model theory notes

7/30/2019 Model Theory Notes

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Model TheorySummer, 2011

1 Introduction

These notes present some of the common notions and theorems of model theory.We begin with ultraproducts and use them to prove the compactness theorem. They

can also be used to give a characterization of elementary equivalence (the Keisler-Shelahtheorem), but we only consider a simplified version of this.

Then we consider the proof method of adding in witnesses (Henkin constants) and useit to prove compactness, completeness, the omitting types theorem, a characterization ofwhen a countable theory has a countable saturated model, and the Craig interopolationtheorem. Beths definability theorem and the Robinson joint consistency theorem fall out ascorollaries.

Next, we consider elementary chains and show how they can be used to construct ele-

mentary extensions with certain desired properties. We show that we can get -saturatedmodels, and -strongly homogeneous models. Also, we give another proof of the Robinson

joint consistency theorem, and get Craig interpolation as a corollary.We examine the properties of -saturated models beyond just existence in the following

section. We show that they are big in the sense that they are +-universal. In fact, -saturated is equivalent to -universal plus -homogeneous. We introduce atomic models asa kind of opposite of saturated models. We show that (countable) atomic models are smallin the sense that they are prime. A characterization of -categorical theories shows whenatomic and saturated collapse into each other in the countable case. Vaughts Never Twotheorem, on the other hand, shows that if they dont collapse into each other (and there aresuch models) then you can get a model strictly in between them.

Next we consider the notion of back and forth equivalence. Back and forth equivalencegives us a structural way of understanding linguistic equivalence. A version of it can be usedto understand elementary equivalence, while the full version matches L-equivalence. Foreach structure A, it turns out there is actually a sentence A of L such that A and B areback and forth equivalent iff B |= A.

Finally, we turn to Lindstroms theorem which yields first order logic as a maximal logicamong compact, skolem logics.

The texts from which the information here was obtained include: Model Theory by Changand Keisler, Model Theory by Hodges, Chapter III ofModel Theoretic Logics by Flum, BasicModel Theory by Doets, and Mathematical Logic by Monk.

2 Ultraproducts

Well define ultraproducts in a moment, but first we ask the question why we would beinterested in them. Well, for starters, they can be used to prove the compactness theorem.

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Further, they can be used to produce -saturated elementary extensions of models in alanguage of cardinality less than (well consider the case = 1). This line of reasoningcan be used to show a simplified version of the Keisler-Shelah theorem which gives an abstractcharacterization of the expressivity of first order logic. The Keisler-Shelah theorem statesthat two models are elementarily equivalent iff they have isomorphic ultrapowers. A corollary

of this is that a class of models K is an elementary class iffK is closed under ultraproductsand isomorphisms and its complement is closed under ultrapowers. Ultraproducts also figureprominently in the discussion of measurable cardinals, and given the interest in Scottstheorem here this provides further motivation. Finally, ultraproducts may be used to definenotions in non-standard analysis.

2.1 The Definition of Ultraproducts and the Fundamental Theo-

rem

Filters A filter on a boolean algebra B is a subset F B such that

1. 0 F (F is proper)

2. Ifx F and y B and x y, then y F (F is upward-closed)

3. Ifx, y F, then x y F (F is closed under meets)

F is called an ultrafilter if in addition for every x B, either x F or x F. In thecontext of ultraproducts the boolean algebra is a power set P(I) of some set I with theboolean operations interpreted as usual, i.e. 0 := , X Y := X Y, and X := I X.

A principal ultrafilter on P(I) is some ultrafilter U such that there exists d I suchthat X U iff d X. The existence of non-principal ultrafilters follows from the axiom of

choice, as well see in the following theorem. Call a subset G of a boolean algebra consistentif it generates a (proper) filter, i.e. for all finite subsets G0 G we have

G0 = 0.

Theorem 1 (Ultrafilter Lemma). Let G be any consistent subset of a boolean algebra B.Then there exists an ultrafilter U extending G, i.e. G U.

Proof. Let B = b0, b1, . . . , bi, . . . | i be a well-ordering of B. We define by transfiniterecursion a sequence Ui | i of subsets of B by:

1. U0 = G

2. Ui+1 = Ui {bi} if consistentUi {bi} otherwise

3. U =

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there are u1, . . . , un in Ui such that u1 un bi = 0. Then introduce v1, . . . , vmbe in Ui such that v1 vm bi = 0, since Ui {bi} is inconsistent. It follows thatu1 un v1 vm = 0, contradicting the fact that Ui is consistent. Finally, checkingthe limit stages just amounts to observing that a finite amount cant be cofinal. The sameobservation yields that U = i


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Finally, consider an n-ary relation symbol R. We say R

iIMi/UfU1 f

Un iff {i I |

RMif1(i) fn(i)} U. We have to check that this doesnt depend on the choice of represen-tative elements. So suppose f1 g1, . . . , f n gn. Suppose {i I | RMif1(i) fn(i)} U.We show {i I | RMig1(i) gn(i)} U. Note that iff1(i) = g1(i), . . . , f n(i) = gn(i) thenof course RMif1(i) fn(i) RMig1(i) gn(i), so

{i I | f1(i) = g1(i)} {i I | fn(i) = gn(i)}

{i I | RMif1(i) fn(i)} {i I | RMig1(i) gn(i)}

Since all of the sets on the left are in U and U is closed under intersection and upward-closed,we get the set on the right in U as well.

Weve finished defining the ultraproduct, but now we want to prove Loss theorem, alsoknown as the fundamental theorem of ultraproducts. This states that the ultraproductthinks a formula holds iff its componentwise truth is in the ultrafilter. Well need a lemmaabout terms.

Componentwise Term Interpretation We show that if t(x1, . . . , xn) is an L-term, andf1, . . . , f n

iIMi, then

t

iIMi/U(fU1 , . . . , f

Un ) = f

U

where f(i) = tMi(f1(i), . . . , f n(i)). We show this by induction on terms. For variables andconstants there isnt much to show. Let F be an m-ary function symbol, and t1, . . . , tmbe terms whose free variables are among x1, . . . , xn and for which our induction hypothesisholds. We show the equation holds true with t = F(t1, . . . , tm) and f1, . . . , f n

iIMi. As

usual, we have

t

iIMi/U(fU

1

, . . . , f U

n

) = F

iIMi/U(t

iI

Mi/U

1

(fU

1

, . . . , f U

n

), . . . , t

iIMi/U

m (fU

1

, . . . , f U

n

))

By the definition of F

iIMi/U, this simplifes to fU where

f(i) = FMi(g1(i), . . . , gm(i))

where gj are representative elements for the t

iIMi/U

j (fU1 , . . . , f

Un ). By the induction hy-

pothesis, we may choose the gj(i) to actually be tMij (f1(i), . . . , f n(i)), obtaining

f(i) = FMi(tMi1 (f1(i), . . . , f n(i)), . . . , tMim (f1(i), . . . , f n(i)))

I.e.,f(i) = tMi(f1(i), . . . , f n(i))

as desired.

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Loss Theorem Now for Loss theorem. We show by induction on formulas (x1, . . . , xn)that, if f1, . . . , f n

iIMi, then

iI

Mi/U |= (fU1 , . . . , f

Un ) {i I | Mi |= (f1(i), . . . , f n(i))} U

Consider first an equation. Let t1 and t2 be two terms whose free variables are amongx1, . . . , xn, and let t1 = t2.

iIMi/U |= t1 = t2(f

U1 , . . . , f

Un ) iff, by our lemma above,

{i I | tMi1 (f1(i), . . . , f n(i)) = tMi2 (f1(i), . . . , f n(i))} U

as desired.Next consider an atomic formula of the form Rt1 tm where R is an m-ary relation

symbol and the free variables of t1, . . . , tm are among x1, . . . , xn.

iIMi/U |= Rt1 tm(f

U1 , . . . , f

Un )

R

iIMi/U(t

iI

Mi/U

1 (fU1 , . . . , f

Un ), . . . , t

iI

Mi/Um (f

U1 , . . . , f

Un ))

R

iIMi/U(gU1 , . . . , g

Um)

where gk(i) = tMik (f1(i), . . . , f n(i)) for k = 1, . . . , m, by the lemma. By the definition of

R

iIMi/U, this is equivalent to

{i I | RMig1(i) gm(i)} U

I.e.,{i I | Mi |= Rt1 tm(f1(i), . . . , f n(i))} U

as desired.Now we begin with the inductive steps. Suppose the statement is true for (x1, . . . , xn),

and we show it for .iI

Mi/U |= (fU1 , . . . , f

Un )

iI

Mi/U |= (fU1 , . . . , f

Un )

{i I | Mi |= (f1(i), . . . , f n(i))} U

{i I | Mi |= (f1, (i), . . . , f n(i)} U

Now assume the statement is true for 1(x1, . . . , xn) and 2(x1, . . . , xn), and we show it

for 1 2. Now, by the induction hypothesis, we haveiI

Mi/U |= (fU1 , . . . , f

Un )

iff{i I | Mi |= 1(f1(i), . . . , f n(i)} U

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and{i I | Mi |= 2(f1(i), . . . , f n(i)} U

This implies{i I | Mi |= 1 2(f1(i), . . . , f 2(i))} U

since U is closed under intersection, and it is implied by it since U is upward-closed.Finally we consider the case x(x, x1, . . . , xn). First we note that

iI

Mi/U |= (fU1 , . . . , f

Un )

iff there exists an f

iIMi such that

A1(f) := {i I | Mi |= (f(i), f1(i), . . . , f n(i))} U

We want to show that this is equivalent to

A2 := {i I | Mi |= x(x, f1(i), . . . , f n(i))} U

First we show there exists an f such that A1(f) U implies A2 U. For this it sufficesto observe that A2 A1(f) and that U is upward-closed. Next we show that A2 Uimplies that there exists an f such that A1(f) U. For each i I such that Mi |=x(x, f1(i), . . . , f n(i)) choose some element f(i) Mi that oversees this, and select f(i)arbitrarily otherwise. We get A1(f) = A2.

Weve completed showing Loss theorem that

iIMi/U |= (f

U1 , . . . , f

Un ) {i I | Mi |= (f1(i), . . . , f n(i))} U

for every first order formula (x1, . . . , xn).

2.2 Compactness via Ultraproducts

The compactness theorem says that if is a finitely satisfiable collection of sentences, then itis satisfiable. Let I be the collection of all finite subsets of . For each i I, let Mi be a modelsuch that Mi |= i. For each , define to be {i I | i}. Now let E := { | }.We note that E is a consistent subset of P(I) since {1, . . . , n} 1 n for any1, . . . , n . Thus, by the Ultrafilter Lemma, there is an ultrafilter U E. We claimthat iIMi/U |= . Let . We show {i I | Mi |= } U. Note that if i, thenMi |= by stipulation, so {i I | Mi |= } U.

2.3 Saturation via Ultraproducts

Its possible to use ultraproducts to get -saturated elementary extensions of models inlanguages of cardinality < . We consider here only the case = 1. We show specifically

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that if |I| = then there exists an ultrafilter U on I such that for every language L ofcountable cardinality and for every collection of L-models Mi | i I we have

iIMi/U

is 1-saturated. This actually works with any non-principal ultrafilter on I. (However, toget larger saturation, more requirements need to be placed on the ultrafilter and I will belarger.) Then, this result applied to ultrapowers yields the desired elementary extensions.

Recall that M is -saturated means that for every subset C of M with cardinality lessthan , we have that M realizes every type over C. I.e., if (x) is a collection of L(C)-formulas (L(C) is the language L of M with new constants added for every element in C)which is finitely realizable in M(C) (M(C) is the natural expansion of M to L(C)) is alsorealizable in M(C).

To show that for every language L of countable cardinality and for any L-structures Mi(i I),

iIMi/U is 1-saturated, it suffices to show that:

For every language L of countable cardinality and for any L-structures Mi (i I), if (x)is a collection of L-formulas finitely realizable in

iIMi/U, then (x) is realizable in

iIMi/U too.This is actually equivalent, as the reverse implication is obvious (apply the saturation hy-pothesis to C = ). To see that this suffices, let L, Mi (i I), and C

iIMi/U be

given. If (x) is some (consistent) type over C, then it is a collection of formulas in L(C)which is finitely realizable in

iIMi/U(C). Since

iIMi/U(C) =

iIMi(Ci)/U (one

just checks that the new constants are given the same interpretation to verify this by theway, the Mi(Ci) are not uniquely determined), we may apply our assumption to L(C) andthe Mi(Ci) as L(C) is still countable to get our conclusion.

So let L be a countable language, Mi (i I) L-structures, and (x) a collection ofL-formulas finitely realizable in

iIMi/U. Recall U is a non-principal ultrafilter over the

countably infinite I. As such, every we may introduce a sequence

I = I0 I1 I2

of elements of U such that

n In = . As the language is countable, we may enumerate(x) as {1(x), 2(x), . . .}. Then for each n 1 we may define

Xn := In {i I | Mi |= x(1(x) n(x))}

Since the same is true for the In, we have Xn Xn+1 for each n, and

n1 Xn = . Also, since(x) is finitely realizable in

iIMi/U, we have that {i I | Mi |= x(1(x) n(x))}

U and so Xn U for each n 1.

Since

n1 Xn = , for each i I with i

n1 Xn there is a largest natural numbern(i) such that i Xn(i). n(i) is defined to be 0 if i

n1 Xn. Now we define an element

f

iIMi as follows, with the aim to show

iIMi/U |= (fU). Ifi

n1 Xn then we

choose an element f(i) Mi such that Mi |= 1 n(i). If i

n1 Xn then we chooseany element f(i) Mi at all.

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To show

iIMi/U |= (fU) let n be given. We show Xn {i I | Mi |=

n(f(i))} U. Let i Xn. Then n n(i) and Mi |= 1 n(i)(f(i)) by the choice off. Thus, in particular, Mi |= n(f(i)).

Weve completed showing that if U is a non-principal ultrafilter on a countably infiniteset I, then any ultraproduct of some models Mi (i I) in some countable language L will

be 1-saturated. In particular, we can take an ultrapower of some given model M (in acountable language) -many times. We know this will also be 1-saturated. The thing thatwe havent check yet is that MU :=

M/U is actually an elementary extension of M. We

check that the map T: M MU defined by T(m) = f(m)U where f(m)(i) = m for all i isan elementary embedding.

First we check that it is injective. Suppose that m1 and m2 are two distinct elements ofM. Then {i I | f(m1)(i) = f(m2)(i)} = U so T(m1) = T(m2).

Next we check that it is an elementary mapping. Let c be a constant symbol. T(cM) = cU

by the definition of cU. Similarly, if F is a function symbol of arity n and m1, . . . , mn areelements of M, then T(FM(m1, . . . , mn)) = F

U(T(m1), . . . , T (mn)) by the definition of FU.

Finally consider an n-ary relation symbol R. RUT(m1) T(mn) iff M |= Rm1 mn byLoss theorem.

2.4 Keisler-Shelah Theorem

Now, Id like to turn our attention to proving a simplified version of the Keisler-Shelahtheorem. This is a theorem which gives a characterization of elementary equivalence interms of ultrapowers. We will prove here that, under the assumption that 1 = 2

, if Lis a language of countable cardinality, then two L-structures A and B of size at most 1are elementarily equivalent iff there is some ultrafilter U such that AU = BU. The fullKeisler-Shelah theorem, which wont be proved here, is that the assumptions on size may be

dropped. I.e., if L is any language and A and B are elementarily equivalent L-structures,then there exists an ultrafilter U such that AU = BU.A corollary of the full theorem is that a class K of L-models is an elementary class (i.e.,

K is definable by a collection of first order sentences) iff K is closed under ultraproductsand isomorphisms and the complement K of K is closed under ultrapowers. This resultmanages to take a linguistic notion (whether there are certain first order sentences with suchand such properties) and converts it to a structural notion (whether a class of structures isclosed under certain structure-building operations). We might say it helps us understandthe expressivity of first order logic.

Lets see how this follows from the Keisler-Shelah theorem. Of course, if K is elementary,then it is closed under these operations. Let K = Mod . If for each i I, Mi |= , then

iIMi/U |= by Loss theorem. If A = B then A B. Also, K is closed underultrapowers, because M MU also by Loss theorem.

Now assume that K is closed under ultraproducts and isomorphisms, and K is closedunder ultrapowers. Let = { | M K, M |= }. We claim Mod = K. Of course ifM K then M |= . So let M |= . We show M K. We claim it suffices to show thatthere is an M K such that M M. Why? Because then we get, by the Keisler-Shelah

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theorem, an ultrafilter U such that MU = MU. It follows that M K by the closureproperties of K.

So lets find an M K with M M. Let T = Th M. Observe that T is finitelysatisfiable among models ofK. I.e. for every finite subset T0 ofT there is some N K suchthat N |= T0. Otherwise, there is some we get, for some finite subset T0, T0 and soM |=

T0 and M |=

T0.

Now, we may use the same argument as in the ultraproduct proof of the compactnesstheorem. Let I consist of the finite subsets of T. For each i I, we may introduce a modelMi K such that Mi |= i. Now, for each sentence T, we may let := {i I | i}.Then E := { | T} is a consistent subset of P(I) since {1, . . . , n} 1 n.So we may introduce an ultrafilter U E on P(I). Then we claim

iIMi/U |= T. Let

T. Then {i I | Mi |= } and so by Loss theorem

iIMi/U |= .In fact, there is a similar result that holds for modal logic, but it doesnt require the

(difficult) Keisler-Shelah theorem to prove. The result is: K is a modally definable classof pointed models (i.e. there is a collection of modal formulas such that (M, w) |= iff (M, w) K) iff K is closed under ultraproducts and bisimulations and the complementof K is closed under ultrapowers. Of course, being closed under bisimulations is a morestringent condition than being closed under isomorphisms, so this characterization at leastmakes sense in so far as it suggests modal definibility implies first order definability, a factwhich we already observed.

Essentially the same proof works, except we dont need the Keisler-Shelah theorem. Notethat ifK is modally definable, then its closed under ultrproducts and the complement underultrapowers: if (Mi, wi) are models for i I then we take the ultraproduct as normal takingthe wi as constants. We know modal formulas are invariant under bisimulations.

Now suppose K has the stated closure properties. Let := { | (M, w) K, (M, w) |=}. We just need to show that if (M, w) |= , then (M, w) K. It suffices to show that there

is an (M

, w

) K such that (M, w) (M

, w

). Let T be the modal theory of (M, w). Asbefore, T is finitely satisfiable among models in K, and so, using the same argument as before,we get (M, w) |= T with (M, w) K. Further, we may take a non-principal ultrafilter Uon , and consider (M, w) := (M, w)U. This model is -saturated, and so M-saturated.Thus, as we have (M, w) modally equivalent to (M, w), we have (M, w) (M, w).

In order to prove our simplified Keisler-Shelah theorem, we will need the following lemma.

Lemma 2. If A and B are saturated models of the same cardinality, then A = B.

Proof. To prove this, we make use of the lemma below, which states that its enough to finda and b tuples which cover A and B respectively such that (A, a) (B, b). To do this, let ciand di (i < ) be enumerations of A and B respectively where = |A| = |B|. We define aiand bi (i < ) by transfinite recursion.

Suppose that ai and bi have been defined for i < < such that (A, ai(i < )) (B, bi(i < )). Now, if is even, i.e. there is some such that = 2, then we considerthe type (x) of c in A with ai (i < ) as parameters. Since B is -saturated, and < (as cardinals too), we may introduce a b B which realizes (x) with the bi (i < ) usedas parameters. We set a = c. Then (A, ai(i )) (B, bi(i )).

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If is odd, i.e. = 2 + 1 for some , then we may do the same thing in reverse. I.e.we let b = d and we introduce a as some element of A which realizes the same type as bover the parameters already chosen.

Lemma 3. If (A, a) (B, b) and a covers A and b covers B, then there is an isomorphism

between A and B. In fact, we only need to assume that (A, a) and (B,b) agree on all atomicsentences.

Proof. Let T = {(a, b) | a and b are corresponding elements of a and b}. Now we claim thatT is the desired isomorphism. First we claim its a function. Let (a, b1) and (a, b2) be inT with corresponding constant symbols c1 and c2. Then (A, a) |= c1 = c2 and so too does(B, b), i.e. b1 = b2.

To show that T is injective, let (a1, b1), (a2, b2) T with a1 = a2. Then (A, a) |= c1 = c2and so too does (B, b) and so b1 = b2. That T is surjective follows from the assumption thatb covers B. That the domain of T is all ofA follows from the assumption that a covers A.

Preserving constants, functions, and relations is not difficult as well. Let d be a constant

symbol of L. We want to show T(dA

) = dB

. Well, we know (A, a) |= c = d for some c (of c,the constants that correspond to a) and so (B, b) |= c = d and thus dA is mapped to dB.Let f be an n-ary function symbol and let a1, . . . , an A. We would like T(fAa1 an) =

fBT(a1) T(an). Let c be such that (A, a) |= c = f c1 cn. Then cB = fBcB1 cBn .

Let R be an n-ary relation symbol, and let a1, . . . , an A. Then RAcA1 cAn

RBcB1 cBn again by the assumption that (A, a) (B, b).

Theorem 4 (Simplified Keisler-Shelah). LetL be a countable language, and let A and B beL-structures of cardinality at most 1. We show that A B iff there is some ultrafilter Usuch that AU = BU. We assume 2 = 1.

Proof. Suppose A

U = B

U

. Then A A

U

B

U

B by two applications of Loss theorem.Now suppose A B. Let U be any non-principal ultrafilter on P(). Then AU and BU

are both 1-saturated. Lets consider the possible cardinalities of AU and BU. Certainly

|AU| |i

A| = |A| (2) = 2 = 1

and similarly BU 1. Since they are both 1 saturated, neither of them can have cardinality. By Loss theorem we have AU BU and as such either both AU and BU are finite andhence isomorphic or they both have cardinality 1. In this case, they must be isomorphicby the uniqueness of saturated structures lemma above.

3 Henkin-style Proofs

In this section we see how a kind of proof method due to Henkin is useful in a number ofsituations. The idea is to add in a bunch of new constant symbols called witnesses to thelanguage and create a structure whose elements are all named by one of these constants.

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Since every element is named by a constant, its easier to control what kind of propertiesthe structure has. For different theorems we enforce different things.

We begin by re-proving the compactness theorem using Henkin constants (the proof ofthis also yields the Skolem theorem). Then we repeat the argument and prove the complete-ness theorem for first order logic. Then we consider the omitting types theorem, the Craig

interpolation theorem (which yields the Beth definability theorem and the Robinson jointconsistency theorem), and finally we give a characterization of which countable theories havea countable saturated model.

3.1 Compactness

For this subsection, a collection of sentences is consistent means that it is finitely satisfiable.Let be finitely satisfiable (= consistent) collection of formulas in some language L. We

wish to show that is satisfiable. Let L be language that results by adding in |L|-manynew constant symbols C to L. is naturally considered as a collection of formulas in L too.

We may extend to a consistent collection which has witnesses in the sense that forevery formula and variable x we have some c C such that x (c) . To see this,we may list these formulas x and then deal with them one at a time, each time using a cthat hasnt occurred so far.

Further, we may extend to a maximal consistent collection T of L-formulas. I.e., ifU + and U + are both not finitely satisfiable, then we can introduce finite subsets U0and U1 of U such that U0 |= and U1 |= . Since U0 + U1 is still a finite subset of U,we see that U is not finitely satisfiable. T still has witnesses.

Next, we introduce an equivalence relation on the L-terms. We say t1 t2 iffT |= t1 =t2 (i.e. t1 = t2 T). This is indeed an equivalence relation since T is maximal consistent.In fact, its a congruence relation so that we get a well-defined structure M whose elements

are equivalence classes [t] of terms, dM

= [d] for each constant symbol d, fM

[t1] [tn] =[f t1 tn] for each function symbol f of arity n, and RM[t1] [tn] Rt1 tn T for

each relation symbol R of arity n.Let v be the valuation for M such that v(x) = [x] for each variable x. By induction

on terms it follows that v(t) = [t] for all terms t. Then, by induction on formulas, one canprove that M, v |= T for all formulas . To do this, one uses the fact that (in

general) if t is substitutable for x in then M, v |= [t/x] M, vv(t)x |= . Also (in our

situation), we have that for every term t there is a witness c with [t] = [c].

3.2 Completeness

We can also prove completeness using the Henkin construction. Although the compactnesstheorem already establishes that if |= , then there is a finite subset 0 of such that0 |= , still we may like to know that if is recursively enumerable, then { | |= }is recursively enumerable too (where the languages have been recursively presented in somenatural way).

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Our proof system is as follows. We first define a set of formulas according to thefollowing inference rules and axioms. Then we define iff prop where prop isthe usual propositional calculus with the atomic formulas and formulas of the form x asproposition letters. The inference rules (for defining ) are:

1. (Gen) From we may infer x for any variable x.2. (MP) From and we may infer

3. (-intro) If x does not occur free in , then from we may infer x

The axioms are as follows:

1. All propositional tautologies are axioms

2. (-dist) x( ) (x x)

3. (-elim) If t is substitutable for x in , x [t/x]

4. t = t, t1 = t2 t2 = t1, (t1 = t2 t2 = t3) t1 = t3

5. (t1 = t1 tn = t

n) (Rt1 tn Rt

1 t

n)

6. (t1 = t1 tn = tn) f t1 tn = f t

1 t

n

We claim that |= iff . (Recall, of course, that |= iff for every (M, v) |= we also have (M, v) |= .) The soundness direction, that , just amounts to checkingthat the elements of are valid (since the propositional calculus is sound), and to show thiswe just have to check that the axioms are valid and that the inferences preserve validity.This isnt so bad.

We have a deduction theorem for our (i.e. + |= = |= ) and so to showthe completeness direction, it suffices to show that if is consistent (in this subsection, thismeans that ), then is satisfiable. Now, may be extended to a maximal consistentcollection T of L-formulas with witnesses in the sense that for every x there is a newconstant c such that [c/x] x T.

We may form the canonical model M whose elements are the L-terms modulo T itcan be shown this is a well-defined model because of the congruence axioms we assumed.Then, we let v be the canonical valuation and it can be shown by induction that M, v |= iff T. Then M, v |= .

3.3 Skolem Theorem

The Skolem theorem says that if T is a satisfiable theory in a countable language, then ithas a countable model. This follows from the proof of the compactness theorem, becausethe canonical model we form of a consistent (=satisfiable) theory has cardinality at most|L|. (IfT contains just sentences, then we can extend it to a maximal consistent collection

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of L-formulas, and then the canonical model M will have M |= T since truth of sentencesdoesnt depend on the valuation.)

The Lowenheim-Skolem theorem is a generalization of this. It follows from compactness.Let T be an L-theory with infinite models. Then for any |L| there is a model M |= Twith |M| = . To see this, we can first add in -many new constant symbols di and let

T = T {di = dj | i = j}. Then T is consistent because there is an infinite model of T.Thus, using the proof of the compactness theorem, we get a model of T of cardinality atmost |L| + = . Yet, its also clear that any model of T must have size at least . Hence,we get a model ofT of size .

We can also present this theorem in terms of elementary extensions and submodels. If Mis any infinite model, and |M| + |L| then there is an elementary extension N M with|N| = . To see this, let T be the elementary diagram of M. Then T is a consistent theorywith infinite models and is greater than the size of the language of T, |M| + |L|. We mayintroduce N |= T with |N| = as above, and N is the desired elementary extension.

Let M be any infinite L-structure, and let X be a subset of the universe of M. Let be a cardinal such that |M| |L| + |X|. Then there exists an elementary submodelN M such that N contains X and |N| = . To see this, first observe that we may assume|X| = . Now, we define by induction a sequence Xn (n N) of subsets of M, such that foreach n, |Xn| = and Xn Xn+1.

We set X0 := X. Assuming Xn has been defined, we define Xn+1. For every formula(x, y) and tuple b from Xn, if M |= x(x, b), then we add in some such element a withM |= (a, b) to Xn+1. The cardinality of Xn+1 is still as for any formula (x, y1, . . . , ym)there are |X|m = many tuples from Xn of length m, and there are only |L|-many formulasand |L|.

We let N :=

nN Xn. Then N is a substructure of M since we can write down formulasx = c for each constant symbol, and the formulas x = f y1 yn for each function symbol.

N is an elementary substructure, since if b N and M |= x(x, b) then there is an a Nsuch that M |= (a, b).

3.4 Omitting Types Theorem

Another theorem other than compactness and completeness that the Henkin method is usefulfor is the omitting types theorem. A type (x) is said to be consistent with a theory T ifthere exists a model M |= T which realizes . We know by the compactness theorem that atype (x) is consistent with T iff it is finitely realizable in T (i.e. for every finite subset 0of there is some M |= T with M realizing 0). However, just because it is consistent withT doesnt mean that every model of T realizes it. This leads us to the question of when it

is possible to have a model M of T that doesnt realize a consistent type. I.e., when doesevery model M |= T realize a type ? IfM doesnt realize , then we say M omits . Sowe may rephrase the question, when is it possible to omit a consistent type (x).

First we observe that if there is a formula (x) consistent with a complete T such thatT |= for all we call such a formula a complete formula for in T and say

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that is locally realizable, or principal, or supported then every model of T realizes .To see this, note that T |= x(x) as T is complete, and so such elements will realize .

In the case where the language is countable, we can also prove the converse. I.e., thatif is realized in every model of T, then there is a complete formula for . We showthat if there is no complete formula, i.e. the type is unsupported or non-principal or locally

omitable, then there is some model which omits . In fact, we shall prove a somewhatstronger result where we may omit countably many types at the same time and T doesntnecessarily have to be complete.

Theorem 5. LetT be a theory in a countable language L. For each m N let m(x) be anunsupported type (not necessarily complete) in nm variables. Then there is a model M of Twhich omits all the types m.

Proof. The proof is a variation on the Henkin construction. First we add countably manynew constant symbols C to the language L obtaining L+. Then we enumerate all the L+-sentences as {0, 1, . . .}. Our goal is to construct a consistent collection of L+-sentences

T

+

T such that the following conditions hold:1. For each L+-sentence , either or is in T+.

2. For each L+-sentence of the form x(x) T+ there are -many witnesses (newconstants) c such that (c) T+.

3. For each m N and for each tuple c (of length nm) of distinct witnesses there is some(x) m such that (c) T

+.

Suppose we have constructed such a T+. Then let M |= T+ and let N = {cM | c C}.We claim that N M. N is a substructure because we may write down the sentencesx(x = d) for each constant symbol d, and so by the second property of T above we havefor some c C, c = d T. Further, we may write down x(x = f c1 cn) so N is closedunder functions. Its an elementary substructure because if M |= x(x, c), then there is ac N such that M |= (c, c). Thus, we have that N |= T+ as well. Further, N omits eachm. Let m(x) be given and let c be a tuple of elements ofN of the appropriate length. Wemay assume that the c are distinct witnesses since there are infinitely many witnesses c suchthat c = ci T

+. Thus, there is some (x) m such that N |= (c) and so N omits m.So now we go about constructing T+. We define an increasing chain of finite L+-theories

= T0 T1 T2 Tn (n )

such that for each n we have T Tn consistent. Our intention is that n Tn = T+. Wemay express the three desired conditions on T+ stated above as countably many tasks toperform, called (1), (2)x(x) for each formula x(x) of this form, and (3)m for each m N,i.e. each type m.

Now, we partition N into infinitely many infinite subsets. We may thus have one infinitesubset ofN for each task we need to perform. We label these subsets as (1), (2)x(x), and(3)m.

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Supposing that Tn has been defined, we define Tn+1 as follows.Case 1: n (1). Suppose that n is the kth element of (1). IfTTn{k} is consistent, we

set Tn+1 = Tn {k}. Otherwise, T Tn {k} is consistent and we set Tn+1 = Tn {k}.Case 2: n (2)x(x). If x(x) Tn then set Tn+1 = Tn. If x(x) Tn then,

noting that Tn is finite, we may find the first witness c that doesnt occur in Tn and set

Tn+1 = Tn {(c)}. Then T Tn+1 is still consistent since c doesnt occur in T or Tn.Case 3: n (3)m. Since C is countably infinite we may introduce a bijection between

(3)m and the tuples c of distinct witnesses of length nm. Suppose that n corresponds to thetuple c. As Tn is finite, we may write

Tn as a sentence (c, d) where d are the witnesses

that appear in Tn other than those of c, and (x, y) is an L-formula. By assumption, T Tnis consistent, and so T {y(x, y)} is consistent. Since m(x) is unsupported, there isa formula (x) m(x) such that y(x, y) (x) is satisfiable among models of T.Since the constants c and d do not occur in T, we may expand any model satisfying T andy(x, y) (x) appropriately so that it satisfies (c, d) (c). Thus, T Tn {(c)}is consistent and we set Tn+1 = Tn {(c)}.

We have completed defining the Tn. Now we check that T+ :=

n

Tn has the desiredproperties. Case 1 takes care of having T+ for every L+-sentence . Case 2 forcesthere to be infinitely many witnesses for existential formulas, since once we get x(x) ina Tn well meet it again infinitely much. Finally, case 3 makes sure that for every m, everytuple of constants of the appropriate length disagrees with m somewhere.

The assumption that the language is countable is used in the proof in case 3. We needTn to be an actual sentence of L

+ to see that the associated formula is not complete. Inthe case where the language is not countable, a similar proof can go through if we relaxwhat a complete formula can be i.e. we allow it to be a complete collection of formulas however, this approach still will not answer the question which countable types may beomitted in an uncountable theory.

3.5 Craig Interpolation Theorem

In this subsection we prove the Craig interpolation theorem using Henkins witness methodand show how Beths definability theorem and the Robinson joint consistency theorem comeout as corollaries. Later we shall see another way to prove the Robinson joint consistencytheorem and see how it implies Craig interpolation.

Theorem 6. LetA and B be sentences such that A |= B. LetL1 be the language consistingof the symbols which occur in A and let L2 be the language consisting of the symbols whichoccur in B. Then there is a sentence in the language L

0= L

1 L

2such that A |= |= B.

Proof. We suppose that there is no such sentence, and prove that A B is satisfiable, acontradiction. Let L := L1 L2. Let C be a countable collection of new constant symbolsand define L := L + C and similarly L0, L

1, and L

2. For T an L

1 theory and U an L

2

theory, we say that (T, U) are separable if there is some L0-sentence such that T |= andU |= .

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We note that ({A}, {B}) are inseparable. Otherwise, we may write a separating L0-sentence as (c) where (x) is an L0-formula. Then we claim that x(x) is an interpolant forA and B. Since the c do not occur in A, and we know A |= (c), we get by generalization onconstants A |= x(x). Also, we have B |= (c), and so B |= x(x), i.e. x(x) |= B.

We set T0 := {A} and U0 := {B}. List the L1-sentences as (0, 1, . . .) and the L2 sen-

tences as (0, 1, . . .). Now we define a sequence of pairs of theories (T0, U0), (T1, U1), . . . , (Tn, Un), . . .(n ) such that for each n:

1. Tn is a finite collection of L1-sentences and Un is a finite collection of L2-sentences.

2. (Tn, Un) is inseparable.

3. Tn Tn+1 and Un Un+1.

4. If (Tn + n, Un) is inseparable, then n Tn+1

5. If (Tn+1, Un + n) is inseparable, then n Un+1

6. If n = x(x) for some variable x and formula , and n Tn+1, then there is aconstant c C which doesnt occur in Tn, Un, n, or n such that (c) Tn+1.

7. If n = x(x) for some variable x and formula , and n Un+1, then there is aconstant c C which doesnt occur in Tn, Un, n, or n such that (c) Un+1.

The construction is as follows. Assuming (Tn, Un) has been defined, we define (Tn+1, Un+1).If (Tn + n, Un) is inseparable, we put n into Tn+1. If we put n in and its of the formx(x), then we find the first constant c C not occuring in Tn, Un, n, or n (possiblesince Tn and Un are finite) and we add (c) to Tn+1. Next, if (Tn+1, Un + n) is inseparable,we add in n to Un+1, and if we add it in and its of the form x(x) then we find the firstconstant c C not occuring in Tn, Un, n, or n and add (c) to Un+1.

Tn+1 and Un+1 are obviously still finite collections of sentences. In fact, all the propertiesare clear except as to whether we have preserved inseparability. The thing to check is thatwhen we add in (c) (into either Tn or Un) then they remain inseparable. We show that if(T, U) are inseparable and x(x) T, and c is a constant not occurring in T or U, then(T + (c), U) is inseparable. (A similar argument works with respect to U.) Well, supposeto get a contradiction that there is some L0-sentence such that T+ (c) |= and U |= .Note that c may occur in . Introduce a variable y that doesnt occur in or and let (y)denote the L0-formula obtained by replacing any occurrences of c by y. By generalizationon constants, we have T |= y((y) (y)) and U |= y(y). Since T |= y(y), we have

T |= y(y). We see that y(y) separates T and U, a contradiction.So we have the sequence (Tn, Un) with the properties as described. Let T =

n Tnand U =

n Un. Then (T, U) are inseparable too, by the compactness theorem. Next,

we claim that T is a maximal consistent L1-theory and U is a maximal consistent L2-

theory. They are clearly both consistent: otherwise they wouldnt be inseparable. To seethe maximality of T, let n be an L1-sentence. Suppose n T and n T to get a

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contradiction. Since n T, we know by our construction that (Tn+n, Un) is separable, sowe may introduce an L0-sentence such that Tn + n |= and Un |= . Thus, T |= n and U |= . Similarly, as n T, we may introduce an L0-sentence

such thatT |= n and U |= . Thus, T |= and U |= ( ) so that

separate T and U, a contradiction. A similar argument shows that U is maximal. In

detail, suppose U and U. Then we may introduce and such that T |= ,U |= , T |= , and U |= . Thus, T |= while U |= ( ).

Let := T U. We claim is a maximal consistent L0-theory. Its clearly consistent.To see that its maximal, note that if is an L0-sentence then T and U cant disagree onthe truth value of , as theyre inseparable. So either is in both T and U, or is inthem both.

Now, let M1 be a model ofT and M2 be a model ofU. Since we have witnesses for eachexistential sentence by construction, the sets N1 := {cM1 | c C} and N2 := {cM2 | c C}are both elementary submodels ofM1 and M2 respectively. Since both N1 and N2 model ,a complete L0-theory, we see by Lemma 3 that N1 L

0

= N2 L0. Thus, we may expandN1 to become an L-model N such that N L1

= N1 and N L2= N2. Thus, N |= A B,

a contradiction.

For the following corollary we need these definitions. Suppose T is a theory in a languageL and R is a new relation symbol (not occuring in L) and (R) is a collection of L + Rsentences. By (R) we mean the collection ofL + R-sentences obtained by replacing everyoccurence of R by R, another new relation symbol. We say that (R) implicitly defines Rwith respect to T if:

T (R) (R) |= x(Rx Rx)

We say that (R) explicitly defines R with respect to T if there is an L-formula (x) suchthat

T (R) |= x(Rx (x))Another way to say that (R) implicitly defines R with repect to T is that for every modelM |= T, if (M, R) and (M, R) both satisfy (R), then R = R.

Corollary 7 (Beth Definability). (R) implicitly defines R with respect to T iff (R) ex-plicitly defines R with respect to T.

Proof. First note that explicit implies implicit is clear. So suppose (R) implicitly definesR with respect to T. Introduce new constants c. By the compactness theorem, we get that

T0 0(R) 0(R) |= Rc Rc

for some finite subsets T0

and 0

of T and respectively. Thus,

T0 0(R) Rc |= 0(R) Rc

Since R only occurs on the left, and R only on the right, by the Craig interpolation theoremwe may introduce an L-formula (x) such that

T0 0(R) Rc |= (c) |= 0(R) Rc

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Since R and R dont occur in (c) we actually also have (c) |= 0(R) Rc. Thus, we have

T (R) |= (c) Rc

and by generalization on constants we obtain

T (R) |= x(Rx (x))

as desired.

One should take care in not overextending the reach of the Beth definability theorem.There are reasonable notions of implicit/explicit definability in the vicinity which do diverge.For example, let L be a signature, and let R be a new relation symbol (of arity n). SetL = L {R}. Let A be an L-structure. We say that an n-ary relation U An is implicitlydefinable if there is some L-sentence such that

(A, U) |= U = U

On the other hand, we say S is explicitly definable if there is some L-formula (x) such that

A |= (a) Ua

It turns out these two notions are not equivalent. E.g., consider the structure A = (N, S, 0).Then the set of even numbers is implicitly definable but not explicitly definable.

Corollary 8 (Robinson Joint Consistency). Let T be a complete theory in a language L.LetL1 and L2 be two languages such that L1 L2 = L. LetT1 T be a consistent L1 theoryand similarly let T2 T be a consistent L2 theory. Then T1 T2 is consistent.

Proof. Suppose its not. Then let T1 and T2 be finite subsets of T1 and T2 respectivelysuch that T1 |= T

2. Then let be an L-sentence such that T

1 |= |= T

2 by the Craig

interpolation theorem. Then as T is complete, either or is in T. If T, then T2 |= and yet T2 |= , a contradiction. If T, then T1 |= and yet T

1 |= , another

contradiction.

3.6 Existence of Countable Saturated Models

We will establish in the next section that there always exists an elementary extension whichis -saturated. However, in the countable case, we have a necessary and sufficient conditionfor when a countable theory has a countable saturated model (i.e. a model which is countableand -saturated).

Given a theory T, we let Sn(T) denote the set of all complete types (x) consistent withT where x has length n.

Proposition 9. Let T be a complete theory in a countable language L. Then T has acountable saturated model iff |Sn(T)| 0 for each n .

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Proof. Suppose T has a countable saturated model M. Of course M, for each n, has onlycountably many tuples of length n. Since T is complete, every type p ofSn(T) is also a typeconsistent with M (of no parameters) and since M is saturated, it is realized in M by sometuple. Since no one tuple may realize two distinct types, we get that there at most countablymany complete types in Sn(T).

Now suppose that |Sn(T)| 0 for each n . Let C be countably many new constantsymbols. For each finite subset Y = {c1, . . . , cn} of C, a type Y(x) in the language L + Yconsistent with T is the same thing as a type (x1, . . . , xn, x) in the language L consistentwith T. As such, for each finite subset of Y of C, there are only countably many completetypes Y(x) consistent with T. Since there are only countably many finite subsets of C,there are at most countably many (x) in any language L + Y, i.e. for any finite subset Yof C. Thus, we may enumerate such types as

0(x), 1(x), . . .

We may also enumerate all the L + C sentences as

0, 1, . . .

Suppose that we have constructed a sequence of consistent L + C-theories Ti (i ) suchthat

1. T0 = T

2. each Ti has only finitely many new constant symbols (i.e. witnesses from C)

3. ifTi i is consistent, then i Ti+1, otherwise i Ti+1

4. ifi Ti+1 and is of the form y(y), then there is a witness c such that (c) Ti+1

5. if i(x) Ti+1 is consistent, then there is a witness c such that i(c) Ti+1

Then we may let T =

i Ti. T is consistent and has a model M. Then we may let, asusual, N = {cM | c C}. Then N M since we have put in witnesses, so N is a model ofT too. Further, N is countable, since C is countable.

Now we check that N is -saturated. Let Y be a finite subset ofN and let p(x) be a typeconsistent with N with parameters from Y. Then we may enlarge p(x) to some complete type(x) which is consistent with N. Then (x) is also consistent with T and so (x) = i(x)for some i. Since i(x) is consistent with N, it is also consistent with Ti+1, and so we mayintroduce a witness c such that i(c) Ti+1. Thus, c realizes i(x) and hence p(x) in N.

Now, it remains to show that we actually can construct such a sequence of theories. Theusual argument of putting in new witnesses works. E.g., we put first put in i or i intoTi, obtaining T

i (and we also add (c) if required). Then, we ask whether i(x) T

i is

consistent or not. If it is, then we add i(c) to Ti where c is the first witness not in Ti or

i(x), obtaining Ti+1. Otherwise, we set Ti+1 = Ti . Now, if i(x) Ti+1 is consistent, then

we know that i(x) Ti is also consistent, and so i(c) Ti+1 for some c.

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4 Elementary Chains

An elementary chain of models is a (transfinite) sequence of models M ( < ) such thatfor each < < we have M M. The main fact about elementary chains is as follows:

Lemma 10 (Elementary Chain Lemma). If M ( < ) is an elementary chain, then

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We define by transfinite recursion a sequence bi (i < ) of elements of M such that for each , bi (i < ) realizes q. Then our result follows because if bi (i < ) realizes q then itrealizes .

So, suppose first that is a limit ordinal and that bi (i < ) is a sequence ofelements such that bi (i < ) realizes q for each < . We claim bi (i < ) realizes q.

Let A = y

F (x, y) be a formula in q. Let k be the largest index such that xi occursfree in A. Then A is also a formula of qk+1 and k + 1 < . Hence the bi satisfy A.

Now suppose that = + 1. Let bi (i < ) realize q . Since

< , we see that thetype q(b, x) is a type of one variable with fewer than parameters. I claim its consistentwith M. I note that q is closed under intersection in the sense that if A1 and A2 are in q,then there is an A3 q such that M |= A3 (A1 A2). Thus, it suffices to show thatfor each A q(b, x), A is satisfiable in M. Given A = y

F (b, x, y) for some finite

subset F of , we may consider the corresponding A = xy

F (b, x, y) of q(b).

Since b realizes q , we may introduce a b so that (b, b) satisfies A. So q(b, x) is consistentin M, and we can cite Ms saturation to get a b such that (b, b) realizes q.

The following lemma gives us the successor step in our elementary chain construction.

Lemma 12. Let be a cardinal. Let L be a language such that |L| , and M an L-structure such that |M| 2. Then there exists an elementary extensionN M suchthat |N| 2 and N is +-saturated over M, i.e. for every subset X of M with |X| and every type (x) with parameters from X consistent with M, there exists b N realizing(x).

Proof. Since |M| 2 there are at most 2 subsets X of size at most ofM. This is becausefor every such subset X there is a distinct map f: 2 whose range is X, and the number

of such functions is (2) = 2. Further, given an X of size at most , the language LX hassize at most , and so there are at most 2 types (x) with parameters from X consistentwith M. Thus, we may introduce a new constant symbol c for each such type, and well beadding in at most 2-many new constant symbols this way. Let T be the theory consistingof the elementary diagram of M, and (c) for each such type. This theory has size at most2 and it is finitely satisfiable in M. Thus, we may introduce a model N of T. This N isthe desired elementary extension.

Now were ready to apply elementary chains to get the saturation.

Lemma 13. Let|L| and letM be an L-structure with |M| 2. Then there exists

an elementary extension N M such that N is +

-saturated and |N| 2

.

Proof. We define an elementary chain Mi (i < 2) with M0 = M such that, first Mi 2

for each i, and second for all i < , Mi+1 is + saturated over Mi. I.e., if X is a set of

parameters from Mi of size at most , and (x) is a type over X consistent with Mi, then is realized in Mi+1.

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Supposing we have constructed such a chain, let N :=

i y. In the lemma we

claimed that 2 has cofinality greater than . This follows from Konigs theorem.

Theorem 14 (Konig). Let I be a set. For each i I, let Ai and Bi be sets such that|Ai| < |Bi|. Then |

iIAi| < |

iIBi|.

Proof. First we construct an injection g :

iIAi

iIBi, and then we show that therecan be no injection h :

iIBi

iIAi.

For each i I we may introduce an injective function fi : Ai Bi. Since each Ai isstrictly smaller than Bi, we know that none of the fi can be surjective. Thus, for eachi I, we may introduce a yi Bi such that yi is not in the range of fi. Now we defineg : iIAi iIBi. We think ofiIAi as the collection of pairs (i, a) such that i Iand a Ai, while

iIBi is the collection of functions w whose domain is I and for each

i I, w(i) Bi. So, we may set g(i, a) = w where w(j) = a if i = j and w(j) = yj if i = j.Is g injective? Well, let (i1, a1) = (i2, a2) and set g(i1, a1) = w1 and g(i2, a2) = w2. Our goalis to show w1 = w2. It suffices to find a j I such that w1(j) = w2(j). Suppose first thati1 = i2. Then let j = i1. We get w1(j) = fj(a1) and w2(j) = xj. Since xj is not in therange of fj , we see that w1(j) = w2(j). Now suppose that i1 = i2 yet a1 = a2. Then set

j = i1 = i2. We get w1(j) = fj(a1) and w2(j) = fj(a2). Since fj is injective, we get thatw1(j) = w2(j). So g is injective.

Now suppose, to get a contradiction, that h :

iIBi

iIAi is injective. Then foreach i I we may define a partial function fi : Ai Bi as follows. We set

fi := {(a, b) Ai Bi | d iI

Bi(h(d) = (i, a) d(i) = b)}

This is indeed a function because if (a, b1), (a, b2) fi, then we have d1 and d2 such thath(d1) = h(d2) and d1(i) = b1 and d2(i) = b2. The injectivity of h yields b1 = b2. Since|Ai| < |Bi|, none of the fi are surjective. Thus, for each i I, we may introduce a w(i) Bi

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such that w(i) is not in the range of fi. But then h(w) = (i, a) for some i I and a Ai.Thus, fi(a) = w(i), a contradiction.

From Konigs theorem we may deduce some facts about cofinality. Let be any infinitecardinal. Then < cf. To see this, let I = cf. Then we may introduce a sequence of

elements Ai , i I, that are unbounded in . As such, we have |

iIAi| = . Further,cf = |

iI|, and for each i, |Ai| < . Thus, we get

= |iI

Ai| < |iI

| = cf

From this observation we may also conclude that, if is infinite, then < cf 2. To see this,suppose, to get a contradiction, that cf 2. Then

2 < (2)cf 2

(2) = 2

Theorem 15. Let M be an L-structure and a cardinal. Then there is an elementary

extension N M such that N is -saturated.

Proof. If M is finite, then it is already -saturated. To see this, let (x) be a type (withsome number of parameters). If (x) were not realized in M, then for each b M we couldintroduce a formula b(x) (x) such that M |= b(b). Then {b | b M} would be afinite subset of not satisfiable in M, making not consistent with M.

So we may assume that M is infinite. Let be some cardinal such that and|L| and |M| 2. Then we may apply Lemma 13 to obtain an elementary extensionN M such that N is +-saturated. Hence N is -saturated too.

Apparently, the existence of fully saturated elementary extensions cannot be proved in

ZF C, although I havent checked this.

4.2 Existence of -strongly homogeneous elementary extensions

We can also use elementary chains to get -strongly homogeneous elementary extensions. Astructure M is called -strongly homogeneous if for every pair of tuples (a, b) of length lessthan such that (M, a) (M, b), there is an automorphism of M with a = b.

A structure is called -homogeneous if for every pair of tuples (a, b) of length less than of the same type, and for every element c, there is an element d such that (ac, bd) have thesame type too. I.e., we can extend correspondences up by one.

Proposition 16. If M is -strongly homogeneous, then M is -homogeneous.Proof. Given (a, b) of the same type and length less than , we may find an automorphism which takes a to b. Then given any element c M, we may set d = (c). This works.

Lemma 17. IfM is a countable structure, then-homogeneous and-strongly homogeneousamount to the same thing.

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Proof. In light of the previous proposition, we may show that M is -strongly homogeneousassuming it is -homogeneous. So let (a, b) be finite tuples having the same type in M. Weare looking for an automorphism : M M such that a = b. List all the elements of Mas (m1, m2, . . .). Given the element m1, we may find a d1 (using homegeneity) such that,letting c1 = m1, we have

(M, ac1) (M, bd1)Similarly, we may find an element c2 such that, letting d2 = m1,

(M, ac1c2) (M, bd1d2)

We may continue in this way until we have (M, ac) (M, bd) with the c and the d bothcovering M. By (the proof of) Lemma 3, we have our desired automorphism.

Lemma 18. There is a countable -homogeneous elementary extension of any countablestructure A in a countable language.

Proof. First, we claim it suffices to find an elementary chain Ai (i ) with the followingproperties:

1. A0 = A

2. Each Ai is countable.

3. Each Ai+1 is homogeneous over Ai in the sense that for every pair of tuples (a, b) ofthe same finite length and the same type over Ai and for every c Ai there is somed Ai+1 such that (ac, bd) have the same type over Ai+1.

For then we can set N = i Ai and we claim its the desired elementary extension. Itsclearly countable, being the countable union of countably many sets. Further, given (a, b)of finite length with (N, a) (N, b) and any c N, we may find an n such that all of aand b and c occur in An. Then (An, a) (N, a) (N, b) (An, b) by the elementary chainlemma, and so by assumption we get d An+1 such that (An+1, ac) (An+1, bd), and hencethe same is true for N, again by the elementary chain lemma.

So, we turn our attention to constructing such a chain. Supposing Ai has been definedappropriately, we define Ai+1. First of all, we may note that there are only countably manythings of the form (ac, b) where a and b are finite tuples in Ai of the same length and the sametype in Ai, and c is some element of Ai. Let consist of such things. For each (ac, b) we may introduce a constant symbol d(ac,b) and the language remains countable. Let T bethe following collection of sentences (allowing parameters from Ai):

{(ac) (bd(ac,b)) | (ac, b) and (xy) is an L-formula}

We claim that the elementary diagram ofAi together with T is finitely satisfiable. It sufficesto show that T is finitely satisfiable on Ai with the parameters named. Let T0 be a finitesubset of T and let 1, . . . , n be the (xy) that occur in T0 that are relevant for d(ac,b).

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Then let J be the subset of {1, . . . , n} such that j J iff Ai |= j(ac). Then let (xy) bethe formula

jJ

j(xy)

j{1,...,n}J

j(xy)

Then Ai |= y(ay) and so by assumption we have Ai |= y(by) and we may let d(ac,b) besuch a y. Then

Ai |= j(ac) j(bd(ac,b))

for each j {1, . . . , n}. We may deal with each d(ac,b) that occurs in T0 separately in thisway.

We get a model Ai+1 of the elementary diagram of Ai together with T. Since it modelsthe elementary diagram of Ai, we may assume Ai Ai+1. Further, Ai+1 is homogeneousover Ai exactly because it models T.

The above results show that we can always get an -strongly homogeneous elementaryextension of a countable structure in a countable language. The next lemma helps us extend

this result beyond the countable situation.

Lemma 19. Let be a cardinal. Let L be a language with |L| 2 and M an L-structurewith |M| 2. Then there is an elementary extension N M such that |N| 2 andN is +-strongly homogeneous.

Proof. We shall actually make use of a generalization of elementary chain called an expandedelementary chain. An expanded elementary chain is a sequence Mi (i < ) each in its ownlanguage Li such that:

1. Li Lj for all i < j <

2. Mi Mj Li for all i < j <

The elementary chain lemma can then be generalized to the expanded elementary chainlemma, so that if Mi (i < ) is an expanded elementary chain, then M :=

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3. |Mi| 2 for each i < 2

4. For each pair (a, b) of tuples of the same L-type in Mi of length at most , there is abinary relation symbol ab Li+1 such that

Mi+1

|= a

b

is an L-automorphism which sends a to b

5. M =

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To show that T is consistent, we note that we may show that T (L(Mi) {}) isconsistent for each new binary relation symbol . To see that this is sufficient, we use theRobinson joint consistency theorem. Enumerate the new relation symbols i (i < 2

). Wefirst show by induction for all 2 that T := T (L(Mi) {i | i < }) is consistent.Assuming that T is consistent and < 2

, lets show that T+1 is consistent. Well, T

(L(Mi) {i | i < }) and T (L(Mi) {}) have L(Mi) as a their common language,and both extend the complete theory of M in this language. Since both are consistent byassumption, their union T is consistent as well by the joint consistency theorem. Nowsuppose 2 is a limit ordinal such that for each < we have T consistent. Then Tmust be consistent too: otherwise we get a finite subset T of T which is inconsistent andthis T would be contained wholly in some T with < .

Finally, one more application of the joint consistency theorem gives us that

T Li(Mi) T (L(Mi) {i | i < 2}) = T

is consistent.

So we just need to show that for each new relation symbol , T (L(Mi) {}) isconsistent. Let T0 be a finite subset of it. Only finitely many symbols from L occur in T0.Let L0 be the language containing these finitely many symbols of L. T0 also contains onlyfinitely many parameters from Mi. In particular, for only finitely many j do we have aj orbj occuring. Let C be finitely many parameters from Mi which contains all the parametersoccuring in T0, and such that if aj C then bj C as well and vice versa. Let a andb be the finite corresponding subtuples of a and b occurring in L0 = L0 + C. Note thatMi L0 |= (a) (b) for all (x) of L0.

Since |L0| = , we may introduce a countable elementary submodel N of Mi L0. By

the previous lemma, we may introduce a (countable) L0-elementary extension N ofN whichis -strongly homogeneous. Since

(N, a) L0 (N, a) L0 (N, b

) L0 (N, b)

we see that a and b have the same L0-type in N. Thus, we may introduce an automorphism of N sending a to b. Then (N, ) is a model ofT0.

Theorem 20. LetM be any structure. Then there is an elementary extension N of M suchthat N is -strongly homogeneous.

Proof. Suppose that M is finite. We show that M is -strongly homogeneous for any .

Another way to express -strongly homogeneous is to say that

(M, a) (M, b) = (M, a) = (M, b)

But always implies = for finite structures, no matter the size of the language. To see this,suppose that M and N have M N and M is finite. Then N is finite too of the same sizeas M, say n. It follows that there are n! bijections between M and N. Let M = {c1, . . . , cn}.

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If M = N, then for each bijection we could introduce a formula h(x1, . . . , xn) such thatM |= h(c) yet N |= h(hc). It follows that

M |= x

i=j

xi = xj

hh(x)

yet N does not model this sentence. (Given a list d of the n distinct elements of N, we mayintroduce a bijection h which sends c to these elements, and so N |= h(d).)

If M is infinite, then we may introduce a cardinal such that |L| 2, |M| 2

and +. By Lemma 19 we get an elementary extension N of M which is +-stronglyhomogeneous. Since +, we get that N is also -strongly homogeneous.

4.3 Robinson Joint Consistency Theorem, again

Now we reprove the Robinson joint consistency theorem using elementary chains.

Theorem 21. LetT be a complete theory in a language L. LetL1 and L2 be two languagessuch that L1 L2 = L. Let T1 T be a consistent L1 theory and similarly let T2 T be aconsistent L2 theory. Then T1 T2 is consistent.

Proof. First, we may introduce A0 |= T1 and B0 |= T2. Since both A0 and B0 model thecomplete theory T, we have that A0 L B0. As such, the union of the L-elementary diagramsof A0 and B0 is consistent, and we may obtain an L-elementary extension B1 of B0 and anL-elementary embedding f1 : A0 B1.

Next, we note that (A0, a (a A0)) L (B1, f1a (a A0)), and so we may repeat thisprocess to obtain an L-elementary extension A1 of A0 and an L-elementary embeddingg1 : B1 A1 which sends f1a to a for each a A0.

Next, we note that (B1, b (b B1)) L (A1, g1b (b B1)) and so we may obtain an L-elementary extension B2 of B1 and an L-elementary embedding f2 : A1 B2 which sendsg1b to b for all b B1.

Continuing on in this way, we obtain two L-elementary chains An (n ) and Bn (n )together with functions fn, gn (n 1) such that

1. fn+1 : An L Bn+1

2. gn : Bn L An+1

3. gn+1fn+1a = a for all a An

4. fn+1gnb = b for all b BnLet A =

n An and B =

n Bn. Of course An L A and Bn L B for each n by the

elementary chain lemma.Let f :=

n1 fn. We claim that f is an L-isomorphism of A and B. It suffices to

show that the domain of f is A, the range is B, and f preserves L (since equality is inthe language).

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Well, given a A, there is some n such that a An, and since a dom fn+1, we see thata dom f. Given b B, there is some n such that b Bn. Since fn+1gnb = b, we see thatb ran fn+1 and so too for f.

Now consider preservation of L. Let (x) be an L-formula, and let a be from A and bfrom B such that (a, b) f. We want to show that

A |= (a) B |= (b)

Well, first note that for each n we have fn+1 fn+2. In detail: let fn+1a = y andwe try to show that fn+2a = y. Well, we know that gn+1y = gn+1fn+1a = a, and sofn+2a = fn+2gn+1y = y, using our two facts about the fn and the gn. It follows fromfn+1 fn+2 that we may introduce an n such that (a, b) fn+1. Thus,

A |= (a) An |= (a)

Bn+1 |= (b)

B |= (b)

Since A =L B, we may expand A to become an (L1 L2)-structure such that A L2 = B.Thus, T1 T2 is satisfiable.

The Robinson joint consistency theorem has the Craig interpolation theorem as a conse-quence just as we saw the reverse direction is true earlier.

Corollary 22 (Craig Interpolation). LetA and B be sentences such that A |= B. LetL1 bethe language consisting of the symbols which occur in A and letL2 be the language consistingof the symbols which occur in B. Then there is a sentence in the language L0 = L1 L2such that A |= |= B.

Proof. Suppose, to get a contradiction, that there is no such . Let := { L0 | A |= }.By our assumption, we know that {B} is finitely satisfiable. Thus, by compactness,we may introduce a model M of {B}. Let T be the complete L0-theory of M. Ofcourse T2 := T {B} is consistent. We claim further that T1 := T {A} is consistent.Otherwise, there would be a finite subset T0 ofT such that A |=

T0, and so M |=

T0

and M |=

T0. Thus, by the Robinson joint consistency theorem, we may conclude thatT1 T2 is consistent. In particular, we have {A} { B} consistent, a contradiction.

5 Big and Small models

In this section we discuss further some properties of models such as saturated, universal,homogeneous, atomic, prime, and categorical. Intuitively saturated models can be thoughtof as big since any locally realizable elements actually do exist in the structure. Furtherwe see that since saturated models are universal, they are big with respect to other modelstoo. Atomic models may be thought of as the opposite of saturated models as they realizeonly the types they have to. Being prime, (countable) atomic models are also small with

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respect to other models too. We consider existence and uniqueness questions for these variousproperties. Finally, we consider the implications of atomic and saturated coinciding or notcoinciding in the countable case.

A model M of a theory T is said to be -universal with respect to T if every modelN of T of cardinality less than elementarily embeds into M. A model M is said to be

-universal if it is -universal with respect to Th M. A model M which is |M|+-universal iscalled universal.

Proposition 23. LetM be-saturated. Then M is +-universal.

Proof. Let N be elementarily equivalent to M with cardinality at most . We need to showthat N elementarily embeds into M. I.e., we would like to find a tuple a of N and a tupleb of M such that (N, a) (M, b) and a covers N. We construct these tuples by transfiniterecursion. We enumerate all of N in a list, and then take the first element not dealt with,realize a type in M over the elements of N already dealt with, so in this way we get amatching element in M.

We say that a model M is -homogeneous if whenever (a, b) is a pair of tuples of lengthless than from M of the same type, then for every element c M there is an elementd M such that (ac, bd) still have the same type. I.e., we can always extend correspondencesup by one. A model which is |M|-homogeneous is called homogeneous.

Proposition 24. If M is -saturated, then it is also -homogeneous.

Proof. If a and b have the same type, and c is another element, then the type of c over ais obviously realized in M. Thus, the corresponding type over b is consistent with M, andso, as the tuple b has length less than , its realized in M by some element d. This is thedesired element.

Proposition 25. A model M in a language of cardinality smaller than is -saturated iffit is -universal and -homogeneous. If the language has cardinality |L| , then we maystill conclude that M is -saturated iff it is +-universal and -homogeneous.

Proof. We have already seen that -saturated implies -universal and -homogeneous. (Ofcourse +-universal implies -universal.) So let M be some model which is both -universaland -homogeneous. Then let (x) be a consistent (complete) type with less than -manyparameters a. Introduce a model B of size less than such that (B, b) |= (d). (In thecase where |L| = , we get that B has size no more than and so we would need to use+-universal in the next step.) As M is -universal, and M B, we may consider B as anelementary substructure ofM. Since (M, a) (B, b), we actually get (M, a) (M, b). Since

M is -homogeneous, we get a c M such that (M, ac) (M, bd). I.e., c realizes (x) withthe a as parameters.

Corollary 26. Let M be an L-structure. Suppose that |L| |M|. Then M is saturated iffit is universal and homogeneous.

Proof. This follows from the previous proposition by letting = |M|.

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Atomic Models A model M is said to be atomic if for every finite tuple a, the type ofa is principal. I.e., there is a formula (x) consistent with M such that M |= forall (x) in the type of a. We can already note that atomic models are in a certain sensesmall and the opposite of saturated models. To see this, we observe that if T is a completetheory, and M an atomic model of T, then only the elements which have to be in M are in

M; i.e., every tuple a realizes a principal type p(x) ofT, and given any other model N ofT,there must be a tuple b which realizes p (let (x) be a complete formula for p and observeT |= x(x) as its complete).

We are actually typically more interested in countable atomic models, as these are nat-urally smaller. The first question we take up is when a theory T has a countable atomicmodel. A sufficient condition (though not necessary Chang and Keisler give real closedordered fields as a counterexample, though I havent checked this) is that for every n < ,|Sn(T)| 0, assuming the language is countable. A necessary and sufficient condition againunder the assumption that the language is countable is that T is atomic. A theory T beingatomic means that for every formula (x) consistent with T there is some complete formula(x) such that T |= (x) (x). (x) is complete means that for every (x), T |= or T |= but not both.

Proposition 27. Suppose that |Sn(T)| 0 for every n , and T is countable andcomplete. Then T has a countable atomic model.

Proof. Since the number of nonprincipal types is at most countable, by the omitting typestheorem, we may introduce a countable model M which omits all of them. Then any tuplefrom M must realize a principal type, i.e. M is atomic.

Proposition 28. Suppose T is a complete, countable theory. T is atomic iff T has a count-able atomic model.

Proof. Suppose that T is atomic. For each n < , introduce n(x1, . . . , xn) as the set of allthe negations of the complete formulas (x) of T. I.e., if(x) is complete, then n.We claim that n is not principal. If it were, then we could introduce some (x) consistentwith T such that T |= for each complete formula , contradicting that there is somecomplete formula with T |= . So, by the omitting types theorem, we obtain a modelM of T which omits each n. Thus, if a is some tuple of M, then there is some completeformula with M |= (a), and so the type of a is principal.

Now suppose that there is a countable atomic model M of T. Let (x) be a formulaconsistent with T. Thus, T |= x(x) and we may introduce a in M such that M |= (a).Then let (x) be a complete formula for the (complete) type of a. Then T |= .

Now we show that countable atomic models are small in the sense that they are prime. Amodel ofT is said to be prime with respect to T if for every model N of T, M elementarilyembeds into N. A model M is said to be prime if it is prime with respect to Th M.

Proposition 29. If M is a countable atomic model, then it is prime. If the language iscountable, then the reverse is also true.

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Proof. Suppose M is a countable atomic model, and let N M. We hope to find anelementary embedding of M into N. Well, list the elements of M as (a0, a1, . . .). Now wedefine a sequence (b0, b1, . . .) of elements of N such that for each n (M, a0, . . . , an1) (N, b0, . . . , bn1). It follows from this that an bn is an elementary embedding. First of all,its an injective function because equality is a relation symbol (an = am bn = bm).

Secondly, given a formula (x) and elements a of M, then there is an n such that all of aoccurs among a0, . . . , an1, and so we get M |= (a) N |= (fa).

So it suffices to describe such a sequence bn. Suppose that b = b0, . . . , bn1 have beendefined so that (M, a) (N, b) (a = (a0, . . . , an1). Then choose a complete formula (x, y)for the type of (a, an) in M. Then M |= y((a, y)) and so by our inductive assumption,N |= y((b, y)). So let bn be such a y. Then given a formula (x, y),

M |= (a, an) M |= (a, y) (a, y)

N |= (b, y) (b, y)

N |= (b, bn)

Now suppose that M is prime and the language is countable. Of course M must becountable because there are countable models of Th M. Let a be a finite tuple from M. Ifthe type of a is not principal, then by the omitting types theorem, we could omit it in somemodel N of Th M. Yet iff: M N is an elementary embedding, then fa must realize thesame type. Note that since the omitting types theorem gives us a countable model N, weactually only need that M elementarily embeds into every countable model of T.

We show now that countable atomic models are unique.

Proposition 30. IfA andB are countable atomic models of the same complete theory, thenthey are isomorphic.

Proof. List the elements of A as (a0, a1, . . .) and the elements of B as (b0, b1, . . .). We shallconstruct a sequence (c0, c1, . . .) of elements of A and a sequence (d0, d1, . . .) of elements ofB such that

1. The cn cover A and the dn cover B

2. (A, c0, . . . , cn1) (B, c0, . . . , cn1) for each n

It follows from this that A = B by Lemma 3. We define (cn, dn) simultaneously and breakit down into cases whether n is even or odd.

Case 1: n = 2k. Then we set cn = ak. We introduce a complete formula (x, y) for

the type of (c, cn) in A, where c is short for (c0, . . . , cn1). Then A |= y((c, y)) and soby inductive hypothesis, B |= y((d, y)), and we may introduce dn as such a y. As in theprevious proof, since is complete, we get as desired (A, c, cn) (B, d, dn).

Case 2: n = 2k +1. Here we set dn = bk and then find a corresponding cn using the sameprocess as the even case except going in the other direction.

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Atomic and Saturated Weve seen that for a countable, complete theory T, the conditionthat |Sn(T)| 0 is a necessary and sufficient condition for the existence of a countablesaturated model. In this section weve also seen that |Sn(T)| 0 is a sufficient condition(though not necessary) for the existence of a countable atomic model. If we strengthenthis condition a bit more, we get a necessary and sufficient condition for the existence of

a countable atomic and saturated model. The condition is |Sn(T)| < 0 (for each n < ).This condition is actually equivalent to a couple of other things of independent interest.

Theorem 31. LetT be a complete theory in a countable language. Then the following areequivalent:

1. T is -categorical (i.e. all countable models of T are isomorphic).

2. T has a countable atomic and saturated model.

3. For eachn < , every type p(x1, . . . , xn) consistent with T is principal.

4. For each n < , there are only finitely many complete types p(x1, . . . , xn) consistentwith T.

5. For eachn < , there are only finitely many formulas (x1, . . . , xn) up to equivalencemodulo T.

6. All models of T are atomic.

Proof. Suppose that T is -categorical. We prove that T has a countable atomic and sat-urated model. Well, let M be the unique (up to isomorphism) model of T of countablesize. Then M elementarily embeds into every countable model of T (via the identity map),and by the proof of Proposition 29 we get that M is atomic. Further, as theres only onecountable model, there are at most countably many types consistent with T, and so there isa countable saturated model, and this must be M.

Now suppose that T has a countable atomic and saturated model M. We prove that foreach n < , every type p(x1, . . . , xn) consistent with T is principal. Well, as M is saturated,we may introduce a which realizes p. Then, since M is atomic, we may introduce a completeformula (x) for the type of a, and this is a complete formula for p too.

Now suppose that for each n < , every type p(x1, . . . , xn) consistent with T is principal.We prove that for each n < there are only finitely many complete types p(x1, . . . , xn)consistent with T. Let n(x1, . . . , xn) consist of all the negations of complete formulas(x1, . . . , xn) of T. We claim that n is inconsistent with T. If not, then it could be

extended to a complete type p(x) consistent with T, and T being principal gives a completeformula for p which is also a complete formula of T. Hence p, a contradiction.So n is inconsistent with T, so we may introduce a finite subset {1, . . . , k} n which isalso inconsistent with T. Hence T |= 1 k and so ifp(x) is a complete type consistentwith T it must contain one of 1, . . . , k, and since the i are complete formulas, there areat most k complete types on x.

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Now suppose that for each n < there are only finitely many complete typesp(x1, . . . , xn)consistent with T. We prove that for each n < there are only finite many formulas(x1, . . . , xn) up to equivalence modulo T. For each formula (x) introduce the set whichcontains the complete types containing . We claim that T |= iff = . Firstsuppose T |= . Then p for any complete type p consistent with T and so

p p. Now suppose = . If T |= , then, without loss of generality,we have consistent with T, and this may be extended to a complete type p. Then pcontains but not . Thus, is an injective function on formulas (on x) modulo T to thepowerset of the set of complete types (on x). Since there are finitely many complete types,the powerset is also finite, and so the number of formulas modulo T is also finite.

Now suppose that for each n < there are only finite many formulas (x1, . . . , xn)modulo T. We prove that all models of T are atomic. Let M be a model of T. Let a bea finite tuple from M. Let 1(x), . . . , m(x) be the finitely many formulas modulo T. LetJ {1, . . . , m} such that i J iff M |= i(a). Then let

(x) := iJ

i(x)

We claim that is complete for the type of a. Let (x) be in the type of a. Then letj {1, . . . , m} such that T |= (x) j(x). In fact, j J. Thus T |= . isconsistent with T because M |= (a).

Now suppose that all models of T are atomic. We prove that T is -categorical. Well,weve already seen that all countable atomic models are isomorphic.

If a countable complete theory T has a countable atomic model, and a countable saturatedmodel, and were not in the case of the preceding theorem where the two are identical, thenin fact we must have a third model. This shows that a countable complete theory can never

have exactly two countable models up to isomorphism.

Theorem 32 (Vaughts Never Two). A countable complete theory T can never have exactlytwo countable models up to isomorphism.

Proof. Suppose, to get a contradiction, that it did. Then there are only countably manycomplete types, and so there is a countable atomic model A and a countable saturated modelB. Now, by the previous theorem, B cannot be atomic, because otherwise there would onlybe one countable model of T. It follows that A = B and so these are the only two countablemodels of T. However, we will construct a countable model which is neither atomic norsaturated.

Since B is not atomic, we may introduce a finite tuple b in B whose type is not principal.We claim that (B, b) is saturated. Well, if C is a finite collection of parameters from (B, b)and (x) is a type consistent with (B, b) then we can also think of (x) as a type consistentwith B with parameters C b. Thus, as B is saturated, (x) is realized in B by someelement s, and this s also realizes as a (B, b)-type. Thus the theory T = Th(B, b) hasa countable saturated model, and so for each n < we have |Sn(T)| 0 and so T has a

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countable atomic model (C, c). Of course the reduct C in the original language is a modelof T. We claim that C is neither atomic nor saturated (thus obtaining a contradiction).

C is not atomic because the type of c is not principal: if (x) were a complete formulafor the type of c in C, then would also be a complete formula for the type of b in B. Indetail: let B |= (b). Then C |= (c) and so C |= , and so B |= .

C is not saturated. As above, it suffices to show that (C, c) is not saturated. BecauseT is not -categorical, by the previous theorem we may introduce an n such that thereare infinitely many formulas (x1, . . . , xn) modulo T. If1(x) and 2(x) are formulas in theoriginal language, then

T |= 1 2 B |= 1 2

(B, b) |= 1 2

T |= 1 2

Thus, if1 and 2 are unequal modulo T, they are also unequal modulo T. Thus, there areinfinitely many formulas (x) modulo T. Thus, T does not have a countable atomic andsaturated model. Thus, (C, c) is not saturated, since it is countable atomic.

6 Back and Forth Equivalence

First we give a notion of back and forth equivalence which matches elementary equivalencefor finite signatures. Then we give a notion of back and forth equivalence which matchesequivalence of the more expressive language L.

6.1 Elementary Equivalence

Dealing With Functions First let me make a preliminary remark about function sym-bols. To make what were about to do work properly, we need some way of dealing withfunction symbols, because they mess things up. The problem is that we will want to be ableto say (in a finite way) that a given finite part of a structure matches a given finite partof another structure up to a certain quantifier depth of formulas. But if theres functionsymbols around, then this finite part could easily become an infinite part by repeated appli-cation of functions. One way to deal with the functions is just to assume that theyre not inthe signature. Another way is to only deal with unnested formulas, which essentially makesfunction symbols into relation symbols, because it doesnt allow us to have function symbolsinside other function symbols. However, since every formula is equivalent to an unnested

formula (though typically with raised quantifier rank), in the end this approach makes goodsense.

Unnested Formulas An unnested atomic formula of L in the free variables x is one ofthe following types of formulas:

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1. f(y) = z where the y and z are some selection of variables from x or constants fromL, and repetitions are ok

2. Ry where the y are some selection of variables from x or constants from L, and repe-titions are ok (note also that R may be the equality symbol)

E.g., f(c,x,y) = x is an unnested atomic formula in the variables x, y, Rc1c2 is an unnestedatomic formula in no free variables, and f(g(x, y), z) = y is not an unnested atomic formula.A formula is said to be unnested if each atomic subformula is unnested. Note that if thereare no function symbols in the language (though constants could be around), then everyformula is unnested.

Lemma 33. Every formula (x) is equivalent to an unnested formula (x).

Proof. First we can prove by induction on terms t, that for every term t, and for everyvariable x not occuring in t, there is an unnested formula w(t, x) such that x = t w(t, x)is valid. We also want to make sure that if z is occurs t, then it occurs free in w(t, x), and

if z does not occur in t, then it does not occur free in w(t, x), unless it is x.If t is a variable or constant d, then we define w(t, x) to be x = d. If t = f(t1, . . . , tn)

where we already have the result for t1, . . . , tn, then, given x, we may introduce distinctvariables y1, . . . , yn not occuring in t or x. Then we define w(t, x) to be

y(x = f(y) ni=1

w(ti, yi))

Does this work? Well, by inductive hypothesis, M, v |= yi = ti w(ti, yi) for any valuationv and for any i. Suppose first that M, v |= x = f(t1, . . . , tn). Then as the yi are all distinct

from x, we have M, v

v(ti)

yi |= x = f(y). Also, as the yi dont occur in the ti, we have for eachi, M, v

v(ti)yi |= yi = ti, and applying the inductive hypothesis yields M, v

v(ti)yi |= w(ti, yi). So

we see that

M, vv(ti)yi |= f(y) = x ni=1

w(ti, yi)

I.e.,

M, v |= y(f(y) = x ni=1

w(ti, yi))

Now assume that M, v |= w(t, x). Then we may introduce d1, . . . , dn M such that

M, v

di

yi |= f(y) = x and M, v

di

yi |= w(ti, yi) for each i. By the inductive hypothesis, we getthat M, vdiyi |= yi = ti. It follows that M, vdiyi

|= f(t) = x, and since the yi dont occur in t orx, we get M, v |=

model theory notes

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