∈ R N for every tuple a of M and every relation symbol R.Observe that a substructure of N is determ**in**ed by its doma**in**, s**in**ce the**in**terpretation of the symbols of function and relation is just the restriction ofthe **in**terpretation of the symbols **in** N. The doma**in**s D of the substructures ofN is a subsets of N ‘closed under application of functions’. That is, satisfy**in**gthe follow**in**g:1. c N ∈ D for every constant symbol c.2. f N (a) ∈ D for every tuple a of D and functional symbol f 1 .Although not every subset A of N is the doma**in** of a substructure, there isa smallest substructure of N connta**in****in**g A, denoted by ⟨A⟩ M. The doma**in** of⟨A⟩ Mis{t N (a) | t = t(x) is a term and a is a tuple of elements of A } (1.2.1)Exercise 1.2.2. Prove that (1.2.1) is the smallest subset of N conta**in****in**g A andclosed under application of functions.We observe that if h : M → N is an embedd**in**g, h(M) is the doma**in** of asubstructure of N, which we will denote also by h(M), and h is an isomorphismfrom M **to** h(M). Conversely, every isomorphism of M on**to** a substructure of Nis an embedd**in**g. In other words, an embedd**in**g of M **in****to** N is an isomorphismof M on**to** a substructure of N. Hence the structures that may be embedded **in**N are the isomorphic copies of substructures of N.Proposition 1.2.3. If h : M → N is an embedd**in**g then there is an extensionM ′ of M (M is a substructure of M ′ ) and an isomorphism h ′ : M ′ → Nextend**in**g h.1 although not stated it should be clear that the arity of f is the same as the length of a

CHAPTER 1. PRELIMINARIES 7Proof. Let A be a set conta**in****in**g M and h ′ an extension of h which is a bijectionfrom A **to** N. Now we def**in**e a structure M ′ with doma**in** A by putt**in**g, foreach tuple a ∈ A:a ∈ R M ′ ⇔ h ′ a ∈ R Nf M ′ (a) = h ′−1 f N (h ′ a) (1.2.2)By def**in**ition, h ′ is an isomorphism extend**in**g h. It rema**in**s **to** prove that M is asubstructure of M ′ . For a tuple a ∈ M, and a relation symbol we have: a ∈ R M ′iff ha = h ′ a ∈ R N iff a ∈ R M , s**in**ce h is an embedd**in**g. For a function symbolf we have: f M ′ (a) = h ′−1 f N (h ′ a) = h ′−1 f N (ha) = h ′−1 hf N (a) = f N (a).Exercise 1.2.4 (homomorphism’s theorem). Let h be an embedd**in**g h : M → N.Prove that:1. For each term t(x) of the language and each a ∈ M it holds that ht M (a) =t N (ha). 22. h preserves all the formulas without quantifiers. That is, if φ(x) is aformula without quantifiers, thenM |= φ(a) iff N |= φ(ha) for all a ∈ M (1.2.3)3. If h is an isomorphism then it preserves all formulas, i.e., (1.2.3) holds forevery formula φ.Exercise 1.2.5. Let h be an embedd**in**g from M **to** N. Prove that:1. If φ(x) is an existential formula ( it is of the form ∃yψ with ψ quantifierfree),a ∈ M and M |= φ(a) then N |= φ(ha).2. If φ(x) is a universal formula ( it is of the form ∀yψ with ψ quantifier-free),a ∈ M and N |= φ(ha) then M |= φ(a).Exercise 1.2.6. Prove the follow**in**g are equivalent:1. h is an embedd**in**g2. h preserves all quantifier-free formulas3. h preserves all a**to**mic formulas4. h preserves the follow**in**g formulas: x = y, x = c for each constant c,Rx 1 , . . . , x n for each relation symbol R, fx 1 , . . . , x n = y for each functionsymbol f.Def**in**ition 1.2.7. **An** elementary embedd**in**g is an embedd**in**g preserv**in**g allformulas, that is, satisfy**in**g (1.2.3) for every formula φ. In the case where theidentity is elementary we will say that M is an elementary substructure ofN and we will denote it by M N.2 In fact this is true for any homomorphism h

CHAPTER 1. PRELIMINARIES 8Po**in**t 3. **in** 1.2.4 says that any isomorphism is an elementary embedd**in**g.Def**in**ition 1.2.8. Two structures M, N are elementarily equivalent, M ≡N **in** symbols, if they satisfy the same sentences, that is, have the same theory.If there is an elementary embedd**in**g of M **in****to** N then M and N are elementarilyequivalent. However not every embedd**in**g between elementarily equivalentstructures is elementary: (2Z,

CHAPTER 1. PRELIMINARIES 9Proof. For each formula φ(y, x) we consider the follow**in**g choice function F φ(y,x) :N n → N (where n is the length of x).{ b, if it exists an element b of N such that N |= φ(b, a);F φ(y,x) (a) =c, otherwise.,where c is a fixed element of N. If M conta**in**s A and is closed under all themaps F φ(y,x) , by Tarski-Vaught’s test 1.2.10 M is an elementary substructureof N. Let M be the closure of A under all the maps F φ(y,x) . This M isbuilt **in** a countable number of steps as follows. We start with A 0 = A and takeA m+1 = A m ∪ ∪ φ(y,x)∈L F φ(y,x)(A n m). Of course at any step |A n | ≤ |A|+|L|+ℵ 0and therefore if we put M = ∪ n A n, the card**in**al of M is also bounded by|A| + |L| + ℵ 0 .As a corollary we obta**in** the usually called downwards Löwenheim-Skolemtheorem. Although it is also true for formulas **in** general (appropriately formulated)we state it for sentences.Theorem 1.2.13 (downwards Löwenheim-Skolem Theorem). If a set of sentenceshas models, it also has models of card**in**ality at most |L| + ℵ 0 .Proof. Let Σ be a set of sentences and N be a models of Σ. By Theorem 1.2.12(with A = ∅) there is some elementary substructure M of N os size at most|L| + ℵ 0 . S**in**ce M ≼ N, M and N must satisfy the same sentences, so M is thedesired model of Σ.A simple consequence of downwards Löwenheim-Skolem theorem us**in**g compactness,is the upwards Löwenheim-Skolem theorem. It says that a theoryhav**in**g **in**f**in**ite models (or arbitrarily large f**in**ite models) also has models **in** anycard**in**al bigger or equal **to** |L| + ℵ 0 .Theorem 1.2.14 (upwards Löwenheim-Skolem Theorem). Let Σ be a set ofsentences such that for each n, Σ has a model of size at least n. Then, for eachcard**in**al λ ≥ |L| + ℵ 0 , Σ has a model of card**in**al λ.Proof. We take a set C of new constants of card**in**al λ and we consider thetheory:Ψ := T ∪ {c ≠ d | c, d are two different constants of C} .Every f**in**ite subset Ψ 0 of Ψ conta**in**s f**in**itely many constants, say n. By hypothesisΣ has a model M of size at least n. Hence Ψ 0 is satisfied **in** a suitableexpansion of M **in**terpret**in**g the new constant as different elements. Becauseof compactness, Ψ has a model. By downwards Löwenheim-Skolem Theorem1.2.13 there is a model M of Ψ of card**in**al at most λ + |L| + ℵ 0 = λ. But as theλ different constant have **to** be **in**terpreted as different elements it follows thatM has card**in**al exactly λ. The reduct of M **to** L is a models of Σ of card**in**alλ.

CHAPTER 1. PRELIMINARIES 10As a consequence, if a theory has only f**in**ite models, there is a f**in**ite bound onthe card**in**ality of its models. The upwards Löwenheim-Skolem theorem says usthat the set of **in**f**in**ite card**in**als where a countable theory has models is either theempty set or all **in**f**in**ite card**in**als. If the language is uncountable we know thateither the set of **in**f**in**ite card**in**als where a theory has models is either the emptyset or conta**in**s {λ | λ is a card**in**al and |L| + ℵ 0 ≤ λ}. However it does no saysanyth**in**g about the card**in**als **in** the **in**terval {λ | λ is a card**in**al and ℵ 0 ≤ λ < |L|}.The follow**in**g examples show that we cannot say much more with absolute generality.Example 1.2.15. 1. Let λ be an uncountable card**in**al and C a set of κ constants.The set {c ≠ d | c, d are two different constants of C} has no modelsof size less than |L| = κ.2. Consider the language L which conta**in**s, for each subset S of N a unarypredicate P S . Let the L-structure N whose doma**in** is N and P S is **in**terpretedas S. Obviously Th(N) has a countable model although |L| = 2 ℵ 0.In the follow**in**g exercise we will see that for f**in**ite card**in**als the situation isradically different.Exercise 1.2.16. The f**in**ite spectrum of a sentence φ is the set of f**in**ite card**in**alswhere the sentence has models: {n ∈ ω | φ has a model of card**in**al n}. Thef**in**ite spectrum can be quite complicated. Construct formulas such that itsf**in**ite spectrum is the follow**in**g one: nN = {nm ∈ N | m ∈ N}, where n a fixednatural number n. The set of all square numbers: { m 2 ∈ N | m ∈ N } . The se**to**f all prime numbers. The set of all non prime numbers. The set of all powersof some prime number. the set of all powers of a fixed prime number p.1.3 Complete theories: the test of ̷Loš-VaughtA theory will be a set of sentences. Many textbooks of mathematical logic def**in**ea theory as a set of sentences closed under logical consequence. Every set ofsentences Σ determ**in**es a unique set closed under logical consequence hav**in**gthe same models: the set of their logical consequencesΣ |= := {φ | φ is a sentence and Σ |= φ} .S**in**ce we are **in**terested on its models this does not make any difference.Exercise 1.3.1. For a class K of L-structures, denoteTh(K) = {φ | φ is a sentence and M |= φ for each M ∈ K} .1. Shows that Σ |= = Th(Mod(Σ)).2. Let Σ 1 , Σ 2 be sets of sentences. Shows that the follow**in**g are equivalent:(a) Σ |= 1 = Σ|= 2 .(b) Σ 1 and Σ 2 have the same models.

CHAPTER 1. PRELIMINARIES 11Def**in**ition 1.3.2. A theory T is said **to** be complete if for every sentence φwe have that T |= φ or T |= ¬φ.The basic example of complete theory is the theory of a structure: Th M ={φ | φ is a sentence and M |= φ}. As a matter of fact, apart from the **in**consistenttheory (that consists of all sentences) there are not more: it is easy **to** verifythat every consistent and complete theory is the theory of any of its models.Exercise 1.3.3. 1. M ≡ N iff N is a model of Th(M).2. A theory is complete iff all its models are elementarily equivalent.Def**in**ition 1.3.4. Let λ be an **in**f**in**ite card**in**al. A theory T is called categorical**in** λ or λ-categorical if any two models of T of card**in**al λ are isomorphic.The λ-categoricity of T says exactly that, except of isomorphism, T hasat most one model. If T has **in**f**in**ite models and λ ≥ |L| + ℵ 0 , by upwardsLöwenheim-Skolem there has **to** be exactly one model up **to** isomorphism.By the homomorphism theorem (exercise 1.2.4) if M is isomorphic **to** N thenM and N are elementarily equivalent. Given any **in**f**in**ite structure M, by theupwards Löwenheim-Skolem theorem Th M has models N of arbitrarily largesize, hence not isomorphic **to** M. In other words, if M is **in**f**in**ite there are structureselementarily equivalent but not isomorphic **to** M. The next propositionshows that the case where M is f**in**ite is completely different.Proposition 1.3.5. 1. If M and L are f**in**ite, there is a sentence φ M thatcharacterizes M up **to** isomorphism, that is, for any N, N is a model ofφ M iff N it is isomorphic **to** M.2. If M is f**in**ite and N ≡ M then M and N are isomorphic.Proof. 1. Let M = {a 1 , . . . , a n }. Consider the sentence φ M := ∃x 1 , . . . , x n ψ M ,where⎛ψ M := ⎝∧( n) ⎞∨x i ≠ x i ∧ ∀y y = x i ∧ ∧ ψ s⎠ ,i=1s∈L1≤i

CHAPTER 1. PRELIMINARIES 122. By 1, the case when L is f**in**ite is obvious. Assume now M ≡ N and Larbitrary. Obviously N is f**in**ite and of the same card**in**alilty as M. Towards acontradiction, assume M ≁ = N. For each bijection g from M **to** M there is asymbol s g not preserved by g. S**in**ce there are f**in**itely many bijections from M **to**N the Language L ′ := {s g | g : M → N bijective} is f**in**ite and M ↾ L ′ ≁ = N ↾ L′whereas M ↾ L ′ ≡ N ↾ L ′ , a contradiction by 1.Observe that, by the previous proposition, if a complete theory only hasf**in**ite models, as a matter of fact it has only one model (up **to** isomorphism). Aquite important question is **to** know when a given theory (without f**in**ite models)is complete. A first simple case is when it is categorical **in** some card**in**al.Theorem 1.3.6 (̷Loš-Vaught’s test). A theory without f**in**ite models and categorical**in** some card**in**al greater or equal **to** |L| + ℵ 0 is complete.Proof. Suppose that T is λ-categorical, with λ ≥ |L| + ℵ 0 and without f**in**itemodels. We will see that any two models M and N of T are elementarilyequivalent. Consider the theories of M and N respectively. Be**in**g M and N**in**f**in**ite we can apply upwards Löwenheim-Skolem and we obta**in** M ′ and N ′elementarily equivalent **to** M and N respectively of card**in**al λ. Then M ′ andN ′ are two models of T of card**in**al λ, therefore isomorphic. F**in**ally M ′ and N ′are elementarily equivalent, and therefore also M and N.For **in**stance, the theory of dense l**in**ear orders without endpo**in**ts is complete,because is ℵ 0 -categorical and has no f**in**ite models. It is easily seen that thistheory is not categorical **in** any other card**in**ality.If λ is an **in**f**in**ite card**in**al number, I(λ, T ) denotes the number of models of Tof card**in**al λ up **to** isomorphism. Each model of card**in**ality λ has an isomorphiccopy with doma**in** λ. Hence I(λ, T ) is at most the number of structures withdoma**in** λ. For each constant symbol we have λ choices, and for each functionsymbol and relation symbol we have 2 λ choices. This shows that there are atmost (2 λ ) |L| = 2 λ+|L| structures with doma**in** λ. This shows that for λ ≥ |L|+ℵ 0one gets I(λ, T ) ≤ 2 λ . If T has **in**f**in**ite models and λ ≥ |L|+ℵ 0 the Löwenheim-Skolem theorem 1.2.14 tells us that I(λ, T ) ≥ 1.Exercise 1.3.7. Prove that:1. The theory of an **in**f**in**ite set is categorical **in** any card**in**al, i.e., I(λ, T ) =1 for each **in**f**in**ite card**in**al λ. This is the theory, **in** the empty languageaxiomatized by {φ ≥n | n ∈ N}, where φ ≥n is the follow**in**g formula:∃x 1 · · · ∃x n( ∧1≤i

CHAPTER 1. PRELIMINARIES 133. If k is a f**in**ite field, the theory of the **in**f**in**ite k-vec**to**r spaces is categorical**in** any **in**f**in**ite card**in**al.4. If k is an **in**f**in**ite field, the theory of nontrivial k-vec**to**r spaces is categoricalonly **in** card**in**ality > |k|. Observe that **in** this case |L| = |k|. Show thatI(λ, T ) = 0 for ℵ 0 ≤ λ < |k|, I(|k| , T ) = |k| and I(λ, T ) = 1 for λ > |k|.5. The theory of all algebraically closed fields of a fixed characteristic iscategorical **in** any non countable card**in**al, however it is not ℵ 0 -categorical.This uses the criterion of Ste**in**itz for isomorphism of algebraically closedfields.6. The theory T DLO of the dense l**in**ear orders (more precisely: dense l**in**earorders without endpo**in**ts) i L = {

CHAPTER 1. PRELIMINARIES 14whenever i ≤ j. When M i is an elementary substructure of M j for every i ≤ jwe will say that it is an elementary cha**in**.When we have a cha**in** of structures, the structure union of the cha**in**,denoted by ∪ i∈I M i, is constructed as follows. Its doma**in** is the union of thedoma**in**s and the symbols are **in**terpreted also tak**in**g the union of their **in**terpretations.For **in**stance, if R is a relation symbol R ∪ i∈α M i= ∪ i∈α. For theRMifunction symbols we do the same with the graph of the function. Equivalently,if f is an n-ari function symbol and a is an n-tuple of elements of ∪ i∈I M i,we take some M j where all the elements of the tuple belong. Then we putf ∪ i∈I M i(a) := f Mj (a). Be**in**g a cha**in**, this does not depend on the chosen M j .We treat the symbols of constant like symbols of function of arity 0. It is easy**to** verify that each M i is a substructure of the union.Exercise 1.4.2. Sow that ∀∃-sentences (sentences of the form ∀x∃yψ with ψquantifier-free) are preserved under unions of cha**in**s. More precisely, if ⟨M i | i ∈ I⟩is a cha**in** of structures and a ∀∃-sentence φ holds **in** any M i then φ also holds**in** the union of the cha**in** ∪ i∈I M i. Use 1.2.4.The follow**in**gs exercise aims **to** show that we can do the same construction**in** a wider situation: **work****in**g wit embedd**in**gs and a non-necessarily ordered set.Exercise 1.4.3. A directed system of embedd**in**gs consists is a collection of structures⟨M i | i ∈ I⟩, where (I, ≤) is a directed (for any i, j ∈ I there is some k ∈ Iwith i ≤ k, j ≤ k) partially ordered set, **to**gether with a collection of embedd**in**gsf i,j : M i → M j for any pair i, j ∈ I with i ≤ j, satisfy**in**g f j,k ◦ f i,j = f i,k .Def**in**e the direct limit lim i M i of the system and embedd**in**gs h i : M i → lim i M iand show that h j ◦ f i,j = h i .Lemma 1.4.4 (cha**in**’s lemma). If the cha**in** ⟨M i | i ∈ I⟩ is elementary, theneach M i is an elementary substructure of the union:M i ∪ j∈IM j .Proof. We prove, by **in**duction on the formula φ(x), that for each i and eachtuple a of M i the follow**in**g holdsM i |= φ(a) iff ∪ i∈IM i |= φ(a)For the a**to**mic formulas we know it because M i is a substructure of the union.The cases of the negation and the conjunction are trivial. Let now φ(a) =∃xψ(x, a). If M i |= ∃xψ(x, a) there is some b ∈ M i with M i |= ψ(b, a). Byhypothesis of **in**duction, ∪ i∈I M i |= ψ(b, a) and thus ∪ i∈I M i |= ∃xψ(x, a).Conversely, if the formula ∃xψ(x, a) holds **in** ∪ i∈I M i we have that ∪ i∈I M i |=ψ(b, a) for a certa**in** b ∈ M j with j ≥ i. By hypothesis of **in**duction M j |= ψ(b, a).Hence M j |= ∃xψ(x, a) and, as the cha**in** is elementary M i |= ∃xψ(x, a).The cha**in**s of structures are usually **in**dexes **in** an ord**in**al α and **in** the limitsare take unions (that is, M i = ∪ j

CHAPTER 1. PRELIMINARIES 15Exercise 1.4.5. A cont**in**uous cha**in** of structures (**in**dexed **in** an ord**in**al) is elementaryif for each i, M i is an elementary substructure of M i+1 .Exercise 1.4.6. This is a cont**in**uation of exercise 1.4.3. Show that the cha**in**lemma holds for a direct limit of structures. More precisely, If (M i | i ∈ I) and(f i,j ) | i ≤ j ∈ I for a directed system of elementary embedd**in**gs (i.e., each f i,jis elementary) then the maps def**in**ed **in** h i : M i → lim i M i def**in**ed **in** 1.4.3 alalso elementary.Exercise 1.4.7. A theory is called model complete iff every embedd**in**g betweenmodels of T is elementary. Prove that T is model complete if given M, N modelsof T and an embedd**in**g f : M → N there is some elementary exetension M ∗of M and an embedd**in**gs g : N → M ∗ such that Id = g ◦ f. H**in**t: given f :M → N build elementary cha**in**s ⟨M n | n ∈ N⟩, ⟨M n | n ∈ N⟩ and embedd**in**gsf n : M n → N n g n : N n → M n+1 with g n ◦ f n = Id f n+1 ◦ g n = Id. Start withM 0 = M, N 0 = N and f 0 = f. Apply the hypothesis **to** obta**in** M 1 = M ∗ andg 0 = g. Apply aga**in** the hypothesis **to** g 0 : N 0 → M 1 **to** obta**in** N 1 and f 1 andso on.1.5 DiagramsWe often **work** with two languages: L and an extension L ′ of it (that is, allsymbols of L are **in** L ′ ), which is denoted by L ⊆ L ′ . In this case any L ′ -structure M ′ becomes an L-structure M naturally forgett**in**g the **in**terpretationof the symbols of L ′ − L. In this case M is called the restriction of M ′ **to** Land denoted by M = M ′ ↾ L . One can also say that M ′ is an expansion ofM **to** L.Now we will consider expansions of a language L by constants. If M is aL-structure, we will expand L by add**in**g a new constant for each element ofM. For simplicity we will use the same symbol **to** denote the constant and theelement of M that designates. We will denote this expansion by L(M). We willdenote by M M the natural expansion of M **to** L(M), where each new constanta is **in**terpreted as the element a of M. If N is another L-structure and h amap from M **to** N the expansion of N **to** L(M) that **in**terprets a as ha will bedenoted by (N, ha) a∈M .Every sentence of L(M) is obta**in**ed by replacement of the variables x of anL-formula φ(x) by constants a ∈ M. This sentence will be denoted by φ(a).With this notation, the Substitution Lemma (1.0.2) becomes:In the same wayM M |= φ(a) iff M |= φ(a). (1.5.1)(N, ha) a∈M |= φ(a) iff N |= φ(ha). (1.5.2)remark 1.5.1. **in** (1.5.1) the expression φ(a) denotes a formula **in** M M |= φ(a)while the same expression φ(a) does not denote a formula **in** M |= φ(a). Thesame applies **to** (1.5.2). The equivalence (1.5.1) says that the notational confusionremarked **in** 1.0.4 is harmless here.

CHAPTER 1. PRELIMINARIES 16The diagram of M is the set of L(M)-formulas without quantifiers hold**in**g**in** M M :Diag(M) = {φ ∈ L(M) | φ is without quantifiers M M |= φ} .The elementary diagram of M is the theory of M M :∆(M) = {φ ∈ L(M) | M M |= φ} = Th(M M ).Next proposition says that every model of the (elementary) diagram of Mconsists of a structure N plus an (elementary) embedd**in**g of M **in** N.Proposition 1.5.2. Let M, N be two L-structures and h a map from M **to** N.1. The map h is an embedd**in**g (of M **in****to** N) iff (N, ha) a∈M is a model ofDiag(M).2. The map h is an elementary embedd**in**g (of M **in****to** N) iff (N, ha) a∈M itis a model of ∆(M).Proof. We beg**in** by prov**in**g 1. Suppose that h is an embedd**in**g. If φ(a) ∈Diag(M) then M M |= φ(a) and thus M |= φ(a). By 1.2.4 N |= φ(ha) and therefore(N, ha) a∈M |= φ(a). Hence (N, ha) a∈M is a model of Diag(M). Conversely,if (N, ha) a∈M is a model of Diag(M), h is **in**jective because a ≠ b ∈ Diag(M)for each couple of elements a, b different of M. Let R is a relation symboland a a tuple of M. If a ∈ R M then R(a) ∈ Diag(M). Therefore(N, ha) a∈M |= R(a) and thus ha ∈ R N . If a /∈ R M then ¬R(a) ∈ Diag(M).Therefore (N, ha) a∈M |= ¬R(a) and ha /∈ R N . If f is a function symbol, a isa tuple of M and b = f M (a) we have that the formula b = f(a) belongs **to** thediagram of M, (N, ha) a∈M |= b = f(a) and therefore h(b) = f N (ha).To prove 2, we suppose that h is an elementary embedd**in**g. Then if φ(a) ∈∆(M) it follows that M M |= φ(a), M |= φ(a), N |= φ(ha) and therefore(N, ha) a∈M |= φ(a). Hence (N, ha) a∈M is a model of ∆(M). Conversely(N, ha) a∈M is model of ∆(M), φ(x) ∈ L and a is a tuple of M, M |= φ(a)iff φ(a) ∈ ∆(M) iff (N, ha) a∈M |= φ(a) iff N |= φ(ha).Exercise 1.5.3. The diagram of a structure could be def**in**ed with a smaller se**to**f formulas. Let us denote diag(M) the follow**in**g set of formulas:• R(a) for each relation symbol R and each tuple a ∈ R M .• ¬R(a) for each relation symbol R and each tuple a /∈ R M .• f(a) = b for each function symbol f, each tuple a ∈ M and each b ∈ Msuch that f M (a) = b.• a ≠ b for each pair of different elements a, b of M.Sow that Diag(M) and diag(M) have the same models.We next see an application of the diagrams.

CHAPTER 1. PRELIMINARIES 17Proposition 1.5.4. Two structure M 1 , M 2 are elementarily equivalent iff thereis a structure N and elementary embedd**in**gs of each M i **in** N.Proof. Suppose that M 1 and M 2 are elementarily equivalent. By proposition1.5.2 everyth**in**g is reduced **to** f**in**d a model of ∆(M 1 ) ∪ ∆(M 2 ). We can supposethat the doma**in**s of M 1 and M 2 are disjo**in**t (otherwise we take an isomorphiccopy of M 1 ). By compactness and the fact that ∆(M i ) is closed under conjunctionwe have **to** f**in**d a model of φ(a) ∧ ψ(b), where φ(a) is from the elementarydiagram of M 1 and ψ(b) is from the elementary diagram of M 2 . Here we assumea ∈ M 1 , b ∈ M 2 and φ(x), ψ(y) are L-formulas.S**in**ce M 2 |= ψ(b) also M 2 |= ∃xψ(x). As M 1 is elementarily equivalent **to**M 2 we have that M 1 |= ∃xψ(x). Therefore M 1 |= ψ(c) for a certa**in** tuple c ofelements of M 1 . Then the expansion M ′ of M 1M1 **in**terpret**in**g b by c satisfiesψ(b) and therefore is a model of φ(a) ∧ ψ(b).remark 1.5.5. In proposition 1.5.4 we can improve the conclusion by stipulat**in**gthat N is an elementary extension of M 1 , i.e. f 1 is the identity. This isdone us**in**g lemma 1.2.3. But we cannot get, **in** general, a common elementaryextension of M 1 and M 2 . For **in**stance, if M 1 ⊆ M 2 the existence of a commonelementary extension implies M 1 ≼ M 2 . Next exercises deal with this problem.Exercise 1.5.6. Let M 1 , M 2 be structures and let A denote the **in**tersectionM 1 ∩ M 2 of their doma**in**s.1. Show that if M 1 and M 2 have a common extension (i.e., there is N withM i ⊆ N) then A is the doma**in** of a common substructure of M 1 and M 2 .2. (**to** be checked) Show that M 1 and M 2 have a common extension iffDiag((M 1 ) A ) ∪ Diag((M 2 ) A ) ∪ {m 1 ≠ m 2 | m 1 ∈ M 1 \ A, m 2 ∈ M 2 \ A}is consistent.3. (**to** be checked) Show that if each pair of f**in**itely generated substructuresN 1 of M 1 and N 2 of M 2 have a common extension then also M 1 and M 2have a common extension. Q: can we do it with ultraproducts????Exercise 1.5.7. Let M 1 , M 2 be structures and let A denote the **in**tersectionM 1 ∩ M 2 of their doma**in**s.1. Show that if M 1 and M 2 have a common elementary extension (i.e., thereis N with M i ≼ N) then A is an algebraically closed subset of M 1 andM 2 .2. (**to** be checked) Show that M 1 and M 2 have a common elementary extensioniff∆((M 1 ) A ) ∪ ∆((M 2 ) A ) ∪ {m 1 ≠ m 2 | m 1 ∈ M 1 \ A, m 2 ∈ M 2 \ A}is consistent.

CHAPTER 1. PRELIMINARIES 183. (**to** be checked) Show that M 1 and M 2 have a common elementary extensioniff (M 1 ) A ≡ (M 2 ) A and A is algebraically closed **in** each M i . H**in**t:use 1 and 2.Exercise 1.5.8. Prove one can generalize proposition 1.5.4: any collection of pairwiseelementarily equivalent models can be embedded elementarily **in** a commonmodel.Corollary 1.5.9. Every structure elementarily equivalent **to** a f**in**ite structureis isomorphic **to** it.Proof. Let M 1 , M 2 be elementary equivalent structures and suppose M 1 is f**in**ite.Let N and f i : M i → N as **in** proposition 1.5.4. S**in**ce the embedd**in**g f 1 iselementary N is f**in**ite with the same card**in**ality as M 1 . This implies f 1 is anisomorphism. Aga**in**, s**in**ce the embedd**in**g f 2 is elementary M 2 is f**in**ite with thesame card**in**ality as N. This implies f 2 is an isomorphism. Now f2 −1 ◦ f 1 is anisomorphism from M 1 **to** M 2 .A theory T is said **to** be closed under substructures if each substructureof each model of T is a model of T . A universal formula is a formula of the form∀xφ, where φ is a quantifier-free formula. The universal formulas are also calledΠ 1 -formulas or sometimes ∀-formulas. One can also def**in**e existential formulas(also Σ 1 -formulas or ∃-formulas) as formulas of the form ∃xφ, where φ is aquantifier-free formula. In fact there is a whole hierarchy: Π n+1 -formulas are ofthe form ∀xφ, where φ is a Σ n formula and Σ n+1 -formulas are of the form ∃xφ,where φ is a Π n formula. A Π n -formula is of the form ∀x 1 ∃x 2 · · · Qx n φ where φis quantifier-free. Also the Σ n -formulas have n blocs of alternat**in**g quantifiers**in** this case start**in**g by al existential block.Let’s see an application of diagrams:iff it can be ax-Proposition 1.5.10. A theory is closed under substructuresiomatized by a set of universal sentences.Proof. In order **to** prove ⇐, it suffices **to** check the follow**in**g: If M is a substructureof N, M satisfies each universal sentence true **in** N. If N |= ∀xφ(x)for some quantifier-free formula φ(x), then for every tuple of elements a ∈ N,N |= φ(a). In particular N |= φ(a) for each tuple a **in** M. As φ is quantifier-free,M |= φ(a) for each tuple a **in** M and thus M |= ∀xφ(x).Conversely, assume T is closed under substructures. We are go**in**g **to** provethat T ∀ = {φ | φ is a universal sentence and T |= φ} is a set of axioms for T ,i.e., Mod(T ) = Mod(T ∀ ). We must see Mod(T ∀ ) ⊆ Mod(T ), the other **in**clusionbe**in**g trivial. Let M be a model of T ∀ . It suffices **to** prove that T ∪ Diag(M)has a model. This will provide us, by proposition 1.5.2, with a model N of Tand an embedd**in**g h : M → N. S**in**ce T is closed under substructures h(M) isa model of T hence also M (as M is isomorphic **to** h(M). By compactness it isenough **to** prove that T ∪ {φ(a)} has a model for any formula φ(a) ∈ Diag(M).If not, for some formula φ(a) ∈ Diag(M) the theory T ∪ {φ(a)} has no models,where a is a tuple of constants a ∈ M and φ(x) is a quantifier-free formula

CHAPTER 1. PRELIMINARIES 19of L. This implies T |= ¬φ(a) and, s**in**ce the constants a do not belong **to** L,T |= ∀x¬φ(x). This means that the formula ∀x¬φ(x) belongs **to** T ∀ . But thisis a contradiction, s**in**ce the M is a model of ∀x¬φ(x) and M |= φ(a).Corollary 1.5.11. A sentence is preserved under substructures iff it is equivalent**to** an existential sentence.Proof. If φ is preserved under substructures, apply**in**g proposition 1.5.10, thetheory {φ} has a set Σ of universal axioms, that is, Mod(Σ) = Mod(φ). Bycompactness there are ψ 1 , . . . , ψ n ∈ Σ such that {ψ 1 , . . . , ψ n } |= φ. Then φ isequivalent **to** ψ 1 ∧. . .∧ψ n . But a conjunction of existential formulas is equivalent**to** an existential formula, by replac**in**g the bound variables is necessary.Given an embedd**in**g f : M → N we say that f is existentially closed ifthe embedd**in**g preserve all existential formulas. More precisely, if (1.2.3) holdsfor all existential formulas. In fact is suffices that for any existential formulaφ(x) and any tuple a ∈ M, if N |= φ(ha) then M |= φ(a). This is equivalent **to**preserve universal formulas **in** the sense of (1.2.3) (and equivalent also **to**: theuniversal formulas transfer from M **to** N). We will denote it by f : M → 1 N**to** suggest f preserves formulas with one block of quantifiers. When M is asubstructure of N one says M is existentially closed **in** N when the identityId M : M → N is an existially closed embedd**in**g. This will be denoted byM ≼ 1 N.Lemma 1.5.12. Let M ⊆ N. Then M is existentially closed **in** Nan extension M ′ of N with M ≼ M ′ .iff there isProof. Assume M is existentially closed **in** N. By proposition 1.5.2, a model of∆(M) ∪ Diag(N) consists of an L-structure M ∗ and embedd**in**gs f : M → M ∗and g : N → M ∗ with f elementary. Observe that g extends f, as we areassum**in**g L(M) ⊆ L(N), i.e., we use the same constants **to** denote an **in**dividualof M **in** ∆(M) and **in** Diag(N). By proposition 1.2.3 there is an extension M ′of N and an isomorphism h : M ′ → M ∗ extend**in**g g. Then M ≼ M ′ , s**in**ceId M = h −1 ◦ f : M → M ′ is the composition of two elementary embedd**in**gs.This shows it is enough then **to** prove that ∆(M) ∪ Diag(N) has a model, orequivalently, by compactness, that ∆(M) ∪ {ψ} has a model for each formulaψ ∈ Diag(N). Separat**in**g the constants from M than those of N \ M, ψ isof the form φ(a, b) for some quantifier-free L-formula φ(x, y) and tuples a ∈M, b ∈ N \ M with N |= φ(a, b). S**in**ce M is existentially closed **in** N andN |= ∃yφ(a, y) we get M |= ∃yφ(a, y). Let c a tuple of elements **in** M suchthat M |= φ(a, c). This implies that M ′′ , the expansion of M M **to** L(Mb) by**in**terpret**in**g the tuple of constants b as the elements c is a model of ψ = φ(a, b).Hence M ′′ is a model of ∆(M) ∪ {ψ}.Exercise 1.5.13. Prove that an embedd**in**g f from M **to** N is existentially closediff there are embedd**in**gs g : M → M ′ and h : N → M ′ , g elementary andg = h ◦ f.

CHAPTER 1. PRELIMINARIES 20A theory T is closed under unions of cha**in**s if any union of models ofT is also a model of T .Theorem 1.5.14 (Chang-̷Los-Suzsko). A theory T is closed under unions ofcha**in**s iff T can be axiomatized by ∀∃-sentences.Proof. We let the reader check that any ∀∃-sentence is preserved under unions ofcha**in**s. This gives the easy part of the proof. To prove the converse, assume T ispreserved under unions of cha**in**s. Wa are go**in**g **to** see that Mod(T ) = Mod(T ∀∃ ),where T ∀∃ denotes the set of ∀∃-consequences of T .Claim Assume M is a model of T ∀∃ . Then there is some N model of T suchthat M ≼ 1 N.Proof. Denote by ∆ 1 (M) the ∀-theory of M M , i.e., the set of all ∀-sentences**in** L(M) true **in** M M . We first show that ∆ 1 (M) ∪ T is consistent. Bycompactness it suffices **to** show ∀xψ(x, a) ∪ T is consistent, where M M |=∀xψ(x, a) and ψ(x, y) is quantifier-free. Otherwise T |= ¬∀xψ(x, a) and thusT |= ∀y∃x¬ψ(x, y). This implies ∀y∃x¬ψ(x, y) ∈ T ∀∃ and thus ∀y∃x¬ψ(x, y) istrue **in** M, a contradiction **to** M M |= ∀xψ(x, a). This ends the proof of theclaim.Let M be a models of T ∀∃ , we must show M is also a model of T . We willbuild a cha**in** of structures (M n | n ∈ ω) start**in**g with M 0 = M. By the claim,there is M 1 ∈ Mod(T ) extend**in**g M 0 with M 0 ≼ 1 M 1 . By Lemma 1.5.12 thereis some extension M 2 of M 1 with M 0 ≼ M 2 . As M 2 is also a model of T ∀∃by the claim aga**in** we obta**in** M 3 ∈ Mod(T ) extend**in**g M 2 with M 2 ≼ 1 M 3 .Aga**in**, by Lemma 1.5.12 there is some extension M 4 of M 3 with M 2 ≼ M 4 .Repeat**in**g this construction ω times obta**in** a cha**in** (M n | n ∈ ω) **in** such a waythat M 2i+1 ∈ Mod(T ) and M 0 ≼ M 2 ≼ M 4 ≼ · · · M 2i ≼ M 2i+2 · · · . As T isclosed under unions of cha**in**s ∪ i∈ω M 2i+1 is a model of T . As (M 2i | n ∈ ω) isan elementary cha**in** M ≼ ∪ i∈ω M 2i = ∪ i∈ω M 2i+1 thus M is a model of T .Example 1.5.15. Very often, nice mathematica theories are axiomatized with∀∃-axioms: groups, r**in**gs, fields, algebraically closed fields, DLO. . .Corollary 1.5.16. A sentence is preserved under unions of cha**in**sequivalent **to** a ∀∃-sentence.iff it isProof. The same proof as corollary 1.5.11 us**in**g theorem 1.5.14 and the factthat a conjunction of ∀∃-formulas is equivalent **to** a ∀∃-formula.Exercise 1.5.17. In this exercise we will prove another preservation theorem: acharacterization of theories preserved under extension. By this we mean thatany extension of any model of the theory is also a model of the theory. As usual,T ∃ denotes the set of all existential sentences which are logical consequences ofT . For an structure M, Th ∀ (M) denotes all the universal sentences true **in** M.1. Prove that any existential sentence is preserved under extensions.2. If M is a model of T ∃ then Th ∀ (M) ∪ T is consistent.

CHAPTER 1. PRELIMINARIES 213. If N is a model of Th ∀ (M) then there is some N ′ extend**in**g N such thatM ≡ N ′ .4. Deduce from 2. and 3. that Mod(T ∃ ) is the class of all structures elementarilyequivalent **to** some extension of a model of T .5. Deduce from 1. and 4. that T is preserved under extensions iff T isaxiomatized by a set of existential sentences.6. Deduce from 5. that a sentence is preserved under extensions iff it isequivalent **to** an existential sentences.Exercise 1.5.18. Prove the follow**in**g:1. Mod(T ∀ ) = {M | M ⊆ N for some M ∈ Mod(T )} == {M | f : M → N for some M ∈ mod (T )}2. Mod(T ∀∃ ) = {M | M ≼ 1 N for some M ∈ Mod(T )} == {M | f : M → 1 N for some M ∈ Mod(T )}Exercise 1.5.19. Prove the converse of exercise 1.4.7. Use diagrams and exercise1.2.9.Exercise 1.5.20. Let M be a given L-structure and T be a theory **in** the languageL. Show that if for any f**in**itely generated substructure M ′ of M there is anembedd**in**g of M ′ **in****to** a model of T then there is an embedd**in**g of M **in****to** amodel of T . H**in**t: use diagrams and compactness.Exercise 1.5.21. **An** Ordered Abelian Group is an Abelian Group **to**gether al**in**ear order ≤ compatible with the group structure, i.e., satisfy**in**g ∀x∀y∀z(x ≤y → x + y ≤ y + z). We assume the natural language for these structures:L = {+, −, 0, ≤}.1. Show that an Ordered Abelian Group is **to**rsion free.2. Show that an Abelian Group can be ordered (**in** a compatible way) ifany f**in**itely generated subgroup can be ordered. H**in**t: Use that OrderedAbelian Group have universal axioms and exercise 1.5.20.3. Prove that any **to**rsion-free Abelian group can be turned **in****to** an OrderedAbelian Group . H**in**t: use that any f**in**itely generatedAbelian group is af**in**ite product (the direct sum of f**in**itely many) cyclic groups.Exercise 1.5.22. Let K be a class of structures **in** a given language closed underisomorphism. Show that:1. K is axiomatizable by universal sentences (i.e., K = Mod(Σ) for some setΣ of universal sentences) iff K is closed under substructures and ultraproducts.2. K is axiomatizable by existential sentences (i.e., K = Mod(Σ) for someset Σ of existential sentences) iff K is closed under ultraproducts and thecomplement is closed under substructures and ultrapowers.

Chapter 2Types and Saturation2.1 TypesHere we will consider aga**in** expansions of the language by add**in**g new constants.Let M be a structure and A be a subset of the doma**in** of M. We will denote byL(A) the set of formulas of expansion of L consist**in**g **in** add**in**g a new constantfor each element of A. There is no harm on tak**in**g the same elements of A asnew constants. Then M expands **in** a natural way **to** an L(A)-structure denotedby M A . Such an A is usually called a set of parameters. The formulas of L(A)are called formulas with parameters **in** A.Until now we have considered languages with a countable amount of variables.Here we will broaden the language by add**in**g an arbitrary amount of newvariables. This set of variables will be considered as a sequence ⟨x i | i ∈ I⟩**in**dexed **in** an a given set I, very often a card**in**al number. This sequence⟨x i | i ∈ I⟩ will be denoted by x and we will call it a tuple of variables.The card**in**ality of this language **in**clud**in**g the variables x is |L| + |x| **in**steadof |L| + ℵ 0 . Here x denotes the number of variables. Moreover, if A is a se**to**f parameters of a model M, the number of formulas **in** this languages withparameters from A is |L| + |x| + |A|.We will denote by L x (A) the set of formulas with parameters **in** A and freevariables **in** x. If φ ∈ L x (A) we put φ = φ(x) **to** emphasize that the formula φhas its free variables among those of x. Obviously φ uses only a f**in**ite numberof those variables. If we have a set π ⊆ L x (A) obviously the free variables ofeach formula of π are among those of the tuple x, and we will write π = π(x)**to** emphasize it.Def**in**ition 2.1.1. Let π ⊆ L x (A), where A is a set of parameters from M. Arealization of π **in** M is a tuple a = ⟨a i | i ∈ I⟩ of elements of M satisfy**in**g allthe formulas of π, denoted by M A |= π(a). This means that for each formulaφ(x) ∈ π it holds M A |= φ(a). A set π which has a realization **in** M is calledrealized **in** M, otherwise is called omitted **in** M. A set π is called f**in**itelysatisfiable **in** M if each f**in**ite subset of π has a realization **in** M. A type of22

CHAPTER 2. TYPES AND SATURATION 23M over A is a subset of L x (A) f**in**itely satisfiable **in** M A .When we say π is a type over A we assume there is some set of variables xand a set of parameters A **in** some model M and π ⊆ L x (A).When x has length n a type (of M over A) with this free variables is called ann-types (of M over A). When we speak about n-types, a fixed set of n variableis presupposed. In this sense L n (A) is used **to** **in**dicate L x (A) for some tuple xof n variables. As a matter of fact the name of the variables is irrelevant, whatreally counts is the number of variables **in** x.Example 2.1.2. We first provide examples of types without parameters.1. Let now L = {+, 0} the language of groups(one of the possible languages).Let π(x) = {∃y(x = ny) | n > 0}. This is a 1-type over ∅ of (Z, +, 0) nonrealized. It is realized **in** (Q, +, 0).2. Let now L = {+, 0} the language of groups (one of the possible languages).Let π(x) = {nx ≠ 0 | n > 0}. This is a 1-type over ∅ realized **in** (Q, +).It is also a non-realized a 1-type of (µ, ·). Here (µ, ·) denotes the set ofcomplex roots of unity with multiplication,i.e. µ = { e πir | r ∈ Q } .3. In the language of graphs (conta**in****in**g a s**in**gle b**in**ary relation R) let π(x, y)denote the follow**in**g set of formulas:{x ≠ y, ¬R(x, y)}∪ { ¬∃z 1 · · · ∃z n(R(x, z1 ) ∧ R(z 1 , z 2 ) ∧ · · · ∧ R(z n , y) ) | n ≥ 1 }In any non-connected graph π is a realized 2-type over the empty set. Inthe graph with doma**in** Z and R(x, y) is **in**terpreted as |x − y| = 1, π is anon-realized type. In fact π is omitted **in** the graph M iff M is connected.4. In the language of order, let x = (x n | n ∈ N) and let π(x) be the follow**in**gset of formulas:{x 0 > x 1 , x 1 > x 2 , x 2 > x 3 , . . .} .In any **in**f**in**ite l**in**early ordered set π is a type. It is omitted **in** Mis well-ordered.Now we provide examples of types with parameters:iff M1. Let M = (Q, ≤) be the rationals with the usual order. The follow**in**g is a1-type of M over Q:{x > r | r ∈ Q, r < √ } {2 ∪ x < r | r ∈ Q, r > √ }2This type is realized **in** (R, ≤) (an elementary extension of (Q, ≤) as wewill see later) and omitted (not realized) **in** (Q, ≤).If N is an elementary extension of M (M Nbut **in** symbols), any subset Aof M is also a subset of N and we can also speak about types of N over A. Asa matter of fact there is no difference between type of M and types of N over

CHAPTER 2. TYPES AND SATURATION 24A provided A is a subset of M. **An**y realization **in** M of an type over A is alsoa realization **in** N of the same type because the extension is elementary. Henceevery type of M over A is a type of N over A. Conversely if π(x) is a type of Nover A, for each f**in**ite subset π 0 (x) of π, as N A |= ∃x ( ∧ π 0 (x)) and M N wehave that M A |= ∃x ( ∧ π 0 (x)) and therefore π is also f**in**itely satisfiable **in** M.When consider**in**g elementary extensions we may omit the superscript.Proposition 2.1.3. Let A ⊆ M and π ⊆ L x (A). The follow**in**g are equivalent:1. π is a type of M.2. π(x) ∪ Th M A is consistent.3. π(x) ∪ ∆(M) is consistent.4. There is an elementary extension of M realiz**in**g π.Proof. 1 ⇒2. Each f**in**ite subset of π(x) ∪ Th M A is a subset of π 0 (x) ∪ Th M Afor some f**in**ite subset π 0 (x) of π(x), thus satisfiable **in** M. By compactnessπ(x) ∪ Th M A is satisfiable.2 ⇒3. By compactness we have **to** show that π(x) ∪ φ(a, b) is satisfiable,where φ(a, b) ∈ ∆(M) and a denotes the parameters of φ conta**in**ed **in** A and bdenotes the parameters not conta**in**ed **in** A. As b has noth**in**g **in** common withπ, the satisfiability of π(x) ∪ φ(a, b) is equivalent **to** that of π(x) ∪ ∃yφ(a, y).S**in**ce ∃yφ(a, y) ∈ Th M A the consistency of this set is ensured by hypothesis.3 ⇒4. Let (N, ha) a∈M be a model of ∆(M) and let b be a tuple of N such that(N, ha) a∈M |= π(b). If a is an enumeration of M, we can write π(x) = π(x, a).By lemma 1.5.2 and exercise 1.2.9 there is some M ′ ≽ M and h ′ ⊇ h suchthat h ′ is an isomorphism from M ′ **to** N. Now (N, ha) a∈M |= π(b, a) impliesN |= π(b, ha) whence, s**in**ce h ′−1 is an isomorphism M ′ |= π(h ′−1 b, a). Thush ′−1 b is a realization of π(x) **in** M ′ .4 ⇒1. π is a type of some elementary extension of M hence a type of M.Def**in**ition 2.1.4. A type π(x) of M over A is said **to** be complete if for eachformula φ ∈ L x (A) we have that either φ ∈ π or ¬φ ∈ π.Facts 2.1.5. Let p(x) be a complete type of M over A. Then1. If π ⊆ L x (A) and π ∪ p is consistent, then π ⊆ p.2. Th(M A ) ⊆ p.3. p is closed under consequence. That is, if φ ∈ L x (A) and p |= φ thenφ ∈ p.4. For all formulas φ 1 , . . . , φ n ∈ L x (A), φ 1 ∧ · · · ∧ φ n ∈ p iff each φ i ∈ p.In particular p is closed under conjunction.5. For all formulas φ 1 , . . . , φ n ∈ L x (A), φ 1 ∨· · ·∨φ n ∈ p iff φ i ∈ p for somei = 1 . . . n.

CHAPTER 2. TYPES AND SATURATION 256. If φ(y, z) ∈ p then ∃zφ(y, z) ∈ p. Here y and z are assumed **to** be subtuplesof x.Proof. 1. Otherwise there is some formula φ ∈ π \ p. Then ¬φ ∈ p, a contradiction**to** the consistency of π ∪ p.2. By proposition 2.1.3 and 1.3. If p |= φ, then p ∪ φ is consistent. By 1. φ ∈ p.4. It follows by 2. and {φ 1 , . . . , φ n } |= φ 1 ∧ · · · ∧ φ n and φ 1 ∧ · · · ∧ φ n |= φ i .5. If for some i φ i ∈ p then p |= φ 1 ∨ · · · ∨ φ n . Conversely, if φ i /∈ p fori = 1 . . . n then ¬φ i ∈ p for i = 1 . . . n and thus p |= ¬(φ 1 ∨ · · · ∨ φ n ). Hence¬(φ 1 ∨ · · · ∨ φ n ) ∈ p and thus φ 1 ∨ · · · ∨ φ n /∈ p.6. Let a be a realization of p(x) is some elementary extension N of M.Then N |= p(a) **in** particular N |= φ(b, c) and N |= ∃zφ(b, z), where b, c are thesubtuples of a correspond**in**g **to** the subtuples y and z of x respectively. Thisimplies p ∪ ∃zφ(y, z) is consistent, as a realizes all these formulas **in** N. By 2.it follows ∃zφ(y, z) ∈ p.Def**in**ition 2.1.6. Let a is a tuple of elements of M and A a subset of M. Wefix a tuple x of variables of the same length as a. The type of a over A **in**M, denoted by tp M (a/A), is the follow**in**g set of formulas:{φ(x) ∈ L x (A) | M A |= φ(a)} .Observe that tp M (a/A) is a complete type (of M over A) with free variables x.Of course, although not explicit **in** the notation tp M (a/A), this depends on thechoice of x, but is ‘unique up **to** replacement of variables’. Moreover, if M ≼ Nand a ∈ M then tp M (a/A) = tp N (a/A). Hence when play**in**g with elementaryextensions, we can omit the superscript.Facts 2.1.7. 1. **An**y type can be completed: given a type π ⊆ L x (A) there issome complete type p (of M over A) with π ⊆ p.2. If p(x) is a complete type of M over A and a is a tuple is M, then:a realizes p iff p = tp M (a/A).Proof. 1. Let, by proposition 2.1.3, a be a realization of π **in** an elementaryextension N of M. Then tp N (a/A) is a complete type of N, hence of M,extend**in**g π.The set of all complete types of M over A with free variables x is denotedby S M x (A). Observe that if N is an elementary extension of M then SM x (A) =S N x (A). Moreover, ∣ ∣ S M x (A) ∣ ∣ ≤ 2 |L|+|A|+|x|+ℵ0 .This is an immediate consequence of the follow**in**g **in**equality:|L x (A)| ≤ |L| + |A| + |x| + ℵ 0 .

CHAPTER 2. TYPES AND SATURATION 26If x and y are tuples of the same card**in**ality, a bijection from x **to** y extends**to** a bijection between L x (A) and L y (A) and thus also **to** Sx M (A) and SM y (A).These spaces of complete types will be **to**pologized later. This bijection turnsout **to** be a homeomorphism between the spaces Sx M (A) and SM y (A). This showsthat what really matters is the number of free variables. When x has n variablesand we do not want **to** put any emphasis **in** the name of the variables it is usuallydenoted by Sn M (A). One could also use the notation Sκ M (A) **to** denote the se**to**f complete types over A with κ free variables for any **in**f**in**ite card**in**al κ.Obviously always|A| ≤ ∣ ∣Sn M (A) ∣ ,s**in**ce the types tp M (a/A) for a ∈ A provides a collection of |A| types: fora, b ∈ A, with a ≠ b, tp M (a/A) ≠ tp M (b/A) s**in**ce the first one conta**in**s theformula x = a while the second conta**in**s x ≠ a.Usually |A| ≥ |L| + ℵ 0 . In this case the above **in**equalities becomes morecompact:|A| ≤ ∣ ∣S M n (A) ∣ ∣ ≤ 2 |A| .The follow**in**g examples show that these extreme values occur **in** practice.Example 2.1.8. 1. Let T the theory of an **in**f**in**ite set. Let M be **in**f**in**ite andA ⊆ M also **in**f**in**ite. Let us see that ∣ ∣Sn M (A) ∣ = |A|. Apart from thetypes tp M (a/A) for a ∈ M there is only one extra type: tp N (b/A) forsome elementary extension N of M and b ∈ N \A. S**in**ce any permutationof N is an au**to**morphism, all elements of N \ A have the same type overA: give a, b ∈ N \ A there is a permutation fix**in**g A po**in**twise and send**in**ga **to** b this implies that a and b must have the same type over A.2. Let T RAND denote the theory of the random graph. This theory is **in**the language of a s**in**gle b**in**ary relation R and express**in**g that (M, R M ) isa graph (i.e.,R is irreflexive and symmetric) with the follow**in**g property:given two f**in**ite and disjo**in**t subsets A, B of M, there is c ∈ M such that(c, a) ∈ R M for each a ∈ A and (c, b) /∈ R M for each b ∈ B. Let us seethat (V ω , ∈) is a model of T RAND , where V ω denotes the ω step **in** theconstruction of the Von-Neumann Universe and ∈ denotes the simetrizedof the membership relation: a ∈ b iff a ∈ b or b ∈ a. Given f**in**ite disjo**in**tsubset X, Y of V ω there are **in**f**in**itely many f**in**ite subset Z of V ω withX ⊆ Z and Z ∩ Y = ∅. Choos**in**g one of those Z such that Z /∈ ∪Y(which exists because ∪Y is f**in**ite) we are done, s**in**ce then x ∈ Z for eachx ∈ X and y /∈ Z, Z /∈ y for each y ∈ Y . S**in**ce Z is f**in**ite, Z ∈ V ω . Let’scheck now that given any model M of T RAND and any **in**f**in**ite subsetA ⊆ M. ∣ ∣S1M (A) ∣ = 2 |A| . Given any subset B ⊆ A, consider the partialtype π B (x) = {xRb | b ∈ B} ∪ {¬xRa | a ∈ A \ B}. This is a type of Mbecause M is a model of T RAND . Let p B (x) ∈ S1M (A) be a completion ofπ B (x) for each B. If B, B ′ are two different subsets of A then some formulax = a belongs **to** one of them and the negation ¬x = a belongs **to** the other(choose a is the symmetric difference). This implies p B ≠ p B ′ and thuswe have an **in**jection from P (A) **to** S1 M (A) show**in**g that ∣ ∣S1 M (A) ∣ = 2 |A| .

CHAPTER 2. TYPES AND SATURATION 27Exercise 2.1.9. Let M be any model. Let m = (m i | i ∈ I) be an **in**jectiveenumeration of M (that is, the map from I **to** M is bijective). Let p(x) =tp M (m/∅), where x = (x i | i ∈ I) is a tuple of variable of the same length asm. Show that a tuple a is a realization of p(x) **in** N iff the map M → N givenby m i ↦→ a i is an elementary embedd**in**g. In general enumerations need not be**in**jective, surjectivity is enough. Show that all **work**s also **in** this case.Exercise 2.1.10. Show that a type of M over A is complete iff it is a maximal(by **in**clusion) **in** the set of all types of M over A with the same free variables.More precisely p ∈ S M x (A) iff it is maximal among subset of L x(A) consistentwith Th(M A ).Exercise 2.1.11. Let M be a structure.1. Let A ⊆ M be a set of parameters and let a = (a i | i ∈ I) be an enumerationof A. Let y = (y i | i ∈ I) be a set of variables of the same length asa. Let q(y) = tp M (a/∅). Given p(x, a) ∈ Sx M (A) let p(x, y) the result ofreplac**in**g each parameter a i by y i **in** all formulas.(a) Show that the map Sx M(A) → SM x,y given by this replacement p(x, a) ↦→p(x, y) is a bijection from Sx M (A) on**to** [q(y)] = { p ∈ Sx,y M | q ⊆ p} .(b) Let b denote another realization of q(x) **in** some structure N, anddenote B = {b i | i ∈ I}. Show that the map Sx M(A) → SN x (B) givenby p(x, a) ↦→ p(x, b) is a bijection.2. Let Now N be a structure realiz**in**g any q ∈ Sy M . Next Lemma 2.1.12 showthat such N exists. For each q ∈ SyM , let a be a realization of q(y) **in** Nand denote A q = {a i | i ∈ I}. Show that∣∣ = ∑ ∣∣Sx N (A q ) ∣ .∣S M x,yq∈S M yNow we will see that we can realize any set of types of M **in** a suitableelementary extension of M. **An**y type over a set A ⊆ M is also a type overM. Same wise, any type π(y) for a certa**in** subtuple y of x can be consideredan type with variables **in** x. For this reason we can consider all types with thesame tuple of free variables.Lemma 2.1.12. Let M be an L-structure and P a set of types of M over Mwith free variables **in** x. Then there is an elementary extension of M of card**in**alat most |M| + |L| + |P | + |x| + ℵ 0 realiz**in**g all the type of P .Proof. For each type π of P take c π an I-tuple of new and different (betweenthem and from those denot**in**g the elements of M) constants. Denote by π(c π )the result of substitut**in**g the free variable of π for the constants of c π . Weconsider now the set of formulasΣ = ∆(M) ∪ ∪π(c π ).π∈P

CHAPTER 2. TYPES AND SATURATION 28As the types are f**in**itely satisfiable **in** M M , every f**in**ite part of Σ is satisfied**in** a suitable expansion of M M . By compactness and downwards Löwenheim-Skolem, Σ has a model of card**in**al at most |M| + |L| + |P | + |x| + ℵ 0 . Thismodel will be of the form (N, ha, c π N ) a∈M, π∈P where h : M → N denotes the**in**terpretation of the constants of M. By proposition 1.5.2 h is an elementaryembedd**in**g from M **in****to** N. Us**in**g exercise 1.2.9 there exists M ′ M and anisomorphism h ′ : M ′ → N that extends h. We f**in**ally verify, for every π ∈ P ,that h ′−1 (c π N ) realizes π **in** M ′ . If φ(x, a) ∈ π then, as (N, ha, c π N ) a∈M, π∈Pis a model of Σ, (N, ha) a∈M |= φ(c π N , a), that is, N |= φ(c π N , ha). As h ′ is anisomorphism extend**in**g h we have M ′ |= φ(h ′−1 (c π N ), a).2.2 SaturationDef**in**ition 2.2.1. Let M a L-structure and κ an **in**f**in**ite card**in**al. The structureM is said **to** be κ-saturated if for each subset A ⊆ M of card**in**al less than κ,M realizes any 1-type of M over A.S**in**ce one can extend every type **to** a complete type, it is enough **to** checkthe realization of complete 1-types.Facts 2.2.2. 1. Every f**in**ite structure is κ-saturated for any card**in**al κ.2. If M is **in**f**in**ite and κ-saturated then |M| ≥ κ.3. If M is κ-saturated and N is the restriction of M **to** a smaller languagethen N is κ-saturated.4. If M is κ-saturated and A ⊆ M with |A| < κ then M A is κ-saturated.Proof. 1. S**in**ce any type of M is realized **in** an elementary extension it is realized**in** the same M.2. Otherwise the type {x ≠ a | a ∈ M} is over a set of less than κ parametersand it is not realized **in** M.3. Obvious.4. **An**y type of M A over B is a type of M over AB.By facts 3. and 4. above, when |A| < κ M is κ-saturated iff M A is κ-saturated.By fact 2 above the maximum degree of saturation that an **in**f**in**ite structureM can have is |M|-saturation. This makes a sense **in** the follow**in**g def**in**ition:Def**in**ition 2.2.3. A model M is called saturated if it is |M|-saturated.Theorem 2.2.4. Let M be a structure and κ be an **in**f**in**ite card**in**al. Then thereis a κ + -saturated elementary extension of M of card**in**al at most 2 |L| |M| κ .Proof. We may assume the structure M **in**f**in**ite. Then M has |M| κ subsets Aof ∣ card**in**al at most κ. Moreover, for each one of these subsets A we have that(A) ∣ ≤ 2 |L|+κ . By lemma 2.1.12 there is an elementary extension M 1 of M∣S M 1

CHAPTER 2. TYPES AND SATURATION 29of card**in**al at most 2 |L| |M| κ realiz**in**g all 1-types of M over a set of at mostκ parameters from M. Apply**in**g aga**in** lemma 2.1.12 **to** M 1 we obta**in** M 2 anelementary extension of M 1 of card**in**ality at most 2 |L| ( 2 |L| |M| κ) κ= 2 |L| |M| κthat realizes all 1-types of M 1 over a set of at most κ parameters from M 1 .Iterat**in**g this procedure we construct a cont**in**uous cha**in** (M α | α < κ + ) withM 0 = M and where M α+1 is an elementary extension of M α of card**in**al at most2 |L| |M| κ that realizes all the types of M α over a set of at most κ parametersfrom M α . Us**in**g exercise 1.4.5 the cha**in** is elementary and, by the lemma of thecha**in** 1.4.4, M is an elementary substructure of the union of the cha**in** N. Byregularity of κ + N is κ + -saturated and its card**in**ality is at most 2 |L| |M| κ .Usually κ ≥ |L|. Then the card**in**al of the extension provided by Theorem2.2.4 is limited by |M| κ . If moreover κ ≥ |M| the upper bound of the card**in**alof the extension becomes 2 κ . Hence if 2 κ = κ + any theory **in** a language with|L| ≤ κ has saturated models of size κ + .remark 2.2.5. If 2 κ = κ + ≥ |L| then any theory has a saturated model ofcard**in**ality κ + .However if 2 κ > κ + ≥ |L| **in** some theories there are not saturated models ofcard**in**ality κ + . For **in**stance, **in** T RAND . If M is a model of T RAND of card**in**alityκ + , pick a subset A of M of card**in**ality κ. Observe the two different completetypes cannot have a common realization: if a ∈ M realizes p and p ∈ ∣ ∣S1M (A) ∣ then p = tp M (a/A). S**in**ce ∣ ∣S1M (A) ∣ = 2 κ and |M| < 2 κ , M cannot realize alltypes **in** S1M (A).Exercise 2.2.6. Prove that every structure M has an ω-saturated elementaryextension of card**in**ality at most 2 |L|+ℵ 0|M|.Exercise 2.2.7. Assume that κ > |L| + ℵ 0 and κ µ = κ for each µ < κ (**in**particular if κ is an strongly **in**accessible card**in**al). Prove that every structureM with |M| ≤ κ has a saturated elementary extension of card**in**ality κ.2.3 Partial elementary mapsIn this section we will consider partial maps from M **to** N. A partial mapf : M → N is an map that goes from a subset of M **to** N. Observe that an**in**jective partial map from M **to** N is a bijection between subset of M and Nrespectively.Def**in**ition 2.3.1. A partial map f : M → N is elementary if for every f**in**itetuple a of the doma**in** of f and every φ(x) ∈ L we haveM |= φ(a) iff N |= φ(fa).Equivalently, for every (f**in**ite) tuple a of the doma**in** of f we have that a and fahave the same type over the empty set (**in** symbols: tp M (a) = tp N (fa)). If a isan enumeration of the doma**in** of f, it is also equivalent **to** tp M (a) = tp N (fa).

CHAPTER 2. TYPES AND SATURATION 30Observe that every partial elementary map is **in**jective and its **in**verse is alsoelementary. If there is a partial elementary map between M and N then M andN have **to** be elementarily equivalent. In this sense it is useful **to** th**in**k thatwhen M and N elementarily equivalent the empty map (between M and N) iselementary.The follow**in**g is easy but important: it will be used several times later.Example 2.3.2. 1. **An**y elementary embedd**in**g h : M → N is a partial elementarymap.2. **An**y subset of a partial elementary map is also a partial elementary map.3. Let M = (Z, s), where s denotes the successor function. The follow**in**gpartial maps {(n, m)}, {(n 1 , n 1 + m), . . . , (n k , n k + m)} are elementarybecause they are subsets of an au**to**morphism of M. But {(1, 8), (3, 5)} isnot a partial elementary map: M |= 3 = s(s(1)) but M ̸|= 5 = s(s(8)). Inother words, y = ss(x) ∈ p(x, y) = tp M ((1, 3)) while y = ss(x) /∈ p(x, y) =tp M ((8, 5)). In this example, the partial elementary maps are the partialmaps that preserve the ‘distance’(the distance between n and m is numberof times you must apply s **to** n **to** obta**in** m, possibly a negative number).Exercise 2.3.3. Let A ⊆ M, let a, b be tuples (of the same length) of M and letc be an enumeration of A. Show that the follow**in**g are equivalent:1. tp M (a/A) = tp M (b/A).2. tp M (ac/∅) = tp M (bc/∅).3. id A ∪ (a, b) is an elementary map.Exercise 2.3.4. Prove that (N, ha) a∈A is a model of T h(M A ) iff the partial map(with doma**in** A) h : M → N is elementary. Hence, a model of T h(M A ) consis**to**n a structure N **to**gether with an elementary partial map h : M → N withdoma**in** A.Def**in**ition 2.3.5. Suppose that f is a partial elementary map from M **to** Nand π is a type of M over dom f. The conjugate of π by f, denoted by π f , isdef**in**ed as{φ(x, fa) | φ(x, a) ∈ π} .Obviously, if π is over A ⊆ dom(f) then π f is over f(A) ⊆ Im(f).Lemma 2.3.6. Let f a partial elementary map from M **to** N.1. If π is a type of M over dom f then π f is a type of N over Im f.2. If a is a tuple of dom f and A ⊆ dom f then fa realizes tp f (a/A). Equivalentlytp(fa/f(A)) = tp f (a/A).3. Let a, b be tuples of the same length of M and N respectively. If b realizestp f (a/ dom f) then f ∪ { (a, b) } is elementary.

CHAPTER 2. TYPES AND SATURATION 31Proof. 1. Observe that π f is a set of formulas with parameters **in** Im f with thesame variables as π. A f**in**ite subset of π f is of the form{φ 1 (x, fa), . . . , φ n (x, fa)} ,where φ i (x, a) ∈ π for each i. As π is a type of M, M |= ∃x ∧ i=1...n φ i(x, a).S**in**ce f is elementary N |= ∃x ∧ i=1...n φ i(x, fa) and therefore π f is f**in**itelysatisfiable **in** N.2. A formula **in** tp f (a/A) is of the form φ(x, fb) for some φ(x, b) ∈ tp(a/A),where φ(x, y) an L-formula and b is a tuple **in** dom(f). If φ(x, b) ∈ tp(a/A)then M |= φ(a, b) and, as f is elementary, N |= φ(fa, fb).3. We have **to** show that for every tuple c of the doma**in** of f and everyformula φ(x, y) ∈ L, M |= φ(a, c) iff N |= φ(b, fc). In fact one of thesetwo implications suffice. If M |= φ(a, c) then φ(x, c) ∈ tp(a/ dom f) and thusφ(x, fc) ∈ tp f (a/ dom f). S**in**ce b realizes tp f (a/ dom f) we get N |= φ(b, fc).Observe that **in** po**in**t 3. of Lemma 2.3.6, by po**in**t 2., the converse alsoholds. Obviously the conjugate by an elementary map of a complete type isalso a complete type. Therefore, if A ⊆ dom(f), conjugation by f is a bijectionbetween Sx M(A) and SN x (f(A)). In fact this bijection will be a homeomorphismof **to**pological spaces.Exercise 2.3.7. Here we will see that partial elementary maps can always beamalgamated is the follow**in**g sense. Assume that for each i ∈ I, f i : M 0 → M iare partial elementary maps. Show that there is N and partial elementary mapsg i : M i → N such that g i ◦ f i makes sense and g i ◦ f i = g j ◦ f j . Let a be anenumeration of ∩ i dom(f i) and x a tuple of variable of the same length. Letp(x) denote tp M 0(a). For each i let b i be an enumeration of dom(f i )\ ∩ i dom(f i)and y i a tuple of variables of the same length. Let q i (x, y i ) denote tp M 0(a, b i ).Show that ∪ i q i(x, y i ) is consistent. Here we assume all variables y i are dist**in**ct.Show that, if a c, (c i | i ∈ I) is realization of ∪ i q i(x, y i ) **in** N the maps f i whichsend g i (a), g i (b i ) **to** c, c i provide a solution.2.4 More about saturationThe follow**in**g shows that a complete theory has at most one saturated model**in** each card**in**ality.Theorem 2.4.1. Two saturated models of the same card**in**ality and elementarilyequivalent are isomorphic.Proof. Suppose that M, N are elementarily equivalent and saturated models,both of card**in**al λ. Let (a i | i ∈ λ) and (b i | i ∈ λ) be enumerations of Mand N respectively. We will construct a cha**in** (f i | i ∈ λ) of partial elementarymaps with a i ∈ dom f i+1 and b i ∈ Im f i+1 and |f i | < κ. The union of the cha**in**will be the searched isomorphism. We start with f 0 = ∅ the empty map and

CHAPTER 2. TYPES AND SATURATION 32**in** the limits we take unions. In order **to** construct f i+1 we take, because ofsaturation, a realization c of tp f i(a i / dom f i ). By lemma 2.3.6 f ′ = f ∪ {(a i , c)}is elementary. Now we take a realization d of tp f ′−1 (b i / Im(f ′ )). By 2.3.6 aga**in**,f i+1 = f i ∪ {(a i , c), (d, b i )} is elementary. Obviously |f i+1 | ≤ |f i | + 2 < κ.This proof **in** fact shows that if M and N are saturated and of the samecard**in**ality, every partial elementary map of card**in**al smaller than |M| = |N|from M **to** N extends **to** an isomorphism between M and N. In particular:Corollary 2.4.2. If M is saturated then every partial elementary map from M**to** M of card**in**al less than |M| extends **to** an au**to**morphism of M.The set of au**to**morphisms of M which are the identity on A (A a subset ofM) is usually denoted by Aut(M/A) and is a group by composition. There is anatural action of Aut(M/A) on M (or on M I for any I) by putt**in**g g ·a := g(a).Assume now that |A| < |M| and let a, b be tuples **in** M of (the same) length< |M|. By the homomorphism theorem 1.2.4, if a and b lie **in** the same orbit ofthe action of Aut(M/A) obviously tp(a/A) = tp(b/A). Conversely, if a, b havethe same type over A, then id A ∪ (a, b) is an elementary map of size < |M|. Bycorollary 2.4.2 there is an au**to**morphism f of M extend**in**g id A ∪ (a, b). Hencea and b lie **in** the same orbit of the action of Aut(M/A). Thus for saturatedmodels M and small A, A-orbits correspond **to** A-types. This holds **in** fact forany strongly homogeneous M (by def**in**ition: each elementary map of size< |M| extends **to** an au**to**morphism of M). Corollary 2.4.2 says that saturatedmodels are strongly homogeneous.Def**in**ition 2.4.3. A structure M is κ-homogeneous if given a partial elementarymap f : M → M of card**in**al < κ and an element a of M there is anextension of f **to** a partial elementary map from M **to** M with a **in** the doma**in**.This is equivalent **to** the existence of an element b **in** M such that f ∪ {(a, b)}is elementary.The follow**in**g exercise is a characterization of κ-saturation **in** terms of extensionsof elementary maps.Exercise 2.4.4. 1. Prove that M is κ-saturated iff given a partial elementarymap f : N → M of card**in**al < κ and an element a of N there is anextension of f **to** a partial elementary map from N **to** M with a **in** thedoma**in**. Equivalently, given a partial elementary map f : N → M ofcard**in**al < κ and an element a ∈ N there exists an element b **in** M suchthat f ∪ {(a, b)} is elementary from N **to** M.2. Conclude that every κ-saturated model is κ-homogeneous.The follow**in**g shows that we can improve the extensibility of elementarymaps **in** saturated models:Proposition 2.4.5. Let M be κ-saturated and let f : N → M be a partialelementary map with |f| < κ. Let B be a subset of N of card**in**ality at mostκ. Then There is a partial elementary map g : N → M, extend**in**g f withB ⊆ dom(g).

CHAPTER 2. TYPES AND SATURATION 33Proof. Let (b i | i ∈ κ) be an enumeration of B. We construct a cha**in** (f i | i ∈ λ)of partial elementary maps with b i ∈ dom f i+1 and |f i | < κ. The union ofthe cha**in** will be the desired elementary map. We start with f 0 = f and**in** the limits we take unions. In order **to** construct f i+1 we take, because ofsaturation, a realization c of tp f i(b i / dom f i ). By lemma 2.3.6 f i+1 = f∪{(a i , c)}is elementary. Obviously |f i+1 | ≤ |f i | + 1 < κ.Corollary 2.4.6. Let M, N be elementarily equivalent structures, where Nis κ-saturated and M is of card**in**al at most κ. Then there is an elementaryembedd**in**g from M **to** N.Proof. Apply proposition 2.4.5 **to** the empty map with B = N.It is usual **to** say that M is κ-universal when for every N of card**in**al lessthan κ and elementarily equivalent **to** M there is elementary embedd**in**g of N**in****to** M. With this term**in**ology, proposition 2.4.7 can be stated as follows: everyκ-saturated model is κ + -universal.The next result shows that we could have replaced 1-types by κ-types **in** thedef**in**ition of κ-saturation.Proposition 2.4.7. Let M be κ-saturated. Then M realizes every type of Mover any sets of parameters of card**in**al less than κ with at most κ variables.Proof. We assume the type is complete. Let p ∈ Sx M (A), where |A| < κ and|x| ≤ κ. Let b ∈ N be a realization of p(x) **in** an elementary extension N ofM. Apply**in**g proposition 2.4.5 **to** the elementary map Id A : N → M and B ={b i | i ∈ I} we get some elementary g with Id A ⊆ g and {b i | i ∈ I} ⊆ dom(g).Now, by lemma 2.3.6, gb realizes p g = p **in** M.Exercise 2.4.8. 1. Assume M and N are κ-saturated. Let A ⊆ M and B ⊆ Nsuch that |A| + |B| ≤ κ and f : M → N a partial elementary map with|f| < κ. Show that there is a partial elementary map g : M → N extend**in**gf with A ⊆ dom(f) and B ⊆ Im(g). H**in**t: pick enumerations (a α | α ∈ κ)and (b α | α ∈ κ) of (a α | α ∈ κ) A and B respectively and build a cha**in**of extensions (f α | α ∈ κ) of f with a α ∈ dom(f α+1 ) and b α ∈ Im(f α+1 )and proceed as **in** the proof of Theorem 2.4.1.2. obta**in** Theorem 2.4.1 from 1.3. obta**in** Corollary 2.4.2 from 1.Exercise 2.4.9. M is said **to** be strongly κ-homogeneous if any partial elementarymap f : M → M with |f| < κ can be extended **to** an au**to**morphism ofM.1. Assume M is κ-homogeneous. Let A, B ⊆ M such that |A| + |B| ≤ κand f : M → M a partial elementary map with |f| < κ. Show that thereis a partial elementary map g : M → M extend**in**g f with A ⊆ dom(f)and B ⊆ Im(g). H**in**t: pick enumerations (a α | α ∈ κ) and (b α | α ∈ κ)of (a α | α ∈ κ) A and B respectively and build a cha**in** of extensions

CHAPTER 2. TYPES AND SATURATION 34(f α | α ∈ κ) of f with a α ∈ dom(f α+1 ) and b α ∈ Im(f α+1 ) and proceed as**in** the proof of Theorem 2.4.1.2. Show that strongly κ-homogeneous implies κ-homogeneous. The converseis not true **in** general. However:3. Show that |M|-homogeneous implies |M|-strongly homogeneous. H**in**t:use 1.Exercise 2.4.10. 1. Let a = (a i | i ∈ I) be an enumeration of M. Letπ(x) = atp M (a) be the a**to**mic type of a, where x = (x i | i ∈ I) is a tupleof variables. Show that a tuple b = (b i | i ∈ I) **in** a structure N realizesπ(x) iff the map from M **to** N given by a i → b i is an embedd**in**g.2. Let N be an |M|-saturated L-structure. Show that if for each f**in**itelygenerated substructure M ′ of M there is an embedd**in**g of M ′ **in****to** N thenthere is an embedd**in**g of M **in****to** N.We know that a κ-saturated models is κ-homogeneous and κ + -universal. Inthe next theorem we will see that the converse also holds **in** a strong form: It isenough (|L| + ℵ 0 ) + -universality plus κ-homogeneity.Theorem 2.4.11. Let M be a model and κ ≥ |L| + ℵ 0 .equivalent:The follow**in**g are1. M is κ-saturated.2. M is κ-homogeneous and (|L| + ℵ 0 ) + -universal.3. M is κ-homogeneous and realizes all types **in** S M n (∅) for each f**in**ite n.Proof. 1. implies 2. is **in** exercise 2.4.4 and corollary 2.4.6.2. implies 3. Let p ∈ S M n (∅). Let, by Löwenheim-Skolem, M ′ ≼ M with|M ′ | ≤ |L| + ℵ 0 and let a be a realization of p(x) **in** an elementary extension Nof M ′ with |N| ≤ |L| + ℵ 0 . By (|L| + ℵ 0 ) + -universality there is an elementaryembedd**in**g f : N → M. Now, by Lemma 2.3.6, fa realizes p f (x) = p(x) **in** M.3. implies 1. Assume M is κ-homogeneous and realizes all types (of M)**in** f**in**itely many variables (over the empty set). We first show that M realizesall types **in** at most κ variables over empty. This will be done by **in**duction onλ, the number of variables. Let p(x) ∈ S M λ (∅), where x = (x α | α ∈ λ) andλ ≤ κ. We will build, by **in**duction on α, a sequence (a α | α ∈ λ) of elementsof M **in** such a way that a ≤α = (a β | β ≤ α) realizes the restriction p ≤α of p **to**(x β | β ≤ α). This will suffice. Assume we have a

CHAPTER 2. TYPES AND SATURATION 35Let A be a subset of M of size less than κ and let p(x) ∈ S1M (A). Let b be arealization of p **in** an elementary extension N of M. Let a = (a i | i ∈ I) be anenumeration of A and let q(y, x) = tp N (a, b). As q is a type (of N hence also ofM) over empty with less than κ variables it has a realization (c, d) **in** M. S**in**cetp N (a, b) = q(y, x) = tp M (c, d) the map (from N **to** M) send**in**g a, b **to** c, d iselementary and thus the map (from N **to** M and also from M **to** M) send**in**g c**to** a is also elementary. By homogeneity there is e ∈ M such that the map (fromM **to** M) send**in**g c, d **to** a, e is elementary. Now the map g : N → M send**in**ga, b **to** a, e is elementary. S**in**ce b realizes p, by 2.3.6, e realizes p g = p.Exercise 2.4.12. Show that 1. and 2. above are equivalent. Show that ifmoreover κ > |L| + ℵ 0 then 1. and 2. are also equivalent **to** 3.1. M realizes all κ-types of M over empty.2. For any N ≡ M and any A ⊆ N with |A| ≤ κ there is a partial elementarymap f : N → M with A ⊆ dom(f).3. M is κ + -universal.Exercise 2.4.13. 1. Let M be κ-homogeneous N ≡ M and assume each f**in**itarytype (with f**in**itely many variables) over empty realized **in** N is alsorealized **in** M.(a) Show that each type **in** at most k variables over empty realized **in**N is also realized **in** M. H**in**t: by **in**duction on λ, the number ofvariables, see the proof of 3. implies 1. **in** Theorem 2.4.11.(b) If |N| ≤ κ, show there is an elementary embedd**in**g from N **to** M.This shows M is k + -universal **in** the class{N | N ≡ M and N omits any f**in**itary type over empty omitted by M} .H**in**t: use (a).(c) Let f : N → M is a partial elementary map with |f| < κ and b ∈ N.Show one can extend f **to** an elementary map g : N → M withb ∈ dom(g). H**in**t: use (a).(d) Let f : N → M is a partial elementary map with |f| < κ andB ⊆ N with |B| ≤ κ. Show one can extend f **to** an elementary mapg : N → M with B ⊆ dom(g). H**in**t: either iterate (c). with a goodenumeration of B or use (a). and exercise 2.4.9.2. Assume M and N are κ-homogeneous elementarily equivalent and realizethe same f**in**itary types over empty.(a) Le f : N → M is a partial elementary map with |f| < κ and B ⊆ N,A ⊆ M with |A| + |B| ≤ κ. Show one can extend f **to** an elementarymap g : N → M with B ⊆ dom(g) and A ⊆ Im(g). H**in**t: iterate1.(c) with good enumerations of A and B.

CHAPTER 2. TYPES AND SATURATION 36(b) Show that if moreover |M| = |N| = κ then M and N are isomorphic.This generalizes Theorem 2.4.1. H**in**t apply 2.(a).Exercise 2.4.14. A theory T is called λ-stable if for any model M of T and anyA ⊆ M with |A| = λ then ∣ SM1 (A) ∣ = λ. Show that if λ ≥ |L| + ℵ0 is regularand T is λ-stable theory with **in**f**in**ite models then T has a saturated model ofcard**in**ality λ. H**in**t: build a cont**in**uous cha**in** of length λ of models of card**in**alityλ each one realiz**in**g all 1-types over the predecessor (if it has). It also true fornon-regular λ, but the proof is much more difficult.Exercise 2.4.15. Suppose that N is λ-saturated (λ + -universal suffice, us**in**g exercise) |M| ≤ λ and each existential sentence hold**in**g **in** M also holds **in** N.Show that there is an embedd**in**g from M **to** N. H**in**t: consider an enumerationa of M and consider atp M (a/∅), the a**to**mic type of a over ∅.Exercise 2.4.16. Let k be an algebraically closed field. Show that1. k is λ-saturated iff the transcendence degree of k over the prime field isat least λ.2. **An**y uncountable algebraically closed field is saturated.

Chapter 3Omitt**in**g Types3.1 A little bit of **to**pologyDef**in**ition 3.1.1. A **to**pological space is a pair (X, τ), where X is a set andτ ⊆ P (X) is a family of subsets of X satisfy**in**g the follow**in**g properties:• For every µ ⊆ τ, ∪µ ∈ τ.• For every f**in**ite µ ⊆ τ, ∩µ ∈ τ.• ∅, X ∈ τThe elements of X are called po**in**ts and the subsets of X sets. The setswhich belong **to** µ will be called open sets and their complements closed sets.The def**in**ition says that the collection of open sets is closed under tak**in**g unions,f**in**ite **in**tersections and conta**in** the empty set and the whole space. Thus thecollection of closed sets is closed under tak**in**g **in**tersections and f**in**ite unions.The empty set and the whole space are also closed. The subsets of X whichare both open and closed are called clopen sets. For **in**stance, ∅ and X areclopen. In order **to** def**in**e a **to**pology **in** X it is quite common **to** provide a nicesubcollection of open sets, **in**stead of provid**in**g the whole collection of all opensets. A base of (X, τ) is a subcollection µ ⊆ τ such that one gets all open setsas unions of sets **in** the base. The elements of µ are called basic open sets. Moreprecisely, each open set is a union of basic open sets i.e., for every V ∈ τ thereis µ ′ ⊆ µ such that V = ∪µ ′ . It is easily checked (and let as an exercise for thereader) that a base µ for a **to**pology **in** X satisfies the follow**in**g two properties:• X = ∪µ.• For every V 1 , V 2 ∈ µ and for every x ∈ V 1 ∩ V 2 there is V 3 ∈ µ such thatx ∈ V 3 ⊆ V 1 ∩ V 2 .Conversely, given a set X and µ ⊆ P (X) satisfy**in**g the two properties abovethe collection τ = {∪µ ′ | µ ′ ⊆ µ} is a **to**pology on X. This is aga**in** left as anexercise.37

CHAPTER 3. OMITTING TYPES 38Def**in**ition 3.1.2. Given a subset A of X, the **in**terior of A, denoted by ◦ A, isthe biggest open set conta**in**ed **in** A:◦A= ∪ {V | V is open and V ⊆ A} .The closure of A is the smallest closed set conta**in****in**g A:A = ∩ {C | C is closed and A ⊆ C} .As tak**in**g complements turns open sets **to** closed sets and reverse the **in**clusions,it is straightforward **to** check that( ◦A ) C= AC .Here we use A C **to** denote X \ A, the complement of A **in** X.A neighborhood of a po**in**t x is any set conta**in****in**g an open set conta**in****in**gx. We let as an exercise **to** show the follow**in**g: x ∈ ◦ A iff there is a neighborhoodof x conta**in**ed **in** A. Also: x ∈ A iff every neighborhood of x **in**tersects (hasnonempty **in**tersection with) A.Def**in**ition 3.1.3. A subset of X is called dense if A = X.Equivalently, a dense set A is a subset of X which has nonempty **in**tersectionwith any nonempty open set. Moreover, A is dense iff A C has empty **in**terior(**in** other words: ∅ is he only open set conta**in**ed **in** the complement of A).Def**in**ition 3.1.4. A **to**pological space (X, τ) is said **to** be Haussdorff iff giventwo different po**in**ts x, y ∈ X there are neighborhoods V x ∋ x and V y ∋ y of xand y respectively such that V x ∩ V y = ∅.Def**in**ition 3.1.5. A subset A of X is called compact (**in** (X, τ)) iff for everyfamily of open sets {V i | i ∈ I} such that A ⊆ ∪ i∈I V i there is a f**in**ite I 0 ⊆ I suchthat A ⊆ ∪ i∈I C i. In words: every open cover**in**g of A has a f**in**ite subcover**in**g.When X is compact we say that the **to**pological space (X, µ) is compact.Tak**in**g complements, one gets the follow**in**g characterization: A is compactiff for every family of closed sets {C i | i ∈ I} such that A ∩ ∩ i∈I C i = ∅ there isa f**in**ite I 0 ⊆ I such that A ∩ ∩ i∈I C i = ∅. Observe that the space X is compactwhenever given any family of closed sets with empty **in**tersection one can f**in**da f**in**ite subfamily with empty **in**tersection. This is usually stated the follow**in**gway: given a collection of closed sets, if any f**in**ite subcollection has nonempty**in**tersection then also the whole family has nonempty **in**tersection.Compact sets behave as f**in**ite sets **in** some sense. For **in**stance, let us as seethat **in** a Haussdorff space we can extend the ‘separation’ of po**in**ts **to** compactsets: given two disjo**in**t compact sets K 1 , K 2 there are disjo**in**t open sets V 1 , V 2with K i ⊆ V i . Let us start with a po**in**t x and a compact set K. For each y ∈ Klet V y , W y be open neighborhoods of x and y respectively with V y ∩ W y = ∅.

CHAPTER 3. OMITTING TYPES 39S**in**ce (W y | y ∈ K) is an open cover**in**g of K, by compactness, there is somef**in**ite F ⊆ K such that (W y | y ∈ F ) also covers K. Now it is easily checked that∩y∈F V y and ∪ y∈F W y are the desired neighborhood of x and K respectively.This implies, **in** particular that K is closed. Repeat**in**g the argument aga**in** wecan separate disjo**in**t compact sets.There is a close relation between ‘compact’ and ‘closed’ subset of a **to**pologicalspace:remark 3.1.6.1. Every closed set of a compact space is compact.2. Every compact subset of a Haussdorff space is closed.Hence, for compact Haussdorff spaces, ‘closed subset’ and ‘compact subset’co**in**cide.Proof. 1. Let C be a closed subset of X. If (V i | i ∈ I) is an open cover**in**g of C,add**in**g C C one gets an open cover**in**g of X. A f**in**ite subcover**in**g of X providesa f**in**ite subcover**in**g of C.2. By the separation argument above.Def**in**ition 3.1.7. A **to**pological space is locally compact if any po**in**t has acompact neighborhood.Obviously any compact space is locally compact: the whole space is a compactneighborhood.remark 3.1.8. In a Haussdorff locally compact space each po**in**t has a base ofcompact neighborhoods: for each po**in**t x and each neighborhood V of x thereis some compact neighborhood K of x conta**in**ed **in** V .Proof. Let K be a compact neighborhood of x an open V neighborhood of x.Let W an open neighborhood of x conta**in**ed **in** K. We will f**in**d a compactneighborhood of x conta**in**ed **in** W ∩ V . By remark 3.1.6 K 1 = K \ (V ∩ W ) iscompact. By separation of x and K 1 let V 1 , V 2 disjo**in**t open sets with x ∈ V 1and K 1 ⊆ V 2 . Now x ∈ V 1 ∩ W ⊆ V2 C ∩ K and thus V2 C ∩ K is a compactneighborhood of x conta**in**ed **in** V ∩ W .Next result, the Baire category theorem also holds for complete metric spaceswith a slight modification on the proof.Theorem 3.1.9 (Baire Category theorem). In a locally compact Haussdorfspace, the **in**tersection of countably many open dense sets is dense.Proof. Let X be locally compact and Haussdorf. Let (V n | n ∈ N) be a collectionof dense open sets. In order **to** show that ∩ n∈N V i is dense we must show thatV ∩ ∩ i∈N V i ≠ ∅ for any nonempty open V . S**in**ce V 1 is dense V 1 ∩ V ≠ ∅ andthus, by remark 3.1.8, we can f**in**d a compact set K 1 with nonempty **in**teriorwith K 1 ⊆ V ∩ V 1 . S**in**ce V 2 is dense and K 1 has nonempty **in**terior, V 2 ∩ K 1 hasalso nonepty **in**terior and thus, by remark 3.1.8 aga**in**, we can f**in**d a compact setK 2 with nonempty **in**terior with K 2 ⊆ K 1 ∩V 1 . Observe that K 2 ⊆ V ∩ V 1 ∩V 2 .Repeat**in**g this procedure we build a decreas**in**g cha**in** of compact sets (K n | n ∈

CHAPTER 3. OMITTING TYPES 40N) with K n ⊆ V ∩ V 1 ∩ · · · ∩ V n . S**in**ce K 1 is compact and each K i is closed,∩Kn∈N is nonempty, hence V ∩ ∩ n∈N V n ≠ ∅.3.2 The spaces of complete types of a theotyIn this section we assume a theory T is given, and x will denote a tuple ofvariables, maybe of **in**f**in**ite length.Def**in**ition 3.2.1. A type of T is a set of formulas π(x) ⊆ L x such that T ∪π(x)is consistent.Hence, a subset of L x is a type of T iff has a realization **in** a model of T .Equivalently, by the compactness theorem, iff every f**in**ite subset of π(x) has arealization **in** a model of T . As **in** section 2.1 we say that a type is completeiff for every formula φ ∈ L x the type conta**in**s exactly one of φ, ¬φ. The set ofall complete types of T with free variables on x is denoted by Sx T .The notion of ‘type of T ’ and ‘type of M’ should not be confused. First,**in** a type of T we do not allow parameters, whereas we allow parameter setsA ⊆ M for the second notion. Moreover, compactness allows us **to** concludethat ‘f**in**itely realizable **in** T ’ (more precisely: **in** a model of T ) is the sameas ‘realizable **in** T ’. Obviously ‘f**in**itely realizable **in** M’ is not the same as‘realizable **in** M’. However both notions are **in**timately related and can be seeneach one as a particular case of the other if T is complete.remark 3.2.2. 1. Let M be a model, A a set of parameters A ⊆ M andπ(x) ⊆ L x (A). Then π(x) is a type of M over A iff it is a type of thetheory Th M A . This is an consequence of 2.1.3. In particular:S M x (A) = S Th(M A)x.2. Let T be a theory. **An**y type π(x) of T is a type of some model of T , s**in**ceπ(x) is realized **in** some model of T . If M is a model of T and π(x) istype of M, π(x) is also s type of T because the consistency of Th(M) ∪ πimplies the consistency of T ∪ π. Hence, π is a type of T iff π is a type ofsome model of T . For complete types, one gets:∪Sx T = Sx M .M∈Mod(T )In the expression above it suffices **to** pick a model for each completion ofT . When T is a complete theory and M a is model of T , then π(x) is atype of M (over ∅) iff it is a type of T . In particular:S T x = S M x .The space S T xis **to**pologized as follows. For every formula φ ∈ L x, consider[φ] = { p ∈ S T x | φ ∈ p } .

CHAPTER 3. OMITTING TYPES 41It is easily checked that [φ] ∩ [ψ] = [φ ∧ ψ], hence {[φ] | φ ∈ L x } forms a baseof a **to**pology. As Sx T \ [φ] = [¬φ], the sets of the base are clopen, hence is a**to**tally disconnected space.As closed sets are **in**tersections of basic clopen sets, any closed set C is ofthe form C = ∩ φ∈π [φ] = { p ∈ Sx T | π ⊆ p} for a certa**in** set π ⊆ L x . Henceclosed sets correspond **to** sets of formulas π. We will denote { p ∈ Sx T | π ⊆ p}by [π]. One also verifies that [π] = ∅ iff T ∪ π is **in**consistent iff π(x) has norealization **in** a model of T . Hence nonempty closed sets correspond **to** types.Proposition 3.2.3. The space S T xis compact.Proof. Suppose we are given a family ([π i ], i ∈ I) of closed sets with empty**in**tersection. S**in**ce ∅ = ∩ i∈I [π i] = [ ∪ i∈I π i] we get that ∪ i∈I π i is **in**consistentwith T . By compactness there is a f**in**ite I 0 ⊆ I such that ∪ i∈I 0π i ∪ T is**in**consistent. This means that ∩ i∈I 0[π i ] = [ ∪ i∈I 0π i ] = ∅.Exercise 3.2.4. Show the follow**in**g properties1. [φ ∨ ψ] = [φ] ∪ [ψ]2. [φ] ⊆ [ψ] iff T |= φ → ψ3. [φ] = [ψ] iff T |= φ ↔ ψ4. [φ] = S T xiff T |= φ.5. [φ] = ∅ iff T ∪ φ is **in**consistent.Exercise 3.2.5. Let T be a theory **in** a language L.equivalence relation ∼ **in** L x : φ ∼ ψ iff T |= φ ↔ ψ.Consider the follow**in**g1. Show that it is an equivalence relation and def**in**e operations such thatL x / ∼ is a boolean algebra. It is called the Tarski-L**in**dembaum algebra.Let us denote it by B T x .2. Given p ∈ Sx T , show that ψ ∼ φ and φ ∈ p then ψ ∈ p. Let us denotep/ ∼ the quotient of p by ∼. Show that p/ ∼ is an ultrafilter of Bx T .3. given an ultrafilter U of B T x , show that ∪ U ∈ S T x .4. Show the maps p ↦→ p/ ∼ and U ↦→ ∪ U are homeomorphisms, one **in**verse**to** the other, between S T x and the s**to**ne space of BT x .3.3 The omitt**in**g types theoremBy def**in**ition, a type of T is realized **in** a model of T . Here we will try **to** answerthe follow**in**g question: it is realized **in** every model? In other words: is there amodel of T not reliz**in**g the type? We beg**in** by nam**in**g th**in**gs:Def**in**ition 3.3.1. We say that a structure M omits a set of formulas π(x) ifM does not realize π(x).

CHAPTER 3. OMITTING TYPES 42If π(x) is **in**consistent with T then obviously every model of T omits π(x).Def**in**ition 3.3.2. **An** n-type π(x) of T is isolated iff there is a formula φ(x),consistent with T , such that T |= φ → ψ for every ψ ∈ π. This is denoted byT |= φ → π, and we say that φ isolates π **in** T .Let us remark that the requirement that the formula isolat**in**g the type has(at most) the same free variables as the type is avoidable. If T |= φ(x, y) → π(x)then also T |= ∃yφ(x, y) → π(x). This notion has a nice **to**pological translation**in** the space Sn T . The consistency of φ with T becomes ∅ ̸= [φ]. The factthat T |= φ → π translates **to** [φ] ⊆ [π]. In other words, an n-type π of Tis isolated iff [π] has nonempty **in**terior **in** the space Sn T . As we have seenabove this characterization does not depend on the n chosen. Even more, onegets the same characterization **in** any space SxT (provided that I conta**in**s thevariables of the type). Tak**in**g complements, one gets that π is non-isolatediff [π] C = Sx T \ [π] is dense. In other words, non-isolated types correspond (**in**the space Sx T ) **to** complements of open dense sets1 .Let’s return **to** the question about if there is a model of T that omits a giventype (or, more generally, a set of types). If the theory T is complete and thetype π is isolated **in** T , it is realized **in** any model of T . For, if φ(x) is consistentwith T , by completeness T |= ∃xφ(x). Hence **in** any model of T there is a tuplesatisfy**in**g φ. If φ isolates π, T |= φ(x) → π(x), hence any tuple satisfy**in**g φ **in**a model of T realizes π. Hence, **in** any model od T we can f**in**d a realizationof π. The follow**in**g theorem provides a sufficient condition **in** order **to** omit acollection of types **in** a countable language.Theorem 3.3.3 (Omitt**in**g types theorem). Let T be a consistent theory **in** acountable language L. Let P be a countable set of non-isolated types, each one**in** a f**in**ite number of variables. There is a model of T that omits all the types**in** P .Proof. We beg**in** by fix**in**g a countable set of variables x = (x n | n ∈ N). Wewill **work** **in** the space SxT of complete types **in** those variables. For every π ∈ Plet us denote by n(π) the (f**in**ite) number of free variables of π. Given a mapσ : {1, . . . , n(π)} → N let p σ denote the type obta**in**ed by replac**in**g the variablesof π accord**in**g **to** σ. More precisely, replac**in**g the i th variable of π by x σ(i) . Letus check that π σ rema**in**s non-isolated (**in** case σ is **in**jective is clear: we are onlyrenam**in**g variables). For, suppose that {y 1 , . . . , y n } are the variables of π andφ(x σ(1) , . . . , x σ(n)) isolates π σ . Then one easily checks that the formulaisolates π **in** T .φ(y 1 , . . . , y n ) ∧∧σ(i)=σ(j)1≤i

CHAPTER 3. OMITTING TYPES 43Now for every formula φ(y, y) ∈ L x let T V φ(y,y) denote the follow**in**g opensetT V φ(y,y) = ∪ [∃yφ(y, y) → φ(x n , y)]n∈NClaim 1. Each set T V φ(y,y) is dense.Proof. We have **to** show that for every ψ ∈ L x consistent with T , [ψ]has nonempty **in**tersection with T V φ(y,y) = [¬∃yφ(y, y)] ∪ ∪ n∈N [φ(x n, y)]. Ifψ ∧ ¬∃yφ(y, y) is consistent with T , [ψ] ∩ [¬∃yφ(y, y)] is nonempty. OtherwiseT |= ψ → ∃yφ(y, y) and thus the formula ψ ∧ φ(x n , y) is consistent with T ,where x n is a variable not occur**in**g **in** ψ ∧ ∃yφ(y, y). In this case [ψ] ∩ [φ(x n , y)]is nonempty. This ends the proof of the claim.Now consider the follow**in**g set∩T V φ(y,y) ∩ ∩ ∩[π σ ] C .φ(y,y)∈L x π∈Pσ∈ n(π) NWe already know that each T V φ(y,y) is dense. Each [π σ ] C is also dense becausethe type π σ is non-isolated. Hence the set above is a countable **in**tersection ofopen dense sets. By the Baire category theorem it conta**in**s a po**in**t p = p(x n |n ∈ N). Let a = (a n | n ∈ N) be a realization of p **in** some model M of T . Weare go**in**g **to** see that {a n | n ∈ N} is the doma**in** of a model of T that omits allthe types of P .Claim 2. The set {a n | n ∈ N} is the doma**in** of an elementary substructureof M.Proof. We use the Tarski-Vaught test, Theorem 1.2.10. Suppose M |=∃yφ(y, a i1 , . . . , a im ) for some formula φ(y, x i1 , . . . , x im ) ∈ L x . This impliesp ∈ [∃yφ(y, x i1 , . . . , x im )]. S**in**ce p also belongs **to** T V φ(y,xi1 ,...,x i m ) then p ∈[∃yφ(y, x i1 , . . . , x im ) → φ(x n , x i1 , . . . , x im )] for some n ∈ N. This implies p ∈[φ(x n , x i1 , . . . , x im )], hence M |= φ(a n , a i1 , . . . , a im ). This ends the proof of theclaim.Let us denote by N this substructure. Obviously N is a model of T . Theproof ends with the follow**in**g:Claim 3. N omits all types **in** P .Proof. We show that given any π = π(y 1 , . . . , y n ) ∈ P , and any σ :{1, . . . , n} → N the subtuple (a σ(1) , . . . , a σ(n) ) of a does not realize π. S**in**cep ∈ [π σ ] C = ∪ φ∈π [¬φσ ], we get ¬φ σ ∈ p for some φ ∈ π. But this impliesM |= ¬φ σ (a). Equivalently M |= ¬φ(a σ(1) , . . . , a σ(n) ). Hence (a σ(1) , . . . , a σ(n) )does no realize π.

Chapter 4Countable modelsIn this chapter the language will be countable.4.1 a**to**mic modelsWe start with a remark about isolated types **in** a theory T . If a complete typep ∈ Sx T is isolated by a consistent formula φ ∈ L x then φ ∈ p. Otherwise ¬φ ∈ pand thus T |= φ → ¬φ, contradict**in**g the consistency of φ with T .Let’s now make some more remarks about isolated types **in** the space Sx M(A),where M is any model and A ⊆ M a parameter set. Recall that Sx M (A) =S Th(M A)x.remark 4.1.1. 1. π ⊆ L x (A) is isolated iff there is some φ(x) ∈ L x (A) withM A |= ∃xφ(x)M A |= ∀x (φ(x) → π(x))2. If π ⊆ L x (A) is isolated then it is realized **in** M A .Proof. 1. If φ(x) is consistent with Th(M A ), s**in**ce Th(M A ) is complete it followsTh(M A ) |= ∃xφ(x). Hence M A |= ∃xφ(x). We use M A |= ∀x (φ(x) → π(x)) **to**denote M A |= ∀x (φ(x) → µ(x)) for any µ ∈ π. Obviously M A |= ∀x (φ(x) → π(x))is equivalent **to** Th(M A ) |= ∀x (φ(x) → π(x)).2. By 1, any tuple of M realiz**in**g φ(x) realizes π(x).If f : M → N is a partial elementary map and A ⊆ dom(f), the mapp ↦→ p f from Sx M(A) **to** SN x (f(A)) is a homeomorphism. We already know itis a bijection. But the image of the basic clopen [φ] is the basic clopen [φ f ].This means this map is open. By the same reason the **in**verse map, given byp ↦→ p f −1is also open. In particular, p ∈ Sx M(A) is isolated iff pf is isolated **in**(f(A)). In fact this is true also for partial (not necessarily complete) types.S N xDef**in**ition 4.1.2. Let M be a structure. We say that M is a**to**micevery f**in**ite tuple a ∈ M, tp M (a) is isolated.iff for44

CHAPTER 4. COUNTABLE MODELS 45Lemma 4.1.3. Let M be a structure and A ⊆ M be a f**in**ite set of parameters.If M is a**to**mic then also M A is a**to**mic.Proof. Let a be a f**in**ite tuple **in** M. Let b be an enumeration of A. S**in**ce Mis a**to**mic, the type p(x, y) of the tuple ab is isolated. Let φ(x, y) ∈ L x,y be aformula that isolates the type: M |= ∃x∃yφ(x, y) ∧ ∀x∀y(φ(x, y) → p(x, y)).Obviously M A |= ∀x(φ(x, b) → p(x, b)). Moreover φ ∈ p implies M |= φ(a, b)and thus M A |= ∃xφ(x, b). This means that tp(a/A) = p(x, b) is isolated byφ(x, b).Next result shows that a complete theory has at most a countable a**to**micmodel.Proposition 4.1.4. Suppose M and N are countable and a**to**mic. Let f : M →N be a f**in**ite partial elementary map. then f extends **to** an isomorphism fromM **to** N. In particular two elementarily equivalent a**to**mic countable models areisomorphic.Proof. Let (a n | n ∈ ω) and (b n | n ∈ ω) be enumerations of M \ dom(f)and N \ Im(f) respectively. We are go**in**g t build a cha**in** of elementary maps(f n | n ∈ ω) from M **to** N with f 0 = f, a n ∈ dom f n+1 , b n ∈ Im f n+1 and f nf**in**ite. The union will be the isomorphism. Assume f n is already constructed.By Lemma 4.1.3 M is a**to**mic over dom(f n ), hence p = tp(a n / dom(f n )) isisolated. Thus p fn is also isolated and thus realized by some b ∈ N. Nowg = f ∪ {(a n , b)} is elementary. Aga**in** by Lemma 4.1.3 N is a**to**mic over Im(g)and thus we can realize tp g−1 (b n / Im(g)) by some a ∈ M. F**in**ally we putf n+1 = f n ∪ {(a n , b), (a, b n )}.Exercise 4.1.5. Show that an a**to**mic model is ω-homogeneous.Next exercise shows that countable a**to**mic models are prime.Exercise 4.1.6. Let M be countable and a**to**mic. Show that M is prime, i.e. forany N ≡ M there is an elementary embedd**in**g from M **to** N.Exercise 4.1.7. Prove the converse of the previous exercise.show that any prime models is a**to**mic and countable.4.2 The Ryll-Nardzewski’s theoremMore precisely:Theorem 4.2.1 (Engeler-Ryll-Nardzewski-Svenonious). Let T be a completetheory **in** a countable language without f**in**ite models. The follow**in**g are equivalent:1. T is ω-categorical.2. for all n all po**in**ts **in** S T n are isolated.3. for all n S T n is f**in**ite.

CHAPTER 4. COUNTABLE MODELS 464. Each model of T is a**to**mic.Proof. 1 implies 2. If some p **in** some Sn T is non-isolated, by the omitt**in**g typesTheorem 3.3.3 T has a countable model omitt**in**g p and also, by proposition2.1.12, a countable model realiz**in**g p. S**in**ce p has no parameters, these twomodels cannot be isomorphic.2 implies 3. The collection of isolated po**in**ts of Sn T is an open cover**in**g ofSn T , which must be f**in**ite by compactness.3 implies 4. Because Sn T is Haussdorff all po**in**ts **in** Sn T are isolated. S**in**ce Tis complete, for any model M of T , Sn T = SnM whence M is a**to**mic.4 implies 1. All countable models of T are a**to**mic hence isomorphic byproposition 4.1.4.Exercise 4.2.2. Let T be a complet theory without f**in**ite models. Prove thatthe follow**in**g are equivalent:1. T is ω-categorical.2. T has a model which is both a**to**mic and saturated.3. T has a countable model which is both a**to**mic and saturated.4. T has a model which is both a**to**mic and ℵ 1 -universal.5. T has a countable model which is both a**to**mic and ℵ 1 -universal.6. T has a model realiz**in**g only f**in**itely many n-types for each n.7. T has a countable model realiz**in**g only f**in**itely many n-types for each n.8. All models of T are ω-saturated.9. All countable models of T are ω-saturated.10. All countable models of T are a**to**mic.11. All models of T realize only f**in**itely many n-types for each n.12. for all n B T n is f**in**ite.13. For any model M of T and any f**in**ite A ⊆ M S M 1 (A) is f**in**ite.14. For any model M of T any f**in**ite A ⊆ M and any n, S M n (A) is f**in**ite.Exercise 4.2.3. Provide some more characterizations of ω-categoricity.Exercise 4.2.4. Let T be a theory without f**in**ite models. Prove that T is ω-categorical iff all its models are partially isomorphic.Exercise 4.2.5. 1. Show that a complete theory with **in**f**in**itely many differentconstants (the language conta**in**s **in**f**in**itely constants and the theory saysthat the are different) cannot be ω-categorical. Even more holds:

CHAPTER 4. COUNTABLE MODELS 472. Show that any model of an ω-categorical theory is locally f**in**ite. Thatis: each f**in**itely generated substructure is f**in**ite. In fact any model of thetheory is uniformly locally f**in**ite: there is a function d : N → N (depend**in**gonly on the theory, not the model) such that |⟨A⟩ M| ≤ d(|A|) for any f**in**itesubset A of any model M of the theory.4.3 countable a**to**mic and countable saturatedmodels of a complete theoryIn the section we well characterize, **in** terms of the spaces S T n , when a givencomplete theory has a countable a**to**mic model and when it has a countablesaturated model.We start with the existence of countable saturated models.remark 4.3.1. Let T be a theory, M a model of T and A ⊆ M a set of parameters.1. If ∣ y = (y i | i ∈ I) a tuple of new variables of the same card**in**ality as A,∣S x M(A)∣ ∣ ≤ ∣ ∣∣SxyT ∣.2. If T is complete ∣ ∣SxT ∣ ≤ ∣ ∣Sx M(A)∣ ∣.Proposition 4.3.2. Let T be a complete theory with **in**f**in**ite models. The follow**in**gare equivalent:1. T has a countable saturated model.2. For all n ∣ ∣SnT ∣ ≤ ℵ 0 .Proof. 1 implies 2. Let M be the countable saturated model of T . S**in**ce T iscomplete, by remark 3.2.2 Sn T = Sn M . S**in**ce M is saturated, any type **in** SnM isrealized **in** M. But the number of possible realizations is countable whence SnMis at most countable.2 implies 1. For any model M of T and any f**in**ite A ⊆ M, by remark 4.3.1,∣∣S1M (A) ∣ ∣≤ ∣S|A|+1T ∣ ≤ ℵ 0 . We build a cha**in** (M n | n ∈ ω) of countable modelsof T as follows. Start**in**g with any countable model, we pick, by 2.1.12 M n+1an elementary extension of M n realiz**in**g all 1-types of M n over a f**in**ite set ofparameters. By hypothesis there are only countably many such types and wecan pick M n+1 **to** be countable. The union of the cha**in** is a countable saturatedmodel of T .The follow**in**g exercise is a generalization of proposition 4.3.2Exercise 4.3.3. Let T be a complete theory and λ a card**in**al with λ

CHAPTER 4. COUNTABLE MODELS 483. S T µ ≤ λ for each µ < λ.4. S M 1 (A) ≤ λ for each model M of T and each A ⊆ M with |A| < λ.Exercise 4.3.4. Let T be any theory (not necessarily complete).1. Show the follow**in**g are equivalent:(a) For all n ∣ ∣SnT ∣ ≤ ℵ 0 .(b) T has at most countably many completions and each completion ofT has a f**in**ite model or a countable saturated model.2. If for all n ∣ ∣SnT ∣ ≤ ℵ 0 , then the number of completions of T co**in**cide withthe number of saturated model of T of card**in**ality at most countable (up**to** isomorphism).3. If T has at most countably many countable models and only one of themis saturate then T is complete.Exercise 4.3.5. Show that if a complete theory has a countable ω 1 -universalmodel then it also has a countable saturated model.Corollary 4.3.6. Let T be any theory with **in**f**in**ite models. If T has at mostcountably many countable models then T has a countable saturated model.Proof. Replac**in**g T by any of its completions we may assume T complete. S**in**ceby 2.1.12 each type **in** S T n is realized **in** a countable model of T , S T n is at mostcountable.Now we go **to** the question about countable a**to**mic models.remark 4.3.7. If N is a**to**mic and N ≼ N then also M is a**to**mic.Proof. It follows from tp M (a) = tp N (a) for every tuple a ∈ M.Hence, the existence of an a**to**mic model of a theory implies the existenceof a countable a**to**mic model. Next proposition characterize the existence ofcountable a**to**mic models.Theorem 4.3.8. Let T be a complete theory. Then T has an a**to**mic modeliff for all n, the set of isolated po**in**ts is dense **in** S T n .Proof. Assume T has a prime model M, and let φ(x) ∈ L n consistent with T .S**in**ce T is complete T |= ∃xφ(x) and thus there is a tuple a **in** M satisfy**in**gφ(x). S**in**ce M is a**to**mic, p = tp(a/∅) is isolated. This implies that p is anisolated po**in**t **in** [φ].Conversely, assume that the set of isolated po**in**ts are dense **in** every spaceS T n . For each n, consider the follow**in**g n-type:π n = { ¬φ | φ ∈ L n and φ isolates a complete type p ∈ S T n}.

CHAPTER 4. COUNTABLE MODELS 49Now we show that each π n is non-isolated **in** T . Otherwise, assume that for somen there is ψ ∈ L n consistent with T such that T |= ψ → π n . Let, by densityof isolated po**in**ts, p ∈ [ψ] isolated **in** T . Let φ ∈ L n a formula isolat**in**g p **in** T .S**in**ce ψ ∈ p it follows that T |= φ → ψ. But ¬φ ∈ π n implies T |= ψ → ¬φ andthus T |= φ → ¬φ, a contradiction with the consistency of φ with T . By theomitt**in**g types theorem there is a model M of T omitt**in**g all π n . We end bycheck**in**g that M is a**to**mic. Given a tuple a of M, a does not realize π n , wheren is the length of a. This implies a satisfies some formula φ(x) isolat**in**g a typep ∈ S T n . Hence p is the type of a, which is isolated.We end this section by show**in**g that the existence of a countable saturatedmodel implies the existence of an a**to**mic one.Theorem 4.3.9. Let T be any theory without f**in**ite models. If T has a countablesaturated model then T has an a**to**mic model. 1Proof. Replac**in**g T by an appropriate completions we may assume T complete.Assume T has not a**to**mic model. By theorem 4.3.8 there is some n such that**in** Sn T the isolated po**in**t are not dense. Let φ ∈ L n such that [φ] does notconta**in** isolated po**in**ts (i.e. [φ] perfect). Let p, q be two different po**in**ts **in** [φ].Let φ o ∈ p with φ o /∈ q and put φ 1 = ¬φ 0 . Observe that [φ] is a disjo**in**tunion [φ] = [φ 0 ] ∪ [φ 1 ] where each [φ i ] is nonempty and perfect aga**in**. We canpartition aga**in** [φ i ] = [φ i0 ] ∪ [φ i1 ] with [φ ij ] nonempty and perfect. Iterat**in**gthis process we build a b**in**ary tree of formulas (φ s | s ∈

CHAPTER 4. COUNTABLE MODELS 50∣theory of M A , where A denotes the set ................ As ∣SmT (a)∣∣ ≤ ∣Sm+|a|T ∣ ≤ ℵ 0 ,by proposition 4.3.2 and theorem 4.3.9 there is a prime model (M 3 ) bof T (a).As tp M 3(b) = p, M 3 is not a**to**mic, hence not isomorphic **to** M 1 . S**in**ce ℵ 0 ≤∣∣SnT ∣ ∣≤ ∣SnT (a) ∣∣, there should be some q ∈ SnT (a) non-isolated. Obviously (M 3 ) bdoes not realize q hence M 3 is not saturated and cannot be isomorphic **to** M 2 .M 3 is our third countable model of T .Exercise 4.4.2. Let L be a language conta**in****in**g countably many unary predicates,L = (P i | i ∈ N). For each each pair A, B of f**in**ite subsets of N considerthe formula φ A,B := ∃x (∧ i∈A P i(x) ∧ ∧ i∈B ¬P i(x) ) . Let T be the follow**in**gtheory {φ A,B | A, B f**in**ite and disjo**in**t subsets of N}. Show the follow**in**g:1. T is consistent and has no f**in**ite models.2. In an ω-saturated model M of T the follow**in**g holds: For each subset Aof N the set ∩ i∈A P M i ∩ ∩ i∈N\A M \ P M i is **in**f**in**ite.3. T is complete and elim**in**ates quantifiers.4. The converse of 2. holds.5. T has no prime model nor countable saturated model nor countable ω 1 -universal model.6. T has 2 ℵ 0countable models.7. for each n, Sn T is perfect (i.e., has no isolated po**in**ts. H**in**t: show that ann-type π is isolated then only f**in**itely may predicates occur **in** π).{2ℵ α, if ℵ α ≤ 2 ℵ 0;8. I(T, ℵ α ) =|α| 2ℵ 0, if ℵ α > 2 ℵ 0.Exercise 4.4.3. Modify previous exercise **to** obta**in** a theory with prime modelbut no countable saturated model (H**in**t: consider T (A) for a countable set A).Exercise 4.4.4. Let T be a complete theory **in** a countable language which isnot ω-categorical. Let π 1 , . . . , π n be a f**in**ite collection of types of T each one **in**a f**in**ite number of variables. Show that T has a countable non-saturated modelrealiz**in**g all the types π 1 , . . . , π n .Exercise 4.4.5. Let L be a language conta**in****in**g countably many unary predicates{P i | i ∈ N} and countably many constants { a i,j | (i, j) ∈ N 2} . Let T be thetheory which says a i,j ≠ a i,k for any i and any pair j < k, ∀x(¬P i (x) ∨ ¬P j (x))for any pair i < j and P i (a i,j ) for any pair i, j. Show the follow**in**g:1. In any ω-saturated model M of T the follow**in**g sets are **in**f**in**ite: M \∪i∈N P M i and P M i \ { a M i,j | j ∈ N} for each i.2. T is complete and elim**in**ates quantifiers.3. The converse of 1. holds.

CHAPTER 4. COUNTABLE MODELS 514. T has a prime model and a countable saturated model. Describe them.5. T has 2 ℵ0 countable models.6. What are the isolated types **in** S T n ?7. I(ℵ α , T ) = (2 + |α|) ℵ 0.Exercise 4.4.6. Proceed as **in** previous exercise with the follow**in**g theory. Thelanguage conta**in**s countably many unary predicates: L = {P s | s ∈

CHAPTER 4. COUNTABLE MODELS 52Let T be a theory (maybe not complete), M a model of T and A ⊆ M a se**to**f parameters. We will denote by T (A) the theory of M A . T (A) is a completetheory **in** L(A) extend**in**g T . In fact any complete theory **in** L(A) extend**in**g Tis of this k**in**d. Recall (exercise 2.3.4) that (N, ha) a∈A is a model of T (A) iff thepartial map (with doma**in** A) h : M → N is elementary. Hence, a model of T (A)consist on a structure N **to**gether with an elementary partial map h : M → Nwith doma**in** A. Observe that, by remark 3.2.2 (or 2.1.3), SnT (A) = Sn M (A). Thea**to**ms of BnT (A) will be called simply a**to**ms of T (A).Def**in**ition 4.5.2. Let M be a model and A ⊆ M a set of parameters. Wesay that M is prime over A if every partial elementary map f : M → N withA = dom(f) extends **to** an elementary embedd**in**g from M **to** N. We say M isprime if M is prime over the empty set, i.e., there is an elementary embedd**in**gfrom M **to** any N elementarily equivalent **to** M.Theorem 4.5.3. Let M be a model and A ⊆ M a countable set of parameters.Then M is prime over A iff M is a**to**mic over A and countable. In particularM is prime iff M is a**to**mic and countable.Proof. If M is prime over A, it is countable, s**in**ce we can embed M **in****to** acountable model of T (A). If M is not a**to**mic over A, let a ∈ M such thattp(a/A) is non-isolated. By the omitt**in**g types theorem, let (N, ha) a∈A be amodel of T (A) omitt**in**g tp(a/A). This means that N omits tp h (a/A). S**in**ce Mis prime over A, h may be extended **to** an elementary embedd**in**g f : M → N.But then f(a) realizes tp h (a/A), a contradiction.Assume now that M is countable and a**to**mic over A and there is an elementarymap f : M → N with doma**in** A. Let (a n | n ≥ 1) be an enumeration ofM \A. We are go**in**g **to** build a cha**in** (f n | n ∈ ω) of elementary maps from M **to**N with dom(f n ) = A ∪ {a 1 , . . . , a n }. Obviously ∪ n∈ω f n will be an elementaryembedd**in**g from M **to** N. We start with f 0 = f. Let us see how **to** build f n+1from f n . S**in**ce M is a**to**mic over A, by lemma 4.1.3, M is a**to**mic over dom(f n ).Hence tp(a n+1 / dom(f n )) is isolated. Obviously tp fn (a n+1 / dom(f n )) is alsoisolated and thus realized by b **in** N. Now we set f n+1 = f n ∪ {(a n+1 , b)}.A Boolean algebra is a**to**mic iff for every 0 ≠ b ∈ B there is some a**to**m awith a ≤ b. Assume BnT is a**to**mic. Let φ ∈ L n consistent with T . Let ψ ∈ L nan a**to**m of BnT with T |= ψ → φ. Let p the type isolated by ψ. Then p ∈ [φ].This shows that the isolated po**in**ts of SnT are dense. Conversely, assume theisolated po**in**ts of SnT are dense. Let φ ∈ L n consistent with T and let p ∈ [φ]an isolated po**in**t. let ψ be the formula that isolates p then ψ is an a**to**m aboveφ. This shows that Bn T is a**to**mic. The same argument shows that a booleanalgebra is a**to**mic iff **in** its s**to**ne space the isolated po**in**ts are dense.The follow**in**g theorem gives a criteria for existence of prime models of T .Exercise 4.5.4. Prove that if φ(x, y) is an a**to**m of T (more precisely, φ is ana**to**m **in** B T n+1, where n denotes the lenght of y) then also ∃xφ is an a**to**m of T(more precisely, ∃xφ is an a**to**m **in** B T n ).

Chapter 5Partial isomorphism5.1 Partial isomorphismIn this chapter we will presuppose a fixed language L and when we speak aboutstructures or models we mean L-structures.Def**in**ition 5.1.1. A partial isomorphism f from M **to** N is a bijection betweensubsets of M and N respectively satisfy**in**g the follow**in**g conditions:1. For every predicate symbol R and every tuple a of the doma**in** of f it holdsthat a ∈ R M iff fa ∈ R N .2. For every function symbol f, every tuple a and element a of the doma**in**of f it holds that f M (a) = a iff f N (fa) = f(a)3. For every constant symbol c and every element a of the doma**in** of f itholds that c M = a iff c M = f(a).Equivalently, we could have described partial isomorphism as partial mapspreserv**in**g, **in** the sense of 1.2.3, the formulas of typex = y, R(x 1 , . . . , x n ), x = f(x 1 , . . . , x n ), x = c.Observe that the **in**verse of a partial isomorphism is also a partial isomorphism.For example, every part(subset) of an isomorphism between substructures ofM and N is a partial isomorphism. It is not true, however, that every partialisomorphism from M **to** N extends **to** an isomorphism f between substructuresof M and N (an embedd**in**g of dom(f) **to** N). If we take M = (Z, s) the map{(1, 4), (3, 7)} is a partial isomorphism from M **to** M which can not be extendedby add**in**g 2 **in** the doma**in**. It is not possible, therefore, extend this map **to** anisomorphism between substructures of M.Exercise 5.1.2. 1. If f : M → N is a partial isomorphism, dom(f) is asubstructure of M iff Im(f) is a substructure of N.53

CHAPTER 5. PARTIAL ISOMORPHISM 542. If f is a partial isomorphism between substructures of M and N then fis an embedd**in**g from dom(f) **to** N.Thus, when the doma**in** of a partial isomorphism f : M → N is a substructureof M then f is an embedd**in**g of dom(f) **in** N. In particular, when thelanguage is relational, a partial isomorphism is the same as an isomorphismbetween substructures.Def**in**ition 5.1.3. A system of partial isomorphism from M **to** N is a nonemptycollection I of partial isomorphisms from M **to** N satisfy**in**g the follow**in**g properties:Back For every f ∈ I and b ∈ N there is a g ∈ I that extends f and such thatb is **in** the image of g.Forth For every f ∈ I and a ∈ M there is a g ∈ I that extends f and suchthat a is **in** the doma**in** of g.When there is a system of partial isomorphism I from M **to** N it is saidthat M and N are partially isomorphic (via I) and this is denoted by M ∼ = p N(I : M ∼ = p N respectively).Example 5.1.4. 1. Let M, N be two dense l**in**ear orders without endpo**in**ts.The collection of isomorphisms between f**in**ite substructures of M and Nis a system of partial isomorphism.2. If f is an isomorphism from M **to** N then {f}, is a system of partialisomorphism.3. If I : M ∼ = p N is a system of partial isomorphisms then{f | f is a partial map and f ⊆ g for some g ∈ I} ,{f | f is a f**in**ite partial map and f ⊆ g for some g ∈ I}are also systems of partial isomorphisms from M **to** N. Hence there is alwaysa system of f**in**ite partial isomorphisms between partially isomorphicstructures.By the above examples two isomorphic structures are partially isomorphic.If the structures are countable the converse holds:Proposition 5.1.5. any two countable and partially isomorphic structures areisomorphic.Proof. Let (a i |∈ ω) and (b i |∈ ω) be enumerations of M and N and let I bea system of partial isomorphism from M **to** N. We will Construct a cha**in** ofpartial isomorphisms (f i | i ∈ ω) such that a i is **in** the doma**in** of f i+1 and b i **in**the image of f i+1 . Then the union of the f i is an isomorphism from M **to** N.Take as f 0 any element of I. Suppos**in**g f i constructed. By forth there is g ∈ Iwith a ∈ dom(g) such that extends f. By back, there is f i+1 that extends g andsuch that b ∈ Im(f i+1 ).

CHAPTER 5. PARTIAL ISOMORPHISM 55If we apply this proposition **to** the example 5.1.4 we obta**in** that there is aunique countable and dense l**in**ear order without endpo**in**ts.Exercise 5.1.6. 1. Prove that any two a**to**mless Boolean algebras are partiallyisomorphic via the set of isomorphisms between f**in**ite subalgebras.2. The Random Graph. The language conta**in**s only a symbol of b**in**aryrelation R. A graph is a symmetrical and antireflexive relation. Theaxioms say that given two f**in**ite sets A 1 , A 2 of vertices (elements of thedoma**in**) there is another vertex related with all the vertices of A 1 and noneof A 2 . Prove that two models any of this theory are partially isomorphicvia the set of isomorphisms between f**in**ite substructures.3. The Random structure **in** a f**in**ite relational language. Let L be relationaland f**in**ite. For any f**in**ite tuple of variables x let ∆(x, y) denote the set offormulas of type R(z) where z is a subtuple of x (any variable of z is **in**x, we allow repetitions) and y occurs **in** z. For any subset Σ of ∆(x, y) letφ Σ denote the formula⎛⎛⎞⎞∀x ⎝‘x different’ → ∃y ⎝‘y /∈ x’ ∧ ∧ ∧ψ ∧ ¬ψ⎠⎠ .ψ∈Σψ∈∆(x,y)\ΣLet T RAND (L) be the theory (of the random structure) axiomatized by allthe sentences φ Σ where Σ is any subset of some ∆(x, y). **An**y two modelsM, N of the theory of the random structure are partially isomorphic viaI f (M, N) the set of all f**in**ite partial isomorphisms from M **to** N.4. Let L be the language conta**in****in**g a s**in**gle b**in**ary relation ≤ and unarypredicates P 1 , . . . , P n . Let T the theory express**in**g that ≤ is a l**in**earorder without endpo**in**ts, the P i ’s form a partition of the doma**in**, andeach P i is dense **in** the doma**in** (between any two po**in**ts there is somepo**in**t of P i ). **An**y two models M, N of this theory are partially isomorphicvia I f (M, N) the set of all f**in**ite partial isomorphisms from M **to** N.remark 5.1.7. If all the models of a countable theory are partially isomorphic,by 5.1.5 then the theory is ω-categorical and therefore complete. We will seelater on that it also holds the converse one. All the examples of the formerexercise are theories ω-categorical.5.2 Karp’s TheoremThe logic L ∞ω is the extension of the logic of first order accept**in**g arbitrarilygreat conjunctions and disjunctions. More precisely, the formulas are constructedfrom the a**to**mic ones and clos**in**g under negation, arbitrary conjunctionand quantify**in**g a s**in**gle variable. Unlike the logic of first order, the formulas ofthis logic always constitute a proper class. We will call also sentences (or closedformulas) the formulas without free variables **in** this language. We are go**in**g

CHAPTER 5. PARTIAL ISOMORPHISM 56**to** see that the partial isomorphism characterizes the elementary equivalence **in**the logic L ∞ω .Theorem 5.2.1 (Karp). Two structures are elementarily equivalents **in** thelogic L ∞ω (that is, they satisfy the same sentences of L ∞ω ) iff they are partiallyisomorphic.Proof. Suppose that I is a system of partial isomorphism between M and N.We have **to** see that M and N are L ∞,ω -equivalent. We will make the proof **in**two steps.Claim 1. Given any a term t = t(x), any f ∈ I and any f**in**ite tuple aof elements of dom(f), there is g ∈ I, f ⊆ g such that t M (a) ∈ dom(g) andg(t M (a)) = t N (g(a)).Proof. By **in**duction on t. The case where t is a variable is obvious. If t =f(t 1 , . . . , t n ), by apply**in**g the hypothesis of **in**duction n times we obta**in** h ∈ I,an extension of f, such that dom(h) conta**in**s t M i (a) and h(t M i (a)) = t N i (h(a))for i = 1 . . . n. Now we take g ∈ I an extension of h with t M (a) ∈ dom(g). Nowg is a partial isomorphism and its doma**in** conta**in**s t M (a) and all t M i (a). S**in**cet M (a) = f M (t M 1 (a), . . . , t M n (a)) we have thatand as that h(t M ig(t M (a)) = f N (g(t M 1 (a)), . . . , g(t M n (a))) =(a)) = t N i (h(a)),= f N (t N 1 (g(a)), . . . , t N n (g(a))) = t N (g(a)).Now we will prove that all the maps of I preserve all the formulas L ∞,ω :Claim 2. Given any formula φ = φ(x) of L ∞,ω , any f ∈ I and any tuple aof elements of dom(f)M |= φ(a) iff N |= φ(fa).Proof. By **in**duction on the formula φ. If the formula is t 1 = t 2 , by claim1, we take g ∈ I an extension of f such that t M i (a) ∈ dom(g) and g(t M i (a)) =t N i (g(a)) for i = 1, 2. Then, as g is **in**jective tM 1 (a) = t M 2 (a) iff t N 1 (g(a)) =t M 2 (g(a)). The case where the formula is R(t 1 , . . . , t n ) is made analogously,apply**in**g the claim 1 n times **in** order **to** achieve an extension g of f witht M i (a) ∈ dom g and g(t M i (a)) = t N i (g(a)) for i = 1 . . . n. Then g preservesR(t 1 , . . . , t n ) s**in**ce it is a partial isomorphism.The cases **in** which the formula is the negation of another formula or theconjunction of a set of formulas come out without any problem apply**in**g the hypothesisof **in**duction. F**in**ally we treat the case **in** which the formula is ∃xφ(x, x).If M |= ∃xφ(x, a) then there is an element a ∈ M such that M |= φ(a, a). Wetake g ∈ I an extension of f with a ∈ dom g. By hypothesis of **in**ductionN |= φ(g(a), g(a)) and therefore N |= ∃xφ(x, g(a)). The converse is done analogouslyus**in**g Back **in**stead of Forth.Now we will prove the converse one. We suppose that M and N satisfy thesame closed formulas of the logic L ∞,ω . We will prove thatI = {f : M → N | f is partial, f**in**ite and preserves all the formulas of L ∞,ω }

CHAPTER 5. PARTIAL ISOMORPHISM 57is a system of partial isomorphism. By hypothesis the empty map is **in** I. Inorder **to** prove Forth, let f ∈ I and a ∈ M. Let a an enumeration of thedoma**in** of f. Everyth**in**g is reduced **to** f**in**d b ∈ M such that the tuples aa andbfa satisfy the same formulas. If there was not certa**in** b we would have thatfor each b ∈ N there is a formula φ b (x, x) of L ∞,ω such that M |= φ b (a, a)but N |= ¬φ b (b, fa). This is an absurd, s**in**ce then M |= ∃x ∧ b∈N φ b(x, a)and N |= ¬∃x ∧ b∈N φ b(x, fa), contradict**in**g the fact that f preserves all theformulas of L ∞,ω . Back is made analogously.Corollary 5.2.2. Two partially isomorphic models are elementarily equivalent.Moreover, every partial isomorphism liv**in**g on a system of partial isomorphismis an elementary map.Exercise 5.2.3. Let M, N be given partially isomorphic structures. Prove that:1. If M is ω-saturated then N is also ω-saturated.2. If M is ω-homogeneous then N is also ω-homogeneous.3. M and N realize the same types **in** a f**in**ite number of variables over theempty set.Exercise 5.2.4. Let M, N be given ω-homogeneous structures. Prove that Mand N are partially isomorphic iff they realize the same types **in** a f**in**ite numberof variables over the empty set.5.3 Partial isomorphism and completenessIn this section we will provide proposition 5.3.2, a quite useful criterion **to** provethat a given theory is complete.We beg**in** by see**in**g that for ω-saturated models, ‘be**in**g elementarily equivalentsand be**in**g partially isomorphic is the same th**in**g.Proposition 5.3.1. Two ω-saturated and elementarily equivalents models arepartially isomorphic via the set of all f**in**ite partial elementary maps.Proof. Suppose that M, N are ω-saturated and elementarily equivalent. Let Idenote set of all partial and f**in**ite elementary maps from M **to** N. I is nonempty s**in**ce then the empty map is **in** I. Everyth**in**g is reduced **to** see that Ihas the Forth property(Back is done the same way). Let f be a f**in**ite partialelementary map and a an element of M. By lemma 2.3.6 everyth**in**g is reduced**to** take, because of saturation, a realization of tp f (a/ dom f).Corollary 5.3.2. A theory T is completepartially isomorphic.iff all its ω-saturated models areProof. The implication from left **to** right follows by apply**in**g the proposition5.3.1. Conversely if M and N are models of T , by the proposition 2.2.4 we cantake M ′ and N ′ elementary and ω-saturated extensions of M and N respectively.As M ′ ∼ = p N ′ then M ≡ M ′ ≡ N ′ ≡ N.

CHAPTER 5. PARTIAL ISOMORPHISM 58Example 5.3.3. 1. The theory T of a **to**tal order with previous and nextelement. A Z-cha**in** of a model of T consists of an element and all itspredecessors and all its successors. A Z-cha**in** is isomorphic **to** Z. Leteasy **to** see that **in** an ω-saturated model of T the Z-cha**in**s are dense andwithout endpo**in**ts. Then, given two ω-saturated models of T the set ofisomorphisms between a f**in**ite number of Z-cha**in**s of M and of N is aset of partial isomorphism. Then this theory is complete and provides anaxiomatization of Th Z,

CHAPTER 5. PARTIAL ISOMORPHISM 59Def**in**ition 5.4.2 (Q.E. second def**in**ition). A theory T has Quantifier Elim**in**ationif for every formula φ = φ(x), where x is a nonempty tuple of variables,there is a formula ψ = ψ(x) without quantifiers such that T |= φ(x) ↔ ψ(x) 2 .Observe that **in** this second def**in**ition we exclude the empty tuple x. Ofcourse any sentence φ has at most one free variable, hence by the second def**in**ition,it is equivalent **to** a quantifier-free formula ψ(x) with (at most) one freevariable x. But we cannot conclude that φ is equivalent **to** a quantifier-freesentence! For **in**stance, if there are not constants **in** the language there are noquantifier-free sentences! But **in** fact this is the only obstruction as po**in**t 2 ofnext proposition shows.Proposition 5.4.3.1. The above def**in**itions of Q.E. are equivalent.2. If T has Q.E. and the language has a constant, then any sentence is equivalentmodulo T **to** a quantifier-free sentence.Proof. We start by prov**in**g that the first def**in**ition implies the second. Letφ(x 1 , . . . , x n ) be any formula with x 1 , . . . , x n a non-empty (i.e. n > 0) tuple ofvariables. By hypothesis there is some quantifier-free formula ψ(x 1 , . . . , x n , y 1 , . . . , y m )equivalent **to** φ mod T (more precisely T |= φ ↔ ψ). It is easily seen that thenφ(x 1 , . . . , x n ) is equivalent mod T **to** the formula ψ(x 1 , . . . , x n , x 1 , . . . , x 1 ).Conversely, it only rema**in**s **to** cover the case where φ is a sentence. Butthen we can apply the second def**in**ition **to** the formula φ ∧ x = x (equivalent **to**φ mod T ).For the proof of 2, let φ be any sentence and let ψ(y 1 , . . . , y m ) be a quantifierfreeformula equivalent **to** φ modulo T . Then it is easily seen that φ is alsoequivalent mod T **to** ψ(c, . . . , c), were c is any constant of the language.The next exercise shows that **in** order **to** elim**in**ates quantifiers it is enough**to** elim**in**ate a s**in**gle quantifier:Exercise 5.4.4. Let T be a theory. Show that the follow**in**g are equivalent:1. T has Q.E.2. Each existential formula with a s**in**gle existential quantifier (i.e., of type∃xψ with ψ quantifier-free) is equivalent, modulo T , **to** quantifier-freeformula.3. For every formula ∃yφ(x, y), where x is a nonempty tuple of variables,and φ(x, y) is quantifier-free, there is a formula ψ(x) without quantifierssuch that T |= φ(x) ↔ ψ(x).Def**in**ition 5.4.5. If a is a tuple of elements of M, the a**to**mic type of a (overthe empty set), that we will denote by atp M (a) (or atp M (a/∅)) is the set ofa**to**mic or negations of a**to**mic formulas that a satisfies:atp(a) = {φ(x) ∈ L | φ is a**to**mic or negation of a**to**mic and M |= φ(a)} .2 this is equivalent **to** T |= ∀x(φ(x) ↔ ψ(x))

CHAPTER 5. PARTIAL ISOMORPHISM 60The follow**in**g criterion will be use later several times.Proposition 5.4.6. A theory T elim**in**ates quantifiers iff given M, N modelsof T and any non empty f**in**ite tuples a, b of M and N respectively such thatatp M (a) = atp N (b) then also tp M (a) = tp N (b).Proof. The implication from left **to** right is easy. We concentrate **in** the converseone. Given φ(x), where x is a non empty tuple, we considerΨ(x) = {ψ(x) | ψ(x) does not have quantifiers and T |= φ(x) → ψ(x)} .By compactness everyth**in**g is reduced **to** prove that T ∪ Ψ(x) |= φ(x). Let Mbe a model of T and a be a tuple of M that realizes Ψ(x). We have **to** showthat M |= φ(a).We claim that T ∪ atp M (a) ∪ {φ(x)} is consistent (we assume that the typeatp M (a) is written us**in**g x as free variables). Otherwise we would have thatT ∪ {φ(x)} |= ¬ ∧ ni=1 ψ i(x) for certa**in** ψ i (x) ∈ atp M (a). This would imply that¬ ∧ ni=1 ψ i(x) ∈ Ψ(x) and therefore M |= ¬ ∧ ni=1 ψ i(a). This is absurd, s**in**ceeach ψ i (x) is satisfied by a.As T ∪ atp M (a) ∪ {φ(x)} is consistent, there is a model N of T and a tuple bof M that realizes atp M (a) ∪ {φ(x)}. Then a and b have the same a**to**mic type(over empty) and by hypothesis they also have the same type (complete andover the empty set). Then M |= φ(a), s**in**ce N |= φ(b).Proposition 5.4.6 has a useful straightforward generalization:Exercise 5.4.7. Let x be a fixed nonempty f**in**ite tuple of variables and let ∆ bea nonempty subset of L x . The ∆-type of a tuple a of a model M is def**in**ed bytp ∆ (a) = {φ(x) | φ ∈ ∆ and M |= φ(a)} Prove that they are equivalent:1. For every given formula φ(x) (where x is a non empty tuple of variables)there exists a boolean comb**in**ation ψ(x) of ∆-formulas such that T |=φ(x) ↔ ψ(x)2. Given a pair M, N of models of T and tuples a, b of M and N respectivelysuch that tp M ∆ (a) = tpN ∆ (b) then also tpM (a) = tp N (b).The follow**in**g exercise shows that Q.E. may be easily achieved by add**in**g,for each formula, a new relation and its def**in**ition.Exercise 5.4.8. Let T be a theory **in** the language L. Consider the follow**in**gexpansion of the language: for each L-formula φ(x) add a new relation R φ ,whose arity is the length of x. Consider the theory T ′ obta**in**ed by add**in**g theaxioms ∀x(φ(x) ↔ R(x)) (for each L-formula φ(x)) **to** T . Prove that1. Each model of T has a unique expansion **to** a model of T ′ .2. If φ is an L-sentence and T ′ |= φ then T |= φ.3. for every φ(x) ∈ L ′ there is ψ(x) ∈ L such that T ′ |= φ(x) ↔ ψ(x)

CHAPTER 5. PARTIAL ISOMORPHISM 614. T ’ elim**in**ates quantifiers.This exercise shows that every theory elim**in**ates quantifiers **in** an appropriatedef**in**itional expansion. A def**in**itional expansion of T is a theory T ′ **in** an expansionL ′ of the language L satisfy**in**g 1, 2 and 3.5.5 Partial isomorphism and Q.E.Now we will give a modification of proposition 5.3.2 that is useful **to** proveelim**in**ation of quantifiers.Let M be a structure. The substructure of M generated by a subset A ofM is the smallest substructure of M conta**in****in**g A. We will denote it by ⟨A⟩ M.A substructure of M is called f**in**itely generated if it is generated by a f**in**itesubset, i.e., it is of the form ⟨A⟩ Mfor some f**in**ite subset A of M. Observethat if a is a (possibly **in**f**in**ite) tuple of elements of M, ⟨a⟩ Mhas as doma**in**{t(a) | t(x) term of L}.Lemma 5.5.1. Let M, N be structures and let a and b be (possibly **in**f**in**ite)tuples **in** M and N respectively. Then the follow**in**g are equivalent:1. a and b have the same a**to**mic type.2. there is an isomorphism from ⟨a⟩ M**to** ⟨ b ⟩ Nsend**in**g a **to** b.Proof. 1 implies 2. Under the assumption atp M (a) = atp N (b) the map thatsends t(a) ↦→ t(b) for every term t is well def**in**ed and an isomorphism betweenthe substructures generated by a and b. This fact may be easily checked by thereader.Conversely, assume there is an isomorphism from ⟨a⟩ M**to** ⟨ b ⟩ send**in**g a **to**Nb. Then, by the homomorphism’s theorem (exercise ??) a and b have the samea**to**mic type.There are two quite **in**terest**in**g candidates **to** be a system of partial isomorphism.The first one is the set of all isomorphisms between f**in**itely generatedsubstructures of M and N respectively, denoted by I fg (M, N). The other oneis I 0 (M, N) = {(a, b) | a, b f**in**ite tuples of M and N respectively with the samea**to**mic type}.Proposition 5.5.2. Given a theory T the follow**in**g are equivalent:1. T is complete and elim**in**ates quantifiers.2. **An**y two ω-saturated models M, N of T are partially isomorphic via theset I 0 (M, N)3. **An**y two ω-saturated models M, N of T are partially isomorphic via theset I fg (M, N)

CHAPTER 5. PARTIAL ISOMORPHISM 62Proof. 1 ⇒ 2. Let M, N be ω-saturated models of T . By elim**in**ates quantifiersany two nonempty tuples liv**in**g **in** models of T have the same type iff they havethe same a**to**mic type. Therefore I 0 (M, N) co**in**cides with the set of all f**in**iteelementary maps from M **to** N. By completeness M and N are elementarilyequivalent. Now proposition 5.3.1 claims that I 0 (M, N) is a system of partialisomorphism.2 ⇒ 3. Tak**in**g **in****to** account lemma 5.5.1, it is easy **to** check that if I 0 (M, N)is a system of partial isomorphism then so is I fg (M, N).3 ⇒ 1. The completeness follows from corollary 5.3.2. For the elim**in**ation ofquantifiers we apply the criterion of proposition 5.4.6. Let M, N be models of Tand a, b be nonempty tuples of M and N respectively. By proposition 2.2.4 wecan suppose that M and N are ω-saturated. As atp M (a) = atp N (b), by lemma5.5.1 it follows that there is an isomorphism f : ⟨a⟩ M→ ⟨a⟩ Nwith f(a) = b.Then f ∈ I fg (M, N), which is a system of partial isomorphism. By corollary5.2.2 f is a partial elementary map and therefore a and b have the same typeover ∅.Example 5.5.3. Apply**in**g 5.5.2, **in** the examples 5.1.4, 5.1.6 and 5.3.3 we haveseen that the follow**in**g theories elim**in**ate quantifiers:1. The dense l**in**ear orders without endpo**in**ts **in** the language L = {≤}.2. The Random graph **in** the language L = {R}.3. The random structure **in** a f**in**ite relational language.4. The a**to**mless Boolean algebras **in** the language L = {∧, ∨, 0, 1}.5. The discrete orders without endpo**in**ts (the theory of (Z, ≤)) **in** the languageL = {≤, s}, where s is the ‘next’ function. There is no Q.E. **in** thepure order language (L = {≤}).6. Inf**in**ite **in**dependent relations **in** the language L = {P i | i ∈ ω}.Exercise 5.5.4. Prove the follow**in**g variation of the proposition 5.5.2. Given atheory T the follow**in**g are equivalent:1. T elim**in**ates quantifiers.2. **An**y two ω-saturated models M, N of T either they are partially isomorphicvia I fg (M, N) or I fg (M, N) is empty.3. **An**y two ω-saturated models M, N of T either they are partially isomorphicvia I 0 (M, N) or I 0 (M, N) is empty.Exercise 5.5.5. Prove that T is complete and elim**in**ates quantifiers iff givenany two (|L| + ℵ 0 ) + -saturated models M, N of T , the set of all isomorphismbetween substructures of M and N of card**in**al at most |L| + ℵ 0 is a systemof partial isomorphism. (H**in**t: for the left **to** right implication observe Q.E.implies that isomorphisms between substructures are elementary maps.)

CHAPTER 5. PARTIAL ISOMORPHISM 63Exercise 5.5.6. State and prove a generalization of exercise 5.5.5 replac**in**g |L|+ℵ 0 by any given card**in**al λ with λ ≥ |L| + ℵ 0 .Exercise 5.5.7. 1. Prove that T elim**in**ates quantifiers iff given any two modelsM, N of T with N (|L| + ℵ 0 ) + -saturated, given M 1 a substructures ofM of card**in**al at most |L| + ℵ 0 , given a ∈ M and given an embedd**in**g fof M 1 **in****to** N we can extend f **to** an embedd**in**g of M 1 ⟨a⟩ **in****to** N. HereM 1 ⟨a⟩ denotes the substructure of M generated by M 1 and a. (H**in**t:observe that this is the forth property for the system provided **in** exercise5.5.5).2. Generalize the criteria of po**in**t 1 replac**in**g |L| + ℵ 0 by any give cad**in**a λwith λ ≥ |L| + ℵ 0 . (H**in**t: use 5.5.6 **in**stead of 5.5.5).Exercise 5.5.8. In the language that conta**in**s a 1-ari function symbol f alone,we consider the theory T that says that f is a permutation (bijective). Provethat T elim**in**ates quantifiers. Give all its completions (h**in**t: for each n describethe number of n-cycles). [To be **work**ed, more h**in**ts]Exercise 5.5.9. Let L = {P i | i ∈ ω} be the language that conta**in**s countablymany unary predicates. We will describe all complete theories is this language.———————————————————————