A Primer on Infinitary Logic

Fall 2007

These lectures are a brief survey of some elements of the model theory ofthe infinitary logics L, and L1,. It is intended to serve as an introductionto Baldwins Categoricity.

1 Basic Definitions

Let L be a language. In the logic L, we build formulas using the symbols ofL, =, connectives ,

and

, quantifiers and , and variables {v : < }.1

We define formulas inductively with the usual formation rules for terms,atomic formulas, negation, and quantifiers and the following rule for

and

.

If X is a set of L, formulas with |X | < , then

X

and

X

are L, formulas.Satisfaction for L,-formulas is defined inductively as usual. If M is an

L-structure and is an assignment of variables, then

M |=

X

M |= for all X

andM |=

X

M |= for some X.

For notational symplicity, we use the symbols and for binary conjunctionand disjunctions.

We can inductively define the notions of subformula, free variable, sentence,theory and satisfiability in the usual way.

Exercise 1.1 If is an L,-sentence and is a subformula of then hasonly finitely many free variables.

We say M, N ifM |= N |=

1For historic reasons we write L1, instead of the more notationally consistent L1, .

1

for all L, sentences .

Definition 1.2 Let is an L,-formula if is an L,-formula for someregular cardinal .

We define M, N in the obvious way.

Exercise 1.3 Give an example of an set T of L1,-sentences such that everyfinite subset is satisfiable but T is not.

Thus the Compactness Theorem fails in infintary languages.2

Exercise 1.4 Show that the following classes are L1,-axiomatizable for ap-propriate languages L.

i) torsion abelian groups;ii) finitely generated fields;iii) linear orders isomorphic to (Z,

Definition 1.8 We say that a set of L,-formulas F is a fragment if there isan infinite set of variables V such that if F , then all variables occuring in are in V and F satisfies the following closure properties:4

i) all atomic formulas using only the constant symbols of L and variablesfrom V are in F ;

ii) F is closed under subformula;iii) If F , v is free in , and t is a term where every variable is in V , then

the formula obtained by substituting all free occurences of v with t is in F ;iv) F is closed under ;v) F is closed under , , , v, and v for v V .

Exercise 1.9 a) Suppose is regular (in particular this holds for L1,). If Tis a set of L,-sentences with |T | < , then there is F a fragment of L, suchthat T F and |F | < .

b) Show that a) may fail if is singular.

We writeMF N andMF N for elementary equivalence and elementarysubmodels with respect to formulas in F .

Theorem 1.10 (Downward Lowenheim-Skolem) Let F be a fragment ofL, such that any formula F has at most finitely many free variables.Let M be an L-stucture with X M . There is N F M with X N and|N | = max(|F|, |X |).

In particular if F is a countable fragment of L1,, then every L-structurehas a countable F-elementary submodel.

The proof is a simple generalization of the proof in first-order logic. It isoutlined in the following Execise.

Exercise 1.11 a) Prove there is L L and F F and M an L-expansionof M such that |L|, |F| = |F| and for each F-formula (v, w) with freevariables from v1, . . . , vn, w, there is an n-ary function symbol f such that

M |= v (w(v, w) (v, f(v))

b) Prove that if N is an L-substructure of M, then N F M.c) Prove that there is an L-substructure N of M with X N and |N | =

max(|F|, |X |).

Exercise 1.12 For L and F define the appropriate notion of built in Skolemfunctions. Prove that for any L and F and F -theory T there are L L,F F and T T with |F| = |T | = |F| where T has built in Skolemfunctions, and every M |= T has an expansion M |= T .

Exercise 1.13 In first order logic, the Upward Lowenheim-Skolem is an easyconsequence of Compactness. In infinitary logics it is generally false. Let L ={U, S,E, c1, . . . , cn, . . .} and let be the conjunction of

4We will usually assume v1, v2, . . . , V .

3

i) x U(x) S(x),ii) xy (E(x, y) U(x) S(y));iii) ci 6= cj , for i 6= j;iv) U(ci) for all i,v) yz ([S(y) S(z) x ((E(x, y) E(x, z))] y = z)vi) x(U(x)

i=1 x = ci.

Prove that every model of has size at most 20 .

L1, and omitting first order types

For countable L1,-theories we can find an equivalent class of models of a firstorder theory omitting a type.

Theorem 1.14 Let T be a countable set of L1,-sentences. There is a count-able language L a first order theory T and a set of types such that: i) ifM |= T and omits all types in , then the L-reduct of M is a model of T ;

ii) every model of T has an L-expansion, that is a model of T omitting alltypes in .

Proof We expand L and F so that for each formula in F with free variablesfrom v1, . . . , vn we have an n-ary relation symbol R we add to T

a sentencesi) if is atomic, add

v R

ii) if is , addv R R

iii) if is

X , addv R R

for all X and let be the type

{R} {R : X}

iv) if is

X, add

v R R

for all in X and let be the type

{R} {R : X}

v) if is w, then add

v R wR.

Let be the collection of types described above.

Exercise 1.15 a) Suppose M |= T and M omits every type in . Prove that

M |= v R

4

for all F.Conclude that the L-reduct of a model of T is a model of T .b) Prove that everyM |= T has an expansion that is a model of T omitting

all types in by interpreting R as .

L,

For completeness, we mention the the logics L,. In these logics we allow amore general quantification rule.

Suppose is a formula and ~v is a sequence of variables freely occuring in with |~v| < , then

~v

is a formula.Thus we are allowed existential and, by taking negations in the usual way,

universal quantification over sequences of variables of cardinality less than .

Exercise 1.16 Show there is L1,1 such that M |= if and only if|M | > 0.

Exercise 1.17 Show there is an L1,1-sentence such that (A,

Theorem 2.4 The following are equivalent:i) M, N ;ii) M=p N ;iii) There is partial isomorphism system between M and N where every

every p P has finite domain;iv) Player II has a winning strategy in G(M,N ).

Proofiii) ii) is clear.

ii) i) Let P be a system of partial isomorphisms. We prove that for all(v1, . . . , vn) L,, f P and a1, . . . , an dom(f), then

M |= (a) N |= (f(a)).

We prove this by inducion on formulas. This is clear for atomic formulasand the induction step is obvious for ,

and

. Suppose (v) is w(v, w).

Suppose M |= w(a, c). There is g P with g f and c dom(P ). Byinduction, N |= (f((a)), f(c)), so N |= (f(a)).

On the other hand, if N |= (f(a), d). There is g P with g f andc dom(g) such that g(c) = d. By induction, M |= (a, c). Thus M |= (a).

This proves the claim. In particular, if is a sentence M |= N |= .

iv) iii) Let be a winning strategy for Player II. Let P be the set of allmaps f(ai) = bi where a1, . . . , an, b1, . . . , bn are the results of some play of thegame where at each stage Player I has played either an or bn and Player II hasresponded using . Since is a winning strategy for Player II, each such f is apartial L-embedding. Since Player I can at any stage play any element from Mor N , P satisfies i) and ii) in the definition of a partial isomorphism system.

i) iv) We need one fact.

Claim Suppose (M, a) , (N , b) and c M , there is d N such that

(M, a, c) , (N , b, d).

Suppose not. Then for all d N there is d such that M |= d((a, c) andN |= d(b, d). But that

M |= v

dN

d(a, v)

andN 6|= v

dN

d(b, v)

a contradicion.

We now describe Player IIs strategy. Player II always has a play to ensure(M, a1, . . . , an) , (N , b1, . . . , bn). As long as Player II does this the resultingmap will be a partial L-embedding.

6

Exercise 2.5 Prove that M, N if and only if there is a forcing extensionV[G] of the universe where M= N .

Exercise 2.6 We say that M=p N if and only if there is a system

P0 P1 . . . P

if and only if for + 1 i) for any f P+1 and a M there is g P with g f and a dom(g);ii) for any f P+1 and b N there is g P with g f and a img(g).Prove thatM=p N if and only ifM |= N |= for all with quantifer

rank less at most .

Exercise 2.7 We say that M=p N if there is a system of partial L-embedingsP such that if f P and C M with |C| < then there is g f with g Pand C dom(g) and if D N and |D| < there is h f with h P andD img(h).

Prove that M=p N if and only if M, N .

Exercise 2.8 Prove that any two 1-like DLO are L,1-equivalent. Hint: Let(A,

where X = { : M |= (a) and is atomic or the negation of an atomicL-formula}. If is a limit ordinal, then

Ma,(v) =

Proof Let = {(a, b) : a, b M l for some l = 0, 1, . . . and (M, a) 6 (M, b)}.Clearly, for < .

Claim 1 If = +1, then = for all > .We prove this by induction on . If is a limit ordinal and the claim holds

for all < , then it also holds for . Suppose that the claim is true for > and we want to show that it holds for +1. Suppose that (M, a) (M, b) andc M . Because (M, a) +1 (M, b), there is d N such that (M, a, c) (M, b, d). By our inductive assumption, (M, a, c) (M, b, d). Similarly, ifd M , then c M such that (M, a, c) (M, b, d). Thus, (M, a) +1 (M, b)as desired.

Claim 2 There is an ordinal < |M |+ such that = +1.Suppose not. Then, for each < |M|+, choose (a, b) +1\. Because

for < , the function 7 (a, b) is one-to-one. Because there areonly |M | finite sequences from M this is impossible.

We conclude this section with Scotts Isomorphism Theorem that everycountable L-structure is described up to isomorphism by a single L1,-sentence.

Let M be an infinite L-structure of cardinality , and let be the Scottrank of M. Let M be the sentence

M,

l=0

aMl

v(Ma,(v) Ma,+1(v)).

Because all of the conjunctions and disjunctions in Ma, are of size , Ma,

L+, for all ordinals < +. Thus M is an L+,-sentence. We call

M theScott sentence of M. If M is countable, then M L1,.

Theorem 2.11 (Scotts Isomorphism Theorem) LetM and N be a count-able L-structures, and let M L1, be the Scott sentence of M. Then,N = M if and only if N |= M.

Proof Because is the Scott rank of M, M |= M. An easy inductionleft to the exercises shows that if N = M, then M and N model the sameL,-sentences.

On the other hand, suppose that N models M. We do a back-and-forthargument to build a sequence of finite partial embeddings f0 f1 . . . fromM to N such that if a is the domain of fi, then

(M, a) (N , fi(a)). ()

Let m0,m1, . . . list M and n0, n1, . . . list N .At stage 0, we let f0 = . Because N |= M,, M N and () holds.Suppose we are at stage n + 1. Let a be the domain of fn. Because

(M, a) (N , f(a)), N |= Ma,(f(a)). Because N |= M, N |= Ma,+1(f(a))

and (M, a) +1 (N , f(a)).If n + 1 = 2i + 1, we want to ensure that mi is in the domain of fn+1. If

mi is in the domain of fn, then fn = fn+1. If not, choose b N such that(M, a,mi) (N , f(a), b) and extend fn to fn+1 by sending mi to b.

9

If n = 2i + 2, we want to ensure that ni is in the image of fn+1. If it isalready in the image of fn, let fn+1 = fn. Otherwise, we can find m M suchthat (M, a,m) (N , f(a), ni) and extend fn to fn+1 by simply sending m toni.

Corollary 2.12 Let be a satisfiable sentence of L1,. The following areequivalent:

a) |= or |= for any L1,;b) |= or |= for any L,;c) is 0-categorical.

If these equivalent conditions hold we say that is complete.

Proof b) a) is clear.c) b) by Scotts Theorem.a) c) by Downward Lowenheim-Skolem.

Exercise 2.13 Give an example of an 1-categorical L1,-sentence that is

References

[1] J. Barwise, Admissible Sets and Structures, Springer-Verlag 1975.

[2] J. Barwise, Back and forth through infinitary logics, Studies in Model The-ory, M. Morley ed., MAA 1973.

[3] J. Barwise and S. Fefferman ed., Handbook of Model Theoretic Logics,Springer-Verlag 1985.

[4] J. Baumgartner, The Hanf number for complete L1, sentences (withoutGCH). J. Symbolic Logic 39 (1974), 575578.

[5] H. J. Keisler,Model Theory of Infinitary Languages, North-Holland 1971.

[6] D. Kueker, Back-and-forth arguments in infinitary languages, InfinitaryLogic: In Memoriam Carol Karp, D. Kueker ed., Lecture Notes in Math72, Springer-Verlag 1975.

[7] G. Hjorth, Knights model, its automorphism group, and characterizing theuncountable cardinals. J. Math. Log. 2 (2002), no. 1, 113144.

[8] D. Marker, Model Theory: an Introduction

[9] M. Nadel, L1, and admissible fragments, in [3]

10