5 Counterexamples to Vaughts Conjecture

A sentence L1, is a counterexample to Vaughts Conjecture, or simply acounterexample, if

0 < I(,0) < 20 ,

i.e., has uncountably many countable models but fewer than continuum manymodels.

Fix a counterexample.A key fact about counterexamples is that there are few types for any count-

able fragment. Let F be a fragment. Let

SF() = {tpMF (a) :M |= , a M

n for some n}

be the set of complete F -types realized in models of . We say that a sentence is scattered if |SF()| 0 for every countable fragment F with F .

Lemma 5.1 If is a counterexample, then is scattered.

Proof Fix F a countable fragment. Then

SF () = {p : M = (N, . . .),M |= , a Mn tpM(a) = p}

is a 11-set (for details see the proof of Theorem 4.4.12 in [10]) . Thus, if it isuncountable, it has cardinality 20 (see for example [5] 14.13) But this is onlypossible if I(,0) = 20 .

Countable Models

We begin by showing a counterexample has exactly 1 countable models. Theproof is exactly the same as the proof for first order theories in S4.4 of [10].

Theorem 5.2 (Morley) If is scattered, then I(,0) 1. Thus if is acounterexample, I(T,0) = 1.

Proof The proof uses an analysis of models is similar to Scotts analysis (butfaster). We build a sequence of countable fragments (F : < 1) such that

F0, if p SF(), then

p

F+1 and F =

Suppose tpN (a) = tpN (b). Call this type p. Suppose (v) F+1 such that

N |= (a) (b). Then

N |= vw[(

p

((v) (w))) (v) (w)

].

Since this sentence is in F+1 is also true inM, contradicting thatM has height.

Let P = {a 7 b : tpM (a) = tpN (b)}. We claim that P is a back-and-forth

system.Suppose tpM (a) = tp

N (b) and c M . Let p(v, w) be the F-type of (a, c).

ThenM |= v, w

p

(v, w).

As this is an F+1-formula, it is true in N . Thus there is (b, d) N realizing

p. Since p extends tpM (a), we must have tpN (b) = tp

N (b

). Since N has height

, tpN+1(b) = tpN+1(b

). But

N |= w

p

(b, w).

As this is an F+1-formula, there is d N such that (b, d) realizes p. Thusa, c 7 b, d P . The other direction is similar.

Thus if M F+1 N are countable models of height they are isomorphic.If M is a countable model of is determined up to isomorphism by its height and Th+1(M) = tpM+1(). As there are at most 1 choices for and, given only countably many choices for its F+1-theory, I(,0) 1.

Our next goal is to show that every counterexample has an uncountablemodel. The proof uses the idea of minimal counterexamples introduced byHarnik and Makkai.

Minimal Counterexamples

Definition 5.3 Suppose is a counterexample. We say that is a minimalcounterexample if for every sentence L1,, either or has atmost countably many countable models.

Lemma 5.4 (Harnik-Makkai) If is a counterexample, then there is a min-imal counterexample with |= .

ProofFix a counterexample to Vaughts Conjecture and suppose there is no min-

imal with |= . The basic idea is that we will build a tree of counterexamples( : 2

i) = ;ii) |= for ;iii) ,0 ,1 is unsatisfiable.

This is easy to do. At any stage , is a non-minimal counterexample. Thusthere is a such that ,0 = and ,1 = are both counterexamples.

We would like to get a contradiction by considering

Tf = {f |n : n }

for f 2. If each Tf is satisfiable, we could easily conclude that has 20

models, a contradiction. The problem is that, in the absence of compactness,we have no guarantee that Tf is satisfiable. We will need to exercise more care.

Add C a countable set of new constant symbols. Let = {s : s is a finite setof L1,-sentences using only finitely many constants from C, such that s{}has uncountably many countable models}.

Claim is a consistency property.

Lets check (C4). Suppose

X

s . Since there are uncountably many

countable models of s. By the Pigeonhole-Principle, there are uncountably manycountable models of s {} for some X . Then s {} .

Suppose s is a finite set of L1,-sentences with only finitely many constantsfrom C. Let (c) be the conjunction of all sentences in s. In any countableL-structure there are only countably many ways to interpret the constants c.Thus s if and only if v(v) is a counterexample.

We will build a sequence of countable fragments F0 F1 . . . and F =Fn. Let 0, 1, . . . , list all F -sentences and let t0, t1, . . . list all basic F -terms,

both lists should have list each item infinitely often.We will also build a tree (s : 2

in the construction, it is easy to build listings 0, 1, . . . and t0, t1, . . .. The onlyadditional condition we need is that n Fn and tn is an Fn-term.

Let s = {}.Suppose we are given s where | | = n. As in the proof of the Model

Existence Theorem, it is easy to find s such that s s :

i) if s {n} , then n s , moreover, in that case ifa) if n =

xX , then s

for some X , and

b) if n = v (v), then (c) s for some c C;ii) tn = c s for some c C.

Let (c) be the conjunction of all formulas in s . As remarked above, v (v) is a counterexample. Since it is not minimal, there is and L1,-sentence (with no constants from C) such that s,0 = s

{} and s1,/

This completes the construction and the proof.

We will need a refinement of this result. Let be a minimal counterexample.Suppose (v) is an L1,-formula, we say that is -large if there are uncount-ably many countable models of v (v). We say that a -large formula (v)is minimal -large if for all L1,-formulas (v) exactly one of and is -large.

Corollary 5.5 If (v) is -large, then there is a minimal -large formula (v)with (v) |= (v).

Proof If (v) has free variables v1, . . . , vn, add constants d1, . . . , dn to L andapply the theorem to (d). There is (d) a minimal counterexample with(d) |= (d). Then (v) is a minimal -large formula with (v) |= (v).

Digression: prime models

We quickly review some material on atomic and prime models from [6]. Fix Fa countable fragment of L1, and T be an F -complete theory.

Definition 5.6 We say (v) is complete if it is satisfiable and for all formulas(v) in F either

T |= (v) (v) or T |= (v) (v).

We say that T is F-atomic if every satisfiable F formula is a T -consequenceof a complete formula.

We say that M |= T is F -atomic if every a Mn satisfies a completeformula.

We say that M |= T is F -prime if there is an F -elementary embeding of Finto any N |= T .

Theorem 5.7 If T has only countably many F-types, then there is M |= Tthat is F-atomic and F-prime.

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Proof Use the Omitting Types Theorem to buildM |= T omitting all types notcontaining a complete formula. Clearly M is F -atomic. The usual proof thatcountable atomic models of first order theories are prime, adapts immediatelyto prove that M is F -prime.

Uncountable models of counterexamples

We now fix a minimal counterexample.Our proof will need fragments with extra closure properties. We say that

F is rich if and for all (v) is -large, there is a -minimal (v) F with |= (v) (v). By Corollary 5.5, if F is a countable fragment we can findF F that is countable and rich. If F is a countable rich fragment of L1,,let

TF = { F : a sentence such that has uncountably many countablemodels}. First note that TF is a complete F -theory. Moreover, is F \ TF ,then has only countably many countable models. Thus TF is satisfiable,indeed it has uncountably many countable models.

If (v) is a -minimal formula, then for any formula (v),

TF |= (v) (v) or TF |= (v) (v).

Thus every -minimal formula is F -complete, and every F -formula is appliedby an -minimal formula. Thus TF is F -atomic.

We now describe our basic construction. For each < 1 we build

F0 F1 . . . F . . . .

Let F0 be a countable rich fragment containing .Let M0 be the prime model of TF0 .Given M and F, let F+1 be a countable rich fragment containing F

and the Scott sentence of M. Then let M+1 be the prime model of TF+1.Note that M F M+1.

For a limit ordinal, let F =

Further Results

The model of size 1 that we constructed above is not li-equivalent to a count-able model.

Theorem 5.9 (Harnik-Makkai) If is a counterexample, then there are mod-els of size 1 that are L,-equivalent to countable models.

The original proof used admissible model theory, Baldwin gave a proof avoid-ing admissibility. The arguments given in Baldwins lecutres show that indeedthere are 1 models of size 1 that are L,-equivalent to countable models butpairwise L,-inequivalent.

Theorem 5.10 (Harrington) For all < 2 there is a model of size 1 withScott rank at least .

I have never seen Harringtons proof.

Baldwin noted that using two deep theorems of Shelahs every first ordercounterexample T has 21 models of size 1. A first order counterexample Tmust be non--stable and every non--stable theory has 21 -models of cardi-nality 1.

Questions 1) Does Harringtons results follow from Sacks recent paper?2) Give a proof of Harringtons result without admissibility.3) Can we prove that I(,1) = 21 for all counterexamples ?4) Can we find 21 models that are L,-equivalent to a countable models?

Theorem 5.11 (Hjorth) There is a counterexample with no models of size2.

Hjorths surprising proof used descriptive set theory of the dynamics of Polishgroup actions.

Question 5 Give a model theoretic proof of Hjorths theorem.

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