Fusing o-minimal structures

Transcription

1 Fusing o-minimal structures A.J. Wilkie Mathematical Institute University of Oxford St Giles Oxford OX1 3LB UK May 3, 2004 Abstract In this note I construct a proper o-minimal expansion of the ordered additive group of rationals. 1 Introduction Following work by van den Dries on expansions of the real field (see [1]) Pillay and Steinhorn layed the foundations for the general theory of o-minimality in [11]. At that time, and very much in the spirit of model theory of the period, Steinhorn asked whether the most atomic of all (interesting) o-minimal structures, namely the ordered, additive group of rationals, could have a proper o-minimal expansion. Although, as we shall show here, the answer to this question is disappointingly positive, much research into o-minimal expansions of ordered, divisible groups was activated by Steinhorn s question (see [12], [2], [6], [7], [8], [5], [14], [10]). The most recent is reported in [9] where many of the natural questions in the case of structures related to R an are settled, but where it is also pointed out that fundamental issues remain unresolved in general. The purpose of this paper is to show that the o-minimal condition places almost no restriction on partial (abelian) group structures that can exist on 1

2 bounded parts of archimedean structures related to R an. The following result gives the flavour of our main theorem. Here, and throughout the paper, we use the convention that if 0 E R and F : E n E, then for r, s E with r < s, F [r,s] : E n E is the function sending x to F ( x) if x E n [r, s] n, and to 0 if x E n \ [r, s] n. Theorem 1. There exists an expansion B; of the ordered ring B := B; 0, 1, <, +, of real algebraic numbers such that the reduct B; 0, 1, <, is isomorphic to the ordered, additive group of rational numbers (with 1 distinguished) and such that both the structures and are o-minimal. B; [r,s] r,s B,r<s B; 0, 1, <, + [r,s], [r,s], r,s B,r<s I should remark that by a result of [6], the domain of an o-minimal structure can support at most two definable ordered groups (up to definable isomorphism) and it follows fairly easily from this that the structure B; itself cannot be o-minimal. However, the second displayed structure in Theorem 1 may be viewed as an expansion of the ordered additive group of rationals, and is certainly a proper expansion (e.g. because its definable closure operator is non-modular), so we do have an answer to Steinhorn s question. In fact, the rationals may be replaced by any countable, archimedean ordered, divisible abelian group and the real algebraics by any countable, archimedean ordered, real closed field. Notice also that Theorem 1 implies that the additive (circle) groups of rationals (mod 1) and real algebraics (mod 1) can coexist o-minimally on the same set. 2 Preliminaries and statements of results A function f : R n R is called restricted analytic if it is of the form F [0,1] : R n R, where F : R n R is analytic at least on some open set U R n with [0, 1] n U. With this notation, if α N n is a multi-index and G : R n R is any function agreeing with the α th derivative F (α) of F on U, then we write f (α) for the restricted analytic function G [0,1]. 2

3 The proof of our main theorem relies heavily on the following result, due to Gabrielov, in which I use R to denote the real ordered field R; 0, 1, <, +,. Proposition 1. Let F be a collection of restricted analytic functions closed under differentiation. Then the structure R; F has a model complete theory and is o-minimal. If we take F to be the collection of all restricted analytic functions, then the corresponding structure R; F is usually denoted R an and was first shown to be model complete by Gabrielov in 1969 (see [3]). The above refinement was established much later in the paper [4]. We can now state our main theorem. Theorem 2. Let F be a countable collection of restricted analytic functions closed under differentiation and set A := R; F. Let B be a countable elementary substructure of A. Then for any countable elementary substructure G of R; 0, 1, <, +, there exists an expansion B, of B such that B; 0, 1, <, = G (where B = dom(b)) and such that the structure B, [r,s] r,s B,r<s is o-minimal. Proof of Theorem 1 assuming Theorem 2. We take F =, B the ordered ring of algebraic numbers and G the ordered additive group of rationals (with 1, as well as 0, distinguished) in Theorem 2. Then all but the very last assertion of Theorem 1 follow immediately. To obtain this as well, we need the following facts and I am grateful to Philip Scowcroft for pointing them out to me (private communication: the crucial result is Theorem 1.1 of [10]): Let D be any expansion of the ordered additive group of rationals (with 0, 1 distinguished) by bounded relations, and let ψ(x) be any formula of L(D). Then (a) there exists q Q such that either p Q with p q, D ψ[p] or p Q with p q, D ψ[p], and (b) for any r, s Q with r < s, there exists a formula θ(x) of L(D), all of whose quantifiers are bounded by closed terms, such that D x[r x s (ψ(x) θ(x))]. Thus, thanks to these results, the o-minimality of the structure B; 0, 1, <, + [r,s], [r,s], r,s B,r<s follows readily from that of B; 0, 1, <, + [r,s], [r,s], [r,s] r,s B,r<s (which, of course, follows immediately from that of B, [r,s] r,s B,r<s ). 3 Towards the proof of the main theorem Let F, A, B, B, G be as in the hypothesis of Theorem 2. Say G = dom(g) so that G is the underlying set of a divisible additive subgroup of R, 0, + 3

4 containing 1. We shall fuse G onto B via the restriction to B of a certain function φ : R R satisfying several properties, the first few of which are as follows:- 3.1 φ is analytic and bijective; 3.2 φ(0) = 0, φ(1) = 1; 3.3 x R φ (1) (x) > 0 (so that φ 1 is also analytic and both φ and φ 1 are monotonically increasing); 3.4 φ maps B bijectively onto G. Now define the function : R 2 R by x y := φ 1 (φ(x) + φ(y)) (for x, y R). Then is analytic and, by 3.4, maps B 2 to B.Also, by , the map φ B is an isomorphism from the structure B; 0, 1, <, to G. In order to establish the o-minimality of the structure B; [r,s] r,s B,r<s consider the expansion A := A;, φ (i) i 1 of A. If 3.5 for all i 1, the i th derivative φ (i) of φ maps B to B, then B; has a (unique) expansion, B say, to a substructure of A. We shall construct φ so that, in addition to , the following holds:- 3.6 B is existentially closed in A. Now before proceeding we need the following Lemma 1. For all multi-indices l, m N 2 with l + m 1, the l, m th derivative x (l,m) y of x y is expressible as a rational function, with integer coefficients, in the quantities φ (i) (x), φ (j) (y) and φ (k) (x y) for 1 i, j, k l + m. A similar result holds for the functions µ N (for N N) defined by N {}}{ µ N (x) := x x x (= φ 1 (N φ(x))) (for x R). Further, the denominators of these expressions are,respectively, powers of φ (1) (x y) and φ (1) (µ N (x)),so by 3.3 they are nowhere vanishing. Proof. Differentiation of the identity φ(x y) = φ(x) + φ(y) yields x (1,0) y = φ(1) (x) φ (1) (x y) and x (0,1) y = 4 φ(1) (y) φ (1) (x y).

5 The result for (both parts) now follows easily by induction on l + m using the chain rule. Similarly µ (1) N (x) = N φ(1) (x) φ (1) (µ N (x)) and so, again by induction, we obtain that µ (1) N (x) is expressible as a rational function, with integer coefficients, in the quantities φ (i) (x), φ (j) (µ N (x)) for 1 i, j l with denominator a power of φ (1) (µ N (x)). Remark: The crucial point in the above expressions is that they do not involve the function φ itself, only its derivatives. Now consider the set { } {µ N : N N} of analytic functions and let H denote its closure under differentiation. Let H res denote the corresponding set {h [0,1] : h H} of restricted analytic functions and let A denote the expansion A; H res of A. Then by Proposition 1, A is o-minimal and has a model complete theory. But by Lemma 1 each function in H res (and H) is certainly existentially definable in A and hence, by 3.6, the structure B has a (unique) expansion to a substructure, B say, of A and, further, B is existentially closed in A. But the theory of A, being model complete, is 2 -axiomatizable so it follows that B is elementarily equivalent to A and hence, by a fundamental result of [11], B is also o-minimal. This obviously implies the o-minimality of its reduct B := B; [0,1], µ N [0,1] N N. However, the presence here of the functions µ N [0,1] clearly implies that all the restricted functions [r,s] (for r, s B, r < s) are definable (with parameters) in B, and hence the structure B; [r,s] r,s B,r<s is o-minimal. Theorem 2 is thus established. So, it remains to construct a function φ : R R satisfying The countably many conditions imposed by 3.5 and 3.6 will be satisfied one by one, and a suitable limit of the inductively defined approximations will then be taken. An appropriate space in which to take such a limit is described in the next section. 4 The Approximation Space I denote by B the set of all entire functions f : C C for which e R M(f; R) is bounded (for R 0) where, as is usual, M(f; R) denotes sup z R f(z), 5

6 and set f := sup e R M(f; R) R 0 for such f. Then B, is easily seen to be a normed linear space. In fact, it is a Banach space. For if f n : n N is any Cauchy sequence in B, then certainly K n N, f n K, so n N, R 0, M(f n ; R) e R K. It follows from Montel s theorem (see [13] page 149 for the exact version that we need here, and see also the subsequent discussion in [13] for why we are working in a complex space even though we are only interested in limits of real functions) that some subsequence f ni : i N of f n : n N converges uniformly on compact subsets of C to an entire function, f say. Let ɛ > 0, and choose N = N(ɛ) large enough so that n, m N f n f m < ɛ/2. Now, for each R > 0 i 0 = i 0 (R) such that M((f f ni0 ), R) < ɛ/2, and we may take i 0 so that n i0 N. It follows that for any n N, M((f f n ), R) M((f f ni0 ), R) + M((f n f ni0 ), R)) < ɛ/2 + e R f n f ni0 < (1 + e R )ɛ/2 and hence f f n < (e R + 1)ɛ/2 < ɛ. Thus the sequence f n : n N converges to f in the space B, as required. Remark: The choice of the dampening factor e R in the definition of is not crucial. We merely require that it dominates every polynomial in R for large R. However, this particular choice does make some later calculations less cumbersome because of the following result. Lemma 2. The map sending f to f (1) is a bounded linear operator on B (with norm e). Proof. Let f B and R > 0. Then for any z C with z R we have, by Cauchy s formula, f (1) (z) = 1 f(w) 2πi C (w z) dw, 2 where C is the circle centred at z and having radius 1. Hence, since w R+1 for w on C, f (1) (z) M(f; R+1) e R+1 f. Thus f (1) e f, as required. We can now state and prove the interpolation property we require of our space B. It is, in fact, a version of the standard Lagrange Interpolation Theorem suitably adapted for present needs. 6

7 Lemma 3. Let n, N be positive integers and K a non-negative integer. Then there exists a positive real number C = C(K, n, N) 1 with the following property: Suppose we are given (i) a 1,..., a n C with a i N and a i a j N 1 for 1 i, j n and i j; (ii) ɛ R with 0 < ɛ < 1; (iii) F B; (iv) non-negative integers K 1,..., K n with K K n = K; (v) b i,j C (for i = 1,..., n, j < K i ) satisfying F (j) (a i ) b i,j < ɛ. Then there exists G B with F G < C ɛ such that G (j) (a i ) = b i,j for each i = 1... n and j < K i. Further, if the a i s and b i,j s are real and if F [R] R, then we may choose G so that G[R] R. Proof. By induction on K. For K = 0 we take C(0, n, N) = 1 and there is nothing to prove. Suppose K > 0 and that C(K 1, n, N) 1 has been found. Set C(K, n, N) := C(K 1, n, N) (2 N 2K (2K 2 + 1) K e N+K + 1). To see that this works, suppose we are given the data as in (i)-(v). Then at least one K i is positive (as K > 0), say K n > 0. By the inductive hypothesis we may choose G 0 B with F G 0 < C(K 1, n, N) ɛ, and such that G (j) 0 (a i ) = b i,j for each i = 1,..., n and j < K i apart from, possibly, the pair i = n and j = K n 1. Now define the functions H, G : C C by and H(z) = n 1 i=1 (z a i) K i n 1 i=1 (a n a i ) K i (z a n) Kn 1, (K n 1)! G(z) = G 0 (z) + H(z) (b n,kn 1 G (Kn 1) 0 (a n )). Since B contains all polynomials (cf, the remark immediately before the statement of Lemma 2) we have that H B and hence G B. Further, for R > 0, e R M(H; R) e R (R+N)K (by (i) and (iv)). But the upper N K 7

8 bound here achieves its maximum at some R 2NK 2, so we obtain that H N 2K (2K 2 + 1) K. Now observe that H (Kn 1) (a n ) = 1 and that H (j) (a i ) = 0 for all other pairs i, j with i = 1,..., n and j < K i and hence, in all cases, G (j) (a i ) = b i,j (by the choice of G 0 ). Notice also that if a i R, b i,j R for i = 1,..., n and j < K i, and if our inductive assumption includes the statement that G 0 [R] R, then G[R] R. So it only remains to calculate the required upper bound for F G. For this, we first note that it follows immediately from Lemma 2 that the map sending f to f (Kn 1) is a bounded linear operator on B of norm e Kn 1. Hence F (Kn 1) G (Kn 1) 0 e Kn 1 F G 0 < e K C(K 1, n, N) ɛ. Thus, using also the lemma hypotheses (v) and (i), we obtain b n,kn 1 G (Kn 1) 0 (a n ) b n,kn 1 F (Kn 1) (a n ) + F (Kn 1) (a n ) G (Kn 1) < ɛ + M(F (Kn 1) G (Kn 1) 0 ; N) ɛ + e N F (Kn 1) G (Kn 1) 0 ɛ(1 + e N+K C(K 1, n, N)). However, by the choice of G 0 and the definition of G, F G 0 (a n ) F G 0 + G G 0 < C(K 1, n, N) ɛ + H b n,kn 1 G (Kn 1) 0 (a n ). Hence, by combining the above inequalities (and also using the fact that C(K 1, n, N) 1 ), we finally obtain that F G < C(K 1, n, N) ɛ + N 2K (2K 2 + 1) K ɛ (1 + e N+K C(K 1, n, N)) C(K 1.n, N)(1 + 2N 2K (2K 2 + 1) K e N+K ) ɛ = C(K, n, N) ɛ, as required. We shall also require the following lemma and definition. Lemma 4. Let V := {f B : C > 1 4, x R, f (1) (x) C e x }. Then V is open in B. Proof. Let f V. Choose C > 1 such that x R, f (1) (x) C e x. 4 Let C = 4C+1, so that C > Let ɛ = (C 1) 1, so that ɛ > e I claim that if g B and g f < ɛ, then g V, thus establishing the lemma. Indeed, we have by Lemma 2 that ɛ > g f e 1 g (1) f (1) and so x R, g (1) (x) f (1) (x) e ɛ e x. Hence, x R, g (1) (x) f (1) (x) e ɛ e x C e x e ɛ e x = C e x, which shows that g V as required. Definition 1. With V as in the statement of Lemma 4, we set V := {f V : f(0) = 0, f(1) = 1 and f[r] R}. 8

9 V is not empty. For example φ 0 V where, for z C, φ 0 (z) := (e e 1 ) 1 (e z e z ). Notice also that if f V and g := f R then g : R R is a real analytic function satisfying g(0) = 0, g(1) = 1 and g (1) (x) 1 for all x R. In 4 particular, g is monotonically increasing and bijective. Further, since convergence with respect to our norm on B certainly implies pointwise convergence, and because of lemma 2, it follows that the same is also true of a function g = f R with f only assumed to lie in the closure (in B) of V. 5 Completion of the proof of Theorem 2 Recall the situation: F is a countable collection of restricted analytic functions (closed under differentiation), A denotes the structure R; F and B is a countable elementary substructure of A. We denote the language of B + by L +, ie. L + is the language of the structure A (or B) together with a new constant symbol c b for each element b of B, where B = dom(b). We are also given a countable elementary substructure G of R; 0, 1, <, + with dom(g) = G, say. Let L + 0 denote the language L + {F i : i 1}, where each F i is a unary function symbol, and let L + 1 denote the language L + 0 {π}, where π is a binary function symbol. If φ : R R is any analytic bijection, then A φ denotes the L + 1 -expansion of A in which F i is interpreted as the i th derivative φ (i) of φ (for i 1) and π by the function φ : R 2 R : x, y φ 1 (φ(x) + φ(y)). Our aim, then, is to construct φ so that are satisfied (for A = A φ and B the (unique) expansion of B to an L + 1 -substructure of A ). As we have already indicated, φ will be the (restriction to R of the) limit in the space B of a Cauchy sequence whose terms lie in V. (Here, and henceforth, V and V denote the sets defined in the statement of Lemma 4 and Definition 1 respectively.) Then 3.1,3.2 and 3.3 will be satisfied by the remarks at the end of the previous section. For 3.4 and 3.5 we require the following definition and lemma. Definition 2. Suppose that ψ B, that Q is a positive integer and that B is a finite subset of B containing 0 and 1. Then ψ is called B, Q -correct if ψ V and, for each b B, ψ(b) G and, for i = 1,..., Q 1, ψ (i) (b) B. Lemma 5. With the notation and assumptions of Definition 2, let ψ be B, Q -correct. Let ɛ > 0 be small enough so that θ B, θ ψ < ɛ 9

10 implies θ V (cf. Lemma 4). Then for any Q N, a B and g G, (i) there is ψ 0 B such that ψ 0 is B {a}, Q -correct, and (ii) there is ψ 1 V and b 0 B such that ψ 1 (b 0 ) = g. Further, for k = 0, 1 we have ψ k ψ < ɛ and, for each b B and j < Q, (b) = ψ(j) (b). ψ (j) k Proof. (i) We may assume that Q > Q and a / B. Say B = {a 1,..., a n 1 } (with the a i s distinct) and write a n for a. Choose N N so that a i N and a i a j N 1 for 1 i, j n with i j. Let C = C(nQ, n, N) (cf Lemma 3). For i = 1,..., n 1, j < Q let b i,j := ψ (j) (a i ) and for Q j < Q choose b i,j B so that ψ (j) (a i ) b i,j < ɛ (using the denseness C of B in R ). Similarly, choose b n,0 G and b n,1,..., b n,q 1 B so that ψ (j) (a n ) b n,j < ɛ for j < C Q. The existence of the required ψ 0 B now follows readily from Lemma 3 and our hypothesis on ɛ. (ii) By the argument at the end of section 4, ψ is surjective, so there exists a R with ψ(a) = g. If a B we may take ψ 1 = ψ. Otherwise, choose N N so large that b N and b b N 1, for all b, b B {a} with b b. Now let C = C(nQ, n, 2N), where n = B + 1, and choose b 0 B to satisfy a b 0 < (2N) 1 and g ψ(b 0 ) < ɛ, and proceed as in (i) to obtain C the desired ψ 1 with ψ 1 (b 0 ) = g. We now state and prove the crucial correction lemma which will allow us to cope with the requirement 3.6. Lemma 6. Let Φ(v 1,..., v m ) be a quantifier-free formula of L + 1 and suppose that B is a finite subset of B containing, as well as 0 and 1, all those elements b of B for which the constant symbol c b occurs in Φ. Suppose further that Q 1 is a natural number greater than all those i 1 such that the function symbol F i occurs in Φ. Now let ψ V be a B, Q -correct function such that the formula Φ(v 1,..., v m ) is satisfiable in the structure A ψ and let ɛ be a positive real number. Then there exists ψ 0 V such that (i) ψ ψ 0 < ɛ, (ii) ψ (j) 0 (b) = ψ (j) (b) for all b B and j < Q, and 10

11 (iii) there exist b 1,..., b m B such that ψ 0 is B {b 1,..., b m }, Q -correct and A ψ0 Φ[b 1,..., b m ]. Proof. We may suppose, by the usual tricks of introducing variables, that Φ has the form (Φ 1 Φ 2 ) Φ 3, where Φ 1 is a quantifier-free L + -formula, Φ 2 is a conjunction of formulas of the form F j (x) = y for 1 j < Q, and Φ 3 is a conjunction of formulas of the form π(x, y) = z, where x, y, z {v 1,..., v m } {c b : b B }. (We are also using the hypotheses on Q and B here.) Notice that such a form of representation of a formula is stable under identification of variables and under substitution of a constant symbol c b (with b B ) for a variable. Hence we may assume (by a suitable inductive hypothesis on m) that Φ(v 1,..., v m ) is satisfied in the structure A ψ by some α 1,..., α m R m with α i / B for i = 1,..., m and α i α j for 1 i, j m, i j. Having fixed such α 1,..., α m, choose N 1 so that a N and a a N 1 for all a, a B {α 1,..., α m } with a a. We now transform Φ 2 (v 1,..., v m ) into a quantifier-free L + -formula in the variables v i, u i,j (for 1 i m, 1 j < Q), Φ 2(v 1,..., v m, ū) say, as follows. Firstly, replace each occurrence of F j (c b ) (for 1 j < Q, b B ) in Φ 2 by the constant symbol c β, where β = ψ (j) (b). (Notice that β B by the B, Q -correctness of ψ.) Secondly, replace each occurrence of F j (v i ) (for 1 j < Q, 1 i m) by the new variable u i,j. Then clearly we have: (*) for any θ V satisfying θ (j) (b) = ψ (j) (b) for all b B and j = 1,..., Q 1, and any γ 1,..., γ m R, A θ (Φ 1 Φ 2 )[γ 1,..., γ m ] A + (Φ 1 Φ 2)[γ 1,..., γ m, δ] where δ denotes the m(q 1)-tuple with i, j -th entry δ i,j := θ (j) (γ i ) (for 1 i m, 1 j < Q), and A + is the natural expansion of A to an L + - structure. In particular, by applying (*) with θ = ψ, γ i = α i and (hence) δ i,j = ψ (j) (α i ) (for 1 i m and 1 j < Q), it follows that for arbitrarily small boxes R mq containing the point α 1,..., α m, δ and having corners in B mq, we have A + v 1,..., v m, ū (Φ 1 Φ 2)(v 1,..., v m, ū). 11

12 But we may transfer this L + -sentence to B + (since B + A + ) and hence obtain a sequence of points α (p) 1,..., α m (p), δ (p) B mq converging to α 1,..., α m, δ as p and each witnessing (in B +, and so in A + ) the formula (Φ 1 Φ 2). Notice also that for each i = 1,..., m, and j = 1,..., Q 1, ψ (j) (α (p) i ) δ (p) i,j 0 as p. As for the formula Φ 3 (v 1,..., v m ), this may be regarded, as it stands, as a formula of the language of the structure R := R; 0, 1, <, +, g g G and we may interpret the function symbol π by + (ie. the usual addition function on R) and the constant symbols c b (for b B ) as ψ(b). (Notice that ψ(b) G for such b by the B, Q -correctness of ψ.) Then, since x ψ y = z if and only if ψ(x) + ψ(y) = ψ(z) (for any x, y, z R), we have (**) for any θ V satisfying θ(b) = ψ(b) for all b B, and any γ 1,..., γ m R, A θ Φ 3 [γ 1,..., γ m ] R Φ 3 [ψ(γ 1 ),..., ψ(γ m )]. Thus,by a similar argument to that following (*) above (except that this time we use the fact that G; g g G R, as given by the hypothesis of Theorem 2), we obtain a sequence of points g (p) 1,..., g m (p) G m converging to ψ(α 1 ),..., ψ(α m ) as p and satisfying in G; g g G, and so also in R, the formula Φ 3. Notice that for each i = 1,..., m, ψ(α (p) i ) g (p) i 0 as p. We now proceed as in the proof of Lemma 5: we let m = B, C = C((m + m ) Q, m + m, 2N) (cf. Lemma 3) and we choose p large enough so that for each i = 1,..., m and j = 1,..., Q 1 we have that ψ (j) (α (p) i ) δ (p) i,j < ɛ and ψ(α(p) C i ) g (p) i < ɛ. We may suppose further C that a 2N and that a a (2N) 1 for a, a B {α (p) 1,..., α m (p) }, a a. Having fixed such a p we now apply Lemma 3, taking {a 1,..., a n } to be B {α (p) 1,..., α m (p) } (keeping the values ψ (j) (b) fixed for b B, j < Q), to obtain a function ψ 0 B satisfying (i) ψ ψ 0 < ɛ, (ii) ψ (j) 0 (b) = ψ (j) (b) for b B, j = 0,..., Q 1, (iii) ψ (j) 0 (α (p) i ) = δ (p) i,j for i = 1,..., m, j = 1,..., Q 1, and (iv) ψ 0 (α (p) i ) = g (p) i for i = 1,..., m. If we now apply (*) and (**) with θ = ψ 0 and γ i = α (p) i for i = 1,..., m, 12

13 we see from (ii), (iii) and (iv) that A ψ0 ((Φ 1 Φ 2 ) Φ 3 )[α (p) 1,..., α m (p) ] and that ψ 0 is B {α (p) 1,..., α m (p) }, Q -correct, as required. Now to complete the proof of Theorem 2 we let E denote the collection of all quantifier-free fromulas of L + 1 and fix an enumeration, Φ n, a n, g n : n 1 say, of the set E B G. Suppose that the variables occuring in the formula Φ n are amongst v 1,..., v mn. Pick φ 0 V (cf. the remark following Definition 1) and choose ɛ 0 > 0 so small that θ V whenever θ B and φ 0 θ < ɛ 0 (cf. Lemma 4). Let B 0 = {0, 1}, Q 0 = 1. Our aim is to construct inductively a sequence φ n : n N of elements of V, finite subsets B 0 B 1 B n of B and integers Q 0 < Q 1 < < Q n <, such that for each n 0 properties (1) n -(7) n below hold. At this point it is convenient to introduce the notation θ 1 θ 2 (mod B, Q), where θ 1, θ 2 B, B B and Q is a positive integer, to mean that θ (j) 1 (b) = θ (j) 2 (b) for each b B and j < Q. (1) n for all b B, if the constant symbol c b occurs in Φ n then b B n ; (2) n for all j 1, if the function symbol F j occurs in Φ n the j < Q n ; (3) n a n B n and φ 1 n (g n ) B n ; (4) n φ n is B n, Q n -correct; (5) n φ n 1 φ n < 2 ɛ 0 3 n ; (6) n φ n φ n 1 (mod B n 1, Q n 1 ); (7) n either for no θ V with φ n θ < ɛ 0 3 n and θ φ n (mod B n, Q n ) is Φ n satisfiable in the structure A θ, or else there exist b 1,..., b mn B n such that A φn Φ n [b 1,..., b mn ]. Now notice that all these conditions hold when n = 0 (mostly vacuously so). So suppose that φ n, B n and Q n have been constructed to satisfy (1) n -(7) n for some n 0. We proceed to construct φ n+1, B n+1 and Q n+1 as follows. Firstly, by repeated use of Lemma 5 (using sufficiently small ɛ s) we may find B n+1 B n, Q n+1 > Q n so that (1) n+1 -(3) n+1 hold, and a function ψ V such that φ n ψ < ɛ 0 3 n 2, ψ is B n+1, Q n+1 -correct and ψ φ n (mod B n, Q n ). (We are using (4) n here.) 13

14 Now if there is no θ V with ψ θ < ɛ 0 3 n 1 and θ ψ (mod B n+1, Q n+1 ) such that Φ n+1 is satisfiable in A θ, then we may set φ n+1 = ψ and (1) n+1 -(7) n+1 are clearly satisfied. Otherwise, choose such a θ. Then we may apply Lemma 6 to obtain a function φ n+1 V such that θ φ n+1 < ɛ 0 3 n 2, φ n+1 θ (mod B n+1, Q n+1 ) and such that there exist b 1,..., b mn B with A φn+1 Φ n [b 1,..., b mn ]. Further, φ n+1 is B n+1 {b 1,..., b mn }, Q n+1 -correct. So if we replace B n+1 by B n+1 {b 1,..., b mn } then (1) n+1 -(4) n+1, (6) n+1 and (7) n+1 are immediate. For (5) n+1 we have φ n+1 φ n φ n θ + θ ψ + ψ φ n < ɛ 0 3 n 2 + ɛ 0 3 n 1 + ɛ 0 3 n 2 < 2 ɛ 0 3 (n+1), which indeed gives (5) n+1 and completes the inductive construction. Now the properties (5) n clearly imply that φ n : n N is a Cauchy sequence in B (consisting of elements of V ) and hence converges to some φ B. By the remarks at the end of Section 4 it follows that the restriction of φ to R satisfies Further, by the properties (6) n these same remarks imply (*) for all n N, φ φ n (mod B n, Q n ). Hence (by the properties (3) n, (4) n ) φ also satisfies 3.5. Finally, to see that φ satisfies 3.6, we require a calculation. For n N let ɛ n := 2 ɛ 0 3 (n+1) φ n φ n+1 so that ɛ n > 0 by (5) n+1. Then for m, n N with m > n we have φ n φ m i=n φ i φ i+1 + φ m+1 φ ɛ n + 2 ɛ 0 m i=n 3 (i+1) + φ m+1 φ (by (5) n+1,..., (5) m+1 ) ɛ n + ɛ 0 3 n (by letting m ). Hence (**) for all n N, φ n φ < ɛ 0 3 n. Notice that by setting n = 0 here it follows, by the definition of ɛ 0, that φ V (so φ V ). Now consider an arbitrary quantifier-free formula of L it will be Φ n (v 1,..., v mn ) for some n - and suppose that it is satisfiable in the structure A φ. By applying (7) n with θ = φ, it follows from (*) and (**) (and the comment following (**)) that A φn Φ n [b 1,..., b mn ] for some b 1,..., b mn B n. But by (1) n, 14

15 (2) n and (*) this clearly implies that A φ Φ n [b 1,..., b mn ]. This establishes 3.6 (recall the comments at the beginning of this section) and completes the proof of the main theorem. References [1] L van den Dries, Remarks on Tarski s problem concerning R, +,, exp, Logic colloquium 1982, North Holland, 1984, [2] M Edmundo, Structure theorems for o-minimal expansions of groups, Annals of Pure and Applied Logic 102 (2000), [3] A Gabrielov, Projections of semianalytic sets, Funct. Anal. Appl. 2 (1968), [4] A Gabrielov, Complements of subanalytic sets and existential formulas for analytic functions, Inv. Math. 125 (1996), [5] D Marker,Y Peterzil and A Pillay, Additive reducts of real closed fields, J Symb Logic 57 (1992), [6] C Miller and S Starchenko, A growth dichotomy for o-minimal expansions of ordered groups, TAMS 350 (1998), [7] Y Peterzil, A structure theorem for semibounded sets in the reals, J. Symb. Logic 57 (1992), [8] Y Peterzil, Reducts of some structures over the reals, J. Symb. Logic 58 (1993), [9] Y Peterzil and S Starchenko, On torsion-free groups in o-minimal structures, preprint (September,2003). [10] A Pillay,P Scowcroft and C Steinhorn, Between groups and rings, Rocky Mountain J of Math. 19(3) 1989, [11] A Pillay and C Steinhorn, Definable sets in ordered structures,i, TAMS 295 (1986), [12] R Poston, Defining multiplication in o-minimal expansions of the positive reals, J. Symb. Logic 60 (1995),

16 [13] R Remmert, Classical topics in complex function theory, Springer, [14] A Strzebonski, One-dimensional groups definable in o-minimal structures, J. Pure and Applied Algebra 96 (1994),