3 The Model Existence Theorem

Although we dont have compactness or a useful Completeness Theorem, Henkin-style arguments can still be used in some contexts to build models. In thissection we describe the general framework and give two applications.

Throughout this section we will assume that L is a countable language andthat C is an infinite set of constant symbols of L (though perhaps not all of theconstant symbols of L). The following idea due to Makkai is the key idea. Ittells us exactly what we need to do a Henkin argument.

In the following a basic term is either a constant symbol (not necessarilyfrom C) or a term of the form f(c1, . . . , cn) where f is a function symbol of Land c1, . . . , cn C. A closed term is a term with no variables.

Definition 3.1 A consistency property is a collection of countable sets ofL1,-sentences with the following properties. Let

C0) and if , with and , then {};C1) if , then 6 ;C2) if , then { } ;C3) if

X , then {} for all X ;

C4) ifX , then {} for some X ;

C5) if v(v) , then {(c)} for all c C;C6) if v(v) , then {(c)} for some c C;C7) let t be a basic term and let c, d C,

a) if c = d , then {d = c} ;b) if c = t, (t) , then {(c)} ;c) for some b C, {b = t} .

Lemma 3.2 Let be a consistency property with , c, d, e C.a) There is {c = c} .b) If c = d, d = e , then {c = e} .c) If , , then {} .

Proof a) By C7b) there is b C such that {b = c} . By C7a)1 = {b = c, c = b} . Let (v) be the formula v = c. Since (b) andc = b are in 1, by C7b) = {c = c} . By C0) {c = c} .

b) Let (v) be the formula v = e. Then (d) and c = d . Thus byC7b) {c = e} .

c) Since , by C4) either {} or {} . By C1)the former is impossible.

The next exerecise shows that we really need only verify C1)-C7).

Exercise 3.3 Suppose 0 satisfies C1)-C7). Let

= {} {0 : 0 0 and , finite }.

Prove that is a consistency property.

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Theorem 3.4 (Model Existence Theorem) If is a consistency propertyand , there is a countable M |= .

Proof Let 0, 1, . . . list all F -sentences. We assume that each sentence islisted infinitely often. Let t0, t1, . . . , list all basic terms.

Using the fact that is a consistency property, we build

= 0 1 . . .

such that each i andA) if n {n} , then n n+1, in this case:

a) if n isX , then n+1 for some X ;

b) if n is v(v), then (c) n+1 for some c C;B) c = tn n+1 for some c C.

Let =n=1 n. We will build a model of .

For c, d C we say c d if c = d .

Claim is an equivalence relationLet c C. The sentence c = c is n for some n. By Lemma 3.2, and A)

c = c n+1 . Thus c c.If c = d , then we can find n such that c = d n and d = c is n. Then

by C7) and condition A) d = c .If c = d, d = e , choose n such that c = d, d = e n and c = e is n.

Then by Lemmma 3.2 and A) c = e .

Let [c] denote the -class of c. Let M = {[c] : c C}. We make M into anL-structure.

Let f be an n-ary function symbol of L and let c1, . . . , cn C. By ii) thereis d C such that d = f(c1, . . . , cn) . Suppose d1 = f(c1, . . . , cn) is also in, using C7b) and A) we see that d = d1 and d d1. Also, note that ifc0 = f(c1, . . . , cn) , and di ci for i = 0, . . . , n then,repeatedly using C7b),d0 = f(d1, . . . , dn) . Thus we can define fM :Mn M by

f([c1], . . . , [cn]) = [d] d = f(c1, . . . , cn) .

For c1, . . . , cn and R an n-ary relation symbol of L, we say

RM([c1], . . . , [cn]) R(c1, . . . , cn) .

Again, this does not depend on the choice of ci.

Claim If , then M |= . We need an annoying analysis of terms.We prove this by induction on complexity.5

i) is s = t, where s and t are closed termsWe build a sequence of formulas s1 = t1, . . . , sm = tm where: s1 = s and t1 = t; sMi = s

Mi+1 and t

Mi = t

Mi+1;

5We need to be slightly careful how we define complexity. For example, v(v) is morecomplex than any (c) and

WX is more complicated than for any X.

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si = ti ; sm and tm are constants in C.

If we accomplish i)-iv) then [sm] = [tm] and, by induction, sM = tM.

Given the formula si = ti unless si, ti C, we can find a basic subterm 6 C of one of them, say si, by B) there is c C such that c = . Obtainsi+1 from si by substituting c for and let ti+1 = ti. An easy induction showsthat sMi = s

Mi+1. By C7) and A) si+1 = ti+1 . si+1 = ti+1.

Thus M |= t1 = t2.

ii) is R(t1, . . . , tn)

Exercise 3.5 Prove that M |= R(t1, . . . , tn). [Hint: Similar to case i).]

iii) isX

By C3) and A), for each X . By induction M |= for all X .Thus M |= .

iv) isX

By C4) and A), there is X such that . By induction M |= .Thus M |= .

v) is v (v)By C5) and A), (c) for all c C. By induction M |= (c) for all

c C. Since every element of M is named by a constant, M |= .

vi) is v (v)By A) there is c C such that (c) . By induction, M |= (c). Thus

M |= .

vii) is By C2) and A), . This now breaks into cases depending on .a) is s = t where s and t are closed termsAs in the proof of i) above we can find a sequence of formulas

s1 6= t1, . . . , sm 6= tm

where each of the formulas is in , sMi = sMi+1, t

Mi = t

Mi+1, and sm, tm C.

Thus M |= .

b) is R(t1, . . . , tm)

Exercise 3.6 Prove M |= R(t1, . . . , tm).

c) is Then . My induction M |= and M |= .

d) isX

ThenX and for some X . By induction, M |= .

Thus M |= .

e) isX

ThenX and for all X . By induction M |= for all

X . Thus M |= .

f) is v (v)

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Then v(v) and (c) for all c C. By induction, M |= (c) forall c C. Thus M |= .

g) is v (v).Then v(v) and (c) for some c C. But then M |= (c) and

M |= .

This completes the induction. Thus M |= .

The next exercise gives a useful extension of the Model Existence Theorem.

Exercise 3.7 [Extended Model Existence Theorem] Suppose is a consistencyproperty and T is a countable set of sentences such that for all and T , {} . Then T has a model for all . [Hint: Consider

1 = { T : }.

The Interpolation Theorem

We give two applications of the Model Existence Theorem. Both results werefirst proved by Lopez-Escobar by different means. The first is the L1, versionof Craigs Interpolation Theorem.

Theorem 3.8 Suppose 1 and 2 are L1,-sentences with 1 |= 2. There isan L1,-sentence such that 1 |= , |= 2 and every relation, function andconstant symbol occurring in occurs in both 1 and 2.

Proof Let C be a countably infinite collection of new constant symbols. LetFi be a countable fragment containing i, where every relation and functionsymbol and every constant symbol not in C occurs in i and every formulacontains only finitely many constants from C. Let F = F1 F2.

Let be the collection of finite = 1 2 where i is a set of Fi-sentencesand if 1, 2 are F -sentences such that 1 |= 1 and 2 |= 2, then 1 2 issatisfiable.

Claim is a consistency property.We verify a couple of properties and leave the rest as an exercise.

C3) SupposeX , where = 1 2 as above. Suppose

X 1. Let 1 = 1 {}. We claim that

1 2 . Suppose

1 |= 1

and 2 |= 2. Then 1 |= 1. Hence 1 2 is satisfiable. This is similar ifX 2.

C4) SupposeX 1. Let 1, = 1 {}. We claim that some

1, 2 . Suppose not. Then for each there are 1,, 2, F suchthat 1, |= 1,, 2 |= 2, and 1, 2, is unsatisfiable. Then 1, |= 2,.Since

1 |=

X

,

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1 |=

X

1,.

But2 |=

X

2,

and

X

1, |=

X

2,

contradicting that .

Exercise 3.9 Finish the proof that is a consistency property.

We now finish the proof of the Interpolation Theorem. Since 1 |= 2, bythe Model Existence Theorem, {1,2} 6 . Thus there are 1 and 2 Fsuch that 1 |= 1, 2 |= 2 and 1 2 is unsatisfiable.

Thus1 |= 1, 1 |= 2, and 2 |= 2.

It follows that1 |= 1 and 1 |= 2.

We would be done except 1 may contain constants from C. Let 1 = (c),where (v) is a F -formula with no constants from C. Then

1 |= v (v) and v (v) |= 2.

Thus we can takev (v)

as the interpolant.

Definition 3.10 We say that K is a PC1,-class of L-structures, L an

expansion of L and L1, such that K is the collection of L-reducts ofmodels of .

Exercise 3.11 Show that the following classes are PC1,.a) incomplete dense linear orders;b) dense linear orderings where |(a, b)| = |(c, d)| for all a < b and c < d;c) bipartite graphs;d) free groups and free abelian groups;e) groups with a proper subgroup of finite index;f) ordered fields with a bounded real closed subfield;

Give other natural examples

Exercise 3.12 Suppose K1 and K2 are disjoint PC1,-classes of L-structures.Prove there is L1, such that M |= for M K1 and M |= forM K2.

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Exercise 3.13 For this exercise we consider only countable structures withuniverse N. Suppose L is a countable language, let X be the set of all L-structures with universe N. Topologize X so that the set of models satisfyingany atomic formula is clopen.

We say that a subset of X is invariant if it is closed under isomorphism.a) Prove that {M :M |= } is an invariant Borel set.b) Prove that any PC1,-class of models is an invariant

11-set.

c) (Scott) Prove that every invariant 11-set is PC1,.d) Prove that every invariant Borel set is {M :M |= } for some L1,.

The Undefinability of Well-Ordering

Let L = {

where b1, b1 + b2, . . . , bm1 + bm.

Let b = bs + . Then bs + b and b+ bs+1 as desired.

Exercise 3.16 Complete the proof that is a consistency property.

We give one more application of the Model Existence Theorem.

Exercise 3.17 [Omitting Types Theorem] Let F be a countable fragment andlet T be a satisfiable set of F -sentences. Suppose Xn(v1, . . . , vmn) is a setof F -formulas with free variables from v1, . . . , vn such that for all n and all(v1, . . . , vmn) F such that

T {x (x)}

is satisfiable, then there is Xn such that

T {x ((x) (x))}

is satisfiable. Prove that

T {x

Xn

(x) : n = 1, 2, . . .}

is satisfiable.Why is this called the Omitting Types Theorem?

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