the model existence theorem

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3 The Model Existence Theorem Although we don’t have compactness or a useful Completeness Theorem, Henkin- style arguments can still be used in some contexts to build models. In this section we describe the general framework and give two applications. Throughout this section we will assume that L is a countable language and that C is an infinite set of constant symbols of L (though perhaps not all of the constant symbols of L). The following idea due to Makkai is the key idea. It tells us exactly what we need to do a Henkin argument. In the following a basic term is either a constant symbol (not necessarily from C) or a term of the form f (c 1 ,...,c n ) where f is a function symbol of L and c 1 ,...,c n C.A closed term is a term with no variables. Definition 3.1 A consistency property Σ is a collection of countable sets σ of L ω1-sentences with the following properties. Let σ Σ C0) ∅∈ Σ and if σ,τ Σ with σ τ and φ τ , then σ ∪{φ}; C1) if φ σ, then ¬φ σ; C2) if ¬φ σ, then σ ∪ {∼ φ}∈ Σ; C3) if φX φ σ, then σ ∪{φ}∈ Σ for all φ X ; C4) if φX φ σ, then σ ∪{φ}∈ Σ for some φ X ; C5) if (v) σ, then σ ∪{φ(c)}∈ Σ for all c C; C6) if (v) σ, then σ ∪{φ(c)}∈ Σ for some c C; C7) let t be a basic term and let c,d C, a) if c = d σ, then σ ∪{d = c}∈ Σ; b) if c = t,φ(t) σ, then σ ∪{φ(c)}∈ Σ; c) for some b C, σ ∪{b = t}∈ Σ. Lemma 3.2 Let Σ be a consistency property with σ Σ, c,d,e C. a) There is σ ∪{c = c}∈ τ . b) If c = d,d = e σ, then σ ∪{c = e}∈ Σ. c) If φ,φ ψ σ, then σ ∪{ψ}∈ Σ. Proof a) By C7b) there is b C such that σ ∪{b = c}∈ Σ. By C7a) σ 1 = σ ∪{b = c,c = b}∈ Σ. Let φ(v) be the formula v = c. Since φ(b) and c = b are in σ 1 , by C7b) τ = σ ∪{c = c}∈ Σ. By C0) σ ∪{c = c}∈ Σ. b) Let φ(v) be the formula v = e. Then φ(d) σ and c = d σ. Thus by C7b) σ ∪{c = e}∈ Σ. c) Since ¬φ ψ σ, by C4) either σ ∪ {¬φ}∈ Σ or σ ∪{ψ}∈ Σ. By C1) the former is impossible. The next exerecise shows that we really need only verify C1)-C7). Exercise 3.3 Suppose Σ 0 satisfies C1)-C7). Let Σ= {∅} ∪ {σ 0 Δ: σ 0 Σ 0 and σ ΣΔ σ, Δ finite }. Prove that Σ is a consistency property. 11

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The Model Existence Theorem

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  • 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.

    11

  • 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.

    12

  • 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)

    13

  • 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

    ,

    14

  • 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.

    15

  • 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?

    17