從模型論的視野裡，我們如何看待現代微積分？

2018年12月6日，我們在“（ε，δ）條件與無窮小方法之比研究”博文小中正式闡明瞭相關學術立場。

2008年，Keisler教授，作為塔爾斯基模型論的傳人，發表研究論文，題為“Quanti?ers in Limits”（極限中的量詞）站在模型論的視角 深入闡述了現代微積分的弊端。

我們的觀點就是從這篇文章中“知識共享”出來的。菲氏極限。論的信徒不知作何感想?

袁萌  陳啟清  12月7日

附件：極限理論中的量詞

Quantiers in Limits

H. Jerome Keisler University of Wisconsin, Madison（2008年發表）

Abstract The standard de?nition of lim z→∞F(z) = ∞ is an ??? sentence. Mostowski showed that in the standard model of arithmetic, these quanti?ers cannot be eliminated. But Abraham Robinson showed that in the nonstandardsetting, thislimitpropertyforastandardfunctionF isequivalent to the one quanti?er statement that F(z) is in?nite for all in?nite z. In general, the number of quanti?er blocks needed to de?ne the limit depends on the underlying structure M in which one is working. Given a structure M with an ordering, we add a new function symbol F to the vocabulary of M and ask for the minimum number of quanti?er blocks neededtode?netheclassofstructures(M,F)inwhichlimz→∞F(z) =∞ holds. We show that the limit cannot be de?ned with fewer than three quanti?erblockswhentheunderlyingstructureMiseithercountable, special, or an o-minimal expansion of the real ordered ?eld. But there are structures M which are so powerful that the limit property for arbitrary functions can be de?ned in both two-quanti?er forms.

1、 Introduction An important advantage of the nonstandard approach to elementary calculus is that it eliminates two quanti?ers in the de?nition of a limit. For example, the standard de?nition of lim z→∞ F(z) =∞ requires three quanti?er blocks, ?x?y?z[y≤z?x≤F(z)]. Mostowski showed that in the standard model of arithmetic, these quanti?ers cannot be eliminated. But Abraham Robinson showed that in the nonstandard setting, this limit property is equivalent to the one quanti?er statement ?z[z∈I ?F(z)∈I], where F is a standard function and I is the set of in?nite elements. Because of thequanti?ers,beginningcalculusstudentscannotfollowthestandardde?nition

1.but have no trouble with the nonstandard de?nition. Since all the basic notions in the calculus depend on limits, students often ?nd the nonstandard approach tothecalculustobeeasiertounderstandthanthestandardapproach(see[Ke2], [Su]). Ingeneral,thenumberofquanti?erblocksneededtode?nethelimitdepends on the underlying structure M in which one is working. Given a structure M with an ordering, we add a new function symbol F to the vocabulary ofMand ask for the minimum number of quanti?er blocks needed to de?ne the class of structures (M,F) in which limz→∞F(z) =∞holds. We show that in the standard setting the limit cannot be de?ned with fewer than three quanti?er blocks when the underlying structureMis not too powerful. We obtain this result in the case thatMis countable, and in the case that Mis an o-minimal expansion of the real ?eld R= (R,≤,+,?). As is usual in the literature, we consider the quanti?er hierarchy which takes intoaccountbothn-quanti?erforms. Foreachn,onecanaskwhetheraproperty is Πn, Σn, both Πn and Σn (called ?n), or Boolean in Πn. Each of Σn and Πn implies Boolean in Πn, which in turn implies ?n+1. The standard de?nition of limit is Π3. In Section 4 we will see that the limit property can never be Boolean in Π1 sentences. However, if M= (R,≤,N,g) where g maps R onto RN, then the limit property is ?2. What happens here is that there is a standard de?nition of limit which uses one less quanti?er block than the usual de?nition, but needs a functionwhichcodessequencesofrealnumbersbyrealnumbers,andistherefore beyond the scope of an elementary calculus course. In Section 5 we show that when M is countable, the limit property is notΣ 3. In Section 6 we prove that the limit property is not Σ3 whenMis the real ordering with an embedded structure with universe N. In Section 7 we prove that the limit property is not Σ3 when M is a saturated or special structure, even when one adds a predicate for the set of in?nite elements. This shows that Robinson’s result for standard functions does not carry over to arbitrary functions. InSection8weconsiderin?nitelylongsentences.

InanorderedstructureM with universe setRand at least a constant symbol for each natural number, the limit property can be expressed naturally by a countable conjunction of countable disjunctions of Π1 sentences. We show that the limit property cannot be expressed by a countable disjunction of countable conjunctions of Σ1 sentences. In Section 9 we prove our main result: If M is an o-minimal expansion of the real ordered ?eld , then the limit property is not Boolean over Π2. We leave open the question of whether one can improve this result by showing that the limit property is not Σ3. Wealsoconsiderthesimilarbutsimplerpropertyofafunctionbeingbounded. The standard de?nition of boundedness is Σ2, or alternatively, a countable disjunction of universal sentences. We show that whenMis countable, special, or an o-minimal expansion of the real ordered ?eld, the boundedness property is not Π2. When M has universe set R, boundedness cannot be expressed by a countable conjunction of existential sentences.

2. Preliminaries We introduce a general framework for the study of quanti?ers required for de?ning limits. We assume throughout that M is an ordered structure with no greatest element. That is, M is a ?rst order structure for a vocabulary L(M) that containsatleasttheorderrelation≤, and≤isalinearorderingwithnogreatest element. We will consider structuresK= (M,f) where f :M→Mis a unary function. The vocabulary of (M,f) is L(M)∪{F} where F is an extra unary function symbol. For a formula ?(~x,~y) of L(M)∪{F}, the notation ?(f,~x,~y) means that (M,f)|= ?(~x,~y). We use the notation |X| for the cardinality of X, ≡ for elementary equiva-lence, ?for elementary substructure, and～ = for isomorphic. Quanti?er-free formulas are called Π0 formulas and also called Σ0 formulas. A Πn+1 formula is a formula of the form?~xθ where θ is a Σn-formula. A Σn+1 formula is one a formula of the form?~xθ where θ is a Πn-formula. The negation of a Πn formula is equivalent to a Σn formula, and vice versa. A formula is said to be Boolean in Πn if it is built from Πn formulas using∧,∨,?. Any formula whichisBooleaninΠn isequivalenttobothaΠn+1 formulaandaΣn+1 formula. De?nition 2.1 In the language L(M)∪{F}, BDD is the Σ2 sentence which says that f is bounded, BDD =?x?yF(y)≤x. LIM is the Π3 sentence which says that limx→∞f(x) =∞, LIM =?x?y?z[y≤z?x≤F(z)]. We will say that a sentence θ of L(M)∪{F} is Πn over M if there is aΠ n sentence of L(M)∪{F}which is equivalent to θ in every structure (M,f). Similarly for Σn. We say that θ is ?n overMif it is both Πn and Σn overM. We will say that θ is Boolean in Πn over M, or Bn over M, if there is a sentence of L(M)∪{F}which is Boolean in Πn and is equivalent to θ in every structure (M,f). With this terminology, there are two quanti?er hierarchies of sentences overM, ?1 ?Π1 ?B1 ??2 ?Π2 ?B2 ??3 ?Π3, ?1 ?Σ1 ?B1 ??2 ?Σ2 ?B2 ??3 ?Σ3. By the quanti?er level of a sentence over M we mean the lowest class in these hierarchies to which a sentence belongs over M. Note that the level of a sentence over an expansion of Mis at most its level overM. In this paper we will consider the following problem. Problem Find the quanti?er level of BDD and LIM over a given structure M.

3.For anyM, BDD is at most Σ2 and LIM is at most Π3 overM. We remark that whenever M is an expansion of an ordered ?eld, the limit property limx→0 f(x) = c will be at the same quanti?er level as LIM. This can be seen by the change of variables z = 1/x. Similar remarks can be made for other limit concepts in the calculus.

3 Results of Mostowski and Robinson In order to compare the results in this paper to earlier results of Mostowski and Robinson, we de?ne the quanti?er level of a sentence over a structure M relative to a set F?MM, where MM is the set of all functions from M into M. We way that a sentence θ of L(M)∪{F} is Πn over M relative to F if there is a Πn sentence of L(M)∪{F}which is equivalent to θ in every structure (M,f) with f ∈F. Similarly for Σn. Thus a sentence is Πn over M in our original sense i? it is Πn overMrelative toMM. Note that if a sentence is Πn overM, then it is Πn overMrelative to everyset F?MM, and similarly for Σn. Thus for ever M and F, BDD is at mostΣ 2 overMrelative toF, and LIM is at most Π3 overMrelative toF. In the paper [Mo2], Mostowski showed that several limit concepts, including BDD and LIM, have the highest possible quanti?er level over the standard model of arithmetic. In fact, over this particular structure, he obtains the stronger result that BDD and LIM have the highest possible quanti?er level relative to the set of primitive recursive functions. Theorem 3.1 (Mostowski [Mo2]). Let F be the set of all primitive recursive functions, and letN be the standard model of arithmetic with a symbol for every function in F. (i) BDD is not Π2 over N relative to F. (ii) LIM is not Σ3 over N relative to F. The proof of Theorem 2 in [Mo2] shows that for every Σ2 formula θ(y) in L(N) there is a primitive recursive function g(x,y) such that θ(y) de?nes the set of y for which g(?,y) is bounded. By the Arithmetical Hierarchy Theorem of Kleene [Kl] and Mostowski [Mo1], θ may be taken to be Σ2 but not Π2 over N, and (i) follows. Similarly, the proof of Theorem 3 in [Mo2] shows that for every Π3 formula ψ(y) in L(N) there is a primitive recursive function h(x,y) such that ψ(y) de?nes the set of y for which limx→∞h(x,y) =∞, and (ii) follows by taking ψ to be Π3 but not Σ3 overN. Abraham Robinson’s characterization of in?nite limits with one universal quanti?er uses an elementary extension of M. De?nition 3.2 In an elementary extension ?M of M, an element is in?nite (over M) if it is greater than every element of M. By a hyperextension of M we mean a structure (?M,I) where ?M is an elementary extension of M with at least one in?nite element, and I is a unary predicate for the set of 4
in?nite elements. A function ?f : ?M→?M is standard if there is a function f :M→M such that (?M,?f) is an elementary extension of (M,f). Theorem 3.3 (A. Robinson) Let (?M,I) be a hyperextension of M. Then BDD and LIM are Π1 over (?M,I) relative to the set standard functions. In fact, Robinson shows that for every standard function ?f : ?M→ ?M,( ?M,?f) satis?es BDD??x[F(x) = 0??I(x)], LIM ??x[I(x)?I(F(x))]. In the nonstandard treatment of elementary calculus, one works in a hyperextension of the system R of real numbers or the system N of natural numbers. The extra predicate I for in?nite elements eliminates one quanti?er block in the de?nition of boundedness and two quanti?er blocks in the de?nition of limit. The question we address here is whether one can also eliminate these quanti?ers in the original ?rst order language L(M)∪{F}. That is, are there any “quanti?er shortcuts” for the statements BDD or LIM in M? This question is completely standard in nature, but is motivated by Robinson’s results in nonstandard analysis.
4 Cases with Low Quanti?er Level In this section we show that when the underlying structure M is su?ciently powerful, the properties BDD and LIM are equivalent to sentences in both two quanti?er forms. All that is needed is a symbol for a particular function which is ?rst order de?nable in the real ?eld with a predicate for the natural numbers. This gives a warning that some restrictions are needed onMin order to show that BDD and LIM are higher than ?2 over M. We also show that BDD and LIM can never be B1 overM. Theorem 4.1 Let M = (R,≤,N,g) be the ordered set of real numbers with a predicate for the natural numbers and a function g : R×N → R such that x 7→ g(x,?) maps R onto RN, that is, for each y ∈RN there exists x ∈R such that y = g(x,?). (i) BDD is ?2 over M. (ii) LIM is ?2 over M. Proof. The language L(M) has the vocabulary{N,≤,G}. (i) BDD is itself a Σ2 sentence. In every structure (M,f), the negation ofBDD is equivalent to the Σ2 sentence ?x(?n∈N)n≤F(G(x,n)). (ii) In every structure (M,f), LIM is equivalent to the Σ2 sentence ?x(?n∈N)?y[G(x,n)≤y?n≤F(y)], 5
and the negation of LIM is equivalent to the Σ2 sentence (?m∈N)?x(?n∈N)[n≤G(x,n)∧F(G(x,n))≤m].
It is well known that in the above theorem, the function g can be taken to be de?nable by a ?rst order formula in the structure (R,≤,+,?,N). One way to do this is to let π :N×N→N be a de?nable pairing function, and for irrational x, let h(x,?)∈NN be the continued fraction expansion of x, and take g(x,n) = z i??mh(z,m) = h(x,π(m,n)). We now show that for an arbitrary M, LIM and BDD cannot be Booleanin Π 1. Theorem 4.2 Let M be an ordered structure with no greatest element. (i) BDD is not Boolean in Π1 over M. (ii) LIM is not Boolean in Π1 over M. Proof. Let ? be a sentence of L(M)∪{F} which is Boolean in Π1. Since conjunctions and disjunctions of Π1 sentences are equivalent to Π1 sentences, and similarly for Σ1, ? is equivalent to a sentence (α1∨β1)∧???∧(αn∨βn) where each αi is Σ1 and each βi is Π1. We may assume that αi and βi have the form αi =?~x?~y[F(~x) = ~y∧αi(~x,~y)], βi =?~x?~y [F(~x) = ~y?βi(~x,~y)] where αi and βi are quanti?er-free formulas of L(M). (This can be proved by induction on the number of occurrences of F). Letussaythatapairoftuples(~a,~ b)inMdecides ? ifforany f,g :M→Msuch that f(~a) = g(~a) = ~ b, ? holds in either both or neither of the models (M,f), (M,g). Claim: There is a pair (~a,~ b) which decides ?. Proof of Claim: The proof is by induction on n. The claim is trivial for n = 0, where the empty conjunction is taken to be always true. Suppose n > 0 and the claim holds for n?1. Then there is a pair (~a,~ b) which decides (α1∨β1)∧???∧(αn?1∨βn?1). A pair (~c, ~ d) is called compatible with (~a,~ b) if bi = dj whenever ai = cj, that is, there exists a function f with f(~a,~c) = (~ b, ~ d). Case 1. There is a pair (~c, ~ d) compatible with (~a,~ b) such that αn(~c, ~ d). Then f(~c) = ~ d implies αn. Case 2. Case 1 fails but there is a pair (~c, ~ d) compatible with (~a,~ b) such that ?βn(~c, ~ d). Then f(~a,~c) = (~ b, ~ d) implies?(αn∨βn). 6
Case 3. For every pair (~c, ~ d) compatible with (~a,~ b), ?αn(~c, ~ d)∧βn(~c, ~ d). Then f(~a) =~ b implies βn. In each case, (~a∪~c,~ b∪~ d) decides ?, completing the induction. Now let (~a,~ b) decide ?. There are functions f,g such that f(~a) = g(~a) =~ b, but BDD holds in (M,f) and fails in (M,g). Therefore ? cannot be equivalent to BDD overM. A similar argument holds for LIM. 5 Countable Structures In this section we consider the case that the universe ofMis countable. In that case, we apply Mostowski’s Theorem 3.1 to show that BDD and LIM have the highest possible quanti?er level. We ?rst observe that Mostowski’s proof of Theorem 3.1 also gives a relativized form of the result. Theorem 5.1 Let α be a ?nite tuple of ?nitary functions on N, letF(α) be the set of all functions which are primitive recursive in α, and letN be the standard model of arithmetic with an extra symbol for each function in the set F(α). (i) BDD is not Π2 over N relative to F(α). (ii) LIM is not Σ3 over N relative to F(α). Corollary 5.2 LetN be an expansion of the standard model of arithmetic with the natural ordering <. (i) BDD is not Π2 over N. (ii) LIM is not Σ3 over N. Theorem 5.3 SupposeMis an ordered structure with no greatest element and the universe of M is countable. (i) BDD is not Π2 over M. (ii) LIM is not Σ3 over M. Proof. WemayassumethatthevocabularyofMis?nite,sinceonly?nitely many symbols occur in a Σ3 formula. We may also assume that the universe of Mis the set N of natural numbers. Let <M be the ordering ofM(which may be di?erent from the natural order < of N). Since M has no greatest element, there is a function h : N → N such that h(n) <M h(n + 1) for each n, and ?x?nx≤M h(n). For each x let λ(x) be the least n such that x≤M h(n). We prove (ii). The proof of (i) is similar. We observe that for each function g : N → N, limx→∞h(g(λ(x))) = ∞ with respect to <M if and only if limx→∞g(x) = ∞ with respect to <. If LIM is not Σ3 over an expansion of M, then it is not Σ3 over M. We may therefore assume that M has symbols for <,+,?,h, and λ. Now suppose that LIM is Σ3 over M, that is, there is a Σ3 sentence ?~x?~y?~z?(~x,~y,~z) of L(M)∪{F} which is equivalent to LIM in all structures (M,f). To complete the proof we will get a contradiction. 7
Let ψ(~x,~y,~z) be the formula obtained from ?(~x,~y,~z) by replacing each term F(u) by h(F(λ(u))). By our observation above, for any function g :N→N, we have limx→∞g(x) =∞ with respect to < if and only if (M,g) satis?es the Σ3 sentence ?~x?~y?~zψ(~x,~y,~z). This contradicts Corollary 5.2 and completes the proof. We also give a second argument, which will be useful later. Instead of Corollary 5.2, this argument uses the following analogous fact from descriptive set theory (see [Kec, Exercise 23.2): In NN, the set L of functions g with limx→∞g(x) =∞ is Π3 but not Σ3 in the Borel hierarchy. We have L =[ ~x∈N\ ~y∈N[ ~z∈N {g : (M,g)|= ψ(~x,~y,~z)}). For each (~x,~y,~z), the set {g : (M,g) |= ψ(~x,~y,~z)} depends only on ?nitely many values of g and thus is clopen in NN. It follows that L is Σ3 in NN, contrary to the preceding paragraph. The next corollary follows by the L¨owenheim-Skolem theorem. Corollary 5.4 Let M be an ordered structure with no greatest element. (i) There is no Π2 sentence of L(M)∪{F} which is equivalent to BDD in all models of Th(M). (ii) There is no Σ3 sentence of L(M)∪{F} which is equivalent to LIM in all models of Th(M). A result in this direction was previously obtained by Kathleen Sullivan in [Su]. She showed that there is no Σ2 sentence of L(M)∪{F}, and no Π2 sentence of L(M)∪{F}, which is equivalent to LIM in all models of Th(M). 6 The Real Line The sentence LIM is a Π3 sentence in the vocabulary with the single relation symbol≤. ItisthereforenaturaltoaskwhetherLIM isΣ3 overtherealordering (R,≤). In this section we show that the answer is no. Given a structureN with universe N, we let (R,≤,N) be the structure formed by adding to (R,≤) a symbol for N and the relations ofN. Theorem 6.1 Let N be a structure with universe N. (i) BDD is not Π2 over (R,≤,N). (ii) LIM is not Σ3 over (R,≤,N). Proof. We prove (ii). The proof of (i) is similar. We may assume that N has a countable vocabulary. Let Q be the set of rational numbers, and let λ(x) = min{n∈N: x≤n}. Claim 1: (Q,≤,N) is an elementary substructure of (R,≤,N).
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Proof of Claim 1. We note that for any two increasing n-tuples ~a,~ b in R such that~a∩N=~ b∩N and λ(~a) = λ(~ b), there is an automorphism of (R,≤,N)which sends ~a to ~ b. Claim 1 now follows from the Tarski-Vaught test ([CK], Proposition 3.1.2). De?ne the function α : NN → RR by (α(g))(x) = g(λ(x)), and let β(g) be the restriction of α(g) to Q. Claim 2: For each function g ∈NN, (Q,≤,N,β(g)) is an elementary substructure of (R,≤,N,α(g)). To see this, note that by Claim 1, (Q,≤,(N,g)) is an elementary substructure of (R,≤,(N,g)). It is easily seen that λ is de?nable in (R,≤,N), and hence α(g) is de?nable in (R,≤,(N,g)). Claim 2 follows. Let L be the set of g ∈ NN such that (M,α(g)) |= LIM. We note thatlim n→∞g(n) =∞in NN if and only if g∈L, and hence that L is not Σ3 in NN. Now suppose that LIM is Σ3 over (R,≤,N), and take a Σ3 sentence θ =?~x?~y?~z?(~x,~y,~z) ofL(R,≤,N)∪{F}whichisequivalenttoLIM inallstructures(R,≤,N,f). By Claim2, foreach g∈NN, (R,≤,N,α(g))satis?es θ ifandonlyif(Q,≤,N,β(g)) satis?es θ. Therefore L =[ ~x∈Q\ ~y∈Q[ ~z∈Q ({g : ?(β(g),~x,~y,~z)}). For each (~x,~y,~z), the set ({g : ?(β(g),~x,~y,~z)}) depends only on ?nitely many values of g and thus is clopen in NN. Since Q is countable, it follows that L is Σ3 in NN. This contradiction proves (ii). We remark that the above theorem also holds, with the same proof, when R is replaced by any dense linear ordering with a co?nal copy of N.
7 Saturated and Special Structures Inthissectionweshowthat BDD and LIM havethehighestpossiblequanti?er level if the underlying structure M is saturated, or more generally, special. This happens even if one adds a symbol for the set I of in?nite elements to the vocabulary. Thus Robinson’s Theorem 3.3 for standard functions cannot be extended to the set of all functions on a special structure. We recall some basic facts (See [CK], Chapter 5). By de?nition, a structure M is κ-saturated if every set of fewer than κ formulas with parameters in M which is ?nitely satis?able in M is satis?able in M. M is saturated if it is |M|-saturated. M is special if it is the union of an elementary chain of λ+-saturated structures where λ ranges over all cardinals less than |M|. Letuscallacardinal κ niceif κ =Pλ<κ 2λ. Forexample, everyinaccessible cardinal is nice, every strong limit cardinal is nice, and every cardinal λ+ = 2λ is nice.
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Facts 7.1 (i) If ω ≤|M| < κ and κ is nice then M has a special elementary extension of cardinality κ. (ii) Any two elementarily equivalent special models of the same cardinality are isomorphic. (iii) Reducts of special models are special. (iv) If M is special, then (M,~a) is special for each ?nite tuple ~a in M. Theorem 7.2 SupposeMis an ordered structure with no greatest element,M is special, |M| is nice, and |L(M)|<|M|. (i) BDD is not Π2 over M. (ii) LIM is not Σ3 over M. Proof. We prove (ii). The proof of (i) is similar. Let T be the set of all Σ3 consequences of Th(M)∪{LIM}. Consider a ?nite subset T0 ? T. The conjunctionofT0 isequivalenttoaΣ3 sentenceθ,andTh(M)∪{LIM}|= θ. By Corollary 5.4, we cannot have Th(M)∪{θ}|= LIM, and therefore we cannot have Th(M)∪T0 |= LIM. By the Compactness Theorem, there is a model (M0,g) of Th(M)∪T ∪{?LIM}. Now let U be the set of all Π3 sentences which hold in (M0,g). By the Compactness Theorem again, there is a model (M1,f) of Th(M)∪U ∪{LIM}. Then every Σ3-sentence holding in (M1,f) holds in (M0,g). Let κ = |M|. Since |L(M)| < κ and κ is nice, it follows from Fact (i) that we may take (M0,g) and (M1,f) to be special models of cardinality κ. By Fact (ii), M0 and M1 are isomorphic to M, so we may take M1 =M0 =M. Then LIM holds in (M,f) and fails in (M,g), but every Σ3 sentence holding in (M,f) holds in (M,g). This proves (ii). The proof of (i) is similar. Given a hyperextension (?M,I) ofM, we will say that a function g : ?M→? M is standard if (?M,g) is an elementary extension of (M,f), or in other words, if (?M,g,I) is a hyperextension of (M,f). Robinson’s Theorem 3.3 shows that BDD and LIM are Π1 for standard functions over any hyperextension (?M,I) ofM. Our next theorem will show that Robinson’s result does not carry over to nonstandard functions. In fact, we will show that Theorem 7.2 holds even when a symbol for the set I of in?nite elements is added to the vocabulary. Thus for nonstandard functions, one cannot lower the quanti?er level of BDD or LIM by adding a symbol for I. Lemma 7.3 LetMbe an ordered structure with no greatest element, such that every element of M is a constant symbol of L(M). Suppose that (?M,I) a hyperextension of M of cardinality κ =|?M| such that |L(M)|< κ, κ is nice, (?M,f,g) is special, and every Πn sentence of L(M)∪{F} true in (?M,f) is true in (?M,g). Then every Πn sentence of L(M)∪{F,I} true in (?M,f,I) is true in (?M,g,I). Similarly for Σn. Proof. The result for Σn follows from the result for Πn by interchanging f and g. Let κ =|?M|. By the Keisler Sandwich Theorem ([CK], Proposition
10
5.2.7 and Exercise 5.2.7), there is a chain (M0,h0)?(M1,h1)????(Mn,hn) such that (M0,h0)≡(M,f), (M1,h1)≡(M,g), and for each m < n?1, (Mm,hm)?(Mm+2,hm+2). By Fact (i) we may take each (Mm,hm) to be a special structure of cardinality κ. For each m ≤ n let Im be the set of elements of Mm greater than each element ofM. We then have (M0,h0,I0)?(M1,h1,I1)????(Mn,hn,In). Since each element of M has a constant symbol in L(M), it follows from Fact (i) that (M0,h0,I0)～ = (?M,f,I) and (M1,h1,I1)～ = (?M,g,I). By remark (iv), (Mm,hm,~a) is still special for each m and ?nite tuple ~a in Mm. Then by Fact (ii), for each m < n?1 and ?nite tuple ~a in Mm, there is an isomorphism from (Mm,hm) onto (Mm+2,hm+2) which ?xes ~a and each element ofM. This isomorphism also sends Im onto Im+2, so (Mm,hm,Im,~a)～ = (Mm+2,hm+2,Im+2,~a). Using the Tarski-Vaught criterion for elementary extensions ([CK], Proposition 3.1.2), it follows that (Mm,hm,Im)?(Mm+2,hm+2,Im+2). By the Keisler Sandwich Theorem again, every Πn-sentence true in (M0,h0,I0) is true in (M1,h1,I1), and hence every Πn-sentence true in (?M,f,I) is true in (?M,g,I). Theorem 7.4 Let M be an ordered structure with no greatest element, and let (?M,I) be a hyperextension of M such that ?M is special and |?M| is uncountable and nice. (i) BDD is not Π2 over (?M,I). (ii) LIM is not Σ3 over (?M,I). Proof. We prove (ii). The proof of (i) is similar. By the proof of Theorem 7.2, there are functions f,g such that (?M,f) and (?M,g) are special, LIM holds in (?M,f) and fails in (?M,g), and every Σ3 sentence true in (?M,f) is true in (?M,g). By Facts (i) and (ii), we may take f and g so that (?M,f,g) is special. Then by the preceding lemma, every Σ3 sentence true in (?M,f,I) is true in (?M,g,I). Therefore LIM is not Σ3 over (?M,I). 11
8 In?nitely Long Sentences In this section we consider the quanti?er levels of BDD and LIM in the in?nitary logic Lω1ω formed by adding countable conjunctions and disjunctions to ?rst order logic. See [Ke1] for a treatment of the model theory of this logic. De?nition 8.1 Let Q be a set of sentences of Lω1ω in the vocabulary L(M). VQ denotes the set of countable conjunctions of sentences in Q.WQ denotes the set of countable disjunctions of sentences in Q. BQ denotes the set of ?nite Boolean combinations of sentences in Q. For example,WVBVΠ1 is the set of sentences of the formWmVn θmn where each θmn is a ?nite Boolean combination of countable conjunctions of universal ?rst order sentences. We say that M has co?nality ω if M has a countable increasing sequencea 0,a1,... which is unbounded, that is, ?xWn x≤an. Every countable M and every structure M with universe R has co?nality ω. In this section our main focus will be uncountableMwith co?nality ω. Suppose that M has universe R and a constant symbol for each natural number. Then BDD isWΠ1 overM, because (M,f)|= LIM ?_ n ?zF(z)≤n. LIM isVWΠ1 overM, because (M,f)|= LIM ?^ m_ n ?z[n≤z?m≤F(z)]. LIM is alsoVΣ2 overM, because (M,f)|= LIM ?^ m ?y?z[y≤z?m≤F(z)]. The next theorem shows that ifMhas co?nality ω, then BDD is notVΣ1over M and LIM is notWVΣ1 over M. Thus if the outer quanti?ers ? and? are replaced byWandV, then Theorem 5.3 holds for uncountable structuresM of co?nality ω. In fact, we get a stronger result with BVΠ1 in place of Σ1. Theorem 8.2 Suppose that M has co?nality ω. (i) BDD is notVBVΠ1 over M. (ii) LIM is notWVBVΠ1 over M. Proof. We prove (ii). The proof of (i) is similar. Let an be an unbounded strictly increasing sequence in M. Since each sentence of Lω1ω has countably many symbols, we may assume without loss of generality that the vocabulary L(M) is countable. We may also assume that L(M) has a constant symbol, say n, for each an, and a symbol for the function
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λ(x) = min{an : x≤an}. AsintheproofofTheorem5.3, welet α :NN→MM be the function de?ned by (α(g))(x) = g(λ(x)), and let L be the set of g ∈NN such that (M,α(g))|= LIM. Claim: Foreach?rstorderquanti?er-freeformula ?(F,~x)inthevocabulary L(M)∪{F}, the set {g∈NN : (M,α(g))|=?~z?(F,~z)} is Σ1 in the Borel hierarchy. Proof of Claim: We may assume without loss of generality that ?~z?(F,~z) has the form ?~x?~y[F(~x) = ~y∧ψ(~x,~y)] where ψ(~x,~y) is a ?rst order quanti?er-free formula of L(M). Since (α(g))(x) = (α(g))(λ(x))forallg andx,(M,α(g))satis?es?~z?(F,~z)ifandonlyifitsatis?es _ ~ p_ ~ q [F(~ p) = ~ q∧?~x[λ(~x) = ~ p∧ψ(~x,~ q)]]. Let U ={(~ p,~ q) :M|=?~x[λ(~x) = ~ p∧ψ(~x,~ q)]}. Then (M,α(g)) satis?es?~z?(F,~z) if and only if g(~ p) = ~ q for some (~ p,~ q)∈U.For each pair ( ~ p,~ q), the set{g : g(~ p) = ~ q}is clopen inNN, so the union of thesesets over U is Σ1 in the Borel hierarchy, as required. Now suppose to the contrary that there is aWVBVΠ1 sentence θ =_ m^ n ?mn(F) which is equivalent to LIM in all models (M,f), where ?mn(F) is BVΠ1 in the vocabulary L(M)∪{F}. We observe that any ?nite conjunction or disjunction ofVΠ1 sentences is equivalent to aVΠ1 sentence, and the negation of aVΠ1 sentence is equivalent to aWΣ1 sentence. It follows that θ is equivalent to a sentence _ m^ n "^ p ?~z?mnp(F,~z)∨_ q ?~zψmnq(F,~z)# where each ?mnp(F,~z) and ψmnq(F,~z) is a ?rst order quanti?er-free formula of L(M)∪{F}. Using the claim, it follows that L =[ m\ n \ p Amnp∪[ q Bmnq! where each Amnp and Bmnq is clopen in NN. By renumbering and rearranging, we can get L =[ m\ n[ q (Amnq ∪Bmnq). Therefore L is Σ3 in NN. This is a contradiction, and proves (ii).
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9 O-minimal Structures In this section we show that BDD has the highest possible quanti?er level, and that LIM is not Boolean in Π2, in the case thatMis an o-minimal expansion of the ordered ?eld of real numbers. We leave open the question whether LIM can be Σ3. We will use two lemmas from [FM] concerning fast and indiscernible functions. An ordered structure is o-minimal if every set de?nable with parameters is a ?nite union of intervals and points. There is an extensive literature on the subject (e.g. see [VDD1], [VDD2]). An example of an o-minimal structure is the ordered ?eld of reals with analytic functions restricted to the unit cube and with the (unrestricted) exponential function. It is easily seen that if M is o-minimal, then every ordered reduct of M is o-minimal, and the expansion of Mformed by adding a symbol for each de?nable relation is o-minimal. Throughout this section we let M be an o-minimal expansion of the real ordered?eldR= (R,≤,+,?). “De?nable”willalwaysmean?rstorderde?nable inMwith parameters fromM. When working with o-minimal structures, one often restricts attention to de?nablefunctions. Itiseasilyseenthat LIM isΠ2 overano-minimalstructure M relative to the set of de?nable functions, since LIM fails for a de?nable function f if and only if?x?y?z[x≤z?f(z)≤y]. However, here we consider arbitrary functions onM. De?nition 9.1 A strictly increasing sequence s of positive integers is M-fast if for each de?nable function f :R→R, there exists N ∈N such that f(s(k)) < s(k +1) for all k > N. For convenience we also require that s(0) = 0. Lemma 9.2 ([FM], 3.3) If L(M) is countable then there exists an M-fast sequence. Given ~u,~v∈Nn and N ∈N, we write ~u =N ~v if min(ui,N) = min(vi,N) fori = 1,...,n, and ~u～~v if ~u,~v are order isomorphic, that is, ui ≤uj i? vi ≤vjfor i,j = 1,...,n. We write ~u～N ~v if ~u =N ~v and ~u～~v. Note that ～N is an equivalence relation on Nn with ?nitely many classes. GivenanM-fastsequencesandatuple~u∈Nn,wewrites(~u)for(s(u1),...,s(un)). Lemma 9.3 ([FM], 3.6). Suppose s is M-fast, h : Rn →R is de?nable, and g :Nn →R is the function given by ~u7→h(s(~u)). Then there exists N ∈N such that g(~u)≤g(~v)?g(w~)≤g(~z) whenever ~u =N ~v and (~u,~v)～N (w~,~z). In [FM], the function g is called almost indiscernible. We will call N an indiscernibility bound for h. Corollary 9.4 Suppose s is M-fast, A?Rn is de?nable, and B ={~u∈Nn : A(s(~u))}. 14
There exists N ∈N such that B(~u)?B(~v) whenever ~u,~v∈Nn and ~u～N ~v. Proof. The characteristic function h of A is de?nable. Let N be an indiscernibility bound for h, and let g be as in Lemma 9.3. Suppose ~u,~v ∈Nn and ~u ～N ~v. Let k ∈ N be greater than N and each ui and vi. By replacing the ui by k + ui whenever N ≤ ui, we obtain a tuple w~ such that w~ ～N ~u and (~u,w~)～N (~v,w~). By Lemma 9.3, g(~u)≤g(w~) i? g(~v)≤g(w~), and g(~u)≥g(w~) i? g(~v)≥g(w~). It follows that g(~u) = g(~v), so B(~u)?B(~v). By an indiscernibility bound for a formula ψ of L(M) with parameters from M we mean an indiscernibility bound for the characteristic function of the set de?ned by ψ. It is clear that if N is an indiscernibility bound for ψ, then any M ≥N is also an indiscernibility bound for ψ. Theorem 9.5 Suppose thatMis an o-minimal expansion of (R,≤,+,?). Then BDD is not Π2 over M. Proof. We may assume that M has a countable vocabulary. By Lemma 9.2 there is an M-fast sequence s. Suppose to the contrary that there is a Π2 sentence?~x?(F,~x) of L(M)∪{F}which is equivalent to BDD in all structures (M,f). We may assume that?~x?(F,~x) has the form ?~x?~y?~z[F(~x,~y) = ~z∧θ(~x,~y,~z)] where θ is a quanti?er-free formula in which F does not occur,|~z|=|~x|+|~y|= j +k, and F(~x,~y) = (F(x1),...,F(xj),F(y1),...,F(yk)). (This can be proved by induction on the number of occurrences of F in ?). Let f : R → R be the function such that f(x) = x if x = s(4i) for some i ∈ N, and f(x) = 0 otherwise. For each m∈N let fm : R→R be the function such that fm(x) = f(x) for x≤s(m) and fm(x) = 0 for x > s(m). Then each fm is bounded but f is unbounded. Therefore?~x?(fm,~x) holds for each m∈N, but ?~x?(f,~x) fails. (Recall that ?~x?(f,~x) means that (M,f) satis?es ?~x?(F,~x).) Hence there exists a tuple ~a such that??(f,~a). The notation w~1 < w~ < w~2 means that each coordinate of w~ is strictly between each coordinate of w~1 and w~2, that is, w~ is in the open box with vertices w~1 and w~2. We will deal with variables which are outside the range of s by putting them into an open box with vertices in the range of s. By Corollary 9.4, there is an N ∈N which is an indiscernibility bound for each formula of the form ?w~[w~1 < w~ < w~2∧θ(~a,~y,~z)] where w~?~y. We also take N so that s(N) > max(~a). By hypothesis, we have ?~y?~z[fN(~a,~y) = ~z∧θ(~a,~y,~z)].
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Choose~ b inMsuch that
?~z[fN(~a,~ b) = ~z∧θ(~a,~b,~z)].
Then θ(~a,~b,fN(~a,~ b)). We wish to ?nd~e inMsuch that θ(~a,~e,f(~a,~e)). This will show that ?(f,~a), contrary to hypothesis, and complete the proof that BDD is not Π2 over M. We cannot simply take ~e = ~ b, because b could have a coordinate bi such that f(bi)6= fN(bi). This happens when bi = s(4m) where 4m > N, so that f(bi) = bi but fN(bi) = 0. Claim: For each ~u ∈Nn there exists ~v ∈Nn such that ~v ～N ~u, f(s(~v)) =f N(s(~v)), and if uj = ui +1 then vj = vi +1. Proof of Claim: Take ~v such that ~v ～N ~u, and vi is not a multiple of 4when vi > N, and vj = vi +1 when uj = ui +1. Then vi = ui below N, and f(s(vi)) = f(s(ui)) = fN(s(vi)) = fN(s(ui)) = 0 when ui > N. It follows that f(s(~v)) = fN(s(~v)). We may rearrange~ b into a k-tuple (~c, ~ d) such that the terms of ~c belong to the range of s and the terms of ~ d do not. Then fN(~ d) =~0 is a tuple of 0’s. Let (~ t,w~) be the corresponding rearrangement of ~y. Then |~c|=|~ t|= n. Let ~ d1, ~ d2 be the vertices of the smallest open box containing ~ d with coordinates in the range of s. That is, ~ d1 < ~d < ~ d2 and the coordinates of ~ d1, ~ d2 are consecutive in the range of s. Then ?w~[~ d1 < w~ < ~ d2∧θ(~a,~c,w~,fN(~a,~c),~0)] holds inM. We have (~c, ~ d1, ~ d2) = s(~u) for some ~u∈Nn. Take ~v as in the claimand let ( ~e0, ~ d3, ~ d4) = s(~v). Then by indiscernibility, ?w~[~ d3 < w~ < ~ d4∧θ(~a,~e0,w~,fN(~a,~e0),~0)]. Note that s(N) > max(~a), so f(~a) = fN(~a). Now let ~e = (~e0,~e1) where ~e1 is a witness for w~ in the above formula. Since the coordinates of ~ d3 and ~ d4 are consecutive in the range of s, the coordinates of ~e1 must be outside the range of s, where f and fN are 0. It follows that f(~a,~e) = fN(~a,~e). Therefore θ(~a,~e,f(~a,~e)), and hence BDD is not Π2 overM. Corollary 9.6 Suppose thatMis an o-minimal expansion of (R,≤,+,?). Then (i) BDD is notVΠ2 over M. (ii) LIM is notVΠ2 over M. Proof. This follows from the proof of Theorem 9.5. For (i), we suppose that there is aVΠ2 sentenceVi?~x?i(F,~x) which is equivalent to BDD in all structures (M,f), choose i ∈N such that ?~x?i(f,~x) fails, and get a contradiction as in the proof of Theorem 9.5. For part (ii), we ?rst observe that all the functions f :M→Mused in the proof of (i) satisfy the Π1 sentence ?x[F(x) = x∨F(x) = 0]. 16
Call this sentence β(F). It follows that β(F)∧BDD(F) is notVΠ2 over M. We also note that β(F)?β(x?F(x)) and β(f)?[BDD(F)?LIM(x?F(x))]. Now suppose that LIM(F) isVΠ2 over M. Then β(F)?LIM(x?F(x)) is alsoVΠ2 overM. Hence β(F)∧BDD(F) isVΠ2 overM, a contradiction. Theorem 9.7 Suppose thatMis an o-minimal expansion of (R,≤,+,?). Then LIM is notWB2 over M. Proof. We may assume that M has a countable vocabulary. By Lemma 9.2 there is anM-fast sequence s. Suppose to the contrary that there is aWB2 sentence ψ which is equivalent to LIM in all structures (M,f). Then ?ψ is equivalent to aVB2 sentence. Since ?nite conjunctions and disjunctions of Π2 sentences are equivalent to Π2 sentences, and similarly for Σ2,?ψ is equivalent to a sentence ^ m (αm∨βm) where each αm is Σ2 and each βm is Π2. We may assume that αm and βm have the form αm =?~x?~y?~z[F(~x,~y) = ~z?αm(~x,~y,~z)], βm =?~x?~y?~z[F(~x,~y) = ~z∧βm(~x,~y,~z)] where αm and βm are quanti?er-free formulas of L(M). We now de?ne a family of functions from N into N which we will later use to build functions from R into R. We begin with a sequence h : N → N with the following properties: (a) For each m∈N, h(m)≤m and h(m) is either 0 or a power of 2. (b) Each power of 2 occurs in?nitely often in the sequence h, (c) If h(m) > 0, them m is halfway between two powers of 2. Notethath(i)is“usually”0,andthatwheneveri < j andh(i) > 0, h(j) > 0, we have h(h(i)) = 0 and 2i≤j. For each ?nite or in?nite sequence of natural numbers σ, let hσ be the function obtained from h by putting hσ(i) = 0 if h(i) = 2n and i > σ(n), and putting hσ(i) = h(i) otherwise. Thus when n is in the domain of σ, h?1 σ {2n} is the ?nite set h?1{2n}∩{0,...,σ(n)}. When σ is?nite and n is outside its domain, h?1 σ {2n}is the in?nite set h?1{2n}.For each σ, de?ne the function fσ :R→R by putting fσ(s(i)) = s(hσ(i)) ifh σ(i) > 0, and fσ(x) = x for all other x. Note that for each in?nite sequence σ we have limx→∞fσ(x) = ∞, so LIM holds in (M,fσ). But for each ?nite sequence σ we have liminfx→∞fσ(x) <∞, so LIM fails in (M,fσ). Wewillnowbuildanin?nitesequenceσ = (σ(0),σ(1),...)suchthat(M,fσ) satis?es ?ψ. This will give us the desired contradiction, since (M,fσ) satis?es LIM. We will simultaneously build a sequence of tuples ~am,m ∈ N, and a
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strictly increasing “growth sequence” g(0) < g(1) < ???. We will form an increasing chain σ[0]?σ[1]???? of ?nite sequences, and take σ to be their union. This chain will have the property that each term of σ[m]\σ[m?1] will be ≥ g(m?1). Whenever possible, σ[m] will be chosen so that αm holds in (M,fσ[m]), and ~am will be a witness for the initial existential quanti?ers of αm. For convenience we let σ[?1] denote the empty sequence and put g(?1) = 1. Suppose m∈N and we already have σ[i], ~ai, and g(i) for each i < m. We have two cases. Case 1. There is a ?nite sequence η ? σ[m?1] such that αm holds in( M,fη), and each term of η\σ[m?1] is ≥ g(m?1). In this case we takeσ [m] to be such an η, and take~am to be a tuple in R which witnesses the initial existential quanti?ers of αm in (M,fσ[m]), that is, ?~y?~z[fσ[m](~am,~y) = ~z?αm(~am,~y,~z)]. Case 2. Otherwise. In this case we take σ[m] to be an arbitrary ?nite sequencesuchthatσ[m]?σ[m?1]andeachtermofσ[m]\σ[m?1]is≥g(m?1), and let ~am be an arbitrary tuple in R. We now de?ne g(m). By Corollary 9.4, there is an N ∈ N which is an indiscernibility bound for each formula of the form ?w~[w~1 < w~ < w~2 ?αm(~am,~y1,w~,~z1,w~)] where w~?~y, ~y1 is the part of ~y outside w~, and ~z1 is the corresponding part of ~z. Let p be the number of variables in the sentences αi,βi,i≤m. Take g(m) so that: (d) g(m)≥N, g(m)≥2p, g(m) > g(m?1), g(m) > max(σ[m]), (e) s(g(m)) > max(~am), (f) The condition stated before Claim 2. Finally, we de?ne σ to be the union σ =Sm σ[m]. To complete the proof we provetwoclaims,Claim1concerningαm andClaim2concerningβm. Condition (f) will not be used in Claim 1, but will be needed later for Claim 2. Claim 1: Suppose αm holds in (M,fσ[m]) (Case 1 above). Then αm holdsin ( M,fσ). Proof of Claim 1: We show that ?~y?~z[fσ(~am,~y) = ~z?αm(~am,~y,~z)]. Suppose not. Then there is a tuple~ b inMsuch that ??~z[fσ(~am,~ b) = ~z?αm(~ai,~b,~z)].
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As in the proof of Theorem 9.5, we may rearrange~ b into a k-tuple (~c, ~ d) such that the terms of ~c belong to the range of s and the terms of ~ d do not. Then fσ(~c) is also in the range of s, and fσ(~ d) = ~ d. Let (~t,w~) be the corresponding rearrangementof~y. Let ~ d1, ~ d2 betheverticesofthesmallestopenboxcontaining ~ d with coordinates in the range of s. Then ??w~[~ d1 < w~ < ~ d2 ?αm(~am,~c,w~,fσ(~am,~c),w~)]. Let ~u be the sequence in N such that (~c, ~ d1, ~ d2) = s(~u). Since the terms of σ beyond σ[m] are greater than g(m), we have hσ(i) = hσ[m](i) for all i ≤ g(m) and fσ(x) = fσ[m](x) for all x≤s(g(m)). For i > g(m) we have either hσ(i) = hσ[m](i), or hσ(i) = 0 and hσ[m](i) > 0. Recall that g(m)≥2p≥2|~u|, where p is the number of variables in αm ∨βm. It follows that the sequence hσ[m] has enough zeros to insure that there is a sequence ~v in Np such that: (g) For each i, (~u,2i)～g(m) (~v,2i), (h) vi = ui whenever hσ(i) > 0, (i) If ui = ui +1 then vi = vi +1, (j) hσ[m](~v) = hσ(~u). Therefore (~v,hσ[m](~v)) ～g(m) (~u,hσ(~u)). Let (~e0, ~ d3, ~ d4) = s(~v). Then by (d), (e), (g), and indiscernibility, we have ??w~[~ d3 < w~ < ~ d4 ?αm(~am,~e0,w~,fσ[m](~am,~e0),w~)]. We may therefore extend ~e0 to a tuple ~e = (~e0,~e1) such that ~ d3 <~e1 < ~ d4∧?αm(~am,~e,fσ[m](~am,~e0),~e1). Thecoordinatesof ~ d1 and ~ d2 areconsecutiveintherangeof s, soby(i)thecoordinates of ~ d1 and ~ d2 are consecutive in the range of s. Therefore the coordinates of ~e1 are outside the range of s, so fσ[m](~e1) =~e1. Then ?αm(~am,~e,fσ[m](~am,~e). This contradicts the fact that ?~y?~z[fσ[m](~ai,~y) = ~z?αm(~am,~y,~z)]. and completes the proof of Claim 1. We now state the postponed condition (f) for the growth sequence g. Recall that p is the number of variables in the sentences αi,βi,i≤m. (f) For each ~u∈Np there exists ~v∈Np with the following properties: (f1) max(~v) < g(m), (f2) If uj = ui +1 then vj = vi +1, (f3) (~v,hσ[m](~v))～m (~u,hσ[m](~u)).
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There is a g(m) with these properties because the equivalence relation ～m has only ?nitely many classes. Claim 2: The sentence?ψ holds in (M,fσ). Proof of Claim 2. We must show that for each m ∈ N, αm ∨βm holds in( M,fσ). Since LIM fails in (M,fσ[m]), αm∨βm holds in (M,fσ[m]). In Case 1 above, by de?nition αm holds in (M,fσ[m]), and by Claim 1, αm holds in (M,fσ). In Case 2, αm fails in (M,fσ[r]) for each r ≥m, and therefore βm holds in( M,fσ[r]) for each r≥m. In this case we prove that βm holds in (M,fσ). We ?x a tuple ~a inMand prove ?~y?~z[fσ(~a,~y) = ~z∧βm(~a,~y,~z)]. By Corollary 9.4 there is an M ∈N which is an indiscernibility bound for each formula of the form ?w~[w~1 < w~ < w~2∧βm(~a,~y1,w~,~z1,w~)] where w~?~y, ~y1 is the part of ~y outside w~, and ~z1 is the corresponding part of ~z. Take r large enough so that r ≥ m and r ≥ M, and s(r) > max(~a). Sinceβ m holds in (M,fσ[r]), we have ?~y?~z[fσ[r](~a,~y) = ~z∧βm(~a,~y,~z)]. We may therefore choose~ b inMso that βm(~a,~b,fσ[r](~a,~ b)). As in the proof of Theorem 9.5, we may rearrange ~ b into a tuple (~c, ~ d) such that the terms of ~c belong to the range of s and the terms of ~ d do not. Then fσ[r](~ d) = ~ d. Let (~t,w~) be the corresponding rearrangement of ~y, and let ~ d1, ~ d2 be the vertices of the smallest open box containing ~ d with coordinates in the range of s. Then ?w~[~ d1 < w~ < ~ d2∧βm(~a,~c,w~,fσ[r](~a,~c),w~)]. Let ~u be the tuple inNsuch that s(~u) = (~c, ~ d1, ~ d2). Then there is a tuple ~v inN such that conditions (f1)–(f3) hold with r in place of m. Here p is the number of variables in αi,βi,i≤r, so βm has at most p variables, and ~u can be taken with length p. Let s(~v) = (~e0, ~ d3, ~ d4). Since r≥M, by (f3) and indiscernibility we have ?w~[~ d3 < w~ < ~ d4∧βm(~a,~e0,w~,fσ[r](~a,~e0),w~)]. Take ~e1 inMsuch that ~ d3 <~e1 < ~ d4∧βm(~a,~e0,~e1,fσ[r](~a,~e0),~e1). 20
By (f2), every coordinate of ~e1 is outside the range of s, so fσ[r](~e1) = ~e1. Putting ~e = (~e0,~e1), we have βm(~a,~e,fσ[r](~a,~e)). It remains to prove that fσ(~a,~e) = fσ[r](~a,~e). Suppose i ≤ g(r) andh σ[r](i) = 2j. If j is in the domain of σ[r], then hσ(i) = 2j because σ(j) = σ[r](j). Otherwise σ(j)≥g(r), hence i≤g(r) < σ(j) and h(i) = 2j, and again hσ(i) = 2j. If hσ[r](k) = 0 then hσ(k) = 0. Therefore hσ(k) = hσ[r](k) for all k≤g(r). By (f1), max(~v) < g(r). Therefore hσ(~v) = hσ[r](~v), and hence fσ(~v) = fσ[r](~v). Since we also have max(~a) < s(r), it follows that fσ(~a,~e) = fσ[r](~a,~e). Therefore βm(~a,~e,fσ(~a,~e)), and we have the required formula ?~y?~z[fσ(~a,~y) = ~z∧βm(~a,~y,~z)]. This establishes Claim 2 and completes the proof. We conclude with a problem which we state as a conjecture. Conjecture 9.8 Suppose that M is an o-minimal expansion of (R,≤,+,?). Then LIM is not Σ3 over M. 10 Conclusion GivenanorderedstructureM,thequanti?erlevelofasentenceθ ofL(M)∪{F} over M is the lowest class in the hierarchies ?1 ?Π1 ?B1 ??2 ?Π2 ... and ?1 ? Σ1 ? B1 ? ?2 ? Σ2 ... which contains a sentence equivalent to θ in all structures (M,f). We investigate the quanti?er levels of the Σ2 sentence BDD, whichsaysthat f isbounded, andtheΠ3 sentence LIM, whichsaysthat limz→∞f(z) = ∞, over a given ordered structure M. This work is motivated by Mostowski’s result that BDD is not Π2 and LIM is not Σ3 relative to the primitive recursive functions over the standard model of arithmetic, and Abraham Robinson’s result which characterizes BDD and LIM for standard functions by Π1 sentences in a language with an added predicate for the set of in?nite elements. We show that BDD and LIM can never be B1 over a structure M, butif M is an expansion of the real ordered ?eld with a symbol for N and each de?nable function, then BDD and LIM are at the lowest possible level ?2 over M. We show that BDD is at its highest possible level, Σ2 but not Π2, and that LIM is at its highest possible level, Π3 but not Σ3, in the following cases: M is countable,Mis the real ordering with an embedded structure on the natural numbers, and M is special of nice cardinality with an extra predicate for the in?nite elements.
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WhenMhas universeR, we obtain analogous results with the outer quanti?ers replaced by countable disjunctions and conjunctions. In that case we show that BDD cannot be expressed by a countable conjunction of existential sentences, and LIM cannot be expressed by a countable disjunction of countable conjunctions of existential sentences. The most interesting case is whereMis an o-minimal expansion of the real ordered ?eld. In that case we show that BDD is at the maximum level, and that LIM is not B2. Moreover, BDD and LIM are notVΠ2, and LIM is not WB2. We leave open the question of whether LIM is at its maximum level, not Σ3, in that case. References [CK] C.C. Chang and H. Jerome Keisler. Model Theory, Third Edition. Elsevier 1990. [VDD1] Lou van den Dries. o-minimal structures. pages 137-185 in: Logic: From Foundations to Applications. W. Hodges et al. eds. Oxford 1996. [VDD2] Lou van den Dries. Tame Topology and o-minimal Structures. Cambridge University Press, 1998. [FM] HarveyFriedmanandChrisMiller.Expansionsofo-minimalstructures by fast sequences. Journal of Symbolic Logic 70:410–418, 2005. [Kec] Alexander Kechris. Classical Descriptive Set Theory. Springer-Verlag 1995. [Ke1] H. Jerome Keisler. Model Theory for In?nitary Logic. North-Holland 1971. [Ke2] H. Jerome Keisler. Elementary Calculus. An In?nitesimal Approach. Second Edition, Prindle, Weber and Schmidt, 1986. Online Edition, http://www.math.wisc.edu/~keisler/, 2000. [Kl] Stephen C. Kleene. Recursive predicates and quanti?ers. Transactions of the American Mathematical Society 53:41–73. 1943. [Mo1] AndrzejMostowski.Onde?nablesetsofpositiveintegers. Fundamenta Mathematicae 34:81–112. 1947. [Mo2] Andrzej Mostowski. Examples of sets de?nable by means of two and three quanti?ers. Fundamenta Mathematicae 42:259–270. 1955. [Ro] Abraham Robinson. Non-standard Analysis. North-Holland 1966. [Su] Kathleen Sullivan. The Teaching of Elementary Calculus: An Approach Using In?nitesimals. Ph. D. Thesis, University of Wisconsin, Madison, 1974.
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