Incomplete version for students of easllc2012 only.
Models and Games
JOUKO VÄÄNÄNEN University of Helsinki University of Amsterdam Incomplete version for students of easllc2012 only.
cambridge university press Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo, Delhi, Tokyo, Mexico City Cambridge University Press The Edinburgh Building, Cambridge CB2 8RU, UK
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© J. Väänänen 2011
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First published 2011
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Contents
Preface page xi 1 Introduction 1 2 Preliminaries and Notation 3 2.1 Finite Sequences 3 2.2 Equipollence 5 2.3 Countable sets 6 2.4 Ordinals 7 2.5 Cardinals 9 2.6 Axiom of Choice 10 2.7 Historical Remarks and References 11 Exercises 11 3 Games 14 3.1 Introduction 14 3.2 Two-Person Games of Perfect Information 14 3.3 The Mathematical Concept of Game 20 3.4 Game Positions 21 3.5 Infinite Games 24 3.6 Historical Remarks and References 28 Exercises 28 4 Graphs 35 4.1 Introduction 35 4.2 First-Order Language of Graphs 35 4.3 The Ehrenfeucht–Fra¨ısse´ Game on Graphs 38 4.4 Ehrenfeucht–Fra¨ısse´ Games and Elementary Equivalence 43 4.5 Historical Remarks and References 48 Exercises 49 Incomplete version for students of easllc2012 only. viii Contents
5 Models 53 5.1 Introduction 53 5.2 Basic Concepts 54 5.3 Substructures 62 5.4 Back-and-Forth Sets 63 5.5 The Ehrenfeucht–Fra¨ısse´ Game 65 5.6 Back-and-Forth Sequences 69 5.7 Historical Remarks and References 71 Exercises 71 6 First-Order Logic 79 6.1 Introduction 79 6.2 Basic Concepts 79 6.3 Characterizing Elementary Equivalence 81 6.4 The Lowenheim–Skolem¨ Theorem 85 6.5 The Semantic Game 93 6.6 The Model Existence Game 98 6.7 Applications 102 6.8 Interpolation 107 6.9 Uncountable Vocabularies 113 6.10 Ultraproducts 119 6.11 Historical Remarks and References 125 Exercises 126 7 Infinitary Logic 139 7.1 Introduction 139 7.2 Preliminary Examples 139 7.3 The Dynamic Ehrenfeucht–Fra¨ısse´ Game 144 7.4 Syntax and Semantics of Infinitary Logic 157 7.5 Historical Remarks and References 170 Exercises 171 8 Model Theory of Infinitary Logic 176 8.1 Introduction 176 8.2 Lowenheim–Skolem¨ Theorem for L ! 176 1 8.3 Model Theory of L!1! 179 8.4 Large Models 184
8.5 Model Theory of L+! 191 8.6 Game Logic 201 8.7 Historical Remarks and References 222 Exercises 223 Incomplete version for students of easllc2012 only. Contents ix
9 Stronger Infinitary Logics 228 9.1 Introduction 228 9.2 Infinite Quantifier Logic 228 9.3 The Transfinite Ehrenfeucht–Fra¨ısse´ Game 249 9.4 A Quasi-Order of Partially Ordered Sets 254 9.5 The Transfinite Dynamic Ehrenfeucht–Fra¨ısse´ Game 258 9.6 Topology of Uncountable Models 270 9.7 Historical Remarks and References 275 Exercises 278 10 Generalized Quantifiers 283 10.1 Introduction 283 10.2 Generalized Quantifiers 284 10.3 The Ehrenfeucht–Fra¨ısse´ Game of Q 296 10.4 First-Order Logic with a Generalized Quantifier 307 10.5 Ultraproducts and Generalized Quantifiers 312 10.6 Axioms for Generalized Quantifiers 314 10.7 The Cofinality Quantifier 333 10.8 Historical Remarks and References 342 Exercises 343 References 353 Index 362 Incomplete version for students of easllc2012 only. 1 Introduction
A recurrent theme in this book is the concept of a game. There are essentially three kinds of games in logic. One is the Semantic Game, also called the Eval- uation Game, where the truth of a given sentence in a given model is at issue. Another is the Model Existence Game, where the consistency in the sense of having a model, or equivalently in the sense of impossibility to derive a con- tradiction, is at issue. Finally there is the Ehrenfeucht–Fra¨ısse´ Game, where separation of a model from another by finding a property that is true in one given model but false in another is the goal. The three games are closely linked to each other and one can even say they are essentially variants of just one basic game. This basic game arises from our understanding of the quantifiers. The purpose of this book is to make this strategic aspect of logic perfectly transpar- ent and to show that it underlies not only first-order logic but infinitary logic and logic with generalized quantifiers alike. We call the close link between the three games the Strategic Balance of Logic (Figure 1.1). This balance is perfectly commutative, in the sense that winning strategies can be transferred from one game to another. This mere fact is testimony to the close connection between logic and games, or, thinking semantically, between games and models. This connection arises from the na- ture of quantifiers. Introducing infinite disjunctions and conjunctions does not upset the balance, barring some set-theoretic issues that may surface. In the last chapter of this book we consider generalized quantifiers and show that the Strategic Balance of Logic persists even in the presence of generalized quanti- fiers. The purpose of this book is to present the Strategic Balance of Logic in all its glory. Incomplete version for students of easllc2012 only. 2 Introduction
Figure 1.1 The Strategic Balance of Logic. Incomplete version for students of easllc2012 only. 2 Preliminaries and Notation
We use some elementary set theory in this book, mainly basic properties of countable and uncountable sets. We will occasionally use the concept of count- able ordinal when we index some uncountable sets. There are many excellent books on elementary set theory. (See Section 2.7.) We give below a simplified account of some basic concepts, the barest outline necessary for this book. We denote the set 0, 1, 2,... of all natural numbers by N, the set of ra- { } tional numbers by Q, and the set of all real numbers by R. The power-set operation is written (A)= B : B A . P { ✓ } We use A B to denote the set-theoretical difference of the sets A and B. \ 1 If f is a function, f 00X is the set f(x):x X and f (X) is the set { 2 } x dom(f):f(x) X . Composition of two functions f and g is denoted { 2 2 } g f and defined by (g f)(x)=g(f(x)). We often write fa for f(a). The notation id is used for the identity function A A which maps every A ! element of A to itself, i.e. id (a)=a for a A. A 2
2.1 Finite Sequences The concept of a finite (ordered) sequence
s =(a0,...,an 1) of elements of a given set A plays an important role in this book. Examples of finite sequences of elements of N are (8, 3, 9, 67, 200, 0) (8, 8, 8) Incomplete version for students of easllc2012 only. 4 Preliminaries and Notation
(24).
We can identify the sequence s =(a0,...,an 1) with the function
s0 : 0,...,n 1 A, { } ! where
s0(i)=ai.
The main property of finite sequences is: (a0,...,an 1)=(b0,...,bm 1) if and only if n = m and ai = bi for all i A B = (a, b):a A, b B . ⇥ { 2 2 } More generally A0 ... An 1 = (a0,...,an 1):ai Ai for all i Finite Sets A set A is finite if it is of the form a0,...,an 1 for some natural number n. { } This means that the set A has at most n elements. If A has exactly n elements we write A = n and call A the cardinality of A. A set which is not finite is | | | | infinite. Finite sets form a so-called ideal, which means that: 1. is finite. ; 2. If A and B are finite, then so is A B. [ 3. If A is finite and B A, then also B is finite. ✓ Further useful properties of finite sets are: 1. If A and B are finite, then so is A B. ⇥ 2. If A is finite, then so is (A). P Incomplete version for students of easllc2012 only. 2.2 Equipollence 5 The Axiom of Choice says that for every set A of non-empty sets there is a function f such that f(a) a for all a A. We shall use the Axiom of Choice 2 2 freely without specifically mentioning it. It needs some practice in set theory to see how the axiom is used. Often an intuitively appealing argument involves a hidden use of it. Lemma 2.1 A set A is finite if and only if every injective f : A A is a ! bijection. Proof Suppose A is finite and f : A B is an injection with B A and ! ⇢ a A B. Let a = a and a = f(a ). It is easy to see that a = a 2 \ 0 n+1 n n 6 m whenever n n The set of all n-element subsets a0,...,an 1 of A is denoted by [A] . { } 2.2 Equipollence Sets A and B are equipollent A B ⇠ 1 if there is a bijection f : A B. Then f : B A is a bijection and ! ! B A ⇠ follows. The composition of two bijections is a bijection, whence A B C = A C. ⇠ ⇠ ) ⇠ Thus divides sets into equivalence classes. Each equivalence class has a ⇠ canonical representative (a cardinal number, see the Subsection “Cardinals” below) which is called the cardinality of (each of) the sets in the class. The cardinality of A is denoted by A and accordingly A B is often written | | ⇠ A = B . | | | | One of the basic properties of equipollence is that if A C, B D and A B = C D = , ⇠ ⇠ \ \ ; then A B C D. [ ⇠ [ Incomplete version for students of easllc2012 only. 6 Preliminaries and Notation Indeed, if f : A C is a bijection and g : B D is a bijection, then ! ! f g : A B C D is a bijection. If the assumption [ [ ! [ A B = C D = \ \ ; is dropped, the conclusion fails, of course, as we can have A B = and \ ; C = D. It is also interesting to note that even if A B = C D = , the \ \ ; assumption A B C D does not imply B D even if A C is assumed: [ ⇠ [ ⇠ ⇠ Let A = N,B = ,C = 2n : n N , and D = 2n +1:n N . However, ; { 2 } { 2 } for finite sets this holds: if A B is finite, [ A B C D, A C, A B = C D = [ ⇠ [ ⇠ \ \ ; then B D. ⇠ We can interpret this as follows: the cancellation law holds for finite numbers but does not hold for cardinal numbers of infinite sets. There are many interesting and non-trivial properties of equipollence that we cannot enter into here. For example the Schroder–Bernstein¨ Theorem: If A B and B C A, then A C. Here are some interesting consequences ⇠ ✓ ✓ ⇠ of the Axiom of Choice: For all A and B there is C such that A C B or B C A. • ⇠ ✓ ⇠ ✓ For all infinite A we have A A A. • ⇠ ⇥ It is proved in set theory by means of the Axiom of Choice that A B | | | | holds in the above sense if and only if the cardinality A of the set A is at | | most the cardinality B of the set B. Thus the notation A B is very | | | | | | appropriate. 2.3 Countable sets A set A which is empty or of the form a0,a1,... , i.e. an : n N , is called { } { 2 } countable. A set which is not countable is called uncountable. The countable sets form an ideal just as the finite sets do. We now prove two important results about countability. Both are due to Georg Cantor: Theorem 2.2 If A and B are countable, then so is A B. ⇥ Proof If either set is empty, the Cartesian product is empty. So let us assume Incomplete version for students of easllc2012 only. 2.4 Ordinals 7 the sets are both non-empty. Suppose A = a ,a ,... and B = b ,b ,... . { 0 1 } { 0 1 } Let (a ,b ), if n =2i3j c = i j n (a ,b ), otherwise. ⇢ 0 0 Now A B = cn : n N , whence A B is countable. ⇥ { 2 } ⇥ Theorem 2.3 The union of a countable family of countable sets is countable. Proof The empty sets do not contribute anything to the union, so let us as- sume all the sets are non-empty. Suppose An is countable for each n N, n 2 say, An = a : m N (we use here the Axiom of Choice to choose an { m 2 } enumeration for each An). Let B = n An. We want to represent B in the i j form bn : n N . If n is given, we consider two cases: If n is 2 3 for some { 2 } i S 0 i and j, we let bn = aj. Otherwise we let bn = a0. Theorem 2.4 The power-set of an infinite set is uncountable. Proof Suppose A is infinite and (A)= bn : n N . Since A is infinite, P { 2 } we can choose distinct elements an : n N from A. (This uses the Axiom of { 2 } Choice. For an argument which avoids the Axiom of Choice see Exercise 2.14.) Let B = a : a / b . { n n 2 n} Since B A, there is some n such that B = b . Is a an element of B ✓ n n or not? If it is, then a / b which is a contradiction. So it is not. But then n 2 n a b = B, again a contradiction. n 2 n 2.4 Ordinals The ordinal numbers introduced by Cantor are a marvelous general theory of measuring the potentially infinite. They are intimately related to inductive def- initions and occur therefore widely in logic. It is easiest to understand ordinals in the context of games, although this was not Cantor’s way. Suppose we have a game with two players I and II. It does not matter what the game is, but it could be something like chess. If II can force a win in n moves we say that the game has rank n. Suppose then II cannot force a win in n moves for any n, but after she has seen the first move of I, she can fix a number n and say that she can force a win in n moves. This situation is clearly different from being able to say in advance what n is. So we invent a symbol ! for the rank of this game. In a clear sense ! is greater than each n but there does not seem Incomplete version for students of easllc2012 only. 8 Preliminaries and Notation to be any possible rank between all the finite numbers n and !. We can think of ! as an infinite number. However, there is nothing metaphysical about the infiniteness of !. It just has infinitely many predecessors. We can think of ! as a tree T! with a root and a separate branch of length n for each n above the root as in the tree on the left in Figure 2.1. Figure 2.1 T! and T!+1. Suppose then II is not able to declare after the first move how many moves she needs to beat II, but she knows how to play her first move in such a way that after I has played his second move, she can declare that she can win in n moves. We say that the game has rank ! +1and agree that this is greater than ! but there is no rank between them. We can think of ! +1as the tree which has a root and then above the root the tree T!, as in the tree on the right in Figure 2.1. We can go on like this and define the ranks ! + n for all n. Suppose now the rank of the game is not any of the above ranks ! + n, but still II can make an interesting declaration: she says that after the first move of I she can declare a number m so that after m moves she declares another number n and then in n moves she can force a win. We would say that the rank of the game is ! +!. We can continue in this way defining ranks of games that are always finite but potentially infinite. These ranks are what set theorists call ordinals. We do not give an exact definition of the concept of an ordinal, because it would take us too far afield and there are excellent textbooks on the topic. Let us just note that the key properties of ordinals and their total order < are: 1. Natural numbers are ordinals. 2. For every ordinal ↵ there is an immediate successor ↵ +1. 3. Every non-empty set of ordinals has a smallest element. 4. Every non-empty set of ordinals has a supremum (i.e. a smallest upper bound). Incomplete version for students of easllc2012 only. 2.5 Cardinals 9 The supremum of the set 0, 1, 2, 3,... of ordinals is denoted by !. An { } ordinal is said to be countable if it has only countably many predecessors, otherwise uncountable. The supremum of all countable ordinals is denoted by !1. Here is a picture of the ordinal number “line”: 0 < 1 < 2 <... Ordinals that have a last element, i.e. are of the form ↵ +1, are called suc- cessor ordinals; the rest are limit ordinals, like ! and ! + !. Ordinals are often used to index elements of uncountable sets. For example, a : ↵ < denotes a set whose elements have been indexed by the ordinal { ↵ } , called the length of the sequence. The set of all such sequences of length of elements of a given set A is denoted by A . The set of all sequences of length < of elements of a given set A is denoted by A< . 2.5 Cardinals Historically cardinals (or more exactly cardinal numbers) are just representa- tives of equivalence classes of equipollence. Thus there is a cardinal number for countable sets, denoted , a cardinal number for the set of all reals, de- @0 noted c, and so on. There is some question as to what exactly are these cardinal numbers. The Axiom of Choice offers an easy answer, which is the prevailing one, as it says that every set can be well-ordered. Then we can let the cardi- nal number of a set be the order-type of the smallest well-order equipollent with the set. Equivalently, the cardinal number of a set is the smallest ordinal equipollent with the set. If we leave aside the Axiom of Choice, some sets need not have have a cardinal number. However, as is customary in current set the- ory, let us indeed assume the Axiom of Choice. Then every set has a cardinal number and the cardinal numbers are ordinals, hence well-ordered. The ↵th in- finite cardinal number is denoted . Thus is the next in order of magnitude @↵ @1 from . The famous Continuum Hypothesis is the statement that = c. @0 @1 For every set A there exists (by the Axiom of Choice) an ordinal ↵ such that the elements of A can be listed as a : < ↵ . The smallest such ↵ is { } called the cardinal number, or cardinality, of A and denoted by A . Thus cer- | | tain ordinals are cardinal numbers of sets. Such ordinals are called cardinals. They are considered as canonical representatives of each equivalence class of equipollent sets. For example, all finite numbers are cardinals, as are ! and !1. The smallest cardinal such that the smaller infinite cardinals can be enumer- ated in increasing order as , < ↵, is denoted ! , or alternatively . If ↵ @↵ Incomplete version for students of easllc2012 only. 10 Preliminaries and Notation = , then is denoted + and is called a successor cardinal. Cardinals @↵ @↵+1 that are not successor cardinals are called limit cardinals. Arithmetic operations + , , for cardinals are defined as follows: · + = , = . | [ | · | ⇥ | Moreover, exponentiation of cardinal numbers is defined as the cardinality of the set of sequences of elements of of length . A certain amount of knowledge about the arithmetic of cardinal numbers in necessary in this book, especially in the later chapters, and Chapters 8 and 9 in particular. The cofinality of an ordinal ↵ is the smallest ordinal for which there is a function f : ↵ such that (1) ⇠ < ⇣ < implies f(⇠) 2.6 Axiom of Choice We have already mentioned the Axiom of Choice. There are so many equiv- alent formulations of this axiom that books have been written about it. The most notable formulation is the Well-Ordering Principle: every set is equipol- lent with an ordinal. The Axiom of Choice is sometimes debated because it brings arbitrariness or abstractness into mathematics, often with examples that can be justifiably called pathological, like the Banach–Tarski Paradox: The unit sphere in three-dimensional space can be split into five pieces so that if the pieces are rigidly moved and rotated they form two spheres, each of the original size. The trick is that the splitting exists only in the abstract world of mathematics and can never actually materialize in the physical world. Con- clusion: infinite abstract objects do not obey the rules we are used to among finite concrete objects. This is like the situation with sub-atomic elementary particles, where counter-intuitive phenomena, such as entanglement, occur. Because of the abstractness brought about by the Axiom of Choice it has received criticism and some authors always mention explicitly if they use it in their work. The main problem in working without the Axiom of Choice is Incomplete version for students of easllc2012 only. 3 Games 3.1 Introduction In this first part we march through the mathematical details of zero-sum two- person games of perfect information in order to be well prepared for the intro- duction of the three games of the Strategic Balance of Logic (see Figure 1.1) in the subsequent parts of the book. Games are useful as intuitive guides in proofs and constructions but it is also important to know how to make the in- tuitive arguments and concepts mathematically exact. 3.2 Two-Person Games of Perfect Information Two-person games of perfect information are like chess: two players set their wits against each other with no role for chance. One wins and the other loses. Everything is out in the open, and the winner wins simply by having a better strategy than the loser. A Preliminary Example: Nim In the game of Nim, if it is simplified to the extreme, there are two players I and II and a pile of six identical tokens. During each round of the game player I first removes one or two tokens from the top of the pile and then player II does the same, if any tokens are left. Obviously there can be at most three rounds. The player who removes the last token wins and the other one loses. The game of Figure 3.1 is an example of a zero-sum two-person game of perfect information. It is zero-sum because the victory of one player is the loss of the other. It is of perfect information because both players know what the other player has played. A moment’s reflection reveals that player II has a way Incomplete version for students of easllc2012 only. 16 Games II 111111 I I I I I 11111 11112 11121 11211 II 12111 II II 21111 II II II 1111 1112 1121 1122 1211 1212 1221 2111 2112 2121 2211 I 111 112 121 122 211 212 221 222 11 12 21 22 1 2 Figure 3.3 cabulary L = W , where W is a four-place predicate symbol. Let be an { } M L-structure2 with M = 1, 2 and { } 4 W M = (a ,b ,a ,b ) M : a + b + a + b =6 . { 0 0 1 1 2 0 0 1 1 } Now we have just proved = x y x y W (x ,y ,x ,y ). (3.1) M| 8 09 08 19 1 0 0 1 1 Conversely, if is an arbitrary L-structure, condition (3.1) defines some M game, maybe not a very interesting one but a game nonetheless: Player I picks an element a M, then player II picks an element b M. Then the same 0 2 0 2 is repeated: player I picks an element a M, then player II picks an element 1 2 b M. After this player II is declared the winner if (a ,b ,a ,b ) W M, 1 2 0 0 1 1 2 and otherwise player I is the winner. By varying the structure we can model M in this way various two-person two-round games of perfect information. This gives a first hint of the connection between games and logic. Games – a more general formulation Above we saw an example of a two-person game of perfect information. This concept is fundamental in this book. In general, the simplest formulation of such a game is as follows (see Figure 3.4): There are two players3 I and II,a domain A, and a natural number n representing the length of the game. Player I starts the game by choosing some element x A. Then player II chooses 0 2 y A. After x and y have been played, and i +1 x0 y0 x1 y1 . . . . xn 1 yn 1 Figure 3.4 A game. and declare that player II wins the game if the sequence formed during the game is in W ; otherwise player I wins. We denote this game by (A, W ). For Gn example, if W = , player II cannot possibly win, and if W = A2n, player ; I cannot possibly win. If W is a set of sequences (x0,y0,...,xn 1,yn 1) where x0 = x1 and if moreover A has at least two elements, then II could not possibly win, as she cannot prevent player I from playing x0 and x1 differently. On the other hand, W could be the set of all sequences (3.2) such that y0 = y1. Then II can always win because all she has to do during the game is make sure that she chooses y0 and y1 to be the same element. If player II has a way of playing that guarantees a sure win, i.e. the opponent I loses whatever moves he makes, we say that player II has a winning strategy in the game. Likewise, if player I has a way of playing that guarantees a sure win, i.e. player II loses whatever moves she makes, we say that player I has a winning strategy in the game. To make intuitive concepts, such as “way of playing” more exact in the next chapter we define the basic concepts of game theory in a purely mathematical way. Example 3.1 The game of Nim presented in the previous chapter is in the present notation ( 1, 2 ,W), where G3 { } n W = (a ,b ,a ,b ,a ,b ) 1, 2 6 : (a + b )=6for some n 2 . 0 0 1 1 2 2 2 { } i i ( i=0 ) X We allow three rounds as theoretically the players could play three rounds even if player II can force a win in two rounds. Example 3.2 Consider the following game on a set A of integers: Incomplete version for students of easllc2012 only. 20 Games Example 3.5 The following game has no moves: III If W = , player II is the winner. If W = , player I is the winner. So this {;} ; is a game with 0 rounds. In practice one of the players would find these games unfair as he or she loses without even having a chance to make a move. It is like being invited to play a game of chess starting in a position where you are already in check-mate. 3.3 The Mathematical Concept of Game Let A be an arbitrary set and n a natural number. Let W A2n. We redefine ✓ the game (A, W ) Gn in a purely mathematical way. Let us fix two players I and II.Aplay of one of the players is any sequence x¯ =(x0,...,xn 1) of elements of A. A sequence (¯x;¯y)=(x0,y0,...,xn 1,yn 1), of elements of A is called a play (of (A, W )). So we have defined the con- Gn cept of play without any reference to playing the game as an act. The play (¯x;¯y) is a win for player II if (x0,y0,...,xn 1,yn 1) W 2 and otherwise a win for player I. Example 3.6 Let us consider the game of chess in this mathematical frame- work. We modify the game so that the number of rounds is for simplicity ex- actly n and Black wins a draw, i.e. if neither player has check-mated the other player during those up to n rounds. If a check-mate is reached the rest of the n-round game is of course irrelevant and we can think that the game is finished with “dummy” moves. Let A be the set of all possible positions, i.e. config- urations of the pieces on the board. A play x¯ of I (White) is the sequence of positions where White has just moved. A play y¯ of II is the sequence of posi- tions where Black has just moved. We let W be the set of plays (¯x;¯y), where either White has not obeyed the rules, or Black has obeyed the rules and White has not check-mated Black. With the said modifications, chess is just the game (A, W ) with White playing as player I and Black playing as player II. Gn Incomplete version for students of easllc2012 only. 3.4 Game Positions 21 A strategy of player I in the game (A, W ) is a sequence Gn =( 0,..., n 1) of functions : Ai A. We say that player I has used the strategy in the i ! play (¯x;¯y) if for all 0 xi = i(y0,...,yi 1) and x0 = 0. The strategy of player I is a winning strategy, if every play where I has used is a win for player I. Note that the strategy depends only on the opponent’s moves. It is tacitly assumed that when the function i+1 is used to determine xi+1, the previous functions 0,..., i were used to determine the previous moves x0,...,xn. Thus a strategy is a winning strategy because of the con- certed effect of all the functions 0,..., n 1. A strategy of player II in the game (A, W ) is a sequence Gn ⌧ =(⌧0,...,⌧n 1) of functions ⌧ : Ai+1 A. We say that player II has used the strategy ⌧ in i ! the play (¯x;¯y) if for all i yi = ⌧i(x0,...,xi). The strategy ⌧ of player II is a winning strategy, if every play where player II has used ⌧ is a win for player II. A player who has a winning strategy in (A, W ) is said to win the game (A, W ). Gn Gn 3.4 Game Positions A position of the game (A, W ) is any initial segment Gn p =(x0,y0,...,xi 1,yi 1) of a play (¯x;¯y), where i n. Positions have a natural ordering: a position p0 extends a position p, if p is an initial segment of p0. Of course, this extension- 4 relation is a partial ordering of the set of all positions, that is, if p0 extends p and p00 extends p0, then p00 extends p, and if p and p0 extend each other, then p = p0. The empty sequence is the smallest element, and the plays (¯x;¯y) are ; 4 See Example 5.7 for the definition of partial order. Indeed this is a tree-ordering. See Example 5.8 for the definition of tree-ordering. Incomplete version for students of easllc2012 only. 22 Games maximal elements of this partial ordering. A common problem of games is that the set of all positions is huge. A strategy of player I in position p =(x0,y0,...,xi 1,yi 1) in the game (A, W ) is a sequence Gn =( 0,..., n 1 i) of functions : Aj A. We say that player I has used strategy after j ! position p in the play (¯x;¯y), if (¯x;¯y) extends p and for all j with i xj = j i(yi,...,yj 1) and xi = 0. The strategy of player I in position p is a winning strategy in position p, if every play extending p where player I has used after position p is a win for player I. A strategy of player II in position p in the game (A, W ) is a sequence Gn ⌧ =(⌧0,...,⌧n 1 i) of functions ⌧ : Aj+1 A. We say that player II has used strategy ⌧ after j ! position p in the play (¯x;¯y) if (¯x;¯y) extends p and for all j with i j yj = ⌧j i(xi,...,xj). The strategy ⌧ of player II in position p is a winning strategy in position p, if every play extending p where player II has used ⌧ after p is a win for player II. The following important lemma shows that if player II has a chance in the beginning, i.e. player I does not already have a winning strategy, she has a chance all the way. Lemma 3.7 (Survival Lemma) Suppose A is a set, n is a natural num- 2n ber, W A and p =(x0,y0,...,xi 1,yi 1) is a position in the game ✓ (A, W ), with i a winning strategy already in position p, as he wins whatever II moves. Let us now make this idea more exact. Suppose there were an xi A such that for all y 2 y A player I has a winning strategy i in position p0 =(x ,y ,...,x ,y ). i 2 0 0 i i We define a strategy =( 0,..., n 1 i) of player I in position p as follows: ( )=x and 0 ; i yi j i(yi,...,yj i)= (yi+1,...,yj i). This is a winning strategy of I in position p, contrary to our assumption that none exists. The following concept is of fundamental importance in game theory and in applications to logic, in particular: Definition 3.8 A game is called determined if one of the players has a win- ning strategy. Otherwise the game is non-determined. Virtually all games that one comes across in logic are determined. The fol- lowing theorem is the crucial fact behind this phenomenon: Theorem 3.9 (Zermelo) If A is any set, n is a natural number, and W A2n, ✓ then the game (A, W ) is determined. Gn Proof Suppose player I has no winning strategy. Then player II has a win- ning strategy based on repeated use of Lemma 3.7. Player II notes that in the beginning of the game, that is, in position , player I does not have a winning ; strategy. Then by the Survival Lemma 3.7 she can, whatever player I moves, find a move such that afterwards player I still does not have a winning strategy. In short, the strategy of player II is to prevent player I from having a winning strategy. After n rounds the game ends and player I still does not have a win- ning strategy. That means player I has lost and player II has won. Let us now make this more precise: We define a strategy ⌧ =(⌧0,...,⌧n 1) of player II in the game (A, W ) as follows: Let a be some arbitrary element Gn of A. By Lemma 3.7 we have for each position p =(x0,y0,...,xi 1,yi 1) in the game (A, W ) such that player I does not have a winning strategy in Gn position p and each x A some y A such that player I does not have a i 2 i 2 winning strategy in position p0 =(x0,y0,...,xi,yi). Let us denote this yi by yi = f(p, xi). Incomplete version for students of easllc2012 only. 24 Games If p =(x0,y0,...,xi 1,yi 1) is a position in which player I does have a win- ning strategy, we let f(p, xi)=a. We have defined a function f defined on po- sitions p and elements xi A. Let ⌧0(x0)=f( ,x0). Assuming ⌧0,...,⌧i 1 2 ; have been defined already, let ⌧i(x0,...,xi)=f(p, xi), where p =(x0,y0,...,xi 1,yi 1) and y0 = ⌧0(x0) yi 1 = ⌧i 1(x0,...,xi 1). It is easy to see that in every play in which player II uses this strategy, every position p is such that player I does not have a winning strategy in position p. It is also easy to see that this is a winning strategy of player II. 3.5 Infinite Games The concept of a game is by no means limited to games with just finitely many rounds. Imagine a chess board which extends the usual board left and right without end. Then the chess game could go on for infinitely many rounds with- out the same configuration of pieces coming up twice. A simple infinite game is one in which two players pick natural numbers each choosing a bigger num- ber, if he or she can, than the opponent. There is no end to this game, since there are infinitely many natural numbers. A third kind of infinite game is the following: Example 3.10 Suppose A is a set of real numbers on the unit interval. We describe a game we denote by G(A). During the game the players decide the decimal expansion of a real number r =0.d0d1 ...on the interval [0, 1]. Player I decides the even digits d2n and player II the odd digits d2n+1. Player II wins if r A. If A is countable, say A = bn : n N , player I has a 2 { 2 } winning strategy: during round n he chooses the digit d so that r = b . 2n 6 n If the complement of A is countable, player II wins with the same strategy. What if A and its complement are uncountable? This is a well-known and much studied hard question. (See e.g. Jech (1997).) Incomplete version for students of easllc2012 only. 3.5 Infinite Games 25 III x0 y0 x1 y1 . . . . Figure 3.5 An infinite game. If A is any set, we use AN to denote infinite sequences (x0,x1,...) of elements of A. We can think of such sequences as limits of an increasing sequence (x0), (x0,x1), (x0,x1,x2),... of finite sequences. Let A be an arbitrary set. Let W AN. We define the game ✓ (A, W ) G! as follows (see Figure 3.5): An infinite sequence (¯x;¯y)=(x0,y0,x1,y1,...), of elements of A is called a play (of (A, W )). A play of one of the players G! is likewise any infinite sequence x¯ =(x0,x1,...) of elements of A. The play (¯x;¯y) is a win for player II if (x ,y ,x ,y ,...) W 0 0 1 1 2 and otherwise a win for player I . A strategy of player I in the game (A, W ) is an infinite sequence G! =( 0, 1,...) of functions : Ai A. We say that player I has used the strategy in the i ! play (¯x;¯y) if for all i N: 2 xi = i(y0,...,yi 1) and x0 = 0. Incomplete version for students of easllc2012 only. 26 Games The strategy of player I is a winning strategy, if every play where I has used is a win for player I. A strategy of player II in the game (A, W ) is an infinite sequence G! ⌧ =(⌧0, ⌧1,...) of functions ⌧ : Ai+1 A. We say that player II has used the strategy ⌧ in i ! the play (¯x;¯y) if for all i yi = ⌧i(x0,...,xi). The strategy ⌧ of player II is a winning strategy, if every play where player II has used ⌧ is a win for player II. A player is said to win the game (A, W ) G! if he or she has a winning strategy in it. A position of the infinite game (A, W ) is any initial segment G! p =(x0,y0,...,xi 1,yi 1) of a play (¯x;¯y). We say that player I has used strategy =( 0, 1,...) after position p in the play (¯x;¯y), if (¯x;¯y) extends p and for all j with i (x0,y0,...,xn 1,yn 1,xn0 ,yn0 ,xn0 +1,yn0 +1,...) W 2 for all x0 ,y0 ,x0 ,y0 ,... A. Respectively, W is closed if AN W n n n+1 n+1 2 \ is open. Finally, W is clopen if it is both open and closed. We call a game G!(A, W ) closed (or open or clopen) if the set W is. We are mainly concerned in this book with closed games. A typical strategy of player II in a closed game is to “hang in there”, as she knows that if player I ends up winning the play p =(x ,y ,...), that is, p/W , there is some n such that player I won the 0 0 2 game already in position (x0,y0,...,xn 1,yn 1). 5 The collection of open subsets of AN is a topology, hence the name. Incomplete version for students of easllc2012 only. 3.5 Infinite Games 27 We can think of infinite games as limits of finite games as follows: Any finite game Gn(A, W ) can be made infinite by disregarding the moves after the usual n moves. The resulting infinite game is clopen (see Exercise 3.31). On the other hand, if G!(A, W ) is an infinite game and n N we can form an 2 n-round game by simply considering only the first n rounds of G!(A, W ) and declaring a play of n rounds a win for player II if any infinite play extending it is in W . Unless W is open or closed, there may be very little connection between the resulting finite games and the original infinite game (see however Exercise 3.32). Lemma 3.11 (Infinite Survival Lemma) Suppose A is a set, W AN, and ✓ p =(x0,y0,...,xi 1,yi 1) is a position in the game !(A, W ), with i N. G 2 Suppose furthermore that player I does not have a winning strategy in position p. Then for every x A there is y A such that player I does not have a i 2 i 2 winning strategy in position p0 =(x0,y0,...,xi,yi). Proof The proof is by contradiction. Suppose there were an xi A such y 2 that for all y A player I has a winning strategy i in position p0 = i 2 (x0,y0,...,xi,yi). We define a strategy =( 0, 1,...) of player I in posi- tion p as follows: ( )=x and for j>i, 0 ; i yi j i(yi,...,yj 1)= (yi+1,...,yj i). This is a winning strategy of player I in position p, contrary to assumption. Theorem 3.12 (Gale–Stewart) If A is any set and W AN is open or closed, ✓ then the game (A, W ) is determined. G! Proof Suppose first W is closed and player I has no winning strategy. We define a strategy ⌧ =(⌧0, ⌧1,...) of player II in the game (A, W ) as follows: Let a be some arbitrary element G! of A. By Lemma 3.11 we have for each position p =(x0,y0,...,xi 1,yi 1) in the game (A, W ) such that player I does not have a winning strategy in G! position p, and each x A, some y A such that player I does not have a i 2 i 2 winning strategy in position p0 =(x0,y0,...,xi,yi). Let us denote this yi by yi = f(p, xi). If p =(x0,y0,...,xi 1,yi 1) is a position in which player I does have a winning strategy, we let f(p, xi)=a. We have defined a function f de- fined on positions p and elements x A. Let ⌧ (x )=f( ,x ). Assuming i 2 0 0 ; 0 ⌧0,...,⌧i 1 have been defined already, let ⌧i(x0,...,xi)=f(p, xi), where Incomplete version for students of easllc2012 only. 28 Games p =(x0,y0,...,xi 1,yi 1) and y0 = ⌧0(x0),yi 1 = ⌧i 1(x0,...,xi 1). It is easy to see that in every play in which player II uses this strategy, every position p is such that player I does not have a winning strategy in position p. It is also easy to see that this is a winning strategy of player II. The proof is similar if W is open. It follows that G!(A, W ) is determined. Theorem 3.12 can been vastly generalized, see e.g. (Jech, 1997, Chapter 33). The Axiom of Determinacy says that the game G!(A, W ) is determined for all sets A and W . However, this axiom contradicts the Axiom of Choice. By using the Axiom of Choice one can show that there are sets A of real numbers such that the game G(A) is not determined (see Exercise 3.37). 3.6 Historical Remarks and References The mathematical theory of games was started by von Neumann and Morgen- stern (1944). For the early history of two-person zero-sum games of perfect in- formation, see Schwalbe and Walker (2001). See Mycielski (1992) for a more recent survey on games of perfect information. Theorem 3.12 goes back to Gale and Stewart (1953). Exercises 3.1 Consider the following game: Player I picks a natural number n. Then player II picks a natural number m. If 2m = n, then II wins, otherwise I wins. Express this game in the form (A, W ). G1 3.2 Consider the following game: Player I picks a natural number n. Then player II picks two natural numbers m and k. If m k = n, then II wins, · otherwise I wins. Express this game in the form (A, W ). G2 3.3 Consider (A, W ), where A = 0, 1, 2 and G3 { } 1. W = (x ,y ,x ,y ,x ,y ) A3 : x = y . { 0 0 1 1 2 2 2 0 2} 2. W = (x ,y ,x ,y ,x ,y ) A3 : y = x or y = x . { 0 0 1 1 2 2 2 0 6 2 2 6 0} 3. W = (x ,y ,x ,y ,x ,y ) A3 : x = y and x = y and x = { 0 0 1 1 2 2 2 0 6 2 1 6 2 2 6 y . 2} Who has a winning strategy? 3.4 Suppose f : R R is a mapping. Express the condition that f is uni- ! formly continuous as a game and as the truth of a first-order sentence in a suitable structure. Incomplete version for students of easllc2012 only. 30 Games 3.10 Examine the game determined by condition (3.1) M = N and W M = (a ,b ,a ,b ) M 4 : a 3.21 Give the winning strategy of player II in Nim (Example 3.1) in the form ⌧ =(⌧0, ⌧1). 3.22 Consider the game of Example 3.3 when f(x)=2x +3, a R, and 2 b =2a +3. Give some winning strategy of player II. 3.23 Consider the game of Example 3.3 when f(x)=2x +3, a =1and b =4. Give some winning strategy of player I. 3.24 Consider the game of Example 3.3 when f(x)=x2, a R, and b = a2. 2 Give some winning strategy of player II. 3.25 A more general version of Nim has m tokens rather than six. Decide who has a winning strategy for each m and give the winning strategy. 3.26 Suppose we have two games (A, W ) and (A0,W0), where A A0 = Gn Gn \ . Let A00 = A A0 and let W 00 be the set of sequences ; [ (x0,y0,...,x2n 1,y2n 1), which satisfy the following condition: (x0,y0,x2,y2 ...,x2n 2,yn 2) W 2 and (x1,y1,x3,y3,...,x2n 1,y2n 1) W 0. 2 Show that: 1. If player I has a winning strategy in (A, W ) or in (A0,W0), Gn Gn then he has one in (A00,W00). G2n 2. If player II has a winning strategy in (A, W ) and in (A0,W0), Gn Gn then she has one in (A00,W00). G2n 3.27 Suppose we have two games (A, W ) and (A0,W0). Let A00 = A Gn Gn ⇥ A0 and let W 00 be the set of sequences (((x0,x0), (y0,y0)),...,((xn 1,xn0 1), (yn 1,yn0 1))), where (x0,y0,...,xn 1,yn 1) W 2 and (x0,y0,...,xn0 1,yn0 1) W 0. 2 Show that: 1. If player I has a winning strategy in (A, W ) or in (A0,W0), Gn Gn then he has one in (A00,W00). Gn 2. If player II has a winning strategy in (A, W ) and in (A0,W0), Gn Gn then she has one in (A00,W00). Gn 5 Models 5.1 Introduction The concept of a model (or structure) is one of the most fundamental in logic. In brief, while the meaning of logical symbols ∧, ∨, ∃,... is always fixed, models give meaning to non-logical symbols such as constant, predicate, and function symbols. When we have agreed about the meaning of the logical and non-logical symbols of logic, we can then define the meaning of arbitrary for- mulas. Depending on context and preference, models appear in logic in two roles. They can serve the auxiliary role of clarifying logical derivation. For example, one quick way to tell what it means for ϕ to be a logical consequence of ψ is to say that in every model where ψ is true also ϕ is true. It is then an almost trivial matter to understand why for example ∀x∃yϕ is a logical consequence of ∃y∀xϕ but ∀y∃xϕ is in general not. Alternatively models can be the prime objects of investigation and it is the logical derivation that is in an auxiliary role of throwing light on properties of models. This is manifestly demonstrated by the Completeness Theorem which says that any set T of first-order sentences has a model unless a contradiction can be logically derived from T , which entails that the two alternative perspec- tives of models are really equivalent. Since derivations are finite, this implies the important Compactness Theorem: If a set of first-order sentences is such that each of its finite subsets has a model it itself has a model. The Compact- ness Theorem has led to an abundance of non-isomorphic models of first-order theories, and constitutes the origin of the whole subject of Model Theory. In this chapter models are indeed the prime objects of investigation and we in- troduce auxiliary concepts such as the Ehrenfeucht–Fra¨ısse´ Game that help us understand models. We use the words “model” and “structure” as synonyms. We have a slight Incomplete version for students of easllc2012 only. 54 Models preference for the word “structure” in a context where absolute generality pre- vails and the structures are not assumed to satisfy any particular axioms. Re- spectively, our preference is to call a structure that satisfies some given axioms a model, so a structure satisfying a theory is called a model of the theory. 5.2 Basic Concepts A vocabulary is any set L of predicate symbols P, Q, R, . . ., function sym- bols f,g,h,..., and constant symbols c, d, e, . . .. Each vocabulary has an arity- function #L : L N ! which tells the arity of each symbol. Thus if P L, then P is a # (P )-ary 2 L predicate symbol. If f L, then f is a # (f)-ary function symbol. Finally, 2 L # (c) is assumed to be 0 for constants c L. Predicate or function symbols L 2 of arity 1 are called unary or monadic, and those of arity 2 are called binary. A vocabulary is called unary (or binary) if it contains only unary (respectively, binary) symbols. A vocabulary is called relational if it contains no function or constant symbols. Definition 5.1 An L-structure (or L-model) is a pair =(M,Val ), M M where M is a non-empty set called the universe (or the domain) of , and M Val is a function defined on L with the following properties: M n 1. If R L is a relation symbol and #L(R)=n, then Val (R) M . 2 M ✓ n 2. If f L is a function symbol and #L(f)=n, then Val (f):M M. 2 M ! 3. If c L is a constant symbol, then Val (c) M. 2 M 2 We use Str(L) to denote the class of all L-structures. We usually shorten Val (R) to RM, Val (f) to f M, and Val (c) to cM. M M M If no confusion arises, we use the notation =(M,RM,...,RM,fM,...,fM,cM,...,cM) M 1 n 1 m 1 k for an L-structure , where L = R ,...,R ,f ,...,f ,c ...,c . M { 1 n 1 m 1 k} Example 5.2 Graphs are L-structures for the relational vocabulary L = E , { } where E is a predicate symbol with #L(E)=2. Groups are L-structures for L = , where is a binary function symbol. Fields are L-structures for { } L = +, , 0, 1 , where +, are binary function symbols and 0, 1 are constant { · } · symbols. Ordered sets (i.e. linear orders) are L-structures for the relational Incomplete version for students of easllc2012 only. 5.2 Basic Concepts 55 vocabulary L = < , where < is a binary predicate symbol. If L = , an { } ; L-structure (M) is a structure with just the universe and no structure in it. If is a structure and ⇡ maps M bijectively onto another set M 0, we can M use ⇡ to copy the relations, functions, and constants of on M 0. In this way M we get a perfect copy 0 of which differs from only in the respect that M M M the underlying elements are different. We then say that 0 is an isomorphic M copy of . For all practical purposes we consider the structures and 0 M M M as one and the same structure. However, they are not the same structure, just isomorphic. This may sound as if isomorphism was a rather trivial matter, but this is not true. In many cases it is a highly non-trivial enterprise to investigate whether two structures are isomorphic or not. In the realm of finite structures the question of deciding whether two given structures are isomorphic or not is a famous case of a complexity question which is between P (polynomial time) and NP (non-deterministic polynomial time) and about which we do not know whether it is NP-complete. In the light of present knowledge it is conceivable that this question is strictly between P and NP. Definition 5.3 L-structures and 0 are isomorphic if there is a bijection M M ⇡ : M M 0 ! such that 1. For all a ,...,a M: 1 #L(R) 2 0 (a ,...,a ) RM (⇡(a ),...,⇡(a )) RM . 1 #L(R) 2 () 1 #L(R) 2 2. For all a ,...,a M: 1 #L(f) 2 0 f M (⇡(a1),...,⇡(a#L(f))) = ⇡(f M(a1,...,a#L(f))). 0 3. For all c L: ⇡(cM)=cM . 2 In this case we say that ⇡ is an isomorphism 0, denoted M ! M ⇡ : = 0. M ⇠ M If also = 0, we say that ⇡ is an automorphism of . M M M Example 5.4 Unary (or monadic) structures, i.e. L-structures for unary L, are particularly simple and easy to deal with. Figure 5.1 depicts a unary structure. Suppose L consists of unary predicate symbols R ,...,R and is an L- 1 n A structure. If X A and d 0, 1 , let Xd = X if d =0and Xd = A X ✓ 2 { } \ Incomplete version for students of easllc2012 only. 56 Models Figure 5.1 A unary structure. otherwise. Suppose ✏ : 1,...,n 0, 1 . The ✏-constituent of is the set { } ! { } A n ✏(i) C ( )= (RA) . ✏ A i i=1 \ A priori, the 2n sets C ( ) can each have any cardinality whatsoever. It is the ✏ A nature of unary structures that the constituents are totally independent of each other. If = , then A ⇠ B C ( ) = C ( ) (5.1) | ✏ A | | ✏ B | for every ✏. Conversely, if two L-structures and satisfy Equation (5.1) for A B every ✏, then = (see Exercise 5.6). We can say that the function ✏ A ⇠ B 7! C ( ) characterizes completely (i.e. up to isomorphism) the unary structure | ✏ A | . There is nothing more we can say about but this function. A A Example 5.5 Equivalence relations, i.e. L-structures for L = such M {⇠} that M is a symmetric (x y y x), transitive (x y z x z), ⇠ ⇠ ) ⇠ ⇠ ⇠ ) ⇠ and reflexive (x x) relation on M can be characterized almost as easily ⇠ as unary structures. Let for every cardinal number M the number of | | equivalence classes of M of cardinality be denoted by EC ( ). If = , ⇠ M A ⇠ B then EC ( )=EC ( ) (5.2) A B for every A . Conversely, if two L-structures and satisfy Equa- | | A B tion (5.2) for every A B , then = (see Exercise 5.12). We can say | [ | A ⇠ B that the function EC ( ) characterizes completely (i.e. up to isomor- 7! A phism) the equivalence relation . There is nothing more we can say about A A but this function. For equivalence relations on a finite universe of size n this Incomplete version for students of easllc2012 only. 62 Models Figure 5.4 A successor structure. If is a successor structure, let Cmp be the set of components of and M M M CCn( )= C Cmp : C is an n-cycle component , M |{ 2 M }| CC ( )= C Cmp : C is a Z-component . 1 M |{ 2 M }| Two successor structures and are isomorphic if and only if CC ( )= M N a M CCa( ) for all a N . N 2 [ {1} 5.3 Substructures The concept of a substructure is in principle a very simple one, especially for relational vocabularies. There are however subtleties which deserve special attention when function symbols are involved. Definition 5.10 An L-structure is a substructure of another L-structure M 0, in symbols 0, if: M M ✓ M 1. M M 0. ✓ 0 n 2. RM = RM M if R L is an n-ary predicate symbol. 0 \ n 2 3. f M = f M M if f L is an n-ary function symbol. 0 2 4. cM = cM if c L is a constant symbol. 2 Substructures are particularly easy to understand in the case that L is re- lational. Then any subset M of an L-structure 0 determines a substruc- M ture the universe of which is M. If L is not relational we have to worry M 0 about the question whether M is closed under the functions f M , f L, 0 2 and whether the interpretations cM of constant symbols c L are in M. 2 For example, if L = f where f is a unary function symbol, then any sub- { } structure of an L-structure which contains an element a has to contain also 0 0 0 f M (a),fM (f M (a)), etc. A substructure of a group need not be a subgroup Incomplete version for students of easllc2012 only. 5.4 Back-and-Forth Sets 63 even when it is closed under the group operation. For example, (N, +) is a substructure of (Z, +) but it is not a group. A substructure of a linear order is again a linear order. Similarly, a substructure of a partial order is again a partial order. A substructure of a tree is a tree if it has a smallest element. Lemma 5.11 Suppose L is a vocabulary, an L-structure, and X M. M ✓ Suppose furthermore that either L contains constant symbols or X = . There 6 ; is a unique L-structure such that: N 1. . N ✓ M 2. X N. ✓ 3. If 0 and X N 0, then 0. N ✓ M ✓ N ✓ N Proof Let X = X cM : c L and inductively 0 [ { 2 } X = X f M(a ,...,a ):a ,...,a X ,f L . n+1 n [ { 1 #L(f) 1 #L(f) 2 n 2 } It is easy to see that the set N = X is the universe of the unique n N n structure claimed to exist in the lemma.2 N S We call the unique structure of Lemma 5.11 the substructure of gen- N M erated by X and denote it by [X] . The following lemma is used repeatedly M in the sequel. Lemma 5.12 Suppose L is a vocabulary. Suppose and are L-structures M N and ⇡ : M N is a partial mapping. There is at most one isomorphism ! ⇡⇤ : [dom(⇡)] [rng(⇡)] extending ⇡. M ! N 5.4 Back-and-Forth Sets One of the main themes of this book is the question: Given two structures M and , how do we measure how close they are to being isomorphic? They may N be non-isomorphic for a totally obvious reason, e.g. two graphs one of which has a triangle while the other does not. They may also be non-isomorphic for an extremely subtle reason which involves the use of the Axiom of Choice (see e.g. Lemma 9.9). One of the basic tools in trying to answer this question is the concept of partial isomorphism. Definition 5.13 Suppose L is a vocabulary and , 0 are L-structures. M M A partial mapping ⇡ : M M 0 is a partial isomorphism 0 if ! M ! M there is an isomorphism ⇡⇤ : [dom(⇡)] [rng(⇡)] extending ⇡.We M ! M0 use Part( , 0) to denote the set of partial isomorphisms 0. If M M M ! M = 0 we call ⇡ a partial automorphism. M M Incomplete version for students of easllc2012 only. 64 Models Note that the extension ⇡⇤ referred to in Definition 5.13 is by Lemma 5.12 necessarily unique. The main topic of this section, the back-and-forth sets, are very useful weaker versions of isomorphisms. To get a picture of this, suppose f : = . Then A ⇠ B f Part( , ) and we can go back and forth between and with f in the 2 A B A B following sense: a A b B(f(a)=b) (5.6) 8 2 9 2 b B a A(f(a)=b). (5.7) 8 2 9 2 We now generalize this to a situation where we do not quite have an isomor- phism but only a set P which reflects the back and forth conditions (5.8) and (5.9) of an isomorphism. Definition 5.14 Suppose and are L-structures. A back-and-forth set for A B and is any non-empty set P Part( , ) such that A B ✓ A B f P a A g P (f g and a dom(g)) (5.8) 8 2 8 2 9 2 ✓ 2 f P b B g P (f g and b rng(g)). (5.9) 8 2 8 2 9 2 ✓ 2 The structures and are said to be partially isomorphic, in symbols A B A 'p , if there is a back-and-forth set for them. B Lemma 5.15 The relation p is an equivalence relation on Str(L). ' Proof The relation p is reflexive, because idA is a back-and-forth set for ' { } 1 and . If P is a back-and-forth set for and , then f : f P is a A B A B { 2 } back-and-forth set for and . Finally, if P is a back-and-forth set for and B A 1 A and P is a back-and-forth set for and , then f f : f P ,f B 2 B C { 2 1 1 2 1 2 2 P2 is a back-and-forth set for and , where we stipulate dom(f2 f1)= }1 A C f1 (dom(f2)). Proposition 5.16 If p , where and are countable, then = . A ' B A B A ⇠ B Proof Let us enumerate A as (an : n This proposition is not true for uncountable structures. Indeed, let L = ; and let and be any infinite L-structures. Then there is a back-and-forth A B set for and (Exercise 5.28). Thus . But = if, for example, A B A 'p B A 6⇠ B A = Q and B = R. The failure of Proposition 5.16 to generalize is a major topic in the sequel. Proposition 5.17 Suppose and are dense linear orders without end- A B points. Then . A 'p B Proof Let P = f Part( , ) : dom(f) is finite . It turns out that this { 2 A B } straightforward choice works. Clearly, P = . Suppose then f P and a A. 6 ; 2 2 Let us enumerate f as (a ,b ),...,(a ,b ) where a <...b and again f (a, b) P . Finally, if a = a , we let b = b and n [{ } 2 i i then f (a, b) = f P . We have proved (5.8). Condition (5.9) is proved [ { } 2 similarly. Putting Proposition 5.16 and Proposition 5.17 together yields the famous result of Cantor (1895): countable dense linear orders without endpoints are isomorphic. See Exercise 6.29 for a more general result. 5.5 The Ehrenfeucht–Fra¨ısse´ Game In Section 4.3 we introduced the Ehrenfeucht–Fra¨ısse´ Game played on two graphs. This game was used to measure to what extent two graphs have sim- ilar properties, especially properties expressible in the first-order language of graphs limited to a fixed quantifier rank. In this section we extend this game to the context of arbitrary structures, not just graphs. Let us recall the basic idea behind the Ehrenfeucht–Fra¨ısse´ Game. Suppose and are L-structures for some relational L. We imagine a situation in A B which two mathematicians argue about whether and are isomorphic or A B not. The mathematician that we denote by II claims that they are isomorphic, while the other mathematician whom we call I claims the models have an intrinsic structural difference and they cannot possibly be isomorphic. The matter would be quickly resolved if II was required to show the claimed isomorphism. But the rules of the game are different. The rules are such that II is required to show only small pieces of the claimed isomorphism. More exactly, I asks what is the image of an element a1 of A that he chooses Incomplete version for students of easllc2012 only. 66 Models Figure 5.5 The Ehrenfeucht–Fra¨ısse´ Game. at will. Then II is required to respond with some element b1 of B so that (a ,b ) Part( , ). (5.10) { 1 1 } 2 A B Alternatively, I might have chosen an element b1 of B and then II would have been required to produce an element a1 of A such that (5.10) holds. The one- element mapping (a ,b ) is called the position in the game after the first { 1 1 } move. Now the game goes on. Again I asks what is the image of an element a2 of A (or alternatively he can ask what is the pre-image of an element b2 of B). Then II produces an element b2 of B (or in the alternative case an element a2 of A). In either case the choice of II has to satisfy (a ,b ), (a ,b ) Part( , ). (5.11) { 1 1 2 2 } 2 A B Again, (a ,b ), (a ,b ) is called the position after the second move. { 1 1 2 2 } We continue until the position (a ,b ),...,(a ,b ) Part( , ) { 1 1 n n } 2 A B after the nth move has been produced. If II has been able to play all the moves according to the rules she is declared the winner. Let us call this game EF ( , ). Figure 5.5 pictures the situation after four moves. If II can win n A B repeatedly whatever moves I plays, we say that II has a winning strategy. Example 5.18 Suppose and are two L-structures and L = . Thus the A B ; structures and consist merely of a universe with no structure on it. In A B this singular case any one-to-one mapping is a partial isomorphism. The only thing player II has to worry about, say in (5.11), is that a1 = a2 if and only if b = b . Thus II has a winning strategy in EF ( , ) if A and B both have 1 2 n A B at least n elements. So II can have a winning strategy even if A and B have different cardinality and there could be no isomorphism between them for the Incomplete version for students of easllc2012 only. 5.5 The Ehrenfeucht–Fra¨ısse´ Game 67 trivial reason that there is no bijection. The intuition here is that by playing a finite number of elements, or even many, it is not possible to get hold of the @0 cardinality of the universe if it is infinite. Example 5.19 Let be a linear order of length 3 and a linear order of A B length 4. How many moves does I need to beat II? Suppose A = a ,a ,a { 1 2 3} in increasing order and B = b ,b ,b ,b in increasing order. Clearly, if I { 1 2 3 4} plays at any point the smallest element, also II has to play the smallest element or face defeat on the next move. Also, if I plays at any point the smallest but one element, also II has to play the smallest but one element or face defeat in two moves. Now in the smallest but one element is the same as the largest A but one element, while in they are different. So if I starts with a , II has to B 2 play b2 or b3, or else she loses in one move. Suppose she plays b2. Now I plays b3 and II has no good moves left. To obey the rules, she must play a3. That is how long she can play, for now when I plays b4, II cannot make a legal move anymore. In fact II has a winning strategy in EF ( , ) but I has a winning 2 A B strategy in EF ( , ). 3 A B We now proceed to a more exact definition of the Ehrenfeucht–Fra¨ısse´ Game. Definition 5.20 Suppose L is a vocabulary and , 0 are L-structures such M M that M M 0 = . The Ehrenfeucht–Fra¨ısse´ Game EFn( , 0) is the game \ ; M M2n (M M 0,W ( , 0)), where W ( , 0) (M M 0) is the set of Gn [ n M M n M M ✓ [ p =(x0,y0,...,xn 1,yn 1) such that: (G1) For all i xi if xi M xi if xi M 0 vi = 2 vi0 = 2 y if y M y if y M 0, ⇢ i i 2 ⇢ i i 2 then fp = (v0,v0),...,(vn 1,vn0 1) { } is a partial isomorphism 0. M ! M We call v and v0 corresponding elements. The infinite game EF ( , 0) i i ! M M is defined quite similarly, that is, it is the game (M M 0,W ( , 0)), G! [ ! M M where W ( , 0) is the set of p =(x ,y ,x ,y ,...) such that for all ! M M 0 0 1 1 n N we have (x0,y0,...,xn 1,yn 1) Wn( , 0). 2 2 M M Note that the game EF! is a closed game. Proposition 5.21 Suppose L is a vocabulary and and are L-structures. A B The following are equivalent: Incomplete version for students of easllc2012 only. 68 Models 1. . A 'p B 2. II has a winning strategy in EF ( , ). ! A B Proof Assume A B = . Let P be first a back-and-forth set for and . \ ; A B We define a winning strategy ⌧ =(⌧ : i Definition 5.22 A back-and-forth sequence (Pi : i n) is defined by the conditions = P ... P Part( , ). (5.13) ;6 n ✓ ✓ 0 ✓ A B f P a A b B g P (f (a, b) g) for i n A 'p B if there is a back-and-forth sequence of length n for and . A B Lemma 5.23 The relation n is an equivalence relation on Str(L). 'p Proof Exactly as Lemma 5.15. Example 5.24 We use (N + N,<) to denote the linear order obtained by putting two copies of (N,<) one after the other. (The ordinal of this order is ! + !.) Now (N,<) 2 (N + N,<), for we may take 'p P = . 2 {;} P1 = (a, b) :0 Proof Let P consist of f Part( , ) with the following property: f = i 2 A B (a0,b0),...,(an i 1,bn i 1) where { } a0 ... an i 1, b0 ... bn i 1, and for all 0 j 1. n . A 'p B 2. II has a winning strategy in EF ( , ). n A B Proof Let us assume A B = . Let (P : i n) be a back-and-forth \ ; i sequence for and . We define a winning strategy ⌧ =(⌧ : i n) for II. A B i Since P = we can fix an element f of P . Condition (5.14) tells us that if n 6 ; n a A, then there are b B and g such that 1 2 1 2 f (a1,b1) g Pn 1. (5.16) [ { } ✓ 2 Let ⌧ (a ) be one such b . Likewise, if b B, then there are a A such that 0 1 1 1 2 1 2 (5.16) holds and we can let ⌧0(b1) be some such a1. We have defined ⌧0(c1) whatever c is. To define ⌧ (c ,c ), let us assume I played c = a A. Thus 1 1 1 2 1 1 2 (5.16) holds with b = ⌧ (a ). If c = a A we can use (5.13) again to find 1 0 1 2 2 2 b = ⌧ (a ,a ) B and h such that 2 1 1 2 2 f (a1,b1), (a2,b2) h Pn 2. [ { } ✓ 2 The pattern should be clear now. As before, the back-and-forth sequence guides II to always find a valid move. Let us then write the proof in more detail: Sup- pose we have defined ⌧i for i For the converse, suppose ⌧ =(⌧ : i n) is a winning strategy of II. Let i Q consist of all plays of EFn( , ) in which player II has used ⌧. Let Pn i A B consist of all possible fp where p =(x0,y0,...,xi 1,yi 1) is a position in the game EF ( , ) with an extension in Q. It is clear that (P : i n) has n A B i the properties (5.13) and (5.14). Note that: P = n {;} Pn 1 = (x0, ⌧0(x0)) : x0 A B { 2 [ } Pn 2 = (x0, ⌧0(x0),x1, ⌧1(x0,x1)) : x0,x1 A B { 2 [ } P0 = (x0, ⌧0(x0),...,xn 1, ⌧n 1(x0,...,xn 1)) : x0,...,xn 1 A B . { 2 [ } 5.7 Historical Remarks and References Back-and-forth sets are due to Fra¨ısse´ (1955). The Ehrenfeucht–Fra¨ısse´ Game was introduced in Ehrenfeucht (1957) and Ehrenfeucht (1960/1961). Back- and-forth sequences were introduced in Karp (1965). Exercise 5.40 is from Ellentuck (1976). Exercise 5.40 is from Ellentuck (1976). Exercise 5.54 is from Barwise (1975). Exercise 5.71 is from Rosenstein (1982). Exercises 5.1 Show that isomorphism of structures is an equivalence relation in the sense that it is reflexive, symmetric, and transitive. 5.2 Suppose L is a finite vocabulary, is a countable L-model, and b : B { n n (2) There is an enumeration an : n = ✓(a ,...,a ) = ✓(b ,...,b ). A| 0 n () B| 0 n 5.3 Suppose L is a vocabulary and is an L-structure. Show that the set M Aut( ) of automorphisms of forms a group under the operation of M M composition of functions. 5.4 Give an example of such that Aut( ) (see the previous exercise) is: M M 1. The trivial one-element group. 2. A non-trivial abelian group (e.g. the additive group of the integers). 3. A non-abelian group (e.g. the symmetric group S3). 5.5 How many automorphisms do the following structures have. 1. A linear order of n elements. 2. (N,<). 3. (Z,<). 4. (Q,<). 5.6 Show that if and are unary structures, then = if and only if for A B A ⇠ B all ✏ : 1,...,n 0, 1 we have C ( ) = C ( ) . Easier version: { } ! { } | ✏ A | | ✏ B | Show that if and are unary structures with a finite universe of size A B n, then = if and only if for all ✏ : 1,...,n 0, 1 we have A ⇠ B { } ! { } C ( ) = C ( ) . | ✏ A | | ✏ B | 5.7 Suppose is a unary structure in which every ✏-constituent has exactly M three elements. How many elements does have? How many automor- M phisms does have? M 5.8 L = P1,...,Pm , where each Pi is unary. Show that the number of { } n+2m 1 non-isomorphic L-structures on the universe 1,...,n is 2m 1 . { } 5.9 Describe the group of automorphisms of a finite unary structure. 5.10 Suppose is an equivalence relation with a finite universe such that M EC ( )=2for each n =1,...,5 and EC ( )=0for other n. n M n M How many elements are there in the universe of ? How many auto- M morphisms does have? M 5.11 Show that for any m N there is m⇤ N such that if n m⇤ then there 2 2 are more than nm non-isomorphic equivalence relations on the universe 1,...,n . Conclude that for any m N there is m⇤ N such that if { } 2 2 n m⇤ then there are more non-isomorphic equivalence relations on the universe 1,...,n than non-isomorphic P ,...,P -structures, { } { 1 m} where each Pi is unary. Incomplete version for students of easllc2012 only. Exercises 73 5.12 Show that if and are equivalence relations, then = if and only if A B A ⇠ B for all A B we have EC ( )=EC ( ). Easier version: Show | [ | A B that if and are equivalence relations with a finite universe of size n, A B then = if and only if for all m n we have EC ( )=EC ( ). A ⇠ B m A m B 5.13 Describe the group of automorphisms of a finite equivalence relation. 5.14 Show that if and are countable dense linear orders, then = M N M ⇠ N if and only if SG( )=SG( ). Demonstrate that this is not true for M N non-dense countable linear orders or for uncountable dense linear orders. 5.15 Show that two well-orders and are isomorphic if and only if M N o( )=o( ). M N 5.16 Prove that two well-founded trees and are isomorphic if and only M N if stp =stp . M N 5.17 Prove that two successor structures and are isomorphic if and only M N if CCa( )=CCa( ) for all a N . Easier version: Prove M N 2 [ {1} that two successor structures and both of which have only finitely M N many components are isomorphic if and only if CC ( )=CC ( ) a M a N for all a N . 2 [ {1} 5.18 Show that any uncountable collection of countable non-isomorphic suc- cessor structures has to contain a successor structure with infinitely many cycle components. 5.19 Describe the group of automorphisms of a successor structure with n Z-components and mi i-cycle components for i =1,...,k. 5.20 Give an example of an infinite structure with no substructures = M N 6 . M 5.21 Consider =(Z, +). What is [X] , if X is M M 1. 0 , { } 2. 1 , { } 3. 2, 2 . { } 5.22 Consider =(Z, +, ). What is [X] , if X is 13, 17 ? M M { } 5.23 Suppose is a successor structure consisting of the standard compo- M nent and two five-cycles. Show that there are exactly four possibilities for the set [X] . M 5.24 Show that the universe of [X] is the intersection of all universes of M substructures of such that X N. N M ✓ 5.25 Prove Lemma 5.12. 5.26 Show that every Boolean algebra is isomorphic to a substructure of M ( (A), ), where A is the set of all ultrafilters of . (This is the so- P ✓ M called Stone’s Representation Theorem.) Incomplete version for students of easllc2012 only. 74 Models 5.27 Show that every tree every element of which has height < ! is isomor- phic to a substructure of the tree (A dom(stp0 ,t)= stp0 ,s : s ImSuc(t) M { M 2 } stp0 ,t(stp0 ,s)=min( 0, s0 ImSuc(t):stp0 ,s =stp0 ,s ). M M @ |{ 2 M M 0 }| Suppose and are well-founded trees such that stp0 =stp0 . M N M N Show that and are partially isomorphic. Give an example of two M N well-founded partially isomorphic trees that are not isomorphic. 5.32 Suppose that and are successor structures, f Part( , ). M N 2 M N Show: 1. f maps elements of the standard component of to elements of the M standard component of . N 2. f maps elements of a cycle component of of size n to elements of M a cycle component of of size n. N 3. f maps elements of a Z-component of to elements of a Z-compo- M nent of . N 5.33 Suppose that two successor structures and satisfy the following M N conditions for all n, m < !: Incomplete version for students of easllc2012 only. Exercises 75 1. CC ( )=m CC ( )=m. n M () n N 2. CC ( )=m CC ( )=m. 1 M () 1 N Show that the successor structures are partially isomorphic. 5.34 Show that Part( , ) is closed under unions of chains, i.e. if f M N 0 ✓ f f ...are in Part( , ), then so is 1 f . 1 ✓ 2 ✓ M N n=0 n 5.35 Suppose (R,<,f) p (R,<,g)), where f : R R is continuous. ' S ! Show that g is also continuous. 5.36 If (M,d), d : M M R, is a metric space, we can think of (M,d) as a ⇥ ! an L-structure =(M,d,R,< ), where L contains a binary function M R symbol, a unary predicate symbol, and a binary relation symbol. Show that there are a separable metric space =(M,d,R,< ) and a non- M R separable metric space 0 =(M 0,d0, R,< ) such that p 0. M R M ' M 5.37 Show that there is a complete separable metric space (Polish space) =(M,d,R,< ) and a non-complete separable metric space 0 = M R M (M 0,d0, R,< ) such that p 0. R M ' M 5.38 Suppose and are structures of the same relational vocabulary L and A B A B = . The disjoint sum of and is the L-structure \ ; A B (A B,(RA RB)R L). [ [ 2 Show that partial isomorphism is preserved by disjoint sums of models. 5.39 Suppose and are structures of the same vocabulary L. The direct A B product of and is the L-structure A B (A B,(RA RB)R L, ⇥ ⇥ 2 (((a0,b0) ...,(an,bn)) (f A(a0,...,an),fB(b0,...,bn)))f L, 7! 2 ((cA,cB))c L). 2 Show that partial isomorphism is preserved by direct products of models. 5.40 Show that if two structures are partially isomorphic, then they are po- tentially isomorphic,2 i.e. there is a forcing extension in which they are isomorphic. Conversely, show that if two structures are potentially iso- morphic, then they are partially isomorphic. 2 5.41 Consider EF2( , ), where =(R 0 ,f), f(x, 0) = (x , 0) M N M ⇥ { } and =(R 1 ,g), g(x, 1) = (x3, 1). Player I can win even without N ⇥ { } looking at the moves of II. How? 3 5.42 Consider EF!( , ), where =(R 0 ,f), f(x, 0) = (x , 0) M N M ⇥ { } and =(R 1 ,g), g(x, 1) = (x5, 1). After a few moves player I N ⇥ { } resigns. Can you explain why? 2 Some authors use the term potential isomorphism for partial isomorphism. Incomplete version for students of easllc2012 only. 76 Models 5.43 Consider EF2( , ), where =(Z, (a, b):a b = 10 ) and = M N M { } N (Q, (a, b):a b =2/3 ). Suppose we are in position ( 8, 1/4) (i.e. { } x = 8 and y = 1/4). Then I plays x = 11/12. What would be a 0 0 1 good move for II? 5.44 Consider EF ( , ), where and are as in the previous exercise. ! M N M N Player I resigns before the game even starts. Can you explain why? 5.45 Suppose M and N are disjoint sets with 10 elements each. Let c M 2 and d N. Who has a winning strategy in EF ( , ) in the following 2 ! M N cases: 1. =(M, (a, b, c):a = b ), =(N, (a, b, d):a = b ), M { } N { } 2. =(M, (a, b, e):a = b ), =(N, (a, b, e):b = e ). M { } N { } 5.46 Who has a winning strategy in EF ( , ) in the following cases: ! M N 1. =(Q,<,1855), =(R,<,1854), M N 2. =(N,<,1855), =(N,<,1854). M N 5.47 Show that ( (X), ) ( (Y ), ), if X and Y are disjoint infi- P ✓ 'p P ✓ nite sets. (Hint: Consider the set of finite partial isomorphisms of the form (A0,B0),...,(Ai 1,Bi 1) , such that (X, A0,...,Ai 1) and { } (Y,B0,...,Bi 1) are partially isomorphic, and then use Exercise 5.29 of Section 5.4.) 5.48 Show that player II has a winning strategy in the game EF ( , ) for ! M N any two atomless (i.e. if 0 for all transitive models M and N of T and for all a ,...,a M we 1 n 2 have M = '(a ,...,a ) M 0 = '(a ,...,a ). | 1 n () | 1 n Show that “x is a vocabulary, y and z are x-structures, and y z” can 'p be defined with a formula '(x, y, z) which is absolute relative to T . 5.55 Suppose is a linear order of length three and a linear order of length A B four. Give a back-and-forth sequence of length two for and . A B 5.56 Suppose is a cycle of four vertices and a cycle of five vertices. Give A B a back-and-forth sequence of length two for and . A B 5.57 Suppose is an equivalence relation of four classes each of size 3 and A B an equivalence relation of three classes each of size 4. Give a back-and- forth sequence of length three for and . A B 5.58 Suppose is an equivalence relation of four classes each of size 2 and A B an equivalence relation of three classes each of size 2. Give a back-and- forth sequence of length three for and . A B 5.59 Suppose is an equivalence relation of four classes each of size 2 plus A one class of size 3, and an equivalence relation of three classes each of B size 2 plus one class of size 4. Give a back-and-forth sequence of length three for and . A B 5.60 Suppose and are successor structures, both consisting of the stan- A B dard component plus some cycle components. Suppose has three five- A cycles and has four five-cycles. Give a back-and-forth sequence of B length three for and . A B 5.61 Show that (7,<) 3 (8,<). 'p 5.62 Show that (Z,<) 3 (Q,<). 6'p 5.63 Show that (N,<) 3 (N + N,<). 6'p 5.64 Show that (Z,<) p (Z + Z,<). 6' 5.65 Show that (N + N,<) 3 (N + N + N,<) 'p 5.66 Finish the proof of Proposition 5.25. 5.67 Prove the claim of Example 5.26. 5.68 Let the game EF⇤ ( , ) be like the game EF ( , ) except that I has ! A B ! A B to play x2n A and x2n+1 B for all n N. Show that player II has a 2 2 2 winning strategy in EF⇤ ( , ) if and only if she has a winning strategy ! A B in EF ( , ). ! A B 5.69 Suppose B = bn : n N . Let the game EF⇤⇤( , ) be like the game { 2 } ! A B EF ( , ) except that I has to play x A and x = b for all ! A B 2n 2 2n+1 n n N. Show that player II has a winning strategy in EF⇤⇤( , ) if and 2 ! A B only if she has a winning strategy in EF ( , ). ! A B Incomplete version for students of easllc2012 only. 78 Models 5.70 Suppose =(A ,< ) and =(A ,< ) are linearly ordered sets. A0 0 0 A1 1 1 Show that if player II has a winning strategy both in EF ( , ) and n A0 B0 in EF ( , ), then she has one in EF ( + , + ). n A1 B1 n A0 A1 B0 B1 5.71 If =(A, <) is a linearly ordered set and a A, then a is the substructure A A of generated by the set x A : x>a. Thus >a is the final A { 2 } A segment of determined by a. Show that if and are ordered sets, A A B then player II has a winning strategy in EF ( , ) if and only if n+1 A B 1. For every a A there is b B such that player II has a winning 2 2 strategy in EF ( 6.1 Introduction We have already discussed the first-order language of graphs. We now de- fine the basic concepts of a more general first-order language, denoted FO, one which applies to any vocabulary, not just the vocabulary of graphs. First- order logic fits the Strategic Balance of Logic better than any other logic. It is arguably the most important of all logics. It has enough power to express inter- esting and important concept and facts, and still it is weak and flexible enough to permit powerful constructions as demonstrated, e.g. by the Model Existence Theorem below. 6.2 Basic Concepts Suppose L is a vocabulary. The logical symbols of the first-order language (or logic) of the vocabulary L are , , , , , , (, ),x ,x ,.... Terms are de- ⇡ ¬ ^ _ 8 9 0 1 fined as follows: Constant symbols c L are L-terms. Variables x ,x ,...are 2 0 1 L-terms. If f L, #(f)=n, and t ,...,t are L-terms, then so is ft ...t . 2 1 n 1 n L-equations are of the form tt0 where t and t0 are L-terms. L-atomic formu- ⇡ las are either L-equations or of the form Rt ...t , where R L, #(R)=n 1 n 2 and t1,...,tn are L-terms. A basic formula is an atomic formula or the nega- tion of an atomic formula. L-formulas are of the form tt0 ⇡ Rt1 ...tn ' ¬ (' ), (' ) ^ _ x ', x ' 8 n 9 n Incomplete version for students of easllc2012 only. 80 First-Order Logic where t, t0,t ,...,t are L-terms, R L with #(R)=n, and ' and are 1 n 2 L-formulas. Definition 6.1 An assignment for a set M is any function s with dom(s) a set of variables and rng(s) . The value tM(s) of an L-term t in under the ✓ M M assignment s is defined as follows: cM(s)=Val (c), xnM(s)=s(xn) and M (ft1 ...tn)M(s)=Val (f)(t1M(s),...,tnM(s)). The truth of L-formulas in M under s is defined as follows: M ✏s Rt1 ...tn iff (t1M(s),...,tnM(s)) Val (R) M 2 M t t iff tM(s)=tM(s) M ✏s ⇡ 1 2 1 2 s ' iff 2s ' M ✏ ¬ M (' ) iff ' and M ✏s ^ M ✏s M ✏s (' ) iff ' or M ✏s _ M ✏s M ✏s x ' iff ' for all a M ✏s 8 n M ✏s[a/xn] 2 M x ' iff ' for some a , M ✏s 9 n M ✏s[a/xn] 2 M a if y = x where s[a/x ](y)= n n s(y) otherwise. ⇢ We assume the reader is familiar with such basic concepts as free variable, sentence, substitution of terms for variables, etc. A standard property of first- order (or any other) logic is that = ' depends only on and the values M| s M of s on the variables that are free in '.Asentence is a formula ' without free variables. Then = ' means = '. In this case we say that ' is true in M| M| ; . M Convention: If ' is an L-formula with the free variables x1,...,xn, we in- dicate this by writing ' as '(x ,...,x ). If is an L-structure and s is an 1 n M assignment for M such that = ', we write = '(a ,...,a ), where M| s M| 1 n ai = s(xi) for i =1,...,n. Definition 6.2 The quantifier rank of a formula ', denoted QR('), is defined as follows: QR( tt0)=QR(Rt ...t )=0, QR( ')=QR('), QR((' ⇡ 1 n ¬ ^ )) = QR((' )) = max QR('), QR( ) , QR( x')=QR( x')= _ { } 9 8 QR(')+1. A formula ' is quantifier free if QR(')=0. The quantifier rank is a measure of the longest sequence of “nested” quan- tifiers. In the first three of the following formulas the quantifiers x and x 8 n 9 n are nested but in the last unnested: x (P (x ) x R(x ,x )) (6.1) 8 0 0 _9 1 0 1 x (P (x ) x R(x ,x )) (6.2) 9 0 0 ^8 1 0 1 x (P (x ) x Q(x )) (6.3) 8 0 0 _9 1 1 Incomplete version for students of easllc2012 only. 6.3 Characterizing Elementary Equivalence 81 ( x P (x ) x Q(x )). (6.4) 8 0 0 _9 1 1 Note that formula (6.3) of quantifier rank 2 is logically equivalent to the for- mula (6.4) which has quantifier rank 1. So the nesting can sometimes be elim- inated. In formulas (6.1) and (6.2) nesting cannot be so eliminated. Proposition 6.3 Suppose L is a finite vocabulary without function symbols. For every n and for every set x ,...,x of variables, there are only finitely { 1 n} many logically non-equivalent first-order L-formulas of quantifier rank Note that Proposition 6.3 is not true for infinite vocabularies, as there would be infinitely many logically non-equivalent atomic formulas, and also not true for vocabularies with function symbols, as there would be infinitely many log- ically non-equivalent equations obtained by iterating the function symbols. 6.3 Characterizing Elementary Equivalence We now show that the concept of a back-and-forth sequence provides an alter- native characterization of elementary equivalence i.e. ' FO( = ' = '). A ⌘ B 8 2 A| () B| This is the original motivation for the concepts of a back-and-forth set, back- and-forth sequence, and Ehrenfeucht–Fra¨ısse´ Game. To this end, let A ⌘n B mean that and satisfy the same sentences of FO of quantifier rank n. A B We now prove an important leg of the Strategic Balance of Logic, namely the marriage of truth and separation: Proposition 6.4 Suppose L is an arbitrary vocabulary. Suppose and are A B L-structures and n N. Consider the conditions: 2 (i) n . A ⌘ Bn (ii) L p L for all finite L0 L. A 0 ' B 0 ✓ We have always (ii) (i) and if L has no function symbols, then (ii) (i). ! $ Proof (ii) (i). If , then there is a sentence ' of quantifier rank n ! A 6⌘n B such that = ' and = '. Since ' has only finitely many symbols, there A| B 6| Incomplete version for students of easllc2012 only. 82 First-Order Logic is a finite L0 L such that L n L . Suppose (Pi : i n) is a back- ✓ A 0 6⌘ B 0 and-forth sequence for L and L . We use induction on i n to prove the A 0 B 0 following Claim If f P and a ,...,a dom(f), then 2 i 1 k 2 ( L ,a1,...,ak) i ( L ,fa1,...,fak). A 0 ⌘ B 0 If i =0, the claim follows from P0 Part( L , L ). Suppose then ✓ A 0 B 0 f P and a ,...,a dom(f). Let '(x ,x ,...,x ) be an L0-formula 2 i+1 1 k 2 0 1 k of FO of quantifier rank i such that L = x0'(x0,a1,...,ak). A 0 | 9 Let a A so that L = '(a, a1,...,ak) and g Pi such that a dom(g) 2 A 0 | 2 2 and f g. By the induction hypothesis, L = '(ga, ga1,...,gak). Hence ✓ B 0 | L = x0'(x0,fa1,...,fak). B 0 | 9 The claim is proved. Putting i = n and using the assumption P = , gives a n 6 ; contradiction with L n L . A 0 6⌘ B 0 (i) (ii). Assume L has no function symbols. Fix L0 L finite. Let P ! ✓ i consist of f : A B such that dom(f)= a0,...,an i 1 and ! { } ( L ,a0,...,an i 1) i ( L ,fa0,...,fan i 1). A 0 ⌘ B 0 We show that (Pi : i n) is a back-and-forth sequence for L and L . A 0 B 0 By (i), P so P = . Suppose f P ,i > 0, as above, and a A. ;2 n n 6 ; 2 i 2 By Proposition 6.3 there are only finitely many pairwise non-equivalent L0- formulas of quantifier rank i 1 of the form '(x, x0,...,xn i 1) in FO. Let them be 'j(x, x0,...,xn i 1), j J. Let 2 J0 = j J : L = 'j(a, a0,...,an i 1) . { 2 A 0 | } Let (x, x0,...,xn i 1)= 'j(x, x0,...,xn i 1) ^ j J ^2 0 'j(x, x0,...,xn i 1). ¬ j J J 2^\ 0 Now L = x (x, a0,...,an i 1), so as we have assumed f Pi, we A 0 | 9 2 have L = x (x, fa0,...,fan i 1). Thus there is some b B with B 0 | 9 2 L = (b, fa0,...,fan i 1). Now f (a, b) Pi 1. The other condi- B 0 | [ { } 2 tion (5.15) is proved similarly. Incomplete version for students of easllc2012 only. 6.3 Characterizing Elementary Equivalence 83 The above proposition is the standard method for proving models elemen- tary equivalent in FO. For example, Proposition 6.4 and Example 5.26 together give (Z, <) (Z + Z, <). The exercises give more examples of partially iso- ⌘ morphic pairs – and hence elementary equivalent – structures. The restriction on function symbols can be circumvented by first using quantifiers to elim- inate nesting of function symbols and then replacing the unnested equations f(x1,...,xn 1)=xn by new predicate symbols R(x1,...,xn). Let Str(L) denote the class of all L-structures. We can draw the following important conclusion from Proposition 6.4 (see Figure 6.1): Corollary Suppose L is a vocabulary without function symbols. Then for all n N the equivalence relation 2 A ⌘n B n divides Str(L) into finitely many equivalence classes Ci , i =1,...,mn, such n n that for each Ci there is a sentence 'i of FO with the properties: 1. For all L-structures : Cn = 'n. A A 2 i () A| i 2. If ' is an L-sentence of quantifier rank n, then there are i ,...,i such 1 k that = ' ('n ... 'n ). | $ i1 _ _ ik n Proof Let 'i be the conjunction of all the finitely many L-sentences of quan- tifier rank n that are true in some (every) model in Cn (to make the con- i junction finite we do not repeat logically equivalent formulas). For the second n n claim, let 'i1 ,...,'ik be the finite set of all L-sentences of quantifier rank n that are consistent with '. If now = ', and Cn, then = 'n. A| A 2 i A| i On the other hand, if = 'n and there is = 'n such that = ', then A| i B| i B| , whence = '. A ⌘n B A| We can actually read from the proof of Proposition 6.4 a more accurate description for the sentences 'i. This leads to the theory of so-called Scott formulas (see Section 7.4). Theorem 6.5 Suppose K is a class of L-structures. Then the following are equivalent (see Figure 6.2): 1. K is FO-definable, i.e. there is an L-sentence ' of FO such that for all L-structures we have K = '. M M 2 () M| 2. There is n N such that K is closed under n. 2 'p As in the case of graphs, Theorem 6.5 can be used to demonstrate that certain properties of models are not definable in FO: Incomplete version for students of easllc2012 only. 84 First-Order Logic Figure 6.1 First-order definable model class K. Figure 6.2 Not first-order definable model class K. Example 6.6 Let L = . The following properties of L-structures are not ; M expressible in FO: 1. M is infinite. 2. M is finite and even. In both cases it is easy to find, for each n N, two models n and n such 2 M N that n , has the property, but does not. Mn 'p Nn M N Example 6.7 Let L = P be a unary vocabulary. The following properties { } of L-structures (M,A) are not expressible in FO: 1. A = M . | | | | 2. A = M A . | | | \ | Incomplete version for students of easllc2012 only. 6.4 The Lowenheim–Skolem¨ Theorem 85 3. A M A . | | | \ | This is demonstrated by the models (N, 1,...,n ), (N, N 1,...,n ), and { } \{ } ( 1,...,2n , 1,...,n ). { } { } Example 6.8 Let L = < be a binary vocabulary. The following properties { } of L-structures =(M,<) are not expressible in FO: M 1. = (Z,<). M ⇠ 2. All closed intervals of are finite. M 3. Every bounded subset of has a supremum. M This is demonstrated in the first two cases by the models n =(Z,<) and M n =(Z + Z,<) (see Example 5.26), and in the third case by the partially N isomorphic models: =(R,<) and =(R 0 ,<). M N \{ } 6.4 The Lowenheim–Skolem¨ Theorem In this section we show that if a first-order sentence ' is true in a structure , M it is true in a countable substructure of , and even more, there are count- M able substructures of in a sense “everywhere” satisfying '. To make this M statement precise we introduce a new game from Kueker (1977) called the Cub Game. Definition 6.9 Suppose A is an arbitrary set. !(A) is defined as the set of P all countable subsets of A. The set (A) is an auxiliary concept useful for the general investigation of P! countable substructures of a model with universe A. One should note that if A 1 is infinite, the set !(A) is uncountable. For example, !(N) = R . The set P |P | | | (A) is closed under intersections and countable unions but not necessarily P! under complements, so it is a (distributive) lattice under the partial order , ✓ but not a Boolean algebra. The sets in (A) cover the set A entirely, but so P! do many proper subsets of (A) such as the set of all singletons in (A) P! P! and the set of all finite sets in (A). P! Definition 6.10 Suppose A is an arbitrary set and a subset of !(A). The C P Cub Game of is the game G ( )=G (A, W ), where W consists of se- C cub C ! quences (a ,a ,...) with the property that a ,a ,... . 1 2 { 1 2 } 2 C 1 Its cardinality is A !. | | Incomplete version for students of easllc2012 only. 6.5 The Semantic Game 93 In particular, for every countable X M there is a countable submodel of ✓ N such that X N and = T . M ✓ N| Proof Let T = ' , ' ,... . By Proposition 6.22 player II has a winning { 0 1 } strategy in G ( ). By Lemma 6.14, player II has a winning strategy in cub C'n Gcub( 1 ' ). If X 1 ' , then [X] = T . n=0 C n 2 n=0 C n M | T T 6.5 The Semantic Game The truth of a first-order sentence in a structure can be defined by means of a simple game called the Semantic Game. We examine this game in detail and give some applications of it. Definition 6.24 Suppose L is a vocabulary, is an L-structure, '⇤ is an M sym L-formula, and s⇤ is an assignment for M. The game SG ( , '⇤) is defined M as follows. In the beginning player II holds ('⇤,s⇤). The rules of the game are as follows: 1. If ' is atomic, and s satisfies it in , then the player who holds (',s) wins M the game, otherwise the other player wins. 2. If ' = , then the player who holds (',s), gives ( ,s) to the other player. ¬ 3. If ' = ✓, then the player who holds (',s), switches to hold ( ,s) or ^ (✓,s), and the other player decides which. 4. If ' = ✓, then the player who holds (',s), switches to hold ( ,s) or _ (✓,s), and can himself or herself decide which. 5. If ' = x , then the player who holds (',s), switches to hold ( ,s[a/x]) 8 for some a, and the other player decides for which. 6. If ' = x , then the player who holds (',s), switches to hold ( ,s[a/x]) 9 for some a, and can himself or herself decide for which. As was pointed out in Section 4.2, = ' if and only if player II has a M| s winning strategy in the above game, starting with (',s). Why? If = ', M| s then the winning strategy of player II is to play so that if she holds ('0,s0), then = '0, and if player I holds ('0,s0), then = '0. M| s0 M 6| s0 For practical purposes it is useful to consider a simpler game which pre- supposes that the formula is in negation normal form. In this game, as in the Ehrenfeucht–Fra¨ısse´ Game, player I assumes the role of a doubter and player II the role of confirmer. This makes the game easier to use than the full game SGsym( , '). M Incomplete version for students of easllc2012 only. 94 First-Order Logic III x0 y0 x1 y1 . . . . Figure 6.11 The game G!(W ). xn yn Explanation Rule (', ) I enquires about ' T . ; 2 (', ) II confirms. Axiom rule ; ('i,s) I tests a played ('0 '1,s) by choosing i ^0, 1 . 2 { } ('i,s) II confirms. -rule ^ ('0 '1,s) I enquires about _ a played disjunction. ('i,s) II makes a choice of i 0, 1 . -rule 2 { } _ (',s[a/x]) I tests a played ( x',s) by choosing a 8 M. 2 (',s[a/x]) II confirms. -rule 8 ( x',s) I enquires about 9 a played existential statement. (',s[a/x]) II makes a choice of a M. -rule 2 9 Figure 6.12 The game SG( ,T). M Definition 6.25 The Semantic Game SG( ,T) of the set T of L-sentences M in NNF is the game (see Figure 6.11) G!(W ), where W consists of sequences (x0,y0,x1,y1,...) where player II has followed the rules of Figure 6.12 and if player II plays the pair (',s), where ' is a basic formula, then = '. M| s In the game SG( ,T) player II claims that every sentence of T is true in M . Player I doubts this and challenges player II. He may doubt whether a M Incomplete version for students of easllc2012 only. 6.5 The Semantic Game 95 certain ' T is true in , so he plays x =(', ). In this round, as in some 2 M 0 ; other rounds too, player II just confirms and plays the same pair as player I. This may seem odd and unnecessary, but it is for book-keeping purposes only. Player I in a sense gathers a finite set of formulas confirmed by player II and tries to end up with a basic formula which cannot be true. Theorem 6.26 Suppose L is a vocabulary, T is a set of L-sentences, and M is an L-structure. Then the following are equivalent: 1. = T . M| 2. Player II has a winning strategy in SG( ,T). M Proof Suppose = T . The winning strategy of player II in SG( ,T) is M| M to maintain the condition =si i for all yi =( i,si), i N, played by M| 2 her. It is easy to see that this is possible. On the other hand, suppose = T , M 6| say = ', where ' T . The winning strategy of player I in SG( ,T) is M 6| 2 M to start with x =(', ), and then maintain the condition = for all 0 ; M 6| si i yi =( i,si), i N, played by II: 2 1. If yi =( i,si), where i is basic, then player I has won the game, because = . M 6| si i 2. If y =( ,s ), where = ✓ ✓ , then player I can use the assumption i i i i 0 ^ 1 = to find k<2 such that = ✓ . Then he plays x = M 6| si i M 6| si k i+1 (✓k,si). 3. If y =( ,s ), where = ✓ ✓ , then player I knows from the as- i i i i 0 _ 1 sumption = that whether II plays (✓ ,s ) for k =0or k =1, the M 6| si i k i condition = ✓ still holds. So player I can play x =( ,s ) and M 6| si k i+1 i i keep his winning criterion in force. 4. If y =( ,s ), where = x', then player I can use the assumption i i i i 8 = to find a M such that = '. Then he plays x = M 6| si i 2 M 6| si[a/x] i+1 (',si[a/x]). 5. If y =( ,s ), where = x', then player I knows from the assumption i i i i 9 = that whatever (',s [a/x]) player II chooses to play, the condi- M 6| si i i tion = ' still holds. So player I can play ( x',s ) and keep his M 6| si[a/x] 9 i winning criterion in force. Example 6.27 Let L = f and =(N,fM), where f(n)=n +1. Let { } M ' = x y fxy. 8 9 ⇡ Clearly, = '. Thus player II has, by Theorem 6.26, a winning strategy in M| the game SG( , ' ). Figure 6.13 shows how the game might proceed. On M { } Incomplete version for students of easllc2012 only. 96 First-Order Logic IIIRule ( x y fxy, ) 8 9 ⇡ ; ( x y fxy, ) Axiom rule ( y fxy, (x, 25) ) 8 9 ⇡ ; 9 ⇡ { } ( y fxy, (x, 25) ) -rule ( y fxy, (x, 25) ) 9 ⇡ { } 8 9 ⇡ { } ( fxy, (x, 25), (y, 26) ) -rule . ⇡ { . } 9 . . Figure 6.13 Player II has a winning strategy in SG( , ' ). M { } IIIRule ( x y fyx, ) 8 9 ⇡ ; ( x y fyx, ) Axiom rule ( y fyx, (x, 0) ) 8 9 ⇡ ; 9 ⇡ { } ( y fyx, (x, 0) ) -rule ( y fyx, (x, 0) ) 9 ⇡ { } 8 9 ⇡ { } ( fyx, (x, 0), (y, 2) ) -rule ⇡ { } 9 (II has no good move) Figure 6.14 Player I wins the game SG( , ). M { } the other hand, suppose = x y fyx. 8 9 ⇡ Clearly, = '. Thus player I has, by Theorem 6.26 and Theorem 3.12, a M 6| winning strategy in the game SG( , ' ). Figure 6.14 shows how the game M { } might proceed. Example 6.28 Let be the graph of Figure 6.15. M and ' = x( y xEy yxEy). 8 9 ¬ ^9 Clearly, = '. Thus player II has, by Theorem 6.26, a winning strategy in M| the game SG( , ' ). Figure 6.16 shows how the game might proceed. On M { } the other hand, suppose = x( y xEy yxEy). 9 8 ¬ _8 Clearly, = '. Thus player I has, by Theorem 6.26 and Theorem 3.12, a M 6| winning strategy in the game SG( , ' ). Figure 6.17 shows how the game M { } might proceed. Incomplete version for students of easllc2012 only. 6.5 The Semantic Game 97 Figure 6.15 The graph . M IIIRule ( x( y xEy yxEy), ) 8 9 ¬ ^9 ; ( x( y xEy yxEy), ) Axiom rule ( y xEy yxEy, (x, d) ) 8 9 ¬ ^9 ; 9 ¬ ^9 { } ( y xEy yxEy, (x, d) ) -rule ( yxEy, (x, d) ) 9 ¬ ^9 { } 8 9 { } ( yxEy, (x, d) ) -rule ( yxEy, (x, d) ) 9 { } ^ 9 { } (xEy, (x, d), (y, c) ) -rule . { . } 9 . . Figure 6.16 Player II has a winning strategy in SG( , ' ). M { } IIIRule ( x( y xEy yxEy), ) 9 8 ¬ _8 ; ( x( y xEy yxEy), ) Axiom rule ( x( y xEy yxEy), ) 9 8 ¬ _8 ; 9 8 ¬ _8 ; ( y xEy yxEy), (x, a) ) -rule ( y xEy yxEy, (x, a) ) 8 ¬ _8 { } 9 8 ¬ _8 { } ( y xEy, (x, a) ) -rule ( xEy, (x, a), (y, d) ) 8 ¬ { } _ ¬ { } ( xEy, (x, a), (y, d) ) -rule ¬ { } 8 Figure 6.17 Player I wins the game SG( , ). M { } Incomplete version for students of easllc2012 only. 98 First-Order Logic 6.6 The Model Existence Game In this section we learn a new game associated with trying to construct a model for a sentence or a set of sentences. This is of fundamental importance in the sequel. Let us first recall the game SG( ,T): The winning condition for II in the M game SG( ,T) is the only place where the model (rather than the set M M M) appears. If we do not start with a model we can replace the winning M condition with a slightly weaker one and get a very useful criterion for the existence of some such that = T : M M| Definition 6.29 The Model Existence Game MEG(T,L) of the set T of L- sentences in NNF is defined as follows. Let C be a countably infinite set of new constant symbols. MEG(T,L) is the game G!(W ) (see Figure 6.11), where W consists of sequences (x0,y0,x1,y1,...) where player II has followed the rules of Figure 6.18 and for no atomic L C-sentence ' both ' and ' are in [ ¬ y ,y ,... . { 0 1 } The idea of the game MEG(T,L) is that player I does not doubt the truth of T (as there is no model around) but rather the mere consistency of T . So he picks those ' T that he thinks constitute a contradiction and offers them 2 to player II for confirmation. Then he runs through the subformulas of these sentences as if there was a model around in which they cannot all be true. He wins if he has made player II play contradictory basic sentences. It turns out it did not matter that we had no model around, as two contradictory sentences cannot hold in any model anyway. Definition 6.30 Let L be a vocabulary with at least one constant symbol. A Hintikka set (for first-order logic) is a set H of L-sentences in NNF such that: 1. tt H for every constant L-term t. ⇡ 2 2. If '(x) is basic, '(c) H and tc H, then '(t) H. 2 ⇡ 2 2 3. If ' H, then ' H and H. ^ 2 2 2 4. If ' H, then ' H or H. _ 2 2 2 5. If x'(x) H, then '(c) H for all c L 8 2 2 2 6. If x'(x) H, then '(c) H for some c L. 9 2 2 2 7. For every constant L-term t there is c L such that ct H. 2 ⇡ 2 8. There is no atomic sentence ' such that ' H and ' H. 2 ¬ 2 Lemma 6.31 Suppose L is a vocabulary and T is a set of L-sentences. If T has a model, then T can be extended to a Hintikka set. Incomplete version for students of easllc2012 only. 6.6 The Model Existence Game 99 xn yn Explanation ' I enquires about ' T . 2 ' II confirms. tt I enquires about an equation. ⇡ tt II confirms. ⇡ '(t0) I chooses played '(t) and tt0 with ' basic and enquires about substituting⇡ t0 for t in '. '(t0) II confirms. 'i I tests a played '0 '1 by choosing i 0, 1 . ^ 2 { } 'i II confirms. '0 '1 I enquires about a played disjunction. _ 'i II makes a choice of i 0, 1 2 { } '(c) I tests a played x'(x) by choosing c C. 8 2 '(c) II confirms. x'(x) I enquires about a played existential statement. 9 '(c) II makes a choice of c C 2 t I enquires about a constant L C-term t. [ ct II makes a choice of c C ⇡ 2 Figure 6.18 The game MEG(T,L). Proof Let us assume = T . Let L0 L such that L0 has a constant M| ◆ symbol c / L for each a M. Let ⇤ be an expansion of obtained by a 2 2 M M interpreting c by a for each a M. Let H be the set of all L0-sentences true a 2 in . It is easy to verify that H is a Hintikka set. M Incomplete version for students of easllc2012 only. 100 First-Order Logic Lemma 6.32 Suppose L is a countable vocabulary and T is a set of L- sentences. If player II has a winning strategy in MEG(T,L), then the set T can be extended to a Hintikka set in a countable vocabulary extending L by constant symbols. Proof Suppose player II has a winning strategy in MEG(T,L). We first run through one carefully planned play of MEG(T,L). This will give rise to a model . Then we play again, this time providing a proof that = T .To M M| this end, let Trmbe the set of all constant L C-terms. Let [ T = 'n : n N , { 2 } C = cn : n N , { 2 } Trm = tn : n N . { 2 } Let (x0,y0,x1,y1,...) be a play in which player II has used her winning strategy and player I has maintained the following conditions: i 1. If n =3, then xn = 'i. 2. If n =2 3i, then x is c c . · n ⇡ i i 3. If n =4 3i 5j 7k 11l, y is t t , and y is '(t ), then x is '(t ). · · · · i ⇡ j k l j n k 4. If n =8 3i 5j, y is ✓ ✓ , and j<2, then x is ✓ . · · i 0 ^ 1 n j 5. If n = 16 3i, and y is ✓ ✓ , then x is ✓ ✓ . · i 0 _ 1 n 0 _ 1 6. If n = 32 3i 5j, y is x'(x), then x is '(c ). · · i 8 n j 7. If n = 64 3i, and y is x'(x), then x is x'(x). · i 9 n 9 8. If n = 128 3i, then x is t . · n i The idea of these conditions is that player I challenges player II in a maximal way. To guarantee this he makes a plan. The plan is, for example, that on round i 3 he always plays 'i from the set T . Thus in an infinite game every element of T will be played. Also the plan involves the rule that if player II happens to play a conjunction ✓ ✓ on round i, then player I will necessarily play 0 ^ 1 ✓ on round 8 3i and ✓ on round 8 3i 5, etc. It is all just book-keeping – 0 · 1 · · making sure that all possibilities will be scanned. This strategy of I is called the enumeration strategy. It is now routine to show that H = y ,y ,... is a { 0 1 } Hintikka set. Lemma 6.33 Every Hintikka set has a model in which every element is the interpretation of a constant symbol. Proof Let c c0 if c0c H. The relation is an equivalence relation ⇠ ⇡ 2 ⇠ on C (see Exercise 6.77). Let us define an L C-structure as follows. [ M Incomplete version for students of easllc2012 only. 6.6 The Model Existence Game 101 We let M = [c]:c C . For c C we let cM =[c]. If f L and { 2 } 2 2 #(f)=n we let f M([ci1 ],...,[cin ]) = [c] for some (any – see Exercise 6.78) c C such that cfc ...c H. For any constant term t there is a c C 2 ⇡ i1 in 2 2 such that ct H. It is easy to see that tM =[c]. For the atomic sentence ⇡ 2 ' = Rt ...t we let = ' if and only if ' is in H. An easy induction 1 n M| on ' shows that if '(x ,...,x ) is an L-formula and '(d ,...,d ) H for 1 n 1 n 2 some d ...,d , then = '(d ,...,d ) (see Exercise 6.79). In particular, 1 n M| 1 n = T . M| Lemma 6.34 Suppose L is a countable vocabulary and T is a set of L- sentences. If T can be extended to a Hintikka set in a countable vocabulary extending L, then player II has a winning strategy in MEG(T,L) Proof Suppose L⇤ is a countable vocabulary extending L such that some Hintikka set H in the vocabulary L⇤ extends T . Let C = cn : n N be { 2 } a new countable set of constant symbols to be used in MEG(T,L). Suppose D = tn : n N is the set of constant terms of the vocabulary L⇤. The { 2 } winning strategy of player II in MEG(T,L) is to maintain the condition that if y is '(c ,...,c ), then '(t ,...,t ) H. i 1 n 1 n 2 We can now prove the basic element of the Strategic Balance of Logic, namely the following equivalence between the Semantic Game and the Model Existence Game: Theorem 6.35 (Model Existence Theorem) Suppose L is a countable vocab- ulary and T is a set of L-sentences. The following are equivalent: 1. There is an L-structure such that = T . M M| 2. Player II has a winning strategy in MEG(T,L). Proof If there is an L-structure such that = T , then by Lemma 6.31 M M| there is a Hintikka set H T . Then by Lemma 6.34 player II has a winning ◆ strategy in MEG(T,L). Suppose conversely that player II has a winning strat- egy in MEG(T,L). By Lemma 6.32 there is a Hintikka set H T . Finally, ◆ this implies by Lemma 6.33 that T has a model. Corollary Suppose L is a countable vocabulary, T a set of L-sentences and ' an L-sentence. Then the following conditions are equivalent: 1. T = '. | 2. Player I has a winning strategy in MEG(T ' ,L). [ {¬ } Proof By Theorem 3.12 the game MEG(T ' ,L) is determined. So [ {¬ } by Theorem 6.35, condition 2 is equivalent to T ' not having a model, [ {¬ } which is exactly what condition 1 says. Incomplete version for students of easllc2012 only. 102 First-Order Logic Condition 1 of the above Corollary is equivalent to ' having a formal proof from T . (See Enderton (2001), or any standard textbook in logic for a definition of formal proof.) We can think of a winning strategy of player I in MEG(T [ ' ,L) as a semantic proof. In the literature this concept occurs under the {¬ } names semantic tree or Beth tableaux. 6.7 Applications The Model Existence Theorem is extremely useful in logic. Our first applica- tion – The Compactness Theorem – is a kind of model existence theorem itself and very useful throughout model theory. Theorem 6.36 (Compactness Theorem) Suppose L is a countable vocabu- lary and T is a set of L-sentences such that every finite subset of T has a model. Then T has a model. Proof Let C be a countably infinite set of new constant symbols as needed in MEG(T,L). The winning strategy of player II in MEG(T,L) is the follow- ing. Suppose (x0,y0,...,xn 1,yn 1) has been played up to now, and then player I plays xn. Player II has made sure that T y0,...,yn 1 is finitely consistent, i.e. each of its finite sub- [ { } sets has a model. Now she makes such a move y that T y ,...,y is still n [ { 0 n} finitely consistent. Suppose this is the case and player I asks a confirmation for ', where ' T . Now T y0,...,yn 1, ' is finitely consistent as it is the 2 [ { } same set as T y0,...,yn 1 . Suppose then player I asks a confirmation for [ { } ✓0, where ✓0 ✓1 = yi for some i consistent. Suppose the contrary. Then there is a finite conjunction of sen- tences in T such that y0,...,yn 1, = '(c). { }| ¬ Hence y0,...,yn 1, = x '(x). { }| 8 ¬ But this contradicts the fact that y0,...,yn 1, has a model in which x'(x) { } 9 is true. Finally, if t is a constant term, it follows as above that there is a constant c C such that T y0,...,yn 1, ct is finitely consistent. 2 [ { ⇡ } It is a consequence of the Compactness Theorem that a theory in a countable vocabulary is consistent in the sense that every finite subset has a model if and only if it is consistent in the sense that T itself has a model. Therefore the word “consistent” is used in both meanings. As an application of the Compactness Theorem consider the vocabulary L = +, , 0, 1 of number theory. An example of an L-structure is the so-called { · } standard model of number theory =(N, +, , 0, 1). L-structures may be N · elementary equivalent to and still be non-standard in the sense that they are N not isomorphic to . Let c be a new constant symbol. It is easy to see that the N theory ' : = ' 1 T czEcz+1 : z Z . [ { 2 } Any finite subset of T 0 mentions only finitely constants cz, so it can be satisfied in the model for a sufficiently large n. By the Compactness Theorem T 0 Mn has a model . Now L = T and the elements cM, z Z, constitute an M M | z 2 infinite cycle in . M As another application of the Model Existence Game we prove the so-called Incomplete version for students of easllc2012 only. 6.8 Interpolation 107 Thus (N, +, , 0, 1,A) 0 . · ⌘ M ⌘ M 2 In general, the significance of the Omitting Types Theorem is the fact that it can be used – as above – to get “standard” models. 6.8 Interpolation The Craig Interpolation Theorem says the following: Suppose = ' , | ! where ' is an L -sentence and is an L -sentence. Then there is an L L - 1 2 1 \ 2 sentence ✓ such that = ' ✓ and = ✓ . Here is an example: | ! | ! Example 6.39 L1 = P, Q, R ,L2 = P, Q, S . Let { } { } ' = x(Px Rx) x(Rx Qx) 8 ! ^8 ! and = x(Sx Px) x(Sx Qx). 8 ! !8 ! Now = ' , | ! and indeed, if ✓ = x(Px Qx), 8 ! then ✓ is an L L -sentence such that 1 \ 2 = ' ✓ and = ✓ . | ! | ! The Craig Interpolation Theorem is a consequence of the following remark- able subformula property of the Model Existence Game MEG(T,L): Player II never has to play anything but subformulas of sentences of T up to a substi- tution of terms for free variables. Theorem 6.40 (Craig Interpolation Theorem) Suppose = ' , where ' | ! is an L -sentence and is an L -sentence. Then there is an L L -sentence 1 2 1 \ 2 ✓ such that = ' ✓ and = ✓ . | ! | ! Proof We assume, for simplicity, that L1 and L2 are relational. This re- striction can be avoided (see Exercise 6.97). Let us assume that the claim of the theorem is false and derive a contradiction. Since = ' , player | ! I has a winning strategy in MEG( ', ,L L ). Therefore to reach { ¬ } 1 [ 2 Incomplete version for students of easllc2012 only. 108 First-Order Logic a contradiction it suffices to construct a winning strategy for player II in MEG( ', ,L L ). If ' alone is inconsistent, we can take any incon- { ¬ } 1 [ 2 sistent L-sentence as ✓. Likewise if alone is inconsistent, we can take any ¬ valid L-sentence as ✓. Let L = L L . Let us consider the following strategy 1 \ 2 of player II. Suppose C = cn : n N is a set of new constant symbols. { 2 } We denote L C-sentences by ✓(c0,...,cm 1) where ✓(z0,...,zm 1) is as- [ sumed to be an L-formula. Suppose player II has played Y = y0,...,yn 1 n n{ } so far. While she plays, she maintains two subsets S1 and S2 of Y such that Sn Sn = Y . The set Sn consists of all L C-sentences in Y , and Sn 1 [ 2 1 1 [ 2 consists of all L C-sentences in Y . Let us say that an L C-sentence ✓ 2 [ [ separates Sn and Sn if Sn = ✓ and Sn = ✓. Player II plays so that the 1 2 1 | 2 | ¬ following condition holds at all times: (?) There is no L C-sentence ✓ that separates Sn and Sn. [ 1 2 Let us check that she can maintain this strategy: (There is no harm in assuming that player I plays ' and first.) ¬ 0 0 Case 1. Player I plays '. We let S1 = ' and S2 = . Condition (?) holds, n { } ; as S1 is consistent. 1 Case 2. Player I plays having already played '. We let S1 = ' and 1 ¬ 1 1 { } S2 = . Suppose ✓(c0,...,cm 1) separates S1 and S2 . Then = ' {¬ } | ! z0 ... zm 1✓(z0,...,zm 1) and = z0 ... zm 1✓(z0,...,zm 1) 8 8 | 8 8 ! contrary to assumption. n+1 Case 3. Player I plays cc, where, for example, c L1 C. We let S1 = n n+1⇡ n 2 [ S0 cc and S2 = S1 cc . Suppose ✓(c0,...,cm 1) separates n+1[ {⇡ }n+1 [ {⇡ } n n S1 and S2 . Then clearly also ✓(c0,...,cm 1) separates S1 and S2 ,a contradiction. n Case 4. Player I plays '0(c1), where, for example, '0(c0), c0c1 S1 .We n+1 n n+1 n ⇡ 2 let S1 = S1 '0(c1) and S2 = S2 . Suppose ✓(c0,...,cm) separates n+1 n+1[ { } n n S1 and S2 . Then as S1 = '0(c1) clearly ✓(c0,...,cm 1) separates S1 n | and S2 , a contradiction. n n+1 Case 5. Player I plays 'i, where, for example, '1 '1 S1 . We let S1 = n n+1 n ^ 2 n+1 S1 'i and S2 = S2 . Suppose ✓(c0,...,cm 1) separates S1 and n+1[ { } n n n S2 . Then, as S1 = 'i, clearly ✓(c0,...,cm 1) separates S1 and S2 ,a | contradiction. n Case 6. Player I plays '0 '1, where, for example, '0 '1 S1 . We claim _ n n _ 2 that for one of i 0, 1 the sets S1 'i and S2 satisfy (?). Otherwise 2 { } [ { } n there is for both i 0, 1 some ✓i(c0,...,cm 1) that separates S1 'i 2 { } [ { } Incomplete version for students of easllc2012 only. 6.8 Interpolation 109 n and S2 . Let ✓(c0,...,cm 1)=✓0(c0,...,cm 1) ✓1(c0,...,cm 1). _ n n n Then, as S1 = '0 '1, clearly ✓(c0,...,cm 1) separates S1 and S2 ,a | _ contradiction. n Case 7. Player I plays '(c0), where, for example, x'(x) S1 . We claim that n n 8 2 the sets S1 '(c0) and S2 satisfy (?). Otherwise there is ✓(c0,...,cm 1) [ { } that separates Sn '(c ) and Sn. Let 1 [ { 0 } 2 ✓0(c1,...,cm 1)= x✓(x, c1,...,cm 1). 8 n n Then, as S1 = x'(x), we have S1 = '(c0), and hence ✓0(c0,c1,...,cm 1) n| 8 n | separates S1 and S2 , a contradiction. n Case 8. Player I plays x'(x), where, for example, x'(x) S1 . Let c C 9 9 2 n 2 be such that c does not occur in Y yet. We claim that the sets S1 '(c) n [ { } and S2 satisfy (?). Otherwise there is some ✓(c, c0,...,cm 1) that separates Sn '(c) and Sn. Let 1 [ { } 2 ✓0(c1,...,cm 1)= x✓(x, c0,...,cm 1). 9 n n Then, as S1 = x'(x) and S1 = '(c) ✓(c, c0,...,cm 1) we clearly have | 9 | n !n that ✓0(c1,...,cm 1) separates S1 and S2 , a contradiction. Example 6.41 The Craig Interpolation Theorem is false in finite models. To see this, let L = R and L = P where R and P are distinct binary 1 { } 2 { } predicates. Let ' say that R is an equivalence relation with all classes of size 2 and let say P is not an equivalence relation with all classes of size 2 except one of size 1. Then = ' holds for finite . If there were a sentence M| ! M ✓ of the empty vocabulary such that = ' ✓ and = ✓ for all M| ! M| ! finite , then ✓ would characterize even cardinality in finite models. It is easy M to see with Ehrenfeucht–Fra¨ısse´ Games that this is impossible. Theorem 6.42 (Beth Definability Theorem) Suppose L is a vocabulary and P is a predicate symbol not in L. Let ' be an L P -sentence. Then the [ { } following are equivalent: 1. If ( ,A) = ' and ( ,B) = ', where is an L-structure, then A = B. M | M | M 2. There is an L-formula ✓ such that ' = x0 ...xn 1(✓(x0,...,xn 1) P (x0,...,xn 1)). | 8 $ If condition 1 holds we say that ' defines P implicitly. If condition 2 holds, we say that ✓ defines P explicitly relative to '. Incomplete version for students of easllc2012 only. 110 First-Order Logic Proof Let '0 be obtained from ' by replacing everywhere P by P 0 (another new predicate symbol). Then condition 1 implies =(' Pc0 ...cn 1) ('0 P 0c0 ...cn 1). | ^ ! ! By the Craig Interpolation Theorem there is an L-formula ✓(x0,...,xn 1) such that =(' Pc0 ...cn 1) ✓(c0,...,cn 1) | ^ ! and = ✓(c0,...,cn 1) ('0 P 0c0 ...cn 1). | ! ! It follows easily that ✓ is the formula we are looking for. Example 6.43 The Beth Definability Theorem is false in finite models. Let ' be the conjunction of 1. “< is a linear order”. 2. x(Px y( xy x Lemma 6.44 Suppose L is a relational vocabulary and P L is a unary 2 predicate symbol. The following are equivalent for all L-formulas ' and all L-structures such that P M = : M 6 ; 1. = '(P ). M| 2. (P M) = '. M | Proof Exercise 6.101. Definition 6.45 Suppose L is a vocabulary. A class K of L-structures is an EC-class if there is an L-sentence ' such that K = Str(L): = ' {M 2 M| } and a PC-class if there is an L0-sentence ' for some L0 L such that ◆ K = L : Str(L0) and = ' . {M M 2 M| } Example 6.46 Let L = . The class of infinite L-structures is a PC-class ; which is not an EC class. (Exercise 6.102.) Example 6.47 Let L = . The class of finite L-structures is not a PC-class. ; (Exercise 6.103.) Example 6.48 Let L = < . The class of non-well-ordered L-structures is { } a PC-class which is not an EC-class. (Exercise 6.104.) Suppose = ' , where ' is an L -sentence and is an L -sentence. Let | ! 1 2 K = (L L ): = ' 1 {M 1 \ 2 M| } and K = (L L ): = . 2 {M 1 \ 2 M| ¬ } Now K and K are disjoint PC-classes. If there is an L L -sentence ✓ 1 2 1 \ 2 such that = ' ✓ and = ✓ , then the EC-class | ! | ! K = : = ✓ {M M| } separates K and K in the sense that K K and K K = . On the 1 2 1 ✓ 2 \ ; other hand, if an EC-class K separates in this sense K1 and K2, then there is an L L -sentence ✓ such that = ' ✓ and = ✓ . Thus the Craig 1 \ 2 | ! | ! Interpolation Theorem can be stated as: disjoint PC-classes can always be separated by an EC-class. Theorem 6.49 (Separation Theorem) Suppose K1 and K2 are disjoint PC- classes of models. Then there is an EC-class K that separates K1 and K2, i.e. K K and K K = . 1 ✓ 2 \ ; Incomplete version for students of easllc2012 only. 112 First-Order Logic Proof The claim has already been proved in Theorem 6.40, but we give here a different – model-theoretic – proof. This proof is of independent interest, being as it is, in effect, the proof of the so-called Lindstrom’s¨ Theorem (Lindstrom¨ (1973)), which gives a model theoretic characterization of first order logic. Case 1: There is an n N such that some union K of n-equivalence classes 2 'p of models separates K1 and K2. By Theorem 6.5 the model class K is an EC-class, so the claim is proved. Case 2: There are, for any n N, L1 L2-models n and n such that 2 \ M N K , K , and there is a back-and-forth sequence (I : i n) Mn 2 1 Nn 2 2 i for and . Suppose K is the class of reducts of models of ', and K Mn Nn 1 2 respectively the class of reducts of models of . Let T be the following set of sentences: 1. '(P1). 2. (P2). 3. (R, <) is a non-empty linear order in which every element with a predeces- sor has an immediate predecessor. 4. z(Rz Q z). 8 ! 0 5. z u ... u v ... v ((Rz Q zu ...u v ...v ) 8 8 1 8 m8 1 8 m ^ n 1 m 1 m ! (✓(u ,...,u ) ✓(v ,...,v ))) for all atomic L L -formulas ✓. 1 m $ 1 m 1 \ 2 6. z u ... u v ... v ((Rz Q zu ...u v ...v ) z0 x((Rz0 8 8 1 8 n8 1 8 m ^ n 1 m 1 m !8 8 ^ z0 For all n N there is a model n of T with (R, <) of length n. The model 2 A n is obtained as follows. The universe An is the (disjoint) union of Mn, Nn, A P An and 1,...,n . The L1-structure ( n L1) 1 is chosen to be a copy of the { } A P An model of '. The L -structure ( L ) 2 is chosen to be a copy of Mn 2 An 2 the model of . The 2i +1-ary predicate Q is interpreted in as the set Nn i An (n i, u1,...,ui,v1,...,vi): (u1,v1),...,(ui,vi) In i . { { } 2 } By the Compactness Theorem, there is a countable model of T with (R, <) M non-well-founded (see Exercise 6.107). That is, there are an, n N, in M such 2 that an+1 is an immediate predecessor of an in for all n N. Let 1 be (P M) M 2 M the L1 L2-structure ( (L1 L2)) 1 . Let 2 be the L1 L2-structure \ (P M)M \ M \ ( (L L )) 2 . Now , for we have the back-and-forth set: M 1 \ 2 M1 'p M2 P = (u1,v1),...,(un,vn) : = Qnanu1 ...unv1 ...vn,n N . {{ } M| 2 } Incomplete version for students of easllc2012 only. 6.9 Uncountable Vocabularies 113 Since and are countable, they are isomorphic. But K and M1 M2 M1 2 1 K , a contradiction. M2 2 2 6.9 Uncountable Vocabularies So far we have concentrated on methods based on the assumption that vocab- ularies are countable. Several key methods work also for uncountable vocabu- laries. A typical application of uncountable vocabularies is the task of finding an elementary extension of an uncountable structure. In this case a new con- stant symbol is added to the vocabulary for each element of the model, and the vocabulary may become uncountable. Strictly speaking, handling uncountable vocabularies does not require deal- ing with ordinals, but since we use the Axiom of Choice anyway, it is more natural to assume our vocabularies are well-ordered as in L = R : ↵ < f : ↵ < c : ↵ < . { ↵ } [ { ↵ } [ { ↵ } We then allow also variable symbols x↵, ↵ < ✏. An important method throughout logic is the method of Skolem functions. Definition 6.50 Suppose L is a vocabulary, is an L-structure, and we M have an L-formula '(x0,...,xn) of first-order logic. A Skolem function for '(x ,...,x ) in is any function f : M n M such that for all elements 0 n M ' ! a0,...,an 1 of M: M = xn'(a0,...,an 1,xn) '(a0,...,an 1,f'(a0,...,an 1)). | 9 ! The following simple but fundamental fact is very helpful in the applications of Skolem functions: Proposition 6.51 (Tarski–Vaught criterion) Suppose L is a vocabulary, M an L-structure, and such that for all L-formulas '(x ,...,x ) the N ✓ M 0 n following holds: If a0,...,an 1 N and = '(a0,...,an 1,an) for some 2 M| (6.7) an M, then = '(a0,...,an 1,an0 ) for some an0 N. 2 M| 2 Then . N M Proof Exercise 6.110. Proposition 6.52 Suppose L is a vocabulary, an L-structure, and a M F family of functions such that every L-formula has a Skolem function in 2 F Incomplete version for students of easllc2012 only. 6.11 Historical Remarks and References 125 formula of the vocabulary Lm. By definition, J = i I i = '(a0(i),...,an 1(i)) D. { 2 M | } 2 Suppose i J E . Again, n m. Now 2 \ m i i = '(a0(i),...,an 1(i)) M | and (6.21) is a position in the game EF ( L , L ) while II uses ⌧ . ni Mi ni Ni ni i Since ⌧i is a winning strategy of II and '(x0,...,xn 1) is a formula of the vocabulary Lni , i = '(b0(i),...,bn 1(i)). N | Thus i I i = '(b0(i),...,bn 1(i)) J Em D, { 2 N | } ◆ \ 2 whence = '(b0,...,bn 1). N| Theorem 6.68 is by no means the best in this direction (see Benda (1969)). A particularly beautiful stronger result is the following result of Shelah (1971): if and only if there are I and an ultrafilter D on I such that /D = M ⌘ N i M ⇠ /D. i N Q Q 6.11 Historical Remarks and References Basic texts in model theory are Chang and Keisler (1990) and Hodges (1993). For the history of model theory, see Vaught (1974). Characterization of ele- mentary equivalence in terms of back-and-forth sequences (Theorem 6.5 and its Corollary) is due to Fra¨ısse´ (1955). The concepts and results of Section 6.4 are due to Kueker (1972, 1977). Theorem 6.23 goes back to Lowenheim¨ (1915) and Skolem (1923, 1970). The idea of interpreting the quantifiers in terms of moves in a game, as in the Semantic Game, is due to Henkin (1961). Hintikka (1968) extended this from quantifiers to propositional connectives and emphasized its role in se- mantics in general. The roots of interpreting logic as a game go back, arguably, to Wittgenstein’s language games. Lorenzen (1961) used a similar game in proof theory. For the close general connection between inductive definitions and games see Aczel (1977). Our Model Existence Game is a game-theoretic rendering of the method of semantic tableaux of Beth (1955a,b), model sets of Hintikka (1955), and consistency properties of Smullyan (1963, 1968). Its roots are in the proof- theoretic method of natural deduction of Gentzen (1934, 1969). A good source Incomplete version for students of easllc2012 only. 126 First-Order Logic for more advanced applications of the Model Existence Game is Hodges (1985). Theorem 6.42 is due to Beth (1953) and the stronger but related Theorem 6.40 to Craig (1957a). For the background of Theorem 6.40 see Craig (2008), and for its early applications Craig (1957b). The failure of Graig Interpolation in fi- nite models was observed in Hajek´ (1976), see also Gurevich (1984), which has this and Example 6.43. The proof of Theorem 6.49 is modeled according to the proof in Barwise and Feferman (1985) of the main result of Lindstrom¨ (1973), the so-called Lindstrom’s¨ Theorem, which characterizes first-order logic as a maximal logic which satisfies the Compactness Theorem and the Lowenheim–¨ Skolem Theorem in the form: every sentence of the logic which has an infinite model has a countable model. The connection to Theorem 6.49 is the follow- ing: Suppose L⇤ were such a logic and ' L⇤. We could treat the class of 2 models of ' and the class of models of ' as we treat the disjoint PC-classes ¬ K1 and K2 in Theorem 6.49. The proof then shows that a first-order sentence ✓ can separate the class of models of ' and the class of models of '. This ¬ would clearly mean that ' would be logically equivalent to ✓, hence first-order definable. Theorem 6.62 is from Keisler and Morley (1968). Ultraproducts were introduced by Łos´ (1955). For a survey of the use of them in model theory see Bell and Slomson (1969). Theorem 6.68 is from Benda (1969). Exercise 6.9 is from Brown and Hoshino (2007), where more information about Ehrenfeucht-Fra¨ısse´ games for paths and cycles can be found. See also Bissell-Siders (2007). Exercise 6.114 is from Morley (1968). Exercises 6.1 A finite connected graph is a cycle if every vertex has degree 2. Write a sentence of quantifier rank 2 which holds in a cycle if and only if the cycle has length 3. Show that no such sentence of quantifier rank 1 exists. 6.2 Write a sentence of quantifier rank 3 which holds in a cycle if and only if the cycle has length 4. Show that no such sentence of quantifier rank 2 exists. Do the same for the cycle of length 5. 6.3 Do the previous Exercise for the cycle of length 6. 6.4 Write a sentence of quantifier rank 4 which holds in a graph if and only if the cycle has length 7. Show that no such sentence of quantifier rank 3 exists. Do the same for the cycle of length 8. 6.5 Write a sentence of quantifier rank 4 which holds in a graph if and only if the cycle has length 9. Show that no such sentence of quantifier rank 3 exists. Incomplete version for students of easllc2012 only. Exercises 127 6.6 Construct a sentence of quantifier rank 2 which is true in an ordered set if and only if has length 2. M M 6.7 Construct a sentence 'n of quantifier rank 3 which is true in an ordered set if and only if has length n, where n is 3, 4, 5, or 6. M M 6.8 Show that there is a sentence of quantifier rank 4 which is true in a graph if and only if the graph is a cycle, but no such sentence of quantifier rank 3 exists. n 1 6.9 Show that if n 3 and and are cycles of length 2 +3, then M N n . M 'p N 6.10 Suppose L = and n N. Into how many classes does n divide ; 2 ⌘ Str(L)? 6.11 Suppose L = c and n N. Into how many classes does n divide { } 2 ⌘ Str(L)? 6.12 Suppose L = P , #(P )=1, and n N. Into how many classes does { } 2 divide Str(L)? ⌘n 6.13 Suppose L = R , #(R)=2. Into how many classes does divide { } ⌘1 Str(L)? 6.14 Suppose L = R , #(R)=2. Show that divides Str(L) into at least { } ⌘2 11 classes. n 6.15 Construct for each n>0 trees and 0 of height 2 such that p 0 n+1 T T T ' T but 0. T 6'p T 6.16 Consider EF ( , 0) where and T 0 are the trees below. Show that 3 T T T player I has a winning strategy. Then write a sentence of quantifier rank 3 which is true in but false in 0. T T 6.17 Suppose is an equivalence relation with n classes of size 1 and n +1 M classes of size 2. Suppose, on the other hand, that is an equivalence N relation with n +1classes of size 1 and n classes of size 2. Show that II has a winning strategy in EF ( , 0) but I has a winning strategy n+1 M M in EF ( , 0). n+2 M M 6.18 Suppose is an equivalence relation with n classes of size k and n +1 M classes of size k+1. Suppose, on the other hand, that is an equivalence N relation with n +1classes of size k and n classes of size k +1. Show that II has a winning strategy in EF ( , 0) but I has a winning n+k M M strategy in EF ( , 0). n+k+1 M M Incomplete version for students of easllc2012 only. 128 First-Order Logic 6.19 Suppose L = c, d . Which of the following properties of L-structures { } can be expressed in FO with a sentence of quantifier rank 1: M (a) M = cM,dM . 6 { } (b) M 2. | | (c) M cM 1. | \{ }| (d) M =2. | | 6.20 Like Exercise 6.19 but L = R , #(R)=2, and the cases are: { } (a) There is a M such that (b, a) RM for all b M a . 2 2 2 \{ } (b) RM is symmetric. (c) RM is reflexive. 6.21 Suppose L = R , #(R)=2. Which of the following properties of L- { } structures can be expressed in FO with a sentence of quantifier rank M 2. (a) is an ordered set. M (b) is a partially ordered set. M (c) is an equivalence relation. M (d) is a graph. M 6.22 Suppose L = < , #(<)=2. Which of the following properties of L- { } structures can be expressed in FO with a sentence of quantifier rank M 3: (a) is a dense linear order. M (b) is an ordered set with at least eight elements. M (c) is a linear order with at least two limit points. (a is a limit point M if a has predecessors but no immediate predecessor.) 6.23 Which of the following sentences are logically equivalent to a sentence of quantifier rank 1: (a) x0 x1( Rx0 Px1). 8 9 ¬ _ (b) x0 x1(Rx1 Rx0). 9 9 ^ (c) x0 x1( x0x1 Px0). 9 9 ¬⇡ ^ (d) x0 x1 x0x1. 8 9 ¬⇡ 6.24 Which of the following sentences are logically equivalent to a sentence of quantifier rank 2: (a) x0 x1 x2(Rx0x2 Sx0x1). 8 9 8 _ (b) x0 x1 x2(Rx0x2 Sx1x2). 9 9 8 _ 6.25 Suppose we are told of an ordered set that 2 (N,<). Can we M M ⌘ conclude that Incomplete version for students of easllc2012 only. Exercises 129 (a) M is infinite? (b) has a smallest element? M (c) Every element has finitely many predecessors? 6.26 Show that if 3 (Q,<), then (Q,<). M ⌘ M ⌘ 6.27 Show that if 3 (Z,<), then (Z,<). M ⌘ M ⌘ 6.28 Show that for all n there are and such that but M N M ⌘n N M 6⌘n+1 . N 6.29 Suppose L is a vocabulary and an L-structure. is !-saturated if A A every type of the expanded structure ( ,a0,...,an 1), where the ele- A ments a0,...,an 1 are from A, is realized in ( ,a0,...,an 1). Sup- A pose and are !-saturated structures such that . Show that A B A ⌘ B . A 'p B 6.30 Let = X !(R):sup(X) = 1000 . What is a good starting move C { 2 P } for player I in Gcub( )? Let 0 = X !(R):inf(X) 1000 . C C { 2 P } What is a good starting move for player II in G ( 0)? cub C 6.31 Let = X !(R):X is dense (meets every non-empty open set) in C { 2 P R . What is a good strategy for player II in Gcub( )? } C 6.32 Let = X !(R):every point in X is a limit point of X . What is C { 2 P } a good strategy for player II in G ( )? cub C 6.33 Let = X N : m N n N(X [n, n + m]= ) . What is a C { ✓ 8 2 9 2 \ ; } good strategy for player I in G ( )? cub C 6.34 Decide which player has a winning strategy in the Cub Game of the following sets: 1. X (A):a X , where a A. { 2 P! 2 } 2 2. X (A):B X is finite , where B (A). { 2 P! \ } 2 P 3. X (A):B X is countable , where B (A). { 2 P! \ } 2 P 4. X !(R):X is bounded . { 2 P } 5. X !(R):X is closed . { 2 P } 6.35 Compute a A a if a = X !(A):a A . 4 2 C C { 2 P 2 } 6.36 Let f : A A. Let = X (A):f(a) X and 0 = X ! Ca { 2 P! 2 } Ca { 2 !(A):f(a) X . Compute a A a and a A a0 . P 62 } 4 2 C 5 2 C 6.37 Suppose is an ordered set. For a M let be the set of X M M 2 Ca ✓ which have an element above a in and let 0 be the set of X M M Ca ✓ which are bounded by a in . Describe the sets a M a and a M a. M 4 2 C 5 2 C 6.38 Let L be a relational vocabulary. Suppose f : = , where and M ⇠ N M and L-structures such that M = N. If A M, there is a unique N ✓ submodel A of with domain A. Show that player II has a M M winning strategy in Gcub( X (M): A = A ). { 2 P! M ⇠ N } Incomplete version for students of easllc2012 only. 130 First-Order Logic 6.39 Suppose is a connected graph. Describe a winning strategy for player G II in Gcub( ), where = X !(G):[X] is connected . C C { 2 P G } 6.40 Suppose (A, <) is an ordered set and X has a last element for a station- ary set of countable X A. Show that (A, <) itself has a last element. ✓ 6.41 Show that the set of sets (A) which contain a cub is a CUBA C ✓ P! countably closed filter (i.e. (1) If and (A), then C 2 CUBA C ✓ D ✓ P! A. (2) If n A for all n N, then n A). D 2 CUB C 2 CUB 2 n N C 2 CUB In fact, is a normal filter (i.e. if for2 all a A, then CUBA Ca 2 CUBAT 2 a A a A). 4 2 C 2 CUB 6.42 Show that if is stationary and cub, then is stationary. D C D \ C 6.43 Show that if = n is stationary, then there is n N such that D n N D 2 is stationary. 2 Dn S 6.44 Show that if = a A a is stationary, then there is a A such that D 5 2 D 2 is stationary. Da 6.45 (Fodor’s Lemma) Suppose is stationary and f(X) X for every D 2 X . Show that there is a stationary such that f is constant on 2 C D ✓ C . (Hint: Let = X : f(X)=a . Assume no is stationary and D Ca { } Ca use Lemma 6.15 to derive a contradiction.) 6.46 Show that if A is an uncountable set, then there is a stationary set C ✓ !(A) such that also !(A) is stationary. Such sets are called bis- P P \C X tationary. Note that then / A. (Hint: Write X = a : n N C 2 CUB { n 2 } whenever X (A). Apply the above Fodor’s Lemma to the func- 2 P! tions f (X)=aX to find for each n a stationary on which f is n n Dn n constant. If each (A) is non-stationary, there is for each n a cub P! \Dn set . Let = and show that can have only one element, Cn ✓ Dn C Cn C which contradicts the fact that is cub.) T C 6.47 Use the previous exercise to conclude that is not an ultrafilter (i.e. CUBA a maximal filter) if A is infinite. 6.48 Show that the set NSA of sets (A) which are non-stationary is a C ✓ P! -ideal (i.e. (1) If NSA and (A), then NSA. (2) D 2 C ✓ D ✓ P! C 2 If NSA for all n , then NSA). In fact, NSA is a n N n N n D 2 2 A 2 D 2 A normal ideal (i.e. if a NS for all a A, then a A a NS ). D 2 S 2 5 2 D 2 6.49 Show that if a sentence is true in a stationary set of countable submodels of a model then it is true in the model itself. More exactly: Let L be a countable vocabulary, an L-model, and ' an L-sentence. Suppose M X !(M):[X] = ' is stationary. Show that = '. { 2 P M | } M| 6.50 In this and the following exercises we develop the theory of cub and stationary subsets of a regular cardinal > !. A set C is closed if it ✓ contains every non-zero limit ordinal < such that C is unbounded \ in , and unbounded if it is unbounded as a subset of . We call C ✓ Incomplete version for students of easllc2012 only. Exercises 131 a closed unbounded (cub) set if C is both closed and unbounded. Show that the following sets are cub (i) . (ii) ↵ < : ↵ is a limit ordinal . { } (iii) ↵ < : ↵ = ! for some . { } (iv) ↵ < : if < ↵ and < ↵, then + < ↵ . { } (v) ↵ < : if ↵ = , then ↵ = or ↵ = . { · } 6.51 Show that the following sets are not cub: (i) . ; (ii) ↵ < ! : ↵ = +1for some . { 1 } (iii) ↵ < ! : ↵ = ! + ! for some . { 1 } (iv) ↵ < ! :cf(↵)=! . { 2 } 6.52 Show that a set C contains a cub subset of !1 if and only if player II wins the game G!(WC ), where WC = (x0,x1,x2,...):supxn C . { n 2 } 6.53 A filter on M is -closed if A for ↵ < , where < , implies F ↵ 2 F A . A filter on is normal if A for ↵ < implies ↵ ↵ 2 F F ↵ 2 F A , where 4T↵ ↵ 2 F A = ↵ < : ↵ A for all < ↵ . 4↵ ↵ { 2 } Note that normality implies -closure. Show that if > ! is regular, then the set of subsets of that contain a cub set is a proper normal F filter on . The filter is called the cub-filter on . F 6.54 A subset of which meets every cub set is called stationary. Equiva- lently, a subset S of is stationary if its complement is not in the cub- filter. A set which is not stationary, is non-stationary. Show that all sets in the cub-filter are stationary. Show that ↵ < ! : cof(↵)=! { 2 } is a stationary set which is not in the cub-filter on !2. 6.55 (Fodor’s Lemma, second formulation) Suppose > ! is a regular car- dinal. If S is stationary and f : S satisfies f(↵) < ↵ for all ✓ ! ↵ S, then there is a stationary S0 S such that f is constant on S0. 2 ✓ (Hint: For each ↵ < let S = < : f( )=↵ . Show that one of ↵ { } the sets S↵ has to be stationary. ) Incomplete version for students of easllc2012 only. 132 First-Order Logic 6.56 Suppose is a regular cardinal > !. Show that there is a bistationary set S (i.e. both S and S are stationary). (Hint: Note that S = ✓ \ ↵ < :cf(↵)=! is always stationary. For ↵ S let : ! ↵ be { } 2 ↵ ! strictly increasing with supn ↵(n)=↵. By the previous exercise there is for each n 6.57 Suppose is a regular cardinal > !. Show that = ↵< S↵ where the sets S are disjoint stationary sets. (Hint: Proceed as in Exercise 6.56. ↵ S Find n !. Show that S is stationary if and only if ✓ every regressive f : S is constant on an unbounded set. ! 6.60 Prove that C ! is in the cub filter if and only if almost all countable ✓ 1 subsets of !1 have their sup in C. 6.61 Suppose S ! is stationary. Show that for all ↵ < ! there is a closed ✓ 1 1 subset of S of order-type ↵. (Hint: Prove a stronger claim by induction on ↵.) 6.62 Decide first which of the following are true and then show how the win- ner should play the game SG( ,T): M 1. (R,<,0) = x y(y Figure 6.22 ... and II has a winning strategy in each SG ( n,T), then II has a 89 M winning strategy in SG ( n1=0 n,T). 89 [ M 6.67 A formula in NNF is universal-existential if it is of the form y ... y x ... x ', 8 1 8 n9 1 9 m where ' is quantifier free. Show that if L is countable and T is a set of universal-existential L-sentences, then = T if and only if player II M| has a winning strategy in the game SG ( ,T). 89 M 6.68 The Positive Semantic Game SG ( ,T) differs from SG( ,T) only pos M M in that the winning condition “If player II plays the pair (',s), where ' is basic, then = '” is weakened to “If player II plays the pair M| s (',s), where ' is atomic, then = '”. Suppose and are L- M| s M N structures. A surjection h : M N is a homomorphism ! M ! N if = '(a ,...,a ) = '(f(a ),...,f(a )) M| 1 n ) N| 1 n for all atomic L-formulas ' and all a ,...,a M. Show that if II 1 n 2 has a winning strategy in SG ( ,T) and h : is a surjective pos M M ! N homomorphism, then II has a winning strategy in SG ( ,T). pos N 6.69 A formula in NNF is positive if it contains no negations. Show that if L is countable and T is a set of positive L-sentences, then = T if and M| only if player II has a winning strategy in the game SG ( ,T). pos M 6.70 The game MEG(T,L) is played with T = Pc, Qfc, x ( Px Qx ), x ( Px Pfx ) . { ¬ 8 0 ¬ 0 _ 0 8 0 ¬ 0 _ 0 } The game starts as in Figure 6.22. How does I play now and win? 6.71 Consider T = x x Rx x , x x Rx x . Now we start the {9 08 1 0 1 9 18 0¬ 0 1} game MEG(T,L) as in Figure 6.23. How does I play now and win? 6.72 Consider T = x ( Px Qx ), x (Qx Px ) . The game {8 0 ¬ 0 _ 0 9 0 0 ^ ¬ 0 } MEG(T,L) is played. Player I immediately resigns. Why? 6.73 The game MEG(T,L) is played with T = x x Ex , x x ( x Ex x Ex ), {8 0¬ 0 0 8 08 1 ¬ 0 1 _ 1 0 x x x Ex , x x x Ex . 8 09 1 0 1 8 09 1¬ 0 1} Incomplete version for students of easllc2012 only. 134 First-Order Logic III x0 x1Rx0x1 9 8 x1Rc0x1 8 x1 x0 Rx0x1 9 8 ¬ x0 Rx0c1 8 ¬ Figure 6.23 Player I immediately resigns. Why? 6.74 Use the game MEG(T,L) to decide whether the following sets T have a model: 1. xP x, y( Py Ry) . {9 8 ¬ _ } 2. xP xx, y x P xy . {8 9 8 ¬ } 6.75 Prove the following by giving a winning strategy of player I in the ap- propriate game MEG(T ' ,L): [ {¬ } 1. x(Px Qx), xP x = xQx. {8 ! 9 }| 9 2. xRxfx = x yRxy. {8 }| 8 9 6.76 Suppose T is the following theory x x 6.93 Suppose p and p0 are types of the theory T . Under which conditions does T have a model which realizes p but omits p0? (Hint: Consider the condition: For every '(x, y) there is (x) p0 such that for no 2 (y),..., (y) p do we have both 1 n 2 T ('(x, y) (y) ... (y)) (x) ` ^ 1 ^ ^ n ! and T x y('(x, y) (y) ... (y)) is consistent.) [ {9 9 ^ 1 ^ ^ n } 6.94 Let T be the theory of dense linear order without endpoints plus the axioms c 6.103 Prove Example 6.47. 6.104 Prove Example 6.48. 6.105 Suppose L is a finite vocabulary, P1 and P2 are unary predicate symbols in L. Show that the class of L-structures such that M (P M) (P M) (P M) (P M) 1 and 2 are well-defined and 1 = 2 M M M ⇠ M is a PC-class. 6.106 Suppose L is a finite vocabulary, P1 and P2 are unary predicate symbols in L. Show that for all n N the class of L-structures such that 2 M (P M) (P M) (P M) n (P M) 1 and 2 are well-defined and 1 2 M M M 'p M is a PC-class. 6.107 Suppose L is a countable vocabulary containing the binary predicate symbol <. Suppose T is a set of L-sentences. Prove that if T has for each n N a model in which n Gn/F . Show that G is disconnected. Q Incomplete version for students of easllc2012 only. 7 Infinitary Logic 7.1 Introduction As the name indicates, infinitary logic has infinite formulas. The oldest use of infinitary formulas is the elimination of quantifiers in number theory: x'(x) '(n) 9 $ n _2N x'(x) '(n). 8 $ n ^2N Here we leave behind logic as a study of sentences humans can write down on paper. Infinitary formulas are merely mathematical objects used to study properties of structures and proofs. It turns out that games are particularly suit- able for the study of infinitary logic. In a sense games replace the use of the Compactness Theorem which fails badly in infinitary logic. 7.2 Preliminary Examples The games we have encountered so far have had a fixed length, which has been either a natural number or ! (an infinite game). Now we introduce a game which is “dynamic” in the sense that it is possible for player I to change the length of the game during the game. He may first claim he can win in five moves, but seeing what the first move of II is, he may decide he needs ten moves. In these games player I is not allowed to declare he will need infinitely many moves, although we shall study such games, too, later. Before giving a rigorous definition of the Dynamic Ehrenfeucht–Fra¨ısse´ Game we discuss some simple versions of it. Incomplete version for students of easllc2012 only. 140 Infinitary Logic Definition 7.1 (Preliminary) Suppose , 0 are L-structures such that L is M M a relational vocabulary and M M 0 = . The Dynamic Ehrenfeucht–Fra¨ısse´ \ ; Game, denoted EFD ( , 0) is defined as follows: First player I chooses a ! M M natural number n and then the game EF ( , 0) is played. n M M Note that EFD ( , 0) is not a game of length !. Player II has a winning ! M M strategy in EFD ( , 0) if she has one in each EF ( , 0). On the other ! M M n M M hand, player I has a winning strategy in EFD ( , 0) if he can envisage a ! M M number n so that he has a winning strategy in EF ( , 0). n M M Example 7.2 If and 0 are L-structures such that M is finite and M 0 M M is infinite, then player I has a winning strategy in EFD ( , 0). Suppose ! M M M = n. Player I has a winning strategy in EF ( , 0). He first plays | | n+1 M M all n elements of M and then any unplayed element of M 0. Player II is out of good moves, and loses the game. Example 7.3 If and 0 are equivalence relations such that has finitely M M M many equivalence classes and 0 infinitely many, then player I has a win- M ning strategy in EFD ( , 0). Suppose the equivalence classes of are ! M M M [a1],...,[an]. The strategy of I is to play first the elements a1,...,an. Then he plays an element from M 0 which is not equivalent to any element played so far. Player II is at a loss. She has to play an element of M equivalent to one of a1,...,an. She loses. Definition 7.4 (Preliminary) Suppose n N. The game EFD!+n( , 0) 2 M M is played as follows. First the game EF ( , 0) is played for n moves. Then ! M M player I declares a natural number m and the game EF ( , 0) is continued ! M M for m more moves. If II has not lost yet, she has won EFD ( , 0). !+n M M Otherwise player I has won. Example 7.5 Suppose and 0 are graphs so that in every vertex has a G G G finite degree while in 0 some vertex has infinite degree. Then player I has G a winning strategy in EFD ( , 0). Suppose a G0 has infinite degree. !+1 G G 2 Player I plays first the element a. Let b G be the response of player II.We 2 know that every element of has finite degree. Let the degree of b be n. Player G I declares that we play n+1 more moves. Accordingly, he plays n+1 different neighbors of a. Player II cannot play n +1different neighbors of b since b has degree n. She loses. Example 7.6 Suppose is a connected graph and 0 a disconnected graph. G G Then player I has a winning strategy in EFD ( , 0). Suppose a and b are !+2 G G elements of G0 that are not connected by a path. Player I plays first elements a and b. Suppose the responses of player II are c and d. Since is connected, G Incomplete version for students of easllc2012 only. 7.2 Preliminary Examples 141 there is a connected path c = c0,c1,...,cn,cn+1 = d connecting c and d in . G Now player I declares that he needs n more moves. He plays the elements c1,...,cn one by one. Player II has to play a connected path a1,...,an in 0. Now d is a neighbor of c in but b is not a neighbor of a in 0 (see G n G n G Figure 7.1). Example 7.7 An abelian group is a structure =(G, +) with + : G G G ⇥ G G satisfying the conditions ! (1) x + (y + z)=(x + y)+ z for x, y, z. G G G G (2) there is an element 0 such that x + 0 =0 + x = x for all x. G G G G G (3) for all x there is x such that x + ( x)=0 . G G (4) for all x and y : x + y = y + x. G G Examples of abelian groups are Figure 7.1 Incomplete version for students of easllc2012 only. 142 Infinitary Logic (Z, +) integers with addition. (Z(n), +) integers modulo n with modular adddition: x +Z(n) y =(x +Z y) mod n. (Q, +) rationals with addition. (R, +) reals with addition. (R+, ) positive reals with multiplication. · Example 7.8 Consider the abelian groups Z =(Z, +) and Z2 =(Z Z, +) ⇥ with (m, n)+(p, q)=(m + p, n + q). 2 It is trivial that II has a winning strategy in EFD1(Z, Z ). But I has a winning 2 strategy already in EFD2(Z, Z ): First he plays x0 =(1, 0) and ↵0 =1. Suppose II responds with y0 Z. Then I plays x1 =(0, 1) and ↵1 =0. 2 Player II responds with y1 Z. Now 2 y1 y0 y0 = y1 i=1 i=1 X X but y1 y0 x =(y , 0) =(0,y )= x 0 1 6 0 1 i=1 i=1 X X unless y1 = y0 =0, in which case II has lost anyway. Example 7.9 Consider the structures (Z, +, 0) and (Z, +, 1). Player II can- not guarantee victory even in a zero-move game, as 0+0=0, but 1+1=1. 6 If instead we have the structures (Z, +, 0) and (Z, , 1), then II wins the zero- · move game, but if I has even just one move, he can play x0 =0in (Z, , 1) · and he wins. Namely, if II plays y0 Z with y0 =0, we have x0 x0 = x0 2 6 · but y + y = y . 0 0 6 0 An element a of an abelian group is a torsion element if there is an n N G 2 such that a + ...+ a =0. In Z(n) every element is a torsion element because n if a Proposition 7.10 If T is a first-order theory in the vocabulary + and { } Z(n) = T for arbitrarily large n N, then T has a model which is not a | 2 torsion group. Proof Let T 0 consist of axioms of abelian groups, T and the axioms c + ...+ c =0 6 n for all n N,n > 0. Any finite| subtheory{z } of T 0 is satisfied by Z(n) for large 2 enough n, if we interpret c as 1. By the Compactness Theorem T 0 has a model . Let cG = a. Now in we have a + ...+ a =0for all n N. Thus a is not G G 6 2 n a torsion element of . G | {z } Lemma 7.11 If is an abelian torsion group and 0 is a non-torsion abelian G G group, then I has a winning strategy in EFD ( , 0). 1 G G Proof We let I play x as a non-torsion element of 0. Suppose II plays 0 G y0 . Now there is n N such that 2 G 2 y0 + ...+ y0 =0 n but | {z } x + ...+ x =0 0 0 6 n so I wins. | {z } We can construe abelian groups also as relational structures. Thus instead of a binary function +:G G G we have a ternary relation R G G G. ⇥ ! + ✓ ⇥ ⇥ Then the axioms of abelian groups are (1) x y zR+xyz. 8 8 9 (2) x y z u((R+xyz R+xyu) z = u). 8 8 8 8 ^ ! (3) x y z u v w((R+xyu R+uzv R+yzw) R+xwv). 8 8 8 8 8 8 ^ ^ ! (4) x y(R+xyy R+yxy z u(R+zux R+uzx)). 9 8 ^ ^8 9 ^ In Ehrenfeucht–Fra¨ısse´ Games abelian groups behave quite differently de- pending on whether they are construed as relational structures or as algebraic structures. Lemma 7.12 If =(G, R+) is an abelian torsion group and 0 =(G0, R+) G G is a non-torsion abelian group, then I has a winning strategy in the game EFD ( , 0). !+1 G G Incomplete version for students of easllc2012 only. 144 Infinitary Logic Figure 7.2 Proof Let I play first x 0 which is not a torsion element. The response 0 2 G y of II is a torsion element, so if we use algebraic notation, we have o 2 G z1,...,zn such that y0 + y0 = z1. z1 + y0 = z2. . . zn + y0 =0. Now I declares there are n+2 moves left, and plays xi = zi for i =1,...,n. Let the responses of II be y1,...,yn. Next I plays xn+1 =0 , and II plays G y 0. Since x 0 is not a torsion element, II cannot have played n+1 2 G 0 2 G yn+1 =0 or else she loses. So there is xn+2 in 0 with xn+2 +yn+1 = xn+2. G0 G 6 Now finally I plays this xn+2, and II plays yn+2. As yn+2 + xn+1 = yn+2, II has now lost. 7.3 The Dynamic Ehrenfeucht–Fra¨ısse´ Game From EFD ( , 0) we could go on to define a game EFD ( , 0) !+n M M !+! M M in which player I starts by choosing a natural number n and declaring that we are going to play the game EFD ( , 0). But what is the general form of !+n M M such games? We can have a situation where player I wants to decide that after n0 moves he decides how many moves are left. At that point he decides that after n1 moves he will decide how many moves now are left. At that point he decides that after n2 moves he . . . until finally he decides that the game lasts nk more moves. A natural way of making this decision process of player I exact is to say that player I moves down an ordinal. For example, if he moves down the ordinal ! + ! +1, he can move as in Figure 7.2. So first he wants n0 moves and after they have been played he decides on n . If he moves down on the ordinal ! ! +1, he first chooses k and wants 1 · Incomplete version for students of easllc2012 only. 7.3 The Dynamic Ehrenfeucht–Fra¨ısse´ Game 145 Figure 7.3 n0 moves and after they have been played he can still make k changes of mind about the length of the rest of the game (see Figure 7.3). Definition 7.13 Let L be a relational vocabulary and , 0 L-structures M M such that 0 = . Let ↵ be an ordinal. The Dynamic Ehrenfeucht– M \ M ; Fra¨ısse´ Game EFD ( , 0) is the game G (M M 0 ↵,W ( , 0)), ↵ M M ! [ [ !,↵ M M where W ( , 0) is the set of !,↵ M M p =(x0, ↵0,y0,...,xn 1, ↵n 1,yn 1) such that (D1) For all i xi if xi M xi if xi M 0 vi = 2 and vi0 = 2 y if y M y if y M 0 ⇢ i i 2 ⇢ i i 2 then fp = (v0,v0), , (vn 1,vn0 1) { ··· } is a partial isomorphism 0. M ! M Note that EFD ( , 0) is not a game of length ↵. Every play in the game ↵ M M EFD ( , 0) is finite, it is just how the length of the game is determined ↵ M M during the game where the ordinal ↵ is used. Compared to EF ( , 0), the ! M M only new feature in EFD ( , 0) is condition (D2). Thus EFD ( , 0) ↵ M M ↵ M M is more difficult for I to play than EF ( , 0), but – if ↵ ! – easier than ! M M any EF ( , 0). n M M Lemma 7.14 (1) If II has a winning strategy in EFD↵( , 0) and ↵, M M then II has a winning strategy in EFD ( , 0). M M (2) If I has a winning strategy in EFD ( , 0) and ↵ , then I has a ↵ M M winning strategy in EFD ( , 0). M M Incomplete version for students of easllc2012 only. 146 Infinitary Logic Proof (1) Any move of I in EFD ( , 0) is as it is a legal move of I in M M EFD ( , 0). Thus if II can beat I in EFD she can beat him in EFD . ↵ M M ↵ (2) If I knows how to beat II in EFD↵, he can use the very same moves to beat II in EFD . Lemma 7.15 If ↵ is a limit ordinal =0and II has a winning strategy in the 6 game EFD ( , 0) for each < ↵, then II has a winning strategy in the M M game EFD ( , 0). ↵ M M Proof In his opening move I plays ↵0 < ↵. Now II can pretend we are actually playing the game EFD ( , 0). And she has a winning strategy ↵0+1 M M for that game! Back-and-forth sequences are a way of representing a winning strategy of player II in the game EFD↵. Definition 7.16 A back-and-forth sequence (P : ↵) is defined by the conditions = P ... P Part( , ) (7.1) ;6 ↵ ✓ ✓ 0 ✓ A B f P a A b B g P (f (a, b) g) for < ↵ (7.2) 8 2 +18 2 9 2 9 2 [ { } ✓ f P b B a A g P (f (a, b) g) for < ↵. (7.3) 8 2 +18 2 9 2 9 2 [ { } ✓ We write ↵ A 'p B if there is a back-and-forth sequence of length ↵ for and . A B The following proposition shows that back-and-forth sequences indeed cap- ture the winning strategies of player II in EFD ( , ): ↵ A B Proposition 7.17 Suppose L is a vocabulary and and are two L-structures. A B The following are equivalent: 1. =↵ . A ⇠p B 2. II has a winning strategy in EFD ( , ). ↵ A B Proof Let us assume A B = . Let (P : i ↵) be a back-and-forth \ ; i sequence for and . We define a winning strategy ⌧ =(⌧i : i N) for A B 2 II. Suppose we have defined ⌧i for i and if xj B, then xj rng(fj). We let ⌧j(x0,...,xj)=fj(xj) if xj A, 2 2 1 2 and ⌧j(x0,...,xj)=fj (xj) otherwise. This ends the construction of ⌧j. This is a winning strategy because every fp extends to a partial isomorphism . M ! N For the converse, suppose ⌧ =(⌧n : n N) is a winning strategy of II. 2 Let Q consist of all plays of EFD ( , ) in which player II has used ⌧. Let ↵ A B P consist of all possible fp where p =(x0, ↵0,y0,...,xi 1, ↵i 1,yi 1) is a position in the game EFD↵( , ) with an extension in Q and ↵i 1 . It is A B clear that (P : ↵) has the properties (7.1) and (7.2). We have already learnt in Lemma 7.14 that the bigger the ordinal ↵ in EFD ( , 0) is, the harder it is for player II to win and eventually, in a ↵ M M typical case, her luck turns and player I starts to win. From that point on it is easier for I to win the bigger ↵ is. Lemma 7.15, combined with the fact that the game is determined, tells us that there is a first ordinal where player I starts to win. So all the excitement concentrates around just one ordinal up to which player II has a winning strategy and starting from which player I has a win- ning strategy. It is clear that this ordinal tells us something important about the two models. This motivates the following: Definition 7.18 An ordinal ↵ such that player II has a winning strategy in EFD ( , 0) and player I has a winning strategy in EFD ( , 0) is ↵ M M ↵+1 M M called the Scott watershed of and 0. M M By Lemma 7.14 the Scott watershed is uniquely determined, if it exists. In two extreme cases the Scott watershed does not exist. First, maybe I has a winning strategy even in EF ( , 0). Here Part( , 0)= . Secondly, 0 M M M M ; player II may have a winning strategy even in EF ( , 0), so I has no ! M M chance in any EFD ( , 0), and there is no Scott watershed. In any other ↵ M M case the Scott watershed exists. The bigger it is, the closer and 0 are M M to being isomorphic. Respectively, the smaller it is, the farther and 0 M M are from being isomorphic. If the watershed is so small that it is finite, the structures and 0 are not even elementary equivalent. M M General problem: Given and 0, find the Scott watershed! M M How far afield do we have to go to find the Scott watershed? It is very natural to try first some small ordinals. But if we try big ordinals, it would be nice to know how high we have to go. There is a simple answer given by the next proposition: If the models have infinite cardinality , and the Scott watershed exists, then it is < +. Thus for countable models we only need to check Incomplete version for students of easllc2012 only. 148 Infinitary Logic Figure 7.4 countable ordinals. For finite models this is not very interesting: if the models have at most n elements, and there is a watershed, then it is at most n. Proposition 7.19 If II has a winning strategy in EFD↵( , 0) for all ↵ < + M M ( M + M 0 ) then II has a winning strategy in EF ( , 0). | | | | ! M M Proof Let = M + M 0 . The idea of II is to make sure that | | | | (?) If the game EF ( , 0) has reached a position ! M M p =(x0,y0,...,xn 1,yn 1) with fp = (v0,v0),...,(vn 1,vn0 1) { } then II has a winning strategy in EFD↵+1(( ,v0,...,vn 1), ( 0,v0,...,vn0 1)) (7.4) M M for all ↵ < +. In the beginning n =0and condition (?) holds. Let us suppose II has been able to maintain (?) and then I plays x in EF ( , 0). Let us look at the n ! M M possibilities of II: She has to play some y and there are possibilities. Let n be the set of them. Assume none of them works. Then for each legal move + yn there is ↵yn < such that Incomplete version for students of easllc2012 only. 7.3 The Dynamic Ehrenfeucht–Fra¨ısse´ Game 149 (1) II does not have a winning strategy in EFD (( ,v ,...,v ), ( 0,v0 ,...,v0 )) ↵yn M 0 n M 0 n where xn if xn M yn if yn M 0 vn = 2 and vn0 = 2 y if x M 0 x if y M . ⇢ n n 2 ⇢ n n 2 + Let ↵ =supy ↵yn . As , we have ↵ < . By the induction n2 | | hypothesis, II has a winning strategy in the game (7.4). So, let us play this game. We let I play x and ↵. The winning strategy of II gives y . Let n n 2 vn and vn0 be determined as above. Now (2) II has a winning strategy in EFD↵(( ,v0,...,vn), ( 0,v0 ,...,v0 )). M M 0 n We have a contradiction between (1), (2), ↵yn < ↵ and Lemma 7.14. The above theorem is particularly important for countable models since countable partially isomorphic structures are isomorphic. Thus the countable ordinals provide a complete hierarchy of thresholds all the way from not be- ing even elementary equivalent to being actually isomorphic. For uncountable models the hierarchy of thresholds reaches only to partial isomorphism which may be far from actual isomorphism. We list here two structural properties of ↵, which are very easy to prove. 'p There are many others and we will meet them later. ↵ ↵ ↵ Lemma 7.20 (Transitivity) If 0 and 0 00, then M 'p M M 'p M M 'p 00. M Proof Exercise 7.14. ↵ ↵ Lemma 7.21 (Projection) If 0, then L 0 L. M 'p M M 'p M Proof Exercise 7.15. We shall now introduce one of the most important concepts in infinitary logic, namely that of a Scott height of a structure. It is an invariant which sheds light on numerous aspects of the model. Definition 7.22 The Scott height SH( ) of a model is the supremum of M M all ordinals ↵ +1, where ↵ is the Scott watershed of a pair ( ,a ,...,a ) ( ,b ,...,b ) M 1 n 6'p M 1 n and a ,...,a ,b ,...,b M . 1 n 1 n 2 Incomplete version for students of easllc2012 only. 150 Infinitary Logic Lemma 7.23 SH( ) is the least ↵ such that if a1,...,an,b1,...,bn M M 2 and ( ,a ,...,a ) ↵ ( ,b ,...,b ) M 1 n 'p M 1 n then ( ,a ,...,a ) ↵+1 ( ,b ,...,b ). M 1 n 'p M 1 n Proof Exercise 7.16. SH( )+! Theorem 7.24 If p M 0, then p 0. M ' M M ' M Proof Let SH( )=↵. The strategy of II in EF ( , 0) is to make sure M ! M M that if the position is (1) p =(x0,y0,...,xn 1,yn 1) then ↵ (2) ( ,v0,...,vn 1) p ( 0,v0,...,vn0 1). M ' M In the beginning of the game (2) holds by assumption. Let us then assume we are in the middle of the game EF ( , 0), say in position p, and (2) ! M M holds. Now player I moves x , say x = v M . We want to find a move n n n 2 y = v0 M 0 of II which would yield n n 2 ↵ (3) ( ,v ,...,v ) ( 0,v0 ,...,v0 ). M 0 n 'p M 0 n ↵+! Now we use the assumption 0. We play a sequence of rounds of M 'p M an auxiliary game G =EFD ( , 0) in which player II has a winning ↵+n+1 M M strategy ⌧. First player I moves the elements v0,...,vn0 1. Let the responses of player II according to ⌧ be u0,...,un 1. We get ↵+1 (4) ( 0,v0,...,vn0 1) p ( ,u0,...,un 1). M ' M By transitivity, ↵ ( ,v0,...,vn 1) p ( ,u0,...,un 1). M ' M See Figure 7.5. By Lemma 7.23, ↵+1 ( ,v0,...,vn 1) p ( ,u0,...,un 1). M ' M Now we apply the definition of ↵+1 and find a M such that 'p 2 ↵ (5) ( ,v0,...,vn 1,vn) p ( ,u0,...,un 1,a). M ' M Finally we play one more round of the auxiliary game G using (4) so that Incomplete version for students of easllc2012 only. 7.3 The Dynamic Ehrenfeucht–Fra¨ısse´ Game 151 Figure 7.5 player I moves a M and II moves according to ⌧ an element y = v0 M 0. 2 n n 2 Again ↵ ( 0,v0,...,vn0 1,vn0 ) p ( ,u0,...,un 1,a), M ' M which together with (5) gives (3). Note, that for countable models we obtain the interesting corollary: Corollary If is countable, then for any other countable 0 we have M M SH( )+! M 0 = 0. M 'p M () M ⇠ M The Scott spectrum ss(T ) of a first-order theory is the class of Scott heights of its models: ss(T )= SH( ): = T . { M M| } It is in general quite difficult to determine what the Scott spectrum of a given theory is. For some theories the Scott spectrum is bounded from above. An extreme case is the case of the empty vocabulary, where the Scott height of any model is zero. It follows from Example 7.29 below that the Scott spectrum of the theory of linear order is unbounded in the class of all ordinals. A gap in a Scott spectrum ss(T ) is an ordinal which is missing from ss(T ). Vaught’s Conjecture: If T is a countable first-order theory, then T has, up to isomorphism, either or exactly 2@0 countable models. @0 Incomplete version for students of easllc2012 only. 152 Infinitary Logic It can be proved that any first-order theory can have only or exactly 2@0 @0 countable models of a fixed Scott height Morley (1970). Thus, since there are Scott heights of countable models, any first-order theory can have or @1 @1 exactly 2@0 countable models, up to isomorphism, all in all. To prove Vaught’s Conjecture it would suffice to prove that for every first-order theory T there is an upper bound ↵ < !1 for the Scott heights of its countable models or else there are 2@0 countable models of some fixed Scott height. This leads to the following concept: a first-order theory is scattered if it has at most countable @0 models of any fixed Scott height. Vaught’s Conjecture now has the following equivalent form: If T is scattered, then the Scott heights of its countable models have a countable upper bound. We now prove that there are for arbitrarily large ↵ models with Scott height ↵. First we prove that for any ↵ there are non-isomorphic models and 0 ↵ M M such that 0. For this we need the following useful concept: M 'p M Definition 7.25 If =(M ,<) and 0 =(M 0,<0) are ordered sets, their M M product 0 is the ordered set (M M 0,<⇤) where M ⇥ M ⇥ (x, x0) <⇤ (y, y0) x0 <0 y0 or (x0 = y0 and x Theorem 7.26 Suppose satisfies the condition ↵ < = !↵ < ) and is any linear order with a first element. Then . M 'p ⇥ M Proof An !✓-interval of is any set of the form I✓ = ↵ : !✓ ⇠ ↵ < !✓ (⇠ + 1) . ⇠ { · · } An !✓-interval of is any set of the form I✓ a , where a M . ⇥ M ⇠ ⇥ { } 2 We shall define a back-and-forth sequence P ... P as follows: A ✓ ✓ 0 partial isomorphism f is put into P✓ if f is a finite subfunction of a partial isomorphism g from to such that ⇥ M Incomplete version for students of easllc2012 only. 7.3 The Dynamic Ehrenfeucht–Fra¨ısse´ Game 153 Figure 7.6 ✓ (1) dom(g) is a union of finitely many ! -intervals I0,...,In of . ✓ (2) rng(g) is a union of finitely many ! -intervals I0 ,...,I0 of . 0 n ⇥ M (3) g(0) = (0, min( )). M (4) g Ij : Ij ⇠= Ij0 . The empty function is in P . If ⌘ < ✓, then P P for every !✓-interval ✓ ✓ ⌘ is a union of !⌘-intervals. To prove the back-and-forth property, suppose f 2 P , where < and (⇠,a) . Suppose f is a finite subfunction of +1 2 ⇥ M g satisfying (1)–(4). If (⇠,a) happens to be in the range of g, it is clear how to proceed: we simply extend f inside g. Let us assume that (⇠,a) is not in the +1 range of g. Let I0,...,In be the ! -intervals in increasing order containing elements of the domain of f. Let the corresponding ! +1-intervals in ⇥ M be I0 ,...,In0 . Let m be the largest m such that (⇠,a) is above the interval Im0 . If m = n, we have Figure 7.6. +1 +1 Let k be an isomorphism between an ! -interval above In0 and an ! - interval of above In0 . Then g k satisfies (1)–(4) and the restriction ⇥ M 1 [ of g k to dom(f) k (⇠,a) is the extension of f in P we are looking [ [ { } Incomplete version for students of easllc2012 only. 154 Infinitary Logic Figure 7.7 for. If on the other hand m Before drawing conclusions from the above important theorem we need to introduce some operations on linear orders. The sum + 0 of two linear orders and 0 is defined as the linear M M M M order consisting of and 0 one after the other, first then 0. More M M M M technically: Definition 7.27 Suppose =(M ,<) and 0 =(M 0,<0) are linear or- M M ders. Their sum + 0 is the linear order (M 00,<00) where M M (1) M 00 = M 0 M 0 1 . ⇥ { } [ ⇥ { } (2) (x, i) <00 (y, j) i The inverse of a linear order =(M ,<) is the linear order ⇤ =(M ,>). M M Note that if is an infinite well-order, then ⇤ is necessarily non-well- M M ordered. Example 7.28 (Z,<) = !⇤ + ! = (Z,<)+1+(Z,<) ⇠ 6⇠ (Q,<) ⇠= (Q,<)+(Q,<) ⇠= (Q,<)+1+(Q,<) (R,<) = (R,<)+1+(R,<) = (R,<)+(R,<). ⇠ 6⇠ Incomplete version for students of easllc2012 only. 7.3 The Dynamic Ehrenfeucht–Fra¨ısse´ Game 155 ↵n Example 7.29 Let ↵0 = !, ↵n+1 = ! , and ✏0 =supn ✏ ✏ 0 ✏ (1 + !⇤). 0 'p 0 ⇥ ↵ More generally, if ↵ =sup ! , then ↵ ↵ (1 + !⇤). <↵ 'p ⇥ The above example shows that there is no ordinal ↵ such that ↵ (( well-order & 0) 0 well-order). 8M M M 'p M ! M This should be compared with the fact (( well-order & 0) 0 = 0). 8M M M 'p M ! M ⇠ M The above example also shows that Scott heights can be arbitrarily large and the Scott spectra of first-order theories can be unbounded in the class of all ordinals. We now prove a result of D. Kueker about the number of automorphisms of countable models. Lemma 7.30 Suppose p 0 where M < M 0 . Then there are a = a0 M ' M | | | | 6 in M and b M 0 such that ( ,a) ( ,a0) ( 0,b). 2 M 'p M 'p M Proof For any b M 0 there is a M such that ( ,a) ( 0,b). Since 2 2 M 'p M there are M 0 many different b but only M many different a, there has to be | | | | one a M such that ( ,a ) ( 0,b ) and ( ,a ) ( 0,b ) for 0 2 M 0 'p M 0 M 0 'p M 1 some b = b . Let a M such that 0 6 1 1 2 ( ,a ,a ) ( 0,b ,b ). M 0 1 'p M 0 1 Thus ( ,a ) ( 0,b ) ( ,a ). M 0 'p M 1 'p M 1 Clearly a = a . 0 6 1 Theorem 7.31 If p 0 where is countable and 0 is uncountable, M ' M M M then has 2@0 automorphisms. M Proof We construct an automorphism ⇡s of for each s : N 2 such that M ! if s = s0, then ⇡s = ⇡s . To this end let M = bn : n N . We define ⇡s 6 6 0 { 2 } as the union of finite partial mappings ⇡s n,n N. Let ⇡ = . Suppose ⇡s n 2 ; ; has been defined and we want to define ⇡s n+1. As an induction hypothesis we assume that if ⇡ = (x ,y ):i then ( ,x0,...,xm 1) p ( ,y0,...,ym 1) M ' M p ( 0,z0,...,zm 1). ' M By Lemma 7.30 there are a = a0 M and b M 0 such that 6 2 2 ( ,x0,...,xm 1,a) p ( ,x0,...,xm 1,a0) M ' M p ( 0,z0,...,zm 1,b). ' M Let c = c0 M such that 6 2 ( ,x0,...,xm 1,a) p ( ,y0,...,ym 1,c) M ' M and ( ,x0,...,xm 1,a,a0) p ( ,y0,...,ym 1,c,c0). M ' M Then ( ,x0,...,xm 1,a) p ( ,x0,...,xm 1,a0) M ' M p ( ,y0,...,ym 1,c0). ' M Let xm = a and c if s(n)=0 y = m c if s(n)=1. ⇢ 0 Let c ,d M such that n n 2 ( ,x ,...,x ,b ,c ) ( ,y ,...,y ,d ,b ) M 0 m n n 'p M 0 m n n and ⇡ = (x ,y ):i m (b ,d ), (c ,b ) . s n+1 { i i } [ { n n n n } Two more applications of the back-and-forth property of guarantee that the 'p induction condition remains valid. Let 1 ⇡s = ⇡s n. n=0 [ If s = s0, say s(n) = s0(n), then ⇡ = ⇡ , so ⇡ = ⇡ . Clearly each 6 6 s n+1 6 s0 n+1 s 6 s0 ⇡ is an automorphism of . s M Corollary If is a countable model with only countably many automor- M phisms, then for all 0 M SH( )+! M 0 = 0. M 'p M () M ⇠ M Proof If 0, then 0 must be countable by the previous theorem. M 'p M M Then = 0 by Proposition 5.16. M ⇠ M Incomplete version for students of easllc2012 only. 7.4 Syntax and Semantics of Infinitary Logic 157 Example 7.32 The following structures have only countably many automor- phisms: (N,<, , 0, 1), (↵,<), (Z,<), (Z, +), (Q, +). · 7.4 Syntax and Semantics of Infinitary Logic The syntax and semantics of the infinitary logic L ! that we now introduce 1 are very much like the syntax and semantics of first-order logic. The logical symbols are , , , , , , (, ),x ,x ,.... Terms and atomic formulas are ⇡ ¬ 8 9 0 1 defined as usual. Formulas of L ! are of the form V W 1 tt0 ⇡ Rt1 ...tn ' ¬ i I 'i, i I 'i 2 2 Vxn', xWn' 8 9 where t, t0,t ,...,t are L-terms, R L with # (R)=n, and ' and all ' , 1 n 2 l i i I, where I is an arbitrary set, are formulas of L !, and the formulas 'i 2 1 have altogether only finitely many free variables.1 We regard ' , ' , ^ _ (' ) and (' ) as abbreviations. ! $ In first-order logic we can think of formulas as finite strings of symbols. In infinitary logic it is customary to consider formulas as sets. Then we have the following more exact albeit more cumbersome definition: Definition 7.33 Suppose L is a vocabulary. The class of L-formulas of L ! 1 is defined as follows: (1) If t and t0 are L-terms, then (0,t,t0) is an L-formula denoted by tt0. ⇡ (2) If t1,...,tn are L-terms, then (1,R,t1,...,tn) is an L-formula denoted by Rt1 ...tn. (3) If ' is an L-formula, so is (2, '), and we denote it by '. ¬ (4) If is a set of L-formulas with a fixed finite set of free variables, then (3, ) is an L-formula and we denote it by ' '. 2 (5) If is a set of L-formulas with a fixed finite set of free variables, then V (4, ) is an L-formula and we denote it by ' '. 2 (6) If ' is an L-formula and n N, then (5, ',n) is an L-formula and we 2 W denote it by x '. 8 n 1 This restriction makes it possible to quantify all free variables in a formula. Incomplete version for students of easllc2012 only. 158 Infinitary Logic (7) If ' is an L-formula and n N, then (6, ',n) is an L-formula and we 2 denote it by x '. 9 n Every formula of L ! is now a finite sequence of sets and the first element 1 of the sequence is one of 0, 1, 2, 3, 4, 5, 6 . With this definition it is easy to { } write exact inductive definitions for various concepts related to infinitary logic. A formula of L ! can be thought of as a tree, too. In this tree the formula 1 itself is the root and the set ISub(') of immediate successors of a node ' of the tree are: (1) ISub((0,t,t0)) = . ; (2) ISub((1,t1,...,tn)) = . ; (3) ISub((2, ')) = ' . { } (4) ISub((3, )) = . (5) ISub((4, )) = . (6) ISub((5, ',n)) = ' . { } (7) ISub((6, ',n)) = ' . { } The tree thus consists of the elements of 1 Sub(')= Subn(') n=0 [ where Sub (')= ' 0 { } Sub (')= ISub( ): Sub (') , n+1 [{ 2 n } and the order is < ✓ ✓ Sub ( ) for some n>0. Sub () 2 n The tree (Sub('), (1) QR( tt0)=0. ⇡ (2) QR(Rt1 ...tn)=0. (3) QR( ')=QR('). ¬ (4) QR( )=sup QR( ): . { 2 } (5) QR( )=sup QR( ): . V { 2 } (6) QR( xn')=QR(')+1. 8W (7) QR( xn')=QR(')+1. 9 Incomplete version for students of easllc2012 only. 7.4 Syntax and Semantics of Infinitary Logic 159 Example 7.34 QR( x ... x ( x x )) = QR( x x )+n+1 = n+1. 9 0 9 n ¬⇡ i j ¬⇡ i j 0 i ✓ (x )= x x QR x1 x1 The truth-definition of L ! is standard: 1 Definition 7.36 The concept of an assignment s : N M satisfying a ! formula ' in a model , = ' is defined as follows: M M| s = t t iff tM(s)=tM(s) M| s ⇡ 1 2 1 2 =s Rt1 ...tn iff (t1M(s),...,tnM(s)) Val (R) M| 2 M = ' iff = ' M| s ¬ M 6| s = ' iff = ' for all i I M| s i I i M| s i 2 = 2 ' iff = ' for some i I M| s Vi I i M| s i 2 = x2' iff = ' for all a M M| s 8W n M| s[a/xn] 2 = x ' iff = ' for some a M . M| s 9 n M| s[a/xn] 2 An alternative definition can be given in terms of games: Definition 7.37 Suppose L is a vocabulary, is an L-structure, '⇤ is an M sym L-formula, and s⇤ is an assignment for M. The game SG ( , '⇤) is defined M as follows. In the beginning player II holds ('⇤,s⇤). The rules of the game are as follows: Incomplete version for students of easllc2012 only. 160 Infinitary Logic III x0 y0 x1 y1 . . . . Figure 7.8 The game G!(W ). 1. If ' is atomic, and s satisfies it in , then the player who holds (',s) wins M the game, otherwise the other player wins. 2. If ' = , then the player who holds (',s), gives ( ,s) to the other player. ¬ 3. If ' = i I 'i, then the player who holds (',s) switches to hold some (' ,s) and2 the other player decides which. i V 4. If ' = i I 'i, then the player who holds (',s) switches to hold some (' ,s) and2 can himself or herself decide which. i W 5. If ' = x , then the player who holds (',s) switches to hold some 8 n ( ,s[a/x ]) and the other player chooses a M. n 2 6. If ' = x , then the player who holds (',s) switches to hold some 9 n ( ,s[a/x ]) and can himself or herself choose a M. n 2 As was pointed out in Section 6.5, = ' if and only if player II has a M| s winning strategy in the above game, starting with (',s). Why? If = ', M| s then the winning strategy of player II is to play so that if she holds ('0,s0), then = '0, and if player I holds ('0,s0), then = '0. M| s0 M 6| s0 The negation normal form NNF is defined for L ! exactly as for first-order 1 logic by requiring that negations occur in front of atomic formulas only. Definition 7.38 The Semantic Game SG( ,T,s) of the set T of L-sentences M of L ! in NNF is the game G!(W )(see Figure 7.8), where W consists of se- 1 quences (x0,y0,x1,y1,...) such that player II has followed the rules of Fig- ure 7.9, and moreover, if i is a basic formula and player II plays the pair ( ,s) then = . i M| s i Proposition 7.39 =s T iff II has a winning strategy in SG( ,T,s). M| M Proof Exercise 7.37. Example 7.40 Let n be the sentence x0 ... xn( 0 i xn yn Explanation Rule (', ) I enquires about ; ' T . 2 (', ) II confirms. Axiom rule ; ('i,s) I tests a played ( i I 'i,s) by choosing i I. 2 2 V ('i,s) II confirms. -rule ^ ( i I 'i,s) I enquires about 2 a played disjunction. W ('i,s) II makes a choice of i I. -rule 2 _ (',s[a/x]) I tests a played ( x',s) by choosing a 8M. 2 (',s[a/x]) II confirms. -rule 8 ( x',s) I enquires about a played 9 existential statement. (',s[a/x]) II makes a choice of a M. -rule 2 9 Figure 7.9 The game SG( ,T,s). M Example 7.41 Let = x x 0 ⇡ 0 1 = x (x Ex x ( x x )). n+1 9 2 0 2 ^9 0 ⇡ 0 2 ^ n Then for graphs we have G = x x ( ) iff is connected. G| 8 08 1 n G n _2N Note that the sentence x x ( ) uses just the variables x ,x , and 8 08 1 n N n 0 1 x . 2 2 W Example 7.42 Consider the vocabulary +, 0 of abelian groups. Let us in- { } troduce the notation x 0=0 i · x (n + 1) = x n + x . i · i · i Incomplete version for students of easllc2012 only. 162 Infinitary Logic Thus x n = x + + x . i · i ··· i n A group is torsion-free iff G | {z } = x ( 0x 0x n) G| 8 0 ⇡ 0 _ ¬⇡ 0 · n>0 ^ and is torsion if G = x ( 0x n). G| 8 0 ⇡ 0 · n 0 _ Example 7.43 Consider the vocabulary +, , 0, 1 of arithmetic. Let xi n { · } · be defined as above. Then for models of Peano’s axioms we have M = (N, +, , 0, 1) iff = x0 x01 n . M ⇠ · M| 8 0 ⇡ · 1 n 0 _ @ A Example 7.44 Suppose (M,d) is a metric space. For each positive rational r let D = (x, y) M M : d(x, y) ( ,f) = x x (D x x D fx fx ) . M | s 8 0 8 1 0 1 ! ✏ 0 1 ✏ ! ^ _ Example 7.45 Consider the formulas ✓↵ of Example 7.35. Then = ✓ iff ( ,s(0))M = ↵ M| s ↵ ⇠ where ( ,x)M =(y M : y f = (x0, ↵0),...,(xn 1, ↵n 1) { } such that for all i ,xi)M ⇠= ↵i. Note that isomorphisms between well-ordered sets are unique. To prove the back-and-forth property for P , suppose first f P and a ( 2 2 ,s(0))M. We play SG(( ,s(0))M, ✓ ,s) such that player I enquires about ↵ ( <↵ ✓ ,s[0/a]). The winning strategy of II yields < ↵ such that she plays (✓ ,s[0/a]). By the induction hypothesis ( ,a)M = . So f (a, ) W ⇠ [{ } 2 P . The other half of back-and-forth is proved similarly. Example 7.46 Let ✓↵ be as above. Then = x ✓ x ✓ iff = ↵. M| 08 0 1 ^ 0 9 0 1 M ⇠ _<↵ ^<↵ @ A @ A The proof is just as above (see Exercise 7.52). We write ! 0 if and 0 satisfy the same L !-sentences and M ⌘1 M M M 1 ↵ 0 if they satisfy the same L !-sentences of quantifier rank ↵. M ⌘ M 1 We now extend an important leg of the Strategic Balance of Logic, namely the equivalence of the Semantic Game and the Ehrenfeucht–Fra¨ısse´ Game, from first-order logic to infinitary logic: Theorem 7.47 The following are equivalent: (i) . A ⌘↵ B (ii) ↵ . A 'p B Proof (ii) (i) Suppose (P : ↵) is a back-and-forth sequence for ! A and . We use induction on ↵ to prove: B Claim: If f P and a1,...,ak dom(f), then 2 2 ( ,a ,...,a ) ( ,fa ,...,fa ). A 1 k ⌘ B 1 k We use induction on ' of quantifier rank to prove the claim ( ,a ,...,a ) = ' ( ,fa ,...,fa ) = '. A 1 k | ) B 1 k | The only non-trivial case is that ' = x (x ) and = QR( ) < . By 9 n n assumption, f P . Since ( ,a ,...,a ) = ', there is a A such that 2 +1 A 1 k | 2 ( ,a ,...,a ,a) = (c), where c is a new constant symbol, a name for a. A 1 k | Since f P , there is b B such that f (a, b) P . By the induction 2 +1 2 [ { } 2 hypothesis ( ,fa ,...,fa ,b) = (c). Thus ( ,fa ,...,fa ) = '. B 1 k | B 1 k | (i) (ii) Let P consist of such finite f Part( , ) that if dom(f)= ! 2 A B a0,...,an 1 , then { } ( ,a0,...,an 1) ( ,fa0,...,fan 1). A ⌘ B Incomplete version for students of easllc2012 only. 164 Infinitary Logic By assumption (i), P , so P = . Certainly < implies P P .To ;2 ↵ ↵ 6 ; ✓ prove the back-and-forth criterion, suppose f P ,a A and there is no 2 +1 2 b B with 2 ( ,a0,...,an 1,a) ( ,fa0,...,fan 1,b). (7.5) A ⌘ B Then for each b B there is some ' of quantifier rank such that 2 b ( ,a0,...,an 1,a) = 'b(c) A | and ( ,fa0,...,fan 1,b) = 'b(c) B | ¬ where c is a name for a in and b in . Thus A B ( ,a0,...,an 1) = x0 'b(x0). A | 9 b B ^2 Since QR( x0 b B 'b(x0)) +1and f P +1 we may conclude 9 2 2 V ( ,fa0,...,fan 1) = x0 'b(x0). B | 9 b B ^2 Let z B with ( ,fa0,...,fan 1,z) = 'b(c). We get the contradic- 2 B | b B tion 2 V ( ,fa0,...,fan 1,z) = 'z(c) 'z(c). B | ¬ ^ Thus a, b B with (7.5) must exist. The other half of the back-and-forth 2 criterion is similar. By combining the above theorem with our previous results about the relation ↵ p , we obtain many interesting facts about L !: ' 1 Proposition 7.48 The following are equivalent for all and : A B 1. ! . A ⌘1 B 2. i.e. there is a back-and-forth set for and . A 'p B A B 3. II has a winning strategy in EF ( , ). ! A B Example 7.49 (1) There is no L !-sentence in the empty vocabulary such 1 that for all : M = iff , M| |M| @0 because all infinite models in this vocabulary are partially isomorphic. (2) There is no L !-sentence in the vocabulary of equivalence rela- 1 {⇠} tions such that any equivalence relation satisfies = iff has only countably many equivalence classes. M| M Incomplete version for students of easllc2012 only. 7.4 Syntax and Semantics of Infinitary Logic 165 (3) There is no L !-sentence of the vocabulary E of graph theory such 1 { } that for all graphs : G = iff has an uncountable clique. G| G The following consequence of Karp’s Theorem (Theorem 7.26) is of funda- mental importance for understanding L !: 1 Corollary There is no L !-sentence of the vocabulary < such that for all 1 { } linear orders : M = iff is a well-order. M| M This should be contrasted with the fact that for all ↵ and all M = ✓ iff is a well-order of type ↵. M| ↵ M If we could take the disjunction of all ✓ , ↵ On, we could characterize ↵ 2 well-order, but On is a proper class, so the disjunction cannot be formed in L !. 1 Definition 7.50 Let L be a vocabulary, an L-structure, and a0,...,an 1 M 2 M. Then we define 0 ,a0,...,an 1 = '(x0,...,xn 1):'(x0,...,xn 1) M { ^is a basic L-formula and = '(a0,...,an 1) M| } ↵+1 = x ↵ x ↵ ,a0,...,an 1 n ,a0,...,an n ,a0,...,an M 8 M ^ 9 M a M ! a M ! n_2 n^2 ⌫ ↵ ,a0,...,an 1 = ,a0,...,an 1 , for limit ⌫ M M ↵<⌫ ↵ ^↵ = , . M M ; = ↵ (a ,...,a ) Lemma 7.51 1. ,a0,...,an 1 0 n 1 . M| M ↵ (x ,...,x ) = ↵ (x ,...,x ) n k +1 2. ,a0,...,ak 0 k ,a0,...,an 1 0 n 1 for . M | M 3. If ↵ < , then (x ,...,x ) = ↵ (x ,...,x ). ,a0,...,an 1 0 n 1 ,a0,...,an 1 0 n 1 M | M Proof Exercise 7.44. Proposition 7.52 The following are equivalent: = ↵ (b ,...,b ) (1) 0 ,a0,...,an 1 0 n 1 . M | M ↵ (2) ( ,a0,...,an 1) p ( 0,b0,...,bn 1). M ' M Incomplete version for students of easllc2012 only. 166 Infinitary Logic ↵ ↵ Proof Note that the quantifier rank of the formula ,a0,...,an 1 is . Since = ↵ (a ,...,a ) M (2) ,a0,...,an 1 0 n 1 by Lemma 7.51, the implication M| M ! (1) follows from Proposition 7.47. Next we prove (1) (2). Intuitively, the ! winning strategy of II in EFD↵ on ( ,a0,...,an 1) and ( 0,b0,...,bn 1) M M ↵ is written into the structure of ,a0,...,an 1 . More exactly, we can define a back-and-forth sequence (P : M ↵) by letting P consist of finite map- pings f = (a0,b0),...,(an 1,bn 1),...,(am,bm) { } such that 0 = (b ,...,b ,...,b ) . ,a0,...,an 1,...,am 0 n 1 m M | M } (P : By the definition of the formulas ,a0,...,an 1 , the sequence M ↵) is indeed a back-and-forth sequence. Note that (1) implies P = , as ↵ 6 ; (a0,b0),...,(an 1,bn 1) P↵. { } 2 Definition 7.53 The Scott sentence of a structure is the L !-sentence M 1 SH( ) SH( ) SH( )+1 = M x ... x ( M M ). , 0 n 1 ,a0,...,an 1 ,a0,...,an 1 M M ; ^ 8 8 M ! M a ,...,a M 0 n 12 n^ 2 N Proposition 7.54 The following are equivalent: (1) 0 = . M | M (2) 0 . M 'p M Proof (2) (1): Lemma 7.51 gives = . The implication follows ! M| M now from Proposition 7.48. (1) (2): Suppose 0 = . We prove 0 p ! M | M M ' by giving a winning strategy for player II in the game EFDSH( )( , 0). M M M M The strategy of II is to make sure that if the position is p =(x0,y0,...,xn 1,yn 1) then SH( ) (?) 0 = v0,...,vMn 1 (v0,...,vn0 1). M | In the beginning of the game (?) holds by assumption. Let us then assume we are in the middle of the game EF ( , 0), say in position p, and (?) holds. ! M M Now player I moves x , say x = v M. Now we use the assumption n n n 2 0 = . It gives M | M SH( )+1 0 = M (v0 ,...,v0 )), ,v0,...,vn 1 0 n 1 M | M Incomplete version for students of easllc2012 only. 7.4 Syntax and Semantics of Infinitary Logic 167 whence ↵ 0 = y ,v0,...,vn 1,a(v0,...,vn0 1,y). M | 9 M a M ^2 By choosing a = v we find a move y = v0 M 0 of II which yields n n n 2 SH( ) 0 = M (v0 ,...,v0 ). M | v0,...,vn 0 n Note that if is a well-ordered set then by the above result it is, up to M isomorphism, the only model of . M Corollary (Scott Isomorphism Theorem) Suppose is a countable model. M Then for all countable 0 M 0 = 0 = . M | M () M ⇠ M This is a remarkable result. It puts countable models on levels of a well- ordered hierarchy according to their Scott height. On each level there is an invariant, the Scott sentence of the model, that characterizes the model up to isomorphism. These invariants need not, of course, be simple in any way, but they have a uniform tree-structure, the differences occurring only at the leaves of the tree. The invariants provide a way to systematize and classify countable models according to the syntactic properties of the Scott sentence. For the next result we have to compute an upper bound for the number of non-equivalent infinitary formulas of a given quantifier rank. As in the fi- nite case (see Propositions 6.3 and 4.15), the upper bound is an exponential tower, only this time we deal with infinite cardinals rather than natural num- bers. These cardinal numbers look very big, but at this point the only relevant thing is that they exist. We want to be sure that there is not a proper class of non-equivalent formulas of a fixed quantifier rank. Note that if we do not limit the quantifier rank, there is a proper class of non-equivalent formulas, namely the Scott sentences of different ordinals. Recall that i0( )= ( )=2i↵( ) 8 i↵+1 < i⌫ ( )=sup↵<⌫ i↵( ). : Lemma 7.55 Suppose L is a vocabulary of size µ and ↵ = ⌫ + n where ⌫ is a limit ordinal. There are at most i⌫+2n+2(µ + 0) non-equivalent formulas @ of L ! of quantifier rank ↵. 1 Incomplete version for students of easllc2012 only. 168 Infinitary Logic Proof There are at most µ+ 0 atomic formulas and therefore at most i2(µ+ @ ) non-equivalent formulas of quantifier rank 0. Suppose then ↵ = ⌫ + n +1 @0 where ⌫ is a limit ordinal. Formulas of quantifier rank ↵ are of the form x ' or of the form x ', where QR(') < ↵, and what can be built from 8 n 9 n them by means of , i I and i I . Thus their number (up to logical equiv- ¬ 2 2 alence) is at most i2(i⌫+2n+2(µ + 0)) = i⌫+2(n+1)+2(µ + 0)). If ⌫ is a V W @ @ limit ordinal, the number of non-equivalent formulas of quantifier rank < ⌫ is sup↵<⌫ i↵(µ + 0)=i⌫ (µ + 0). Therefore, the number of non- @ @ equivalent formulas of quantifier rank ⌫ is at most i2(i⌫ (µ + 0)) = @ i⌫+2(µ + 0). @ Thus, for example, for any ↵ there is only a set of non-equivalent sentences ↵ , while there is a proper class of non-equivalent sentences . M M Corollary Suppose L is a vocabulary. Then for all ordinals ↵ the equivalence relation A ⌘↵ B divides the class Str(L) of all L-structures into a set of equivalence classes C↵,i I, such that if we choose any representatives C↵, then: i 2 Mi 2 i ↵ ↵ 1. For all L-structures : Ci = . M M 2 () M| Mi 2. If ' is an L-sentence of L ! of quantifier rank ↵, then there is a set 1 I I = ' ↵ 0 such that i I i . ✓ | $ 2 0 M Proof For any L-structure W let Th↵( ) be the set of L !-sentences of M M 1 quantifier rank ↵ (up to logical equivalence) which are true in . Thus M 0 Th ( )=Th ( 0). M ⌘↵ M () ↵ M ↵ M Let Th ( ),i I, be a complete list of all Th ( ). The claim follows. ↵ Mi 2 ↵ M For the second claim let I consist of such i I that = '. If = ' 0 2 Mi | M| Th ( )=Th( ) i I = ↵ . and ↵ ↵ i , then 0 and i Conversely, if ↵ M M 2 M| M = +i,i I0, then ↵ i and = ' follows. M| M 2 M ⌘ M M| Note again that if we tried to prove the above corollary for the finer relation , we would run into the difficulty that there is a proper class of equivalence 'p classes. Corollary Suppose L is an arbitrary vocabulary and K is a class of L- structures. Then the following are equivalent: (i) K is definable in L !. 1 (ii) K is closed under ↵ for some ↵. 'p Incomplete version for students of easllc2012 only. 7.4 Syntax and Semantics of Infinitary Logic 169 Figure 7.10 Model class K definable in L !. 1 Figure 7.11 Model class K not definable in L !. 1 These equivalent conditions are strictly stronger than (iii) K is closed under . 'p The above theorem gives a kind of normal form for sentences of L !: every 1 sentence is a disjunction of sentences ↵ , which in turn have a very canonical form. For finite ↵ and finite relationalM vocabulary the formulas ↵ are first- order. M Definition 7.56 Suppose is a regular cardinal. L! is the fragment of L ! 1 which obtains if in the definition of the syntax of L ! we modify condition 1 (4) and (5) by requiring that I < . | | First-order logic is in this notation L!!. The most important non-first-order case is L!1!, the extension of first-order logic obtained by allowing countable Incomplete version for students of easllc2012 only. 170 Infinitary Logic disjunctions and conjunctions. Note that in a countable vocabulary the Scott sentence of a countable model is in L!1!. Proposition 7.57 Suppose is a countable model in a countable vocabu- M lary and P M n. Then the following are equivalent: ✓ (i) P is closed under automorphisms of . M (ii) There is a formula '(x ,...,x ) in L such that for all a ,...,a 0 n !1! 0 n 2 M (a ,...,a ) P = '(a ,...,a ). 0 n 2 () M| 0 n Proof (ii) (i) is trivial because automorphisms preserve truth. To prove ! (i) (ii) consider ! '(x0,...,xn)= ( ,a ,...,a )(x0,...,xn):(a0,...,an) P , { M 0 n 2 } _ where ( ,a ,...,a )(x0,...,xn) denotes the formula obtained from the sen- M 0 n tence ( ,a ,...,a ) by replacing the name of ai by the variable symbol xi. M 0 n Since M (and hence P ) is countable, '(x ,...,x ) L . If we now 0 n 2 !1! have (a0,...,an) P , then = ( ,a ,...,a )(a0,...,an). Thus = 2 M| M 0 n M| '(a0,...,an). Conversely, suppose = ( ,a ,...,a )(b0,...,bn) and (a0,...,an) P. M| M 0 n 2 Then ( ,b ,...,b ) ( ,a ,...,a ). Thus there is an automorphism ⇡ M 0 n 'p M 0 n of such that ⇡(b )=a for i n. Since P is closed under automorphisms, M i i (b ,...,b ) P . 0 n 2 If we want to show that a relation on a countable structure is not definable in L!1!, a natural approach is to show that the relation is not preserved by auto- morphisms of the structure. The above theorem demonstrates that this natural approach is as good as any other. Example 7.58 Let =(Z,<). The only subsets of Z that are closed un- M der automorphisms of are and Z. Thus they are the only subsets of M ; M definable in L!1!. Corollary If is a rigid countable model in a countable vocabulary, then M every relation on M is L -definable on . !1! M 7.5 Historical Remarks and References Infinitary languages were introduced in propositional calculus in Scott and Tarski (1958) and in predicate logic in Tarski (1958). An early book on infinitary Incomplete version for students of easllc2012 only. Exercises 171 languages is Karp (1964). More recent books are Keisler (1971), Dickmann (1975), and Barwise (1975). A good source is the survey article Makkai (1977). The back-and-forth sets, and thereby in effect the Ehrenfeucht–Fra¨ısse´ Game was introduced to infinitary logic in Karp (1965), where Proposition 7.48 and Proposition 7.26 appear. A good survey article on back-and-forth sets is Kueker (1975). Propositions 7.54 and 7.57 and their corollaries are from Scott (1965). Definition 7.16 is from Karp (1965). Theorem 7.31 is from Kueker (1968). Exercises 7.1 Show that if II has a winning strategy in EFD ( , 0) and M is finite, ! M M then = 0. M ⇠ M 7.2 Let =(Z,<) and 0 =(Z + Z,<) (i.e. two copies of one M M M after the other). For which n does I have a winning strategy in the game EFD ( , 0), and for which does II? !+n M M 7.3 Suppose and 0 are graphs such that player II has a winning strategy G G in EFD ( , 0). Show that if has a cycle path, then so does 0. ! G G G G 7.4 Suppose and 0 are graphs and player II has a winning strategy in G G EFD ( , 0). Show that if has infinitely many edges, also 0 has. ! G G G G 7.5 Suppose where =(N, +, , 0, 1) but = . Show that I M ⌘ N N · M 6⇠ N has a winning strategy in EFD ( , ). 1 M N 7.6 Player I wants to play EFD ( , 0) but cannot decide which ↵ to ↵ M M choose. He wants to play as follows: 1. First I wants to play 10 moves. 2. Then, depending on how II has played, I wants to play 2n moves for some n that he chooses. 3. Then I wants to play five additional moves. 4. Then, depending on how II has played, I wants to play 5n +1moves for some n that he chooses. 5. Finally I wants to play 15 additional moves, whereupon the game should end. Can you help him choose ↵? 7.7 Show that player I (II) has a winning strategy in EF ( , 0) iff he n M M (she) has a winning strategy in EFD ( , 0). n M M 7.8 Let the game EFD⇤ ( , ) be like the game EFD ( , ) except that I ↵ A B ↵ A B has to play x2n A and x2n+1 B for all n N. Show that if ⌫ is a 2 2 2 limit ordinal, then player II has a winning strategy in EFD⇤ ( , ) ⌫+2n A B if and only if she has a winning strategy in EF ( , ). ⌫+n A B Incomplete version for students of easllc2012 only. 172 Infinitary Logic 7.9 Suppose B = bn : n N . Let the game EFD⇤⇤( , ) be like the { 2 } ↵ A B game EFD ( , ) except that I has to play x A and x = b ↵ A B 2n 2 2n+1 n for all n N. Show that if ⌫ is a limit ordinal, then player II has a 2 winning strategy in EFD⇤⇤ ( , ) if and only if she has a winning ⌫+2n A B strategy in EFD ( , ). ⌫+n A B 7.10 Find the Scott watershed for (a) (N, +, , 0, 1) and (Q, +, , 0, 1). · · (b) (Z + Z,<) and (Z + Z + Z,<). 7.11 What is the Scott watershed of (Z2, +) and (Z3, +)? 7.12 What is the Scott watershed of (Q, +) and (R, +)? 7.13 Prove Z(15) = Z(3) Z(5). ⇠ ⇥ 7.14 Prove Lemma 7.20. 7.15 Prove Lemma 7.21. 7.16 Prove Lemma 7.23. SH( ) 7.17 Show that if ( ,v0,...,vn 1) p M ( ,v0,...,vn0 1), then M ' M ( ,v0,...,vn 1) p ( ,v0,...,vn0 1). M ' M 7.18 A model is -homogeneous if the following holds for all v ,...,v M @0 0 n and v0,...,vn0 1 in M: If ( ,v0,...,vn 1) ( ,v0,...,vn0 1) M ⌘ M then there is v0 in M such that ( ,v ,...,v ) ( ,v0 ,...,v0 ). n M 0 n ⌘ M 0 n Show that the Scott height of an -homogeneous model is !. Show @0 that if is a countable -homogeneous model and M @0 ( ,v0,...,vn 1) ( ,v0,...,vn0 1), M ⌘ M then there is an automorphism of which maps each v to v0. M i i 7.19 Show that there are, up to isomorphism, exactly three countable - @0 homogeneous models such that ! (!,<). M M 'p 7.20 Show that if and 0 are well-orderings, then so is 0. M M M ⇥ M 7.21 Prove ↵ = ↵ starting from the inductive definition of multiplica- · ⇠ ⇥ tion in Exercise 2.22. 7.22 Prove ↵ < = !↵ < for uncountable cardinals . Recall the ) inductive definition of exponentiation in Exercise 2.26. 7.23 Prove ( 0 00) = ( 0) 00 for linear orders , 0 M ⇥ M ⇥ M ⇠ M ⇥ M ⇥ M M M and 00. Show also that it is possible that 0 0 . M !1 M!⇥1 M M ⇥ M 7.24 Show that !1 p !1 2, and (Q,<) !1 p (R,<) (!1 2). ' 0 · ✏0 ⇥ 0 ' ⇥ · 7.25 Show that ✏0 (R ,<) p !1 (Q ,<). ⇥ ' ⇥ ↵ 7.26 Find for all ↵ a scattered and a non-scattered 0 such that M M M 'p 0. M 7.27 Prove the claims of Example 7.28. Incomplete version for students of easllc2012 only. Exercises 173 7.28 Suppose =(M,<) is a linear order and 0 is the set of initial M M segments of ordered by proper inclusion. Show that = 0. M M 6⇠ M 7.29 Show that there is a countable model such that has 2@0 automor- M M phisms but there is no uncountable 0 such that 0. (Hint: M M 'p M Let =(M,!,<,R) where < is the usual ordering of !, M = ! M [ (n, i):n N,i 0, 1 and R = (n, (n, i)) : n N,i 0, 1 .) { 2 2 { }} { 2 2 { }} 7.30 Write a sentence of L !, as simple as possible, which holds in a finite 1 graph iff (a) the number of vertices is even. (b) the number of edges is even. (c) the graph has a cycle path. 7.31 Write a sentence of L !, as simple as possible, which holds in a finite 1 graph iff the graph is 3-colorable. Use this to prove that there is a sen- tence of L ! which holds in a graph iff the graph is 3-colorable. (Hint: 1 use the Compactness Theorem of propositional logic to reduce the sec- ond part to the finite case.) 7.32 An ordered field (K, +, , 0, 1,<) is Archimedian if for all r > 0 and r · 1 2 in K there is a natural number n so that r + + r >r. Show that 1 ··· 1 2 n the Archimedian property can be expressed in L !. 1 n | {z } 7.33 Let L ! denote the fragment of L ! consisting of formulas in which 1 1 only variables x0,...,xn 1 occur. Show that if and 0 are graphs G G so that player II has a winning strategy in the n-Pebble Game, then n they satisfy the same sentences of L !. Hence, if and 0 satisfy the 1 G G n extension axiom En, then they satisfy the same sentences of L !. (See 1 Exercise 4.17 for the definition of the n-Pebble Game.) 7.34 Use Exercise 7.33 to conclude that if two graphs satisfy E for all n n 2 N, then the graphs are partially isomorphic. (See Exercise 4.16 for the definition of En.) Conclude also that if ' is a first-order sentence, then En = ' for some n N, or else En = ' for some n N. | 2 | ¬ 2 7.35 Show that there is no L !-sentence of the vocabulary of linear order 1 such that for all linear orders : = iff has cofinality ! (i.e. M M| M has a countable unbounded subset.) 7.36 Show that there is no L !-sentence of the vocabulary P, Q of two 1 { } unary predicates such that (M,PM,QM) = iff P M = QM . | | | | | 7.37 Prove Proposition 7.39. Incomplete version for students of easllc2012 only. 174 Infinitary Logic 7.38 Show that there is no L !-sentence of the vocabulary < such that 1 { } for all linear orders : M = iff M is countable. M| | | Show that such a exists if “linear order” is replaced by “well-order”. 7.39 Let (M,d) be a metric space and =(M,(D ) ) as in Exam- M r r>0 ple 7.44. Write a sentence ' of L ! such that 1 (1) ( ,p0,p1,...) = ' iff the sequence p0,p1,...converges in (M,d). M | (Expand the vocabulary to include names for the points pn.) (2) ( ,f0,f1,...,f) = ' iff the sequence f0,f1,...of functions f : M | M M converges uniformly to f. (Expand the vocabulary to ! include names for the functions fn and for the function f.) (3) ( ,A) = ' iff the set A is closed. (Expand the vocabulary to in- M | clude a name for A.) 7.40 Let (M,d) and be as above. Is there a sentence ' of L ! such that M 1 = ' iff (M,d) is compact? M| 7.41 Let V be a Q-vector space. Let V =(V,+V , 0V , (fr)r 0), where for M each non-negative rational r, f (v)=r v. r ·V Write a sentence ' of L ! such that 1 (1) V = ' iff dim(V )=n. M | (2) V = ' iff dim(V ) is infinite. M | (3) ( V ,f) = ' iff f : V V V is a linear mapping. (Expand the M | ⇥ ! vocabulary to include a name for f.) 7.42 Let V be an R-vector space with a norm V : V R. Let V = || · || ! N (V,+V , 0V ,DV , (fr)r 0) where DV = v V : v < 1 and for { 2 || || } non-negative rational r, fr is as above. Write a sentence ' of L ! such 1 that (1) ( V ,f) = ' iff f : V V is continuous. N | ! (2) ( V ,f) = ' iff f : V V is differentiable. N | ! (In both (1) and (2), expand the vocabulary to include a name for f.) 7.43 Let =(R, +, , 0, 1,<) and L a vocabulary which extends the vocab- M · ulary of by a name for a function f : M M. Write a sentence ' M ! of L ! such that ( ,f) = ' iff 1 M | (1) f [0, 1] has bounded variation, i.e. there exists an M such that n f(x ) f(x ) M for all 0=x (2) f is homogeneous, i.e. there is n N such that f(ax)=anf(x) for 2 all a R. 2 (3) f [0, 1] is Riemann integrable, i.e. for each ✏ > 0 there are 0= x 8.1 Introduction The model theory of L!1! is dominated by the Model Existence Theorem. It more or less takes the role of the Compactness Theorem which can be right- fully called the cornerstone of model theory of first-order logic. The Model Existence Theorem is used to prove the Craig Interpolation Theorem and the important undefinability of the concept of well-order. When we move to the stronger logics L+!, > !, the Model Existence Theorem in general fails. However, we use a union of chains argument to prove the undefinability of well-order. In the final section we introduce game quantifiers. Here we cross the line to logics in which well-order is definable. Game quantifiers permit an approximation process which leads to the Covering Theorem, a kind of Inter- polation Theorem. 8.2 Lowenheim–Skolem¨ Theorem for L ! 1 In Section 6.4 we saw that if a first-order sentence is true in a model it is true in “almost” every countable approximation of that model. We now extend this to L ! but of course with some modification because L ! has consistent sen- 1 1 tences without any countable models. We show that if a sentence ' of L ! is 1 true in a structure , a countable ”approximation” of ' is true in a countable M ”approximation” of , and even more, there are this kind of approximations M of ' and in a sense “everywhere”. To make this statement precise we em- M ploy the Cub Game introduced in Definition 6.10. We say ...X... for almost all X (A) 2 P! Incomplete version for students of easllc2012 only. 8.2 Lowenheim–Skolem¨ Theorem for L ! 177 1 if player II has a winning strategy in G ( (A)). cub P! Recall the following facts: 1. If X (A), then X X for almost all X (A). 0 2 P! 0 ✓ 2 P! 2. If X for almost all X (A) and 0, then X 0 for almost 2 C 2 P! C ✓ C 2 C all X (A). 2 P! 3. If for all n N we have X n for almost all X !(A), then X 2 2 C 2 P 2 for almost all X (A). n N Cn 2 P! 4. If for2 all a A we have X for almost all X (A), then X T 2 2 Ca 2 P! 2 a A a for almost all X !(A). 4 2 C 2 P In other words, the set of subsets of (A) which contain almost all X P! 2 (A) is a countably complete filter. P! Now that approximations extend not only to models but also to formulas we assume that models and formulas have a common universe V , which is supposed to be a transitive1 set. As the following lemma demonstrates, the exact choice of this set V is not relevant: Lemma 8.1 Suppose = A V and !(A). Then the following are ;6 ✓ C ✓ P equivalent: 1. X for almost all X (A). 2 C 2 P! 2. X A for almost all X (V ). \ 2 C 2 P! Proof (1) implies (2): Let a A. Player II applies her winning strategy in 2 G ( ) in the game G ( X (V ):X A ) as follows: If I plays cub C cub { 2 P! \ 2 C} his element in A, player II interprets it as a move in G ( ), where she has a cub C winning strategy. If I plays xn outside A, player II plays yn = a. (2) implies (1): player II interprets all moves of I in A as his moves in V and then uses her winning strategy in G ( X (V ):X A ). cub { 2 P! \ 2 C} Definition 8.2 Suppose ' L ! and X is a countable set. The approxima- 2 1 tion 'X of ' is defined by induction as follows: X (1) ( tt0) = tt0. ⇡ X⇡ (2) (Rt1 ...tn) = Rt1 ...tn. (3) ( ')X = 'X . ¬ ¬ (4) ( )X = 'X : ' X . { 2 \ } (5) ( )X = 'X : ' X . V X V{ X 2 \ } (6) ( xn') = xn(' ). 8W W8 1 A set A is transitive if y x A implies y A for all x and y. 2 2 2 Incomplete version for students of easllc2012 only. 178 Model Theory of Infinitary Logic X X (7) ( xn') = xn(' ). 9 9 X Note that ' is always in L!1!, whatever countable set X is. Example 8.3 Suppose X '↵ : ↵ < !1 = '↵ , '↵ ,... . Then \ { } { 0 1 } X x ' (x ) = x 'X (x ) 8 0 ↵ 0 8 0 ↵n 0 ↵ Definition 8.6 Suppose L is a vocabulary and an L-structure. Suppose ' M is a first-order formula in NNF and s an assignment for the set M the domain of which includes the free variables of '. We define the set of countable D',s subsets of M as follows: If ' is basic, contains as an element any count- D',s able X V such that X M is the domain of a countable submodel of ✓ \ A M such that rng(s) A and: ✓ If ' is tt0, then tA(s)=t0A(t). • ⇡ If ' is tt0, then tA(s) = t0A(t). • ¬⇡ 6 If ' is Rt ...t , then (tA(s),...,t A(t)) RA. • 1 n 1 n 2 If ' is Rt ...t , then (tA(s),...,t A(t)) / RA. • ¬ 1 n 1 n 2 For non-basic ' we define ,s = ' ',s. •D 4 2 D V ,s = ' ',s •D 5 2 D Wx',s = a M ',s[a/x]. •D8 4 2 D x',s = a M ',s(a/x). •D9 5 2 D Incomplete version for students of easllc2012 only. 8.3 Model Theory of L!1! 179 If ' is a sentence, we denote by . If ' is not in NNF, we define D',s D' D',s and by first translating ' into a logically equivalent NNF formula. D' Intuitively, is the collection of countable sets X, which simultaneously D' give an L -approximation 'X of ' and a countable approximation X of !1! M such that X = 'X . M M | Proposition 8.7 Suppose is an L-structure and X ',s. Then [X X A 2 D \ A] =t ' . A | Proof This is trivial for basic '. For the induction step for suppose X 2 ,s. Suppose ' X . Then X ',s. By the induction hypothesis D X 2 \ 2X D V [XV A] =t ' . Thus [X] =t ( ) . The other cases are as in the proof \ A | A | of Proposition 6.21. V Proposition 8.8 Suppose L is a countable vocabulary and an L-structure M such that = '. Then player II has a winning strategy in G ( ). M| cub D' Proof We use induction on ' to prove that if = ', then II has a winning M| s strategy in G ( ). Most steps are as in the proof of Proposition 6.22. Let cub D',s us look at the induction step for . We assume = '. It suffices to M| s prove that II has a winning strategy in G ( ) for each ' . But this V cub D',s V 2 follows from the induction hypothesis. Theorem 8.9 (Lowenheim–Skolem¨ Theorem) Suppose L is a countable vo- cabulary, an arbitrary L-structure, and ' an L !-sentence of vocabulary M 1 L, and V a transitive set containing and ' such that M TC(')= . M \ ; Suppose = '. Let M| X = X !(V ):[X M] = ' . C { 2 P \ M | } Then player II has a winning strategy in the game G ( ). cub C Proof The claim follows from Propositions 8.7 and 8.8. X X Theorem 8.10 1. ! if and only if = for almost all X. 1 ⇠ M ⌘ NX X M N 2. ! if and only if = for almost all X. M 6⌘1 N M 6⇠ N 8.3 Model Theory of L!1! The Model Existence Game MEG(T,L) of first-order logic (Definition 6.35) can be easily modified to L!1!. Incomplete version for students of easllc2012 only. 180 Model Theory of Infinitary Logic xn yn Explanation ' I enquires about '. ' II confirms. tt I enquires about an equation. ⇡ tt II confirms. ⇡ '(t) I chooses played '(c) and ct with ' basic and enquires about substituting⇡ t for c in '. '(t) II confirms. 'i I tests a played i I 'i by choosing i I. 2 2 V 'i II confirms. i I 'i I enquires about a played disjunction. 2 W 'i II makes a choice of i I. 2 '(c) I tests a played x'(x) by choosing c C. 8 2 '(c) II confirms. x'(x) I enquires about a played existential statement. 9 '(c) II makes a choice of c C. 2 t I enquires about a constant L C-term t. [ ct II makes a choice of c C. ⇡ 2 Figure 8.1 The game MEG(T,L). Definition 8.11 The Model Existence Game MEG(',L) for a countable vo- cabulary L and a sentence ' of L!1! is the game G!(W ) where W consists of sequences (x0,y0,x1,y1,...) where player II has followed the rules of Fig- ure 8.1 and for no atomic L C-sentence both and are in y ,y ,... . [ ¬ { 0 1 } We now extend the first leg of the Strategic Balance of Logic, the equiva- Incomplete version for students of easllc2012 only. 8.3 Model Theory of L!1! 181 lence between the Semantic Game and the Model Existence Game, from first- order logic to infinitary logic: Theorem 8.12 (Model Existence Theorem for L!1!) Suppose L is a count- able vocabulary and ' is an L-sentence of L!1!. The following are equivalent: (1) There is an L-structure such that = '. M M| (2) Player II has a winning strategy in MEG(',L). Proof The implication (1) (2) is clear as II can keep playing sentences ! that are true in . For the other implication we proceed as in the proof of M Theorem 6.35. Let C = cn : n N and Trm = tn : n N . Let { 2 } { 2 } (x0,y0,x1,y1,...) be a play in which player II has used her winning strategy and player I has maintained the following conditions: 1. If n =0, then xn = '. 2. If n =2 3i, then x is c c . · n ⇡ i i 3. If n =4 3i 5j 7k 11l, y is c t , and y is '(c ), then x is '(c ). · · · · i ⇡ j k l j n i 4. If n =8 3i 5j and y is ' , then x is ' . · · i m N m n j 5. If n = 16 3i and y is 2' , then x is ' . · i mV N m n m N m 6. If n = 32 3i 5j, y is x'2(x), then x is '(c ).2 · · i W8 n Wj 7. etc. The rest of the proof is exactly as in the proof of Theorem 6.35. Our success in the above proof is based on the fact that even if we deal with infinitary formulas we can still manage to let player I list all possible formulas that are relevant for the consistency of the starting formula. If even one uncountable conjunction popped up, we would be in trouble. It suffices to consider in MEG(',L) such constant terms t that are either constants or contain no other constants than those of C. Moreover, we may assume that if player I enquires about tt, then t = cn for some n N. ⇡ 2 Corollary Let L be a countable vocabulary. Suppose ' and are sentences of L!1!. The following are equivalent: (1) ' = . | (2) Player I has a winning strategy in MEG(' ,L). ^ ¬ The proof of the Compactness Theorem does not go through, and should not, because there are obvious counter-examples to compactness in L!1!. In many proofs where one would like to use the Compactness Theorem one can instead use the Model Existence Theorem. The non-definability of well-order in L ! was proved already in Theorem 7.26 but we will now prove a stronger 1 version for L!1!: Incomplete version for students of easllc2012 only. 182 Model Theory of Infinitary Logic Theorem 8.13 (Undefinability of Well-Order) Suppose L is a countable vo- cabulary containing a unary predicate symbol U and a binary predicate sym- bol <, and ' L . Suppose that for all ↵ < ! there is a model of ' 2 !1! 1 M such that (↵,<) (U M, Proof Let D = dr : r Q be a set of new constant symbols. Let us call { 2 } them d-constants. Let ✓ = r yi = 'i(c0,...,cm,dr1 ,...,drl ), where dr ,...,dr are the d-constants appearing in y0,...,yn 1 except in 1 l { } ✓. She maintains the following condition: (?) For all ↵ < !1 there is a model of ' and b1,...,bl U M !1 such M 2 ✓ that = x ... x ' (x ,...,x ,b ,...,b ) M| 9 0 9 m i 0 m 1 l i ↵ b1,b1 + ↵ b2,...,bl 1 + ↵ bl. We show that player II can indeed maintain this condition. For most moves of player I the move of II is predetermined and we just have to check that (?) remains valid. For a start, if I plays ', condition (?) holds by assumption. If I enquires about substitution or plays a conjunct of a played conjunction, no new constants are introduced, so (?) remains true. Also, if I tests a played x'(x) or enquires about a played x'(x), no new constants of 8 9 D are introduced, so (?) remains true. We may assume that I enquires about tt only if t = c and so (?) holds by the induction hypothesis. Let us then ⇡ n Incomplete version for students of easllc2012 only. 8.3 Model Theory of L!1! 183 Figure 8.2 assume (?) holds and I enquires about a played disjunction i I i. For each ↵ < ! we have a model as in (?) and some i I such that2 = . 1 M↵ ↵ 2 W M↵ | i↵ Since I is countable, there is a fixed i I such that for uncountably many 2 ↵ < ! : = . If II plays this , condition (?) is still true. 1 M↵ | i i The remaining case is that I enquires about a constant term t. We may as- sume t = dr as otherwise there is nothing to prove. The constants of D oc- curring so far in the game are dr1 ,...,drl . Let us assume ri The following corollary is due to Lopez-Escobar (1966b). Corollary If ' is a sentence of L!1! in a vocabulary which contains the unary predicate U and the binary predicate <, and (U M, Corollary The class of well-orderings is not a PC-class of L!1!. The undefinability of well-ordering as a PC-class of L ! will be estab- 1 lished later. We now prove the Craig Interpolation Theorem for L!1!. There are several different proofs of this theorem, some of which employ the above corollary directly. Our proof is like the original proof by Lopez-Escobar, ex- cept that we operate with Model Existence Games instead of Gentzen systems. Theorem 8.14 (Separation Theorem) Suppose L1 and L2 are vocabularies. Suppose ' is an L1-sentence of L!1! and is an L2-sentence of L!1! such Incomplete version for students of easllc2012 only. 184 Model Theory of Infinitary Logic that = ' . Then there is an L L -sentence ✓ of L such that = ' | ! 1 \ 2 !1! | ! ✓ and = ✓ . | ! Proof This is similar to the proof of Theorem 6.40. We assume, w.l.o.g., that L and L are relational. Let L = L L . We describe, assuming that no 1 2 1 \ 2 such ✓ exists, a winning strategy of II in MEG(' ,L L ). We now ^ ¬ 1 [ 2 follow closely the proof of Theorem 6.40, where the strategy of II was to n n n divide the set of her moves into two parts S1 and S2 such that S1 consists of the L C-sentences of and Sn consists of the L C-sentences of . 1 [ 2 2 [ In addition it is assumed that (*) There is no L C-sentence ✓ that separates Sn and Sn. [ 1 2 There are two new cases over and above those of Theorem 6.40: n n+1 Case 50. Player I plays 'i where for example i I 'i S1 . Let S1 = n n+1 n n+1 2 n+12 S1 'i and S2 = S2 . If ✓ separates S1 and S2 , then clearly ✓ also [ { } n n V separates S1 and S2 . n Case 60. Player I plays 'i, where for example 'i S . We claim i I i I 2 1 that for some i I the sets2 Sn ' and Sn satisfy (*).2 Otherwise there is 2 W 1 [ { i} 2 W for each i I some ✓ that separates Sn ' and Sn. Let ✓ = ✓ . 2 i 1 [ { i} 2 i I i Then ✓ separates Sn and Sn contrary to assumption. 2 1 2 W 8.4 Large Models Suppose is an L-fragment of L + of size and is an L-structure of size ! M > . If we define on M a b for every '(x) : = '(a) = '(b) ⇠ () 2 M| () M| then by the Pigeonhole Principle there is a subset I of M of size > such that for a, b I and '(x) : 2 2 = '(a) = '(b). M| () M| We say that the set I is indiscernible in with respect to . If we want M indiscernibility relative to formulas with more than one free variable, we have to use Ramsey theory. Definition 8.15 Suppose L is a vocabulary, is an L-structure, and is an M L-fragment. A linear order (I,<), where I M, is -indiscernible in if ✓ M for all a1 <... 0 AAb bi0 i ⌧ - 6 qq a a0 i ⌧ i 6 qq ? ? xi yi xi yi 8 9 8 9 n m n m WVi i i VW i 6 6 Figure 8.9 Strategy ⇢. Although elementary equivalence relative to L G coincides with elemen- 1 tary equivalence relative to L !, there are properties expressible in L G 1 1 which are not expressible even in L , as we shall prove in Proposition 9.38. 11 8.7 Historical Remarks and References Good sources for the model theory of infinitary logic are Keisler (1971) and Makkai (1977). Theorem 8.10 is from Kueker (1972, 1977). Theorem 8.12 is from Keisler (1971), where the method of consistency properties is first presented in the context of infinitary logic. Subsequently it was extensively used in infinitary logic in Makkai (1969b,a), Harnik and Makkai (1976), and Green (1975). Theorem 8.13 and its two vorollaries are from Lopez-Escobar (1966b,a). The strong formulation of Theorem 8.13 is from Keisler (1971). Theorem 8.14 is from Lopez-Escobar (1965). By means of the Model Existence Game (or consistency properties) many variations of the Craig Interpolation Theorem can be proved for L!1!, as demonstrated convincingly in Keisler (1971). We have collected some of them in Exercises 8.10–8.17. Theorem 8.22 is from Morley (1968). Theorem 8.31 goes back to Morley (1968), but the present proof is due to Shelah. Proposition 8.32 is from Lopez- Escobar (1966b,a). Proposition 8.41 is essentially due to Keisler (1965). A useful source for game quantifiers is Burgess (1977). Approximations of game Incomplete version for students of easllc2012 only. Exercises 223 formulas were first considered in set theory in Kuratowski (1966). For more on Souslin–formulas, see Burgess (1978) and Green (1978). Exercises 8.10–8.12 and 8.16 are from Lopez-Escobar (1965). Exercises 8.13-8.14 are from Malitz (1969). Exercise 8.15 is from Barwise (1969). Ex- ercise 8.17 is from Makkai (1969b), which is a good source for all the Exer- cises 8.10–8.17. Exercise 8.30 is from Sierpinski´ (1933). Exercise 8.46 is from Burgess (1978), where also related results are proved. Exercise 8.47 is from Green (1979). Exercises 8.1 Find an uncountable model such that there is no countable with M N ! . N ⌘1 M 8.2 Let L be the vocabulary E cn : n N , where each cn is a constant { }[{ 2 } symbol and E is a binary relation symbol. Let for A N the sentence ✓ 'A be c c x x x Ex c x . ¬⇡ n m ^9 08 1 1 0 $ ⇡ n 1 n countably many countable models. Show that it is definable in L!1!. 8.6 Show that the class of models ( ,P), where P is L !-definable on M 1 , is a closed class. M 8.7 Show that the class of countable well-orders is the union of two disjoint closed classes. 8.8 Suppose (A, <) is an uncountable linear order. Show that (A, <) con- tains a copy of !1, a copy of !1⇤, or a copy of the rationals. (Hint: As- sume not. Prove first that there is a A such that both ( ,a] and [a, ) 2 ! are uncountable in (A, <).) 8.9 Show that if ' is a sentence of L!1! in a vocabulary which contains the unary predicate U and the binary predicate < and ' has a model with M Incomplete version for students of easllc2012 only. 224 Model Theory of Infinitary Logic (U M, 8.10 (Lyndon Interpolation Theorem) Suppose ' and are sentences in L!1! and = ' . Show that there is ✓ in L such that = ' ✓, | ! !1! | ! = ✓ , every relation symbol occurring positively (negatively) in ✓ | ! occurs positively (negatively) in ' and . 8.11 Assume in Exercise 8.10 the sentences ' and have no function or constant symbols, and no identity. Assume also = ' and = . Show 6| ¬ 6| that ✓ can be chosen so that it does not contain identity. 8.12 Suppose ' and are sentences of L such that if and are models !1! M N of ', is a homomorphic image of , and = , then = . N M M| N| Show that there is a positive L - sentence ✓ such that ' = ✓. !1! | $ 8.13 Suppose L1 and L2 are vocabularies which contain no function symbols. Let ' be an L1-sentence and an L2-sentence of L!1! such that is universal and = ' . Show that there is a universal L L -sentence | ! 1 \ 2 ✓ of L such that = ' ✓ and = ✓ . !1! | ! | ! 8.14 Suppose ' and are sentences of L such that if and are models !1! M N of ', is a submodel of , and = , then = . Show that there N M M| N| is a universal sentence ✓ of L such that ' = ✓. !1! | $ 8.15 Suppose ' and are sentences of L!1!. Show that every countable model of ' can be embedded in some countable model of if and only if every universal logical consequence of in L!1! is a logical conse- quence of '. 8.16 Suppose ' and are sentences of L!1!. Show that every homomorphic image of a model of ' is a model of if and only if there is a positive sentence ✓ in L such that = ' ✓ and = ✓ . !1! | ! | ! 8.17 The class K of countable structures such that is isomorphic to a B B substructure of some model of ' L is identical to the class of all A 2 !1! countable models of the set of universal sentences ✓ L such that 2 !1! ' = . | 8.18 Consider the following game G ( ) where is a set of subsets of M cub C C of cardinality . Players I and II play as in G ( ) but the game goes cub C on for rounds producing a set X = x : ↵ < y : ↵ < . { ↵ } [ { ↵ } Player II wins if X . Show that if II has a winning strategy in 2 C G ( ) and G ( ), then she has a winning strategy in G ( ). cub C cub D cub C \ D 8.19 Show that if is a set of finitary functions on M, = , and is the F |F| C set of X M of cardinality closed under each function in , then II ✓ F has a winning strategy in G ( ). Use this to conclude that if is an cub C M L-structure, L , ' L + , = ' and is the set of domains of | | 2 ! M| C with = ', then II has a winning strategy in G ( ). M0 ✓ M M0 | cub C Incomplete version for students of easllc2012 only. Exercises 225 8.20 Show that we can extend each vocabulary L to L⇤, translate each L+!- sentence ' in the vocabulary L to an L+!-sentence '⇤ in the vocabu- lary L⇤, expand each L-structure to an L⇤-structure ⇤, and extend M M any L-fragment to a set ⇤ of cardinality of L + -sentences in T T ! the vocabulary L⇤ such that for all L-structures , L⇤- structures , M N L-fragments and L + -sentences ' in the vocabulary L: T ! 2 T (1) = ' = ⇤ = ⇤ '⇤ . M| ) M | T [ { } (2) = ⇤ '⇤ = L = '. N| T [ { } ) N | (3) '⇤ is quantifier-free. (4) ⇤ is a set of universal sentences. T 8.21 Suppose L is a vocabulary of cardinality , ' is a sentence of L + in ! the vocabulary L, is an L-structure such that = ', and µ M M| M . Show that there is such that = ' and M = µ. | | M0 ✓ M M0 | | 0| 8.22 Find a counter-example to the following claim: Suppose L is a vocabu- lary of cardinality , ↵ < +, and is an L-structure. Then there is M such that ↵ and M . M0 ✓ M M0 'p M | 0| 8.23 Prove that the class of models (M, A, B), where A, B M and A < ✓ | | B is not RPC in L !. In fact, prove the following statement: If ' | | 1 2 L + has a model such that U M < V M , then ' has a model ! M | | | | such that U N = V N . N | | | | 8.24 Show that the class of linear orders with uncountable cofinality is not PC- definable in L !. In fact, prove the following statement: If ' L+! 1 + 2 has a model such that (U M, 2 8.29 Show that there is a sentence in L+! which has a model of size 2 but none of size > 22 . ! 2 8.30 Show 2 (!1) . (Hint: Take a well-ordering of R. Then define 6! 2 a coloring of pairs x, y of reals by reference to and the standard { } ordering of R.) 8.31 Show that if ()2, then is regular. ! 2 8.32 Prove directly 6 (3)2. ! 2 8.33 Show 15 (4)2 (it is probably just as easy to show 17 (4)2.) 6! 2 6! 2 8.34 Deduce the undefinability of well-order in L!1! directly from Theo- rem 8.22. (Hint: Assume well-order could be defined in L!1!. Show that then there is a sentence in L!1! with a model of size i!1 but none bigger.) 8.35 Show that for every vocabulary L, every L-fragment of L ! and every 1 L-structure there is L0 L, an L0-fragment 0 such that 0 , and M ◆ ◆ an expansion 0 of to an L0-structure such that L0 = L + , M M | | | | @0 0 = + and 0 has Skolem functions for all ' 0. | | | | @0 M 2 8.36 Construct for each infinite a linear order with 2 automorphisms. 8.37 An element a of a Boolean algebra is an atom if for all b M :0 M 2 b a implies 0=b or b = a. Show that if is a Boolean algebra and M (I,<) is a linear order such that I is a set of atoms of , then (I,<) is M -indiscernible in for any fragment of L !. M 1 8.38 Suppose is an equivalence relation and (I,<) is a linear order such M that I is included in one of the equivalence classes of . Show that M (I,<) is -indiscernible in for all fragments of L !. Is the same M 1 true if I is a set of non- -equivalent elements of M? M 8.39 Suppose L is a vocabulary, and 0 L-structures, an L-fragment M M of L !, (I, Proof Suppose T is a bottleneck. Let ↵ < + such that T V [G ]. Let 2 ↵ A↵ be the Cohen subset of added at stage ↵. Note that A↵ is a bistationary subset of . We first show that T T (A ). Suppose 6 ↵ p fˆ : T (A ) T is strictly increasing. ↵ ! When we force with A , calling the forcing notion 0, an uncountable branch ↵ P appears in T (A ), hence also in T . The product forcing ? 0 contains a ↵ P↵ P -closed dense set (Exercise 9.45). Hence it cannot add a branch of length to T . We have shown that T (A ) T in V [G]. Since T is a bottleneck, ↵ 6 T T (A ). By repeating the same with A we get T T ( A ). In ↵ ↵ ↵ sum, T T (A ) T ( A ) (see Exercise 9.44 for the definition of ). But ↵ ⌦ ↵ ⌦ T (A ) T ( A ) T (Exercise 9.46). Hence T T . ↵ ⌦ ↵ p p It is also known (Todorceviˇ c´ and Va¨an¨ anen¨ (1999)) that if V = L, then there 1 are no bottlenecks in the class 1, 1 above Tp@ . T@ @ 9.5 The Transfinite Dynamic Ehrenfeucht–Fra¨ısse´ Game In this section we introduce a more general form of the Ehrenfeucht–Fra¨ısse´ Game. The new game generalizes both the usual Ehrenfeucht–Fra¨ısse´ Game and the dynamic version of it. In this game player I makes moves not only in the models in question but also moves up a po-set, move by move. The game goes on as long as I can move. This game generalizes at the same time the games EF↵( 0, 1) and EFD ( 0, 1). Therefore we denote it by EF A A A A P rather than by EFD . P If is a po-set, let b( ) denote the least ordinal so that does not have P P P an ascending chain of length . Definition 9.66 Suppose 0 and 1 are L-structures and is a po-set. A A P The Transfinite Dynamic Ehrenfeucht–Fra¨ısse´ Game EF ( 0, 1) is like the P A A game EF ( , ) except that on each round I chooses an element c A0 A1 ↵ 2 0, 1 , an element x A , and an element p . It is required that { } ↵ 2 c↵ ↵ 2 P p0 < ...< p↵ < .... P P P Finally I cannot play a new p anymore because is a set. Suppose I has ↵ P played z¯ = (c ,x ): < ↵ and II has played y¯ = y : < ↵ . If p is h i h i z,¯ y¯ a partial isomorphism between and , II has won the game, otherwise I A0 A1 has won. Incomplete version for students of easllc2012 only. 9.5 The Transfinite Dynamic Ehrenfeucht–Fra¨ısse´ Game 259 Thus a winning strategy of I in EF ( 0, 1) is a sequence ⇢ = ⇢↵ : ↵ < P A A h b( ) and a strategy of II is a sequence ⌧ = ⌧ : ↵ < b( ) . Note that P i h ↵ P i EF ( , ) is the same game as EF ( , ), ↵ A0 A1 (↵,<) A0 A1 and EFD ( , ) is the same game as EF ( , ). ↵ A0 A1 (↵,>) A0 A1 Naturally, if ↵ is finite, the games EF ( , ) and EF ( , ) (↵,<) A0 A1 (↵,>) A0 A1 are one and the same game. But if ↵ happens to be infinite, there is a big difference: The first is a transfinite game while the second can only go on for a finite number of moves. The ordering 0 of po-sets has a close connection to the question who P P wins the game EF ( 0, 1), as the following two results manifest: P A A Lemma 9.67 If II wins the game EF ( 0, 1) and 0, then II wins P0 A A P P the game EF ( 0, 1). If I wins the game EF ( 0, 1) and 0, then P A A P A A P P I wins the game EF ( 0, 1). P0 A A Proof Exercise 9.50. Proposition 9.68 Suppose II wins EF ( 0, 1) and I wins EF ( 0, 1). P A A P0 A A Then 0. P P Proof Suppose II wins EF ( 0, 1) with ⌧ and I wins EF ( 0, 1) with P A A P0 A A ⇢. We describe a winning strategy of I in G( 0, ), and then the claim follows P P from Lemma 9.60. Suppose ⇢ ( )=(c ,x ,p0 ). The element p0 is the first 0 ; 0 0 0 0 move of I in G( 0, ). Suppose II plays p . Let P P 0 2 P y0 = ⌧0(((c0,x0,p0))), (c1,x1,p10 )=⇢1((y0)). The element p0 is the second move of I in G( 0, ). More generally the equa- 1 P P tions y = ⌧ ( (c ,x ,p ): ) h i (c ,x ,p0 )=⇢ ( y : < ↵ ) ↵ ↵ ↵ ↵ h i define the move p0 of I in G( 0, ) after II has played p : < ↵ . The ↵ P P h i game can only end if II cannot move p↵ at some point, so I wins. Suppose = . Then there is a least ordinal A0 6⇠ A1 Card(A ) + Card(A ) 0 1 such that II does not win EF ( , ). Thus for all ↵ +1 < there is a A0 A1 Incomplete version for students of easllc2012 only. 260 Stronger Infinitary Logics winning strategy for II in EF ( , ). Let K(= K( , )) be the set ↵+1 A0 A1 A0 A1 of all winning strategies of II in EF ( , ) for ↵ +1< . We can make ↵+1 A0 A1 K a tree by letting ⌧ : ⇠ ↵ ⌧ 0 : ⇠ ↵0 h ⇠ ih ⇠ i if and only if ↵ ↵0 and ⇠ ↵(⌧ = ⌧ 0 ). 8 ⇠ ⇠ Definition 9.69 We call K, as defined above, the canonical Karp tree of the pair ( , ). A0 A1 Note that even when is a limit ordinal K does not have a branch of length , for otherwise II would win EF ( , ). A0 A1 Lemma 9.70 Suppose is a po-set. Then P wins EF ( 0, 1) 0 K. 9 P A A () P Proof Suppose II wins EF ( 0, 1) with ⌧. If s = s⇠ : ⇠ ↵ 0 , ) P A A h i2 P we can define a strategy ⌧ 0 of II in EF ( , ) as follows ↵+1 A0 A1 ⌧ 0 ( (c ,x ):⌘ ⇠ )=⌧ ( (c ,x ,s ):⌘ ⇠ ). ⇠ h ⌘ ⌘ i ⇠ h ⌘ ⌘ ⌘ i Since K does not have a branch of length , ↵ < , and hence ⌧ 0 K. The 2 mapping s ⌧ 0 is an order-preserving mapping 0 K. 7! P ! Suppose f : 0 K is order-preserving. We can define a winning ( P ! strategy of II in EF ( 0, 1) by the equation P A A ⌧ ( (c ,x ,s ):⌘ ⇠ )=f( s : ⇠ ↵ )( (c ,x ,s ):⇠ ↵ ). ↵ h ⇠ ⇠ ⇠ i h ⇠ i h ⇠ ⇠ ⌘ i Proposition 9.71 Suppose is a limit ordinal and II wins EF↵( 0, 1) for A A all ↵ < . The following are equivalent: (i) II wins EF ( , ). A0 A1 (ii) II wins EF ( 0, 1) for every po-set with no branches of length . P A A P Proof To prove (ii) (i), suppose II does not win EF ( , ). Let = ! A0 A1 P K( , ). Then does not have branches of length , hence by (ii) II A0 A1 P wins EF ( 0, 1) and we get from Lemma 9.70, a contradiction P A A P P with Lemma 9.55. The other direction (i) (ii) is trivial. ! Note Suppose = Card(A0)+Card(A1). Then we can compute Card(K) ↵ ↵ sup ( )↵ =sup . If GCH and is regular, then Card(K) +. ↵< ↵< Furthermore, if we assume GCH, we can assume Card( ) in (ii) above P (Hyttinen). For = ! this does not depend on GCH. II wins EF ( 0, 1) if P A A Incomplete version for students of easllc2012 only. 9.5 The Transfinite Dynamic Ehrenfeucht–Fra¨ısse´ Game 261 and only if II wins EF ( 0, 1). So from the point of view of the existence 0P A A of a winning strategy for II we could always assume that is a tree. P Corollary II never wins EF K ( 0, 1). A A Definition 9.72 A po-set is a Karp po-set of the pair ( 0, 1) if II wins P A A EF ( 0, 1) but not EF ( 0, 1). If a Karp po-set is a tree, we call it a P A A P A A Karp tree. By Lemma 9.70 and the above corollary, there are always Karp trees for every pair of non-isomorphic structures. Suppose I wins EF ( 0, 1) with the strategy ⇢. Let S⇢ be the set of se- P A A quences y¯ = y : ⇠ ↵ dom(⇢) such that h ⇠ i2 p Part( , ). ⇢ ↵+1,y 2 A0 A1 Thus S⇢ is the set of sequences of moves of II before she loses EF ( 0, 1), P A A when I plays ⇢. We can make S⇢ a tree by ordering it as follows y : ⇠ ↵ y0 : ⇠ ↵0 h ⇠ ih ⇠ i if and only if ↵ ↵0 and ⇠ ↵(y = y0 ). 8 ⇠ ⇠ Lemma 9.73 I wins EF S ( 0, 1). ⇢ A A Proof The following equation defines a winning strategy ⇢0 of I in the game EF ( , ): S⇢ A0 A1 ⇢0 y : ⇠ < ↵ )= c ,x , (y : ⇠ : < ↵ , ↵h ⇠ i h ↵ ↵ h ⇠ i i where ⇢ ( y : ⇠ < ↵ )=(c ,x ,p ). ↵ h ⇠ i ↵ ↵ ↵ Lemma 9.74 S⇢ . P Proof Suppose s = y : ⇠ : < ↵ S , where hh ⇠ i i2 ⇢ 0 < 1 <...< ⌘ <...(⌘ < ↵). Let =sup⌘<↵ ⌘ and ⇢ ( y : ⇠ < )=(c ,x ,p ). h ⇠ i We define f(s)=p . Then f : S is order-preserving. ⇢ ! P Incomplete version for students of easllc2012 only. 262 Stronger Infinitary Logics Note that Lemma 9.74 implies S . In particular, if I wins EF ( , ) P 6 ⇢ A0 A1 with ⇢, then S⇢ is a tree with no branches of length . Suppose 0 is such that 0 and I wins EF . So 0 could be P P P P0 P S . Suppose furthermore that there is no such that and I wins ⇢ P1 P1 P0 EF . Lemma 9.57 implies that this assumption can always be satisfied. P1 Lemma 9.75 I does not win EF ( 0, 1). P0 A A Proof Suppose I wins EF 0 ( 0, 1) with ⇢0. Then I wins EF S ( 0, 1) P A A ⇢0 A A and S , contrary to the choice of . ⇢0 P0 P0 Definition 9.76 A po-set is a Scott po-set of ( 0, 1) if I wins the game P A A EF ( 0, 1) but not the game EF ( 0, 1). If a Scott po-set is a tree, P A A P A A we call is a Scott tree. If is both a Scott and a Karp po-set, it is called a P determined Scott po-set. By Lemma 9.73 and Lemma 9.75, S is always a Scott tree of ( , ), so ⇢ A0 A1 Scott trees always exist. Note that ↵ Card(S⇢) sup (Card(A0) + Card(A1)) . ↵ Lemma 9.77 Suppose I wins EF ( 0, 1) with ⇢ and K = K( 0, 1). P A A A A Then K S . ⇢ Proof Suppose ⌧ K. Let II play ⌧ against ⇢ in EF ( 0, 1). The result- 2 P A A ing sequence y¯ of moves of II is an element of S . The mapping ⌧ y¯ is ⇢ 7! order-preserving. Suppose II wins EF ( 0, 1) and I wins EF ( 0, 1) with ⇢. Fig- P0 A A P1 A A ure 9.7 shows the resulting picture. In summary, we have proved: Theorem 9.78 Suppose II wins EF ( 0, 1) and I wins EF ( 0, 1). P0 A A P1 A A Then there are trees T0 and T1 such that (i) 0 T T . P0 0 1 P1 (ii) II wins EF ( , ) but not EF ( , ). T0 A0 A1 T0 A0 A1 (iii) I wins EF ( , ) but not EF ( , ). T1 A0 A1 T1 A0 A1 Example 9.79 Suppose I wins EF!( 0, 1). By Proposition 7.19 there A A is a unique = ( , ) such that II wins EF ( , ) and I wins A0 A1 ( ,>) A0 A1 EF ( , ). Then ( ,>) is both a Karp and a Scott po-set for and ( +1,>) A0 A1 A0 . A1 Incomplete version for students of easllc2012 only. 9.5 The Transfinite Dynamic Ehrenfeucht–Fra¨ısse´ Game 263 Figure 9.7 The boundary between II winning and I winning. Example 9.80 Suppose II wins EF↵( 0, 1) but not EF↵+1( 0, 1). Then A A A A (↵,<) is a Karp tree (in fact a Karp well-order) of and . This follows A0 A1 from the fact that (↵,<) (↵ +1,<). ⌘ Example 9.81 Suppose I wins EF↵+1( 0, 1) but not EF↵( 0, 1). Then A A A A (↵,<) is a Scott tree (in fact a Scott well-order) of and . A0 A1 If T is a tree, T +1is the tree which is obtained from T by adding a new element at the end of every maximal branch of T . Note that T +1may be uncountable even if T is countable. Lemma 9.82 Suppose S !1 is bistationary, 0 = (S), 1 = ( ), and ✓ A A ; = T (!1 S)+1. Then I wins EF ( 0, 1). P \ P A A Proof Suppose I has already played (c ,x ,p ) and II has played y for < ↵. Suppose I now has to decide how to play (c↵,x↵,p↵) in EF ( 0, 1). P A A We assume that I has played in such a way that 1. p = : : < ( (T (! S) + 1). hh i i 2 1 \ 2. x N(f,↵)= g ! : f ↵ = g ↵ , { 2 } where ↵ < , form the basis of the topology. Let us denote this general- ized Baire space by . Now properties of models of size correspond N to subsets of . In particular, modulo coding, isomorphism of structures of N cardinality becomes an “analytic” property in this space. One of the basic questions about models of size that we can try to attack with methods of logic is the question which of those models can be identified up to isomorphism by means of a set of invariants. Shelah’s Main Gap Theorem gives one answer: If is any structure of cardinality ! in a countable M 1 vocabulary, then the first-order theory of is either of the two types: M Structure Case All uncountable models elementary equivalent to can be M characterized in terms of dimension-like invariants. Non-structure Case In every uncountable cardinality there are non-isomorphic models elementary equivalent to that are extremely difficult to dis- M tinguish from each other by means of invariants. The game-theoretic methods we have developed in this book help us to an- alyze further the non-structure case. For this we need to develop some basic topology of . A set A is dense if A meets every non-empty open set. N ✓ N The space has a dense subset of size < consisting of all eventually con- N stant functions. If the Generalized Continuum Hypothesis GCH is assumed, then < = for all regular and < = + for singular . Theorem 9.87 (Baire Category Theorem) Suppose A↵, ↵ < , are dense open subsets of . Then A is dense. N ↵ ↵ Proof Let f andT↵ < be arbitrary. If f and ↵ for ⇠ < have 0 2 N 0 ⇠ ⇠ been defined so that ↵ < ↵ and f N(f , ↵ ) ⇣ ⇠ ⇠ 2 ⇣ ⇣ for ⇣ < ⇠ < , then we define f and ↵ as follows: Choose some g 2 N such that g N(f , ↵ ) for all ⇠ < and let ↵ =sup ↵ . Since A 2 ⇠ ⇠ ⇠< ⇠ is dense, there is f A N(g,↵ ). When all f and ↵ for ⇠ < have 2 \ ⇠ ⇠ Incomplete version for students of easllc2012 only. 9.6 Topology of Uncountable Models 271 been defined, we let f be such that f N(f , ↵ ) for all ⇠ < . Then f 2 ⇠ ⇠ 2 A N(f , ↵ ). ↵ ↵ \ 0 0 1 DefinitionT 9.88 A subset A of is said to be ⌃1 (or analytic) if it is a N 1 projection of a closed subset of . A set is ⇧1 (or co-analytic) if its N ⇥ N 1 1 1 complement is analytic. Finally, a set is 1 if it is both ⌃1 and ⇧1. Example 9.89 Examples of analytic sets relevant if is a regular cardinal > !, are CUB = f : ↵ < : f(↵)=0 contains a club { 2 N { } } and NS = f : ↵ < : f(↵) =0 contains a club . { 2 N { 6 } } The set of ↵-sequences of elements of for various ↵ < form a tree N< under the subsequence relation. Any subset T of which is closed under N< subsequences is called a tree in this section. A -branch of such a tree is any linear subtree (branch) of height . Let us denote g( ): < ↵ by g¯(↵). h i Lemma 9.90 A set A is analytic iff there is a tree T < < ✓ N ✓ N ⇥ N such that for all f: f A T (f) has a -branch, (9.7) 2 () where T (f)= g¯(↵):(¯g(↵), f¯(↵)) T . Such a tree is called a tree repre- { 2 } sentation of A. Proof Suppose first A is analytic and B is a closed set such that ✓ ⇥ f A g((f,g) B). 2 () 9 2 Let T = (f¯(↵), g¯(↵)) : (f,g) B,↵ < . { 2 } Clearly now f A if and only if T (f) has a -branch. Conversely, suppose 2 such a T exists. Let B be the set of (f,g) such that (f¯(↵), g¯(↵)) T for all 2 ↵ < . The set B is closed and its projection is A. Respectively, a set is co-analytic if and only if there is a tree T ✓ N< ⇥ such that for all f: N< f A T (f) has no -branches. (9.8) 2 () Let denote the class of all trees without -branches. Let denote the T T , set of subtrees of < of cardinality without any -branches. Incomplete version for students of easllc2012 only. 272 Stronger Infinitary Logics Proposition 9.91 Suppose B is a co-analytic subset of and T is as in N (9.8). For any tree S let 2 T B = f B : T (f) S . S { 2 } Then B = BS, S 2[T , where = <. Proof Clearly B B if S . Conversely, suppose f B. Then of S ✓ 2 T 2 course f B . It remains to observe that T (f) <. 2 T (f) | | Suppose A B is analytic and S is a tree as in (9.7). Let ✓ T 0 = (f¯(↵), g¯(↵), h¯(↵)) :g ¯(↵) T (f), h¯(↵) S(f) . (9.9) { 2 2 } < Note that T 0 and T 0 has no -branches, for such a branch would give | | rise to a triple (f,g,h) which would satisfy f A B. Note also that if f A, 2 \ 2 then there is a -branch h¯(↵):↵ < in S(f), and hence the mapping { } g¯(↵) (f¯(↵), g¯(↵), h¯(↵)) 7! witnesses T (f) T 0. We have proved: Proposition 9.92 (Covering Theorem for ) Suppose B is a co-analytic N subset of and S is as in (9.8). Suppose A B is analytic. Then N ✓ A B ✓ T for some T , where = <. 2 T , The idea is that the sets B , T , cover the co-analytic set B com- T 2 T , pletely, and moreover any analytic subset of B can be already covered by a single B . Especially if B happens to be 1, then there is T such that T 1 2 T , B = BT . 1 Corollary (Souslin–Kleene Theorem for ) Suppose B is a subset of N 1 . Then N B = BT for some T , where = <. 2 T , Incomplete version for students of easllc2012 only. 9.6 Topology of Uncountable Models 273 Corollary (Luzin Separation Theorem for ) Suppose A and B are disjoint N analytic subsets of . Then there is a set of the form C for some co-analytic N T set C and some T , where = <, that separates A and B, i.e. A C 2 T , ✓ and C B = . \ ; In the case of classical descriptive set theory, which corresponds to assuming = !, the sets BT are Borel sets. If we assume CH, then CUB and NS cannot be separated by a Borel set. < Proposition 9.93 If = , then the sets BT are analytic. If in addition T 1 is a strong bottleneck, then BT is 1. Let us call a family of elements of universal if for every T there B T , 2 T is some S such that T S. If has a universal family of size µ, and 2 B T , < = , then by the above results every co-analytic set in is the union N of µ analytic sets. By results in Mekler and Va¨an¨ anen¨ (1993) it is consistent + relative to the consistency of ZFC that + , 2 = , has a universal family + T of size ++ while 2 = +++. Definition 9.94 The class of Borel subsets of is the smallest class con- N taining the open sets and the closed sets which is closed under unions and intersections of length . Note that every closed set in is the union of < open sets (Exer- N cise 9.57). So if < = , then the definition of Borelness can be simplified. < Theorem 9.95 Assume = > !. Then has two disjoint analytic N sets that cannot be separated by Borel sets. Proof Note that is a regular cardinal. Every Borel set A has a “Borel code” c such that A = Bc. Let us suppose A = Bc separates the disjoint analytic sets and NS defined in Example 9.89. For example, A and CUB CUB ✓ A NS = . Let =(2<, ) be the Cohen forcing for adding a generic \ ; P subset for . Let G be -generic and g = . Now either g A or g/A. P P 2 2 Let us assume, w.l.o.g., that g A. Let p gˇ Bcˇ. Let M (H(µ), ,<⇤) 2 S 2 < 2 for a large µ such that ,p, ,TC(c) M, M M, and <⇤ is a well- P 2 ✓ order of H(µ). Since < = > !, we may also assume M = . Since | | P is < -closed, it is easy to construct a -generic G0 over M in V such that P ↵ < : M = “(ˇg) (↵) = 0” contains a club. (9.10) { | G0 6 } It is easy to show that Bc =(Bcˇ)G0 . Since M =“p gˇ B ”, | 2 cˇ whence (ˇg) B and therefore (ˇg) / NS . This contradicts (9.10). G0 2 c G0 2 Incomplete version for students of easllc2012 only. 274 Stronger Infinitary Logics Example 9.96 Suppose is a structure with M = . We call the analytic M set : N = and = {N N ⇠ M} the orbit of . Let = . Now player I has an obvious winning strategy M N 6⇠ M ⇢ in EF ( , ): he simply makes sure that all elements of both models are M N played. Obviously there are many ways to play all the elements but any of them will do. Let us consider the co-anaytic set B = f : N = and = { N N 6⇠ . Let S( ) be the Scott tree S of the pair ( , ). Let us choose a tree M} N ⇢ M N representation T of B in such a way that for all with N = , T (f )= N N S( ). If now f BT , then player I wins EFT ( , ). N N 2 0 0 M N Recall that if is a countable structure and ↵ is the Scott height of , then M M I wins EFD ( , ) whenever = and N is countable. Equivalently, ↵+! M N M 6⇠ N using the notation of Example 9.54, player I wins EF ( , ) whenever B↵+! M N = and N is countable. We now generalize this property of B to M ⇠6 N ↵+! uncountable structures. Definition 9.97 Suppose is an infinite cardinal and is a structure of M cardinality . A tree T is a universal Scott tree of a structure if T has no M branches of length and player I wins EF ( , ) whenever = and T M N M 6⇠ N N = M . | | | | The idea of the universal Scott tree is that the tree T alone suffices as a clock for player I to win all the 2 different games EF ( , ) where = and T M N M 6⇠ N N = M . Universal Scott trees exist: there is always a universal Scott tree | | | | of cardinality 2 as we can put the various Scott trees of the pairs ( , ), M N = , M = N , each of them of the size <, together into one tree. M 6⇠ N | | | | So the question is: How small universal Scott trees does a given structure have? If < = and has a universal family of size µ, then every structure T , of size has a universal Scott tree of size µ. If we allowed T to have a branch of length , any such tree would be a universal Scott tree of any structure of cardinality . We ask whether I wins EF ( , ) rather than in EF ( , ) in order T M N T M N to preserve the analogy with the concept of a Scott tree. A universal Scott tree T in our sense would give rise to a universal Scott tree T in the latter sense. Note that T = T <, so this is the order of magnitude of a difference in the | | | | size of universal Scott trees in the two possible definitions. Proposition 9.98 Suppose < = and is a structure with M = . The M following are equivalent: (1) The orbit of is 1. M 1 Incomplete version for students of easllc2012 only. 9.7 Historical Remarks and References 275 (2) has a universal Scott tree of cardinality . M Proof Suppose first (2) is true. Then = player I wins EF ( , ). M 6⇠ N () T M N 1 The existence of a winning strategy of I can be written in ⇧1 form since we as- sume < = . Assume then (1). Let ⇢ be a strategy of player I in EF ( , ) M N in which he simply enumerates the universes. Note that this is independent of . Let S( ) be the Scott tree S of the pair ( , ). Let us consider the N N ⇢ M N co-anaytic set B = f : N = and = . Let us choose a tree rep- { N N 6⇠ M} resentation T of B as in Example 9.96. If now f BT , then player I wins N 2 0 EF ( , ). By the above Souslin–Kleene Theorem, (1) implies the exis- T 0 M N tence of a tree T 0 such that B = B0 . Thus for any with N = , = T N M 6⇠ N implies that player I wins EFT 0 ( , ). Thus T 0 is a universal Scott tree of < M N . Moreover, T 0 = = . M | | The question whether the orbit of is 1 is actually highly connected to M 1 stability-theoretic properties of the first-order theory of , see Hyttinen and M Tuuri (1991) for more on this. 9.7 Historical Remarks and References Excellent sources for stronger infinitary languages are the textbook Dickmann (1975), the handbook chapter Dickmann (1985), and the book chapter Kueker (1975). The Ehrenfucht-Fra¨ısse´ Game for the logics L appeared in Benda 1 (1969) and Calais (1972). Proposition 9.32, Proposition 9.45, and the corollary of Proposition 9.45 are due to Chang (1968). The concept of Definition 9.40 and its basic properties were isolated independently by Dickmann (1975) and Kueker (1975). Theorem 9.31 is from Shelah (1990). Looking at the origins of the transfinite Ehrenfeucht–Fra¨ısse´ Game, one can observe that the game plays a role in Shelah (1990), and is then systematically studied, first in the framework of back-and-forth sets in Karttunen (1984), and then explicitly as a game in Hyttinen (1987), Hyttinen (1990), Hyttinen and Va¨an¨ anen¨ (1990) and Oikkonen (1990). The importance of trees in the study of the transfinite Ehrenfeucht–Fra¨ısse´ Game was first recognized in Karttunen (1984) and Hyttinen (1987). The cru- cial property of trees, or more generally partial orders, is Lemma 9.55 part (ii), which goes back to Kurepa (1956). A more systematic study of the quasi- order 0 of partial orders, with applications to games in mind, was P P started in Hyttinen and Va¨an¨ anen¨ (1990), where Lemma 9.57, Definition 9.58, Incomplete version for students of easllc2012 only. 276 Stronger Infinitary Logics Lemma 9.59, and Lemma 9.60 originate. The important role of the concept of persistency (Definition 9.63) gradually emerged and was explicitly isolated and exploited in Huuskonen (1995). Once it became clear that trees may be incomparable by , the concept of bottleneck arose quite naturally. Defini- tion 9.64 is from Todorceviˇ c´ and Va¨an¨ anen¨ (1999). The relative consistency of the non-existence of non-trivial bottlenecks (Theorem 9.65) was proved in Mekler and Va¨an¨ anen¨ (1993). For more on the structure of trees see Todorceviˇ c´ and Va¨an¨ anen¨ (1999) and Dzamonjaˇ and Va¨an¨ anen¨ (2004). The point of studying trees in connection with the transfinite Ehrenfeucht– Fra¨ısse´ Game is that there are two very natural tree structures behind the game. The first tree that arises from the game is the tree of sequences of moves, as in Lemma 9.73. This tree originates in Karttunen (1984). The second, and in a sense more powerful tree is the tree of strategies of a player, as in Defini- tion 9.69 and the subsequent Proposition 9.71. This idea originates from Hyt- tinen (1987). The “transfinite” analogues of Scott ranks are the Scott and Karp trees, in- troduced in Hyttinen and Va¨an¨ anen¨ (1990). Because of problems of incom- parability of some trees, the picture of the “Scott watershed” is much more complicated than in the case of games of length !, as one can see by compar- ing Figure 7.4 and Figure 9.7. Proposition 9.85 and Theorem 9.86 are from Tuuri (1990). There is a form of infinitary logic the elementary equivalence of which cor- responds exactly to the existence of a winning strategy for II in EF↵, in the spirit of the Strategic Balance of Logic. These infinitary logics are called in- finitely deep languages. Their formulas are like formulas of L but there are infinite descending chains of subformulas. Thus, if we think of the syntax of a formula as a tree, the tree may have transfinite rank. These languages were introduced in Hintikka and Rantala (1976) and studied in Karttunen (1979), Rantala (1981), Karttunen (1984), Hyttinen (1990), and Tuuri (1992). See Va¨an¨ anen¨ (1995) for a survey on the topic. There is also a transfinite version of the Model Existence Game, the other leg of the Strategic Balance of Logic, with applications to undefinability of (gen- eralized) well-order and Separation Theorems, see Tuuri (1992) and Oikkonen (1997). It was recognized already in Shelah (1990) that the roots of the problem of extending the Scott Isomorphism Theorem to uncountable cardinalities lie in stability theoretic properties of the models in question. This was made explicit in the context of transfinite Ehrenfeucht–Fra¨ısse´ Games in Hyttinen and Tuuri (1991). It turns out that there is indeed a close connection between the structure of Scott and Karp trees of elementary equivalent uncountable models and the Incomplete version for students of easllc2012 only. 9.7 Historical Remarks and References 277 stability theoretic properties such as superstability, DOP, and OTOP, of the (common) first-order theory. For more on this, see Hyttinen (1992), Hyttinen et al. (1993), and Hyttinen and Shelah (1999). A good testing field for the power of long Ehrenfeucht–Fra¨ısse´ Games turned out to be the area of almost free groups, where it seemed that the applica- bility of the infinitary languages L had been exhausted. For results in this direction, see Mekler and Oikkonen (1993), Eklof et al. (1995), Shelah and Vais¨ anen¨ (2002), and Vais¨ anen¨ (2003). An alternative to considering transfinite Ehrenfeucht–Fra¨ısse´ Games is to study isomorphism in a forcing extension. Isomorphism in a forcing extension is called potential isomorphism. The basic reference is Nadel and Stavi (1978). See also Huuskonen et al. (2004). Early on it was recognized that the trees T (S) (see Example 9.61) are very useful and in some sense fundamental in the area of transfinite Ehrenfeucht– Fra¨ısse´ Games. The question arose, whether there is a largest such tree for S ! bistationary. Quite unexpectedly the existence of a largest such tree ✓ 1 turned out to be consistent relative to the consistency of ZF. The name “Ca- nary trees” was coined for them, because such a tree would indicate whether some stationary set was killed. See Mekler and Shelah (1993) and Hyttinen and Rautila (2001) for results on the Canary tree. While the Ehrenfeucht–Fra¨ısse´ Game of length ! is almost trivially deter- mined, the Ehrenfeucht–Fra¨ısse´ Game of length !1 (and also of length ! +1) can be non-determined, see Hyttinen (1992), Mekler et al. (1993), and Hytti- nen et al. (2002). This has devastating consequences for attempts to use trans- finite Ehrenfeucht–Fra¨ısse´ Games to classify uncountable models. It is a phe- nomenon closely related to the incomparability of non-well-founded trees by the relation . This non-determinism is ultimately also the reason why the simple picture in Figure 7.4 becomes Figure 9.7. Some of the complexities of uncountable models can be located already on the topological level, as is revealed by the study of the spaces . These spaces N were studied under the name of -metric spaces in Sikorski (1950), Juhasz´ and Weiss (1978), and Todorceviˇ c´ (1981b). Their role as spaces of models, in the spirit of Vaught (1973), was emphasized in Mekler and Va¨an¨ anen¨ (1993). For more on the topology of uncountable models, see Va¨an¨ anen¨ (1991), Va¨an¨ anen¨ (1995), and Shelah and Va¨an¨ anen¨ (2000). See Va¨an¨ anen¨ (2008) for an informal exposition of some basic ideas. Theorem 9.95 is from Shelah and Va¨an¨ anen¨ (2000). Exercise 9.22 is from Nadel and Stavi (1978). Exercises 9.29 and 9.30 are from Hyttinen (1987). Exercise 9.35 is from Hyttinen and Va¨an¨ anen¨ (1990). Incomplete version for students of easllc2012 only. 278 Stronger Infinitary Logics Exercise 9.40 is from Kurepa (1956). Exercise 9.41 is from Huuskonen (1995). Exercise 9.47 is from Todorceviˇ c´ (1981a). Exercise 9.56 is due to Lauri Hella. Exercises 0 9.1 Show that player II wins EF@ ( , 0) if and only if she has a winning ! M M strategy in EF!( , 0). M M! 9.2 Show that I wins EFD 1 ( , ) if =(Q,<) and =(R,<). 2 M N M N 9.3 Show that in Example 9.2 player I has a winning strategy already in !1 EFD ( , 0). 2 M M 9.4 Show that , where and are as in Example 9.4. M 'p N M N 9.5 Prove the claim of Example 9.5. 9.6 Prove the claim of Example 9.19. 9.7 Give necessary and sufficient conditions for player I to have a winning strategy in EFD ( , 0), when and 0 are L-structures for a ↵ M M M M unary vocabulary L. 9.8 Show that if =(M,d,R, = A , RAi . Ai 0 i 1 i I i I i I !R L ]2 [2 [2 2 @ A Show that if : i I and : i I are families of L-structures {Ai 2 } {Bi 2 } for a relational vocabulary L and for each i , Ai ' Bi Incomplete version for students of easllc2012 only. Exercises 279 then . Ai ' Bi i I i I ]2 ]2 9.14 Suppose i : i I is a family of L-structures. The direct product of the family{A : i2 I} is the L-structure {Ai 2 } i i i = Ai, RA , (prodi I f A )f L, ((c : i I))c L A 0 2 2 2 2 1 i I i I i I !R L Y2 Y2 Y2 2 @ A where i i (prodi I f A )((a : i I)) = (fiA (ai):i I). 2 2 2 Show that if : i I and : i I are families of L-structures {Ai 2 } {Bi 2 } and for each i I 2 , Ai ' Bi then . Ai ' Bi i I i I Y2 Y2 9.15 Suppose : i I is a family of L-structures in a vocabulary L {Ai 2 } containing a distinguished constant symbol 0L. The direct sum i I i 2 A of the family i : i I is the substructure of i I i consisting of {A 2 } 2 A L (a : i I) such that a =0Ai for all but finitely many i I. Show that i 2 i L Q 2 if : i I and : i I are such families and for all i I {Ai 2 } {Bi 2 } 2 , Ai ' Bi then . Ai ' Bi i I i I M2 M2 9.16 Suppose is an L-structure for a relational vocabulary L. Let I J M ✓ be sets of size . Show that . M ' M i I i J M2 M2 9.17 Consider Z as an abelian group. Show that for any set I: Z p Z. ' i I i I M2 Y2 9.18 Show that “has a clique of size ” is not definable in L . ↵ 1 9.19 Prove Exercise 9.13 for !. ⌘1 Incomplete version for students of easllc2012 only. 280 Stronger Infinitary Logics 9.20 Prove Theorem 9.29. 9.21 Prove Proposition 9.32. 9.22 Let us write ( PI) if there is a forcing notion which does not M N add new sets of cardinality < (such forcing is called < -distributive) which forces and to be isomorphic. This is a form of “poten- M N tial isomorphism”, i.e. isomorphism in a forcing extension. Show that ( PI) is not a transitive relation among structures and thereby M N does not correspond to elementary equivalence relative to any logic. (Hint: Use the models (A) of Definition 9.8.) 9.23 Show that if is -homogeneous, then for any sequences ~a and ~b of M the same length from M: ( ,~a) ( ,~b) ( ,~a) s ( ,~b). M ⌘ M ) M ' M 9.24 Let H be the lexicographically ordered set of sequences s !↵ 0, 1 ↵ 2 { } (i.e. s< s0 if s(⇠) 9.33 Show that (!1,<) and (R,<) are incomparable by the quasi-order of po-sets. 9.34 Prove 0 0 0. P P () P P 9.35 Suppose S ! . Let T (S) be the tree of closed ascending sequences ✓ 1 of elements of S. Choose disjoint bistationary sets S1 and S2. Then T (S ) T (S ) and T (S ) T (S ) (see Example 9.61). 1 6 2 2 6 1 9.36 Prove the claims of Example 9.54. 9.37 Suppose T is a tree. Show that T has no infinite branches if and only if there is an ordinal ↵ so that T (↵,>). ⌘ 9.38 Prove that if a po-set is a union of countably many antichains, it satis- P fies (Q,<). P 1 9.39 Show that F 1 and Tp@ are special trees. @ 9.40 Prove the claim in Example 9.56 that 0Q is special but Q non-special. 9.41 Show that T is the -smallest persistent tree in . p T 9.42 Prove that T is a strong bottleneck in the class p T 9.43 If T , i I, is a family of trees, let T be the tree which consists i 2 i I i of a union of disjoint copies of T , i I,2 identified at the root. Show that i 2L i I Ti is the supremum of Ti : i I in the sense that Ti i I Ti 2 { 2 } 2 for all i I and if Ti T for all i I, then i I Ti T . L 2 2 2 L 9.44 If Ti, i I, is a family of trees, let i I Ti be the product tree 2 2 L T = s : dom(s)=QI, i I(s(i) T ) . i { 8 2 2 i } i I Y2 s s0 i I(s(i) s0(i)). () 8 2 Ti Let i I Ti be the subtree 2 N T = s T : i I j I(ht (s(i)) = ht (s(j)) . i 2 i 8 2 8 2 Ti Tj i I ( i I ) O2 Y2 We denote i 0,1 Ti by T0 T1. Prove that i I Ti is the infimum 2{ } ⌦ 2 of Ti0 : i I , that is, i I Ti Ti for each i I, and if T Ti for { 2N} 2 N2 all i I, then T i I Ti. 2 2N 9.45 Show that ? 0 in the proof of Theorem 9.65 contains a -closed dense P↵ P N set. (Hint: Suppose (s, s0) ? 0. Thus s : 0, 1 , s < , and 2 P↵ P ! { } | | s forces that s0 is a closed sequence of length < in A↵. Consider the sets of (s, s0) for which sup : s( )=1 = max(s0).) { } 9.46 Suppose A and B are disjoint stationary subsets of a regular cardinal . Show that T (A) T (B) T . (Hint: Show that II has a winning ⌦ p strategy in the game G(T (A) T (B),T).) ⌦ p 9.47 Suppose S !1. Prove that T (S) Q if and only if S is non-stationary. ✓ Incomplete version for students of easllc2012 only. 282 Stronger Infinitary Logics 9.48 Suppose S ! is bistationary and T is Aronszajn. Show that T (S) ✓ 1 6 T and T T (S). 6 9.49 Prove b((Q,<)) = b((R,<)) = !1. 9.50 Prove Lemma 9.67. 9.51 Prove Lemma 9.83. 9.52 Show that if I wins EF ( , ) for all i I, then II does not win Ti A0 A1 2 EF ( , ) i I Ti 0 1 . 2 A A 9.53 Show that the family of Scott trees of ( , ) is closed under suprema. N A0 A1 9.54 Show that the family of Karp trees of ( , ) is closed under suprema. A0 A1 9.55 Suppose is a Scott po-set of ( , ), where Card(A ), Card(A ) P A0 A1 0 1 2@0 . Show that there is a Scott tree of ( , ) such that T and A0 A1 P Card(T ) 2@0 . + 9.56 Show that if 2 =2 , then + + has an upper bound in + . T , T2 , 9.57 Show that every closed set in is the union of < open sets. N 9.58 Show that if cf()!, then the intersection of countably many open sets in is again open. Topological spaces with this property are called N -additive. 9.59 Show has a basis consisting of clopen sets. Topological spaces with N this property are called zero-dimensional. 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Walkoe, Jr., Wilbur John. 1970. Finite partially-ordered quantification. Journal of Symbolic Logic, 35, 535–555. Cited on page 207. Incomplete version for students of easllc2012 only. Index A B,4 -back-and-forth sequence, see back-and-forth A<⇥ ,9 sequence A!, 334 G8 (A), 24 Qeven, 286 cf !(A, W ), 25 Q! , 333 G LO n(A, W ), 17, 20 , 333 G Q idA,3 Q,3 -branch, 271 R,3 -Scott height, 240 -additive, 282 -Scott watershed, 240 -ideal, 130 Incomplete version for students of easllc2012 only. Index 363 1 ⌃1, see set closed, see set Str(L), 54 closed class, 223 !-saturated, see structure co-analytic, see set absolute, see formula CO-PLU, see generalized quantifier all-but finite, see generalized quantifier cofinal, 333 almost all, 176 cofinality, 10, 225, 333 almost free groups, 277 cofinality model, 333 analytic, see set cofinality quantifier, see generalized quantifier anti-isolated, 36 Compactness Theorem, 102, 116, 123, 322, anti-symmetric, see relation 332, 342 antichain, 61, 290 complement, see generalized quantifier Aronszajn tree, see tree Completeness Theorem, 331, 341 assignment, 35, 80 component, 61 at-least-one-half, see generalized quantifier cycle component, 61 atom, see generalized quantifier standard component, 61 atomic, see formula composition, 3 atomless, see Boolean algebra confirmer, 93 automorphism, 55 connected, see graph axiom consistent, 103 KA, 328 constituent, 56 Keisler’s axiom, 328 context SA, 335 countable, 286 Shelah’s axiom, 335 finite, 286 Axiom of Choice, 5 Continuum Hypothesis, 10 Axiom of Determinacy, 28 corresponding vertex, 40 back-and-forth sequence, 69, 146 countable, see set -back-and-forth sequence, 238 countable context, see context back-and-forth set, 64, 275 countable ordinal, see number -back-and-forth set, 237 countable-like, see formula strong -back-and-forth set, 245 countably complete, see filter basic, see formula countably incomplete, see filter basis, see generalized quantifier counting quantifier, see generalized quantifier Beth Definability Theorem, 109 cover, 225 BI-PLU, see generalized quantifier covering theorem, 272 bijection closed, see generalized quantifier Craig Interpolation Theorem, 107, 209, 214, bijective, see game 226 binary, see vocabulary cub, see set bistationary, see set Cub Game, see game Boolean algebra cycle, 50, see graph atomless, 76 cycle component, see component Borel, see set definable, 50, 83, 199, 310 bottleneck, 257 degree, 36 bounded, see generalized quantifier dense, see linear order, see set branch, see tree dependence logic, 205 Canary tree, see tree descriptive set theory, 209 canonical Karp tree, see tree determined, see game, see Scott po-set Cantor Normal Form, 13 diagonal intersection, 88 Cantor ternary set, 33 diagonal union, 89 cardinal, see number diagram, 134 cardinality, 9 direct product, 75 Cartesian product, 4 discrete, 69 categorical, 248 disjoint sum, 75, 278 CH, 10 distance, 43 chain, see graph, see tree distributive, 280 chess, see game Distributive Normal Form, 52 clique, see graph doubter, 93 Incomplete version for students of easllc2012 only. 364 Index Downward Lowenheim–Skolem¨ Theorem, Souslin-formula, 210 114 true, 80 dual, see generalized quantifier universal-existential, 133 duality, 288 fully compact, 342 Dynamic, see game Gale–Stewart Theorem, 27 EC class, 111 game, 16 edge, see graph bijective, 350 Ehrenfeucht–Fra¨ısse´ Game, see game chess, 20 elementary chain, 325 clopen, 26 elementary equivalent, 325 closed, 26 elementary extension, 324 Cub, 85 elementary submodel, 324 determined, 23, 277 empty sequence, 4 Dynamic, 145, 258 enumeration strategy, 100 dynamic, 230 equipollent, 5 Ehrenfeucht–Fra¨ısse,´ 40, 67, 229, 250, 258, equivalence relation, see relation 275–277, 296 even-cardinality, see generalized quantifier Existential Semantic, 132 eventually counting, see generalized quantifier Model Existence, 98, 180, 276, 336 existential formula, see formula Monotone Model Existence, 317 existential quantifier, see generalized Nim, 14 quantifier non-determined, 23 Existential Semantic Game, see game open, 26 expansion, see structure Pebble, 50 explicitly definable, 109 perfect information, 14 expressible, 50 play, 20, 25 extension axiom, 50 position, 21, 26, 66 Positive Semantic, 133 fan, see tree Semantic, 36, 94, 160 FIL, see generalized quantifier Transfinite Dynamic, 258 filter, 59, 177 Universal-Existential Semantic, 132 countably complete, 177 zero-sum, 14 countably incomplete, 124 game logic, see logic generator, 59 game quantifier, see generalized quantifier normal, 130 GCH, 270 principal, 59 generalized Baire space, 270 regular, 238 Generalized Continuum Hypothesis, 270 ultrafilter, 59, 130 generalized quantifier, 291 finite, see set all-but finite, 288 finite context, see context at-least-one-half, 286 finite sequence, 3 atom, 290 finitely consistent, 102 basis of, 289 first-order, see logic BI-PLU, 317 first-order axiomatization, 205 bi-plural, 317 first-order language, 79 bijection closed, 291 Fodor’s Lemma, 130, 131 bounded, 295 forcing, 75, 257, 273, 277, 280 CO-PLU, 317 formula co-plural, 317 absolute, 76 cofinality, 333, 343 atomic, 35 complement, 288 basic, 79 counting, 285 countable like, 327 dual, 288 existential, 132 even-cardinality, 286 Henkin formula, 204, 206 eventually counting, 295 positive formula, 133 existential, 285 quantifier free, 80, 309 FIL, 315 relativization, 110 finiteness, 285 sentence, 80 game, 201 Incomplete version for students of easllc2012 only. Index 365 Hartig,¨ 351 Keisler’s axiom, see axiom Henkin, 204 Kripke–Platek, 76 IDE, 315 lattice, 58 infinite failure, 289 length, 4 infinity, 285 level, 59 MON, 314 lexicographic order, see linear order monotone, 289 limit cardinal, see number most, 287 limit ordinal, see number NON-TRI, 317 Lindstrom’s¨ Theorem, 112, 126 non-trivial, 317 linear order, 57 permutation closed, 291 dense, 57 PLU, 317 inverse, 154 plural, 317 lexicographic, 280 postcomplement, 288 sum, 58 self-dual, 294 well-order, 57 smooth, 343, 345 LO, 333 unbounded, 295 Los´ Lemma, 122 universal, 285 logic weak, 284 first-order, 79 generalized quantifier logic, see logic game logic, 201 generated, see structure generalized quantifier, 307 generator, see filter infinitary, 139 graph, 35 infinite quantifier, 228 chain, 43 Luzin Separation Theorem, 273 clique, 50 Lyndon Interpolation, 136 connected, 43 cycle, 126 metric space, 75 edge, 35 model, 54, see also structure vertex, 35 non-standard, 103, 106 standard, 103 Hartig¨ quantifier, see generalized quantifier Model Existence Game, see game Hamiltonian, 50 Model Existence Theorem, 101, 115, 181, height, 59 318, 336 Henkin formula, see formula MON, see generalized quantifier Henkin quantifier, see generalized quantifier monadic structure, see structure Hintikka set, 98, 319 Monotone Model Existence Game, see game homomorphism, 133 monotone quantifier, see generalized quantifier IDE, see generalized quantifier most, see generalized quantifier ideal, 4 natural number, see number identity function, 3 negation normal form, 91 immediate predecessor, 58 negative occurrence, 136 immediate successor, 58 Nim, see game implicitly definable, 109 NLE, 334 infinitary logic, see logic NNF, 91 infinite, see set non-decreasing, 345 infinite failure, see generalized quantifier non-determined, see game infinite quantifier logic, see logic non-standard, see model Infinite Survival Lemma, 27 NON-TRI, see generalized quantifier infinitely deep language, 276 normal, 130, see filter infinity quantifier, see generalized quantifier number inverse, see linear order cardinal, 9 involution, 205 limit cardinal, 10 isomorphic, 55, 324 limit ordinal, 9 isomorphism, 55 successor cardinal, 10 KA, see axiom successor ordinal, 9 Karp po-set, 261 cardinal number, 9 Karp tree, see tree countable ordinal, 9 Incomplete version for students of easllc2012 only. 366 Index natural number, 3 relational, see structure ordinal, 8 relativization, see formula, see structure rational number, 3 root, see tree real number, 3 SA, see axiom regular cardinal, 10 scattered, 152 singular cardinal, 10 Scott height, 149 uncountable ordinal, 9 Scott Isomorphism Theorem, 167, 276 number of variables, 37 Scott po-set, 262 number triangle, 292 determined, 262 omits, 104 Scott sentence, 166 Omitting Types Theorem, 104, 118, 323 Scott spectrum, 151 orbit, 274 Scott tree, see tree ordered, 57 Scott watershed, 147, 276 ordinal, see number, 58 self-dual, see generalized quantifier ordinal addition, 12 Semantic Game, see game ordinal exponentiation, 13 semantic proof, 102 ordinal multiplication, 12 sentence, see formula partial isomorphism, 63 Separation Theorem, 111, 183, 276 partially isomorphic, 64 set 1 partially ordered, see set 1, 271 1 path, 49 ⇧1, 271 Pebble Game, see game 1 ⌃1, 271 perfect, see set analytic, 271 perfect information , see game bistationary, 130, 277 permutation closed, see generalized quantifier Borel, 273 persistent, see tree closed, 90, 130 play, see game co-analytic, 271 PLU, see generalized quantifier countable, 6 position, see game cub, 90, 131 positive formula, see formula dense, 270 positive occurrence, 136 finite, 4 Positive Semantic Game, see game infinite, 4 postcomplement, see generalized quantifier partially ordered, 58 potential isomorphism, 75, 277, 280 perfect, 34 power set, see set power set, 3 precise extension, 330, 341 stationary, 91, 131 predecessor, 58 transitive, 177 principal, see filter, 104 unbounded, 90, 130 product, 152 uncountable, 6 quantifier free, see formula Shelah’s axiom, see axiom quantifier rank, 37, 80, 309 singular, see number rank, 60 Skolem expansion, 114 rational number, see number Skolem function, 113 real number, see number Skolem Hull, 114 realizes, 104 Skolem Normal Form, 207 recursively saturated, see structure smooth, see generalized quantifier reduced product, see structure Souslin–Kleene Theorem, 272 reduct, see structure Souslin-formula, see formula reflexive, see relation special, see tree regular, see number, see filter stability theory, 242, 275–277 relation standard component, see component anti-symmetric, 58 standard model, see model equivalence relation, 56 stationary, see set reflexive, 56, 58 stationary logic, 343 symmetric, 56 Strategic Balance of Logic, 1, 2, 14, 79, 81, transitive, 56, 58 101, 163, 180, 238, 276, 309 Incomplete version for students of easllc2012 only. Index 367 strategy type, 104 in a position, 22 ultrafilter, see filter of player I, 21, 25, 251 ultraproduct, see structure of player II, 21, 26, 251 unary, see vocabulary, see structure used, 21, 25, 26 unbounded, see set, see generalized quantifier used after a position, 22 uncountable, see set winning, 15, 21, 26, 66, 251 uncountable ordinal, see number winning in a position, 22 Union Lemma, 325, 338 strong -back-and-forth set, see union of a chain, 325, 338 back-and-forth set universal, 273 strong bottleneck, 257 universal quantifier, see generalized quantifier structure, 54 universal Scott tree, see tree -homogeneous, 246 universal-existential formula, see formula -saturated, 247 Universal-Existential Semantic Game, see !-saturated, 129 game expansion, 110 Upward Lowenheim–Skolem¨ Theorem, 117 generated, 63 monadic, 55 Vaught’s Conjecture, 151 recursively saturated, 203 vector space, see structure reduced product, 122 vertex, see graph reduct, 110 vocabulary, 54 relational, 54 binary, 54 relativization, 110 unary, 54 substructure, 62 Weak Compactness Theorem, 338 ultraproduct, 122 Weak Omitting Types Theorem, 338 unary, 55 weak quantifier, see generalized quantifier vector space, 174, 185, 231 well-founded, see tree subformula property, 107 well-order, see linear order substructure, see structure win, 20, 21, 25, 26, 251 successor, 58 winning strategy, see strategy successor cardinal, see number Zermelo–Fraenkel axioms, 10 successor ordinal, see number zero-dimensional, 282 successor structure, 61 zero-sum, see game successor type, 60 Zorn’s Lemma, 12 sum, see linear order Survival Lemma, 22 symmetric, see relation Tarski–Vaught criterion, 113 threshold function, 294 Transfinite Dynamic, see game transitive, see relation, see set tree, 59, 271, 275 Aronszajn, 61 Canary, 277 universal Scott, 274 branch, 60 canonical Karp, 260 chain, 60 fan, 257 Karp, 261, 276 persistent, 257, 276 root, 59 Scott, 262, 276 special, 61 well-founded, 60 tree represenation, 271 true, see formula