O-MINIMAL EXPANSIONS OF REAL CLOSED FIELDS AND COMPLETENESS IN THE SENSE OF SCOTT

OLIVIER FRÉCON

Abstract. We consider an o-minimal expansion M_{0} = (R_{0}, <, +, ⋯) of a real closed field, and a real closed field R, complete in the sense of D. Scott, containing R_{0} as a dense subfield. We show that M_{0} has an elementary extension M = (R, <, +, ⋯) with domain R. Moreover, such a structure M with domain R is unique.

Note

In an unpublished article, Antongiulio Fornasiero proved a more general result than the main theorem of this paper. Indeed, he showed a similar result for dminimal expansions of a real closed field [1, Proposition 11.6]. However our proofs are very different.

1. Introduction

By the Compactness Theorem, it is easy to elementarily embed any expansion M_{0} = (R_{0}, <, +, ⋯) of a real closed field in an expansion M = (R, < +, ⋯) of a Scott-complete real closed field, that is complete in the sense of Dana Scott (Definition 1.1). However, we have little control on the size of the elementary extension obtained. For instance, if R_{0} is countable, it is possible that the field R has no countable dense subfield. The main result of this paper shows that any o-minimal expansion of a real closed field R_{0} is elementarily embedded in an expansion of a Scott-complete real closed field, in which R_{0} is dense (Theorem 1.2). We note that, since we consider an expansion of a real closed field, not just the field structure, the model completeness of the theory of real closed fields will not help us.

We recall the definition of a complete real closed field in the sense of [2].

Definition 1.1.. – If K and L are ordered fields and K ⊆ L, then K is dense in L if between any two distinct elements of L there lies an element of K.

A given ordered field is called Scott-complete if it has no proper extension to an ordered field in which the given field is dense.

The main result of this paper is the following.

Theorem 1.2.. – Any o-minimal expansion of a real closed field R_{0} has an elementary extension of domain a Scott-complete field R in which R_{0} is dense. Moreover, for a fixed field R, this elementary extension is unique.

Date: May 15, 2015.

2010 Mathematics Subject Classification. 03C64.

Key words and phrases. Real closed field, o-minimal structure.

2. Notations and preliminary results

For the rest of this paper, we fix an o-minimal expansion M_{1} = (R_{0}, <, ⋯) of a real closed field R_{0}, and we denote by R a Scott-complete field in which R_{0} is dense: its existence is ensured by Fact 2.1 below.

Fact 2.1.. – [2, Theorem 1] Given any ordered field K, there is a Scott-complete ordered field K̂ in which K is dense. Any other Scott-complete ordered field in which K is dense is isomorphic to K̂ by a unique isomorphism that is the identity on K.

In the following, we say that a set X is definable if it is definable in the structure M_{1}.

For each integer k and each subset A of R^{k}, we denote by \overline{A} its (topological) closure in R^{k}. For each subset X of R_{0}^{k}, we denote by X̆ the union of subsets \overline{F} for F a R_{0}-closed definable subset contained in X.

For clarity, we use a very different notation for the closure in R_{0}^{k} of a subset X of R_{0}^{k}: we denote it by cl X. Moreover, we denote by ∂X its frontier: ∂X = cl X \ X.

For any element x = (x_{1}, …, x_{k}) of R^{k}, we consider

|x| = max{|x_{i}| | i ∈ {1, …, k}}

Remark 2.2.. – • For any subset X of R_{0}^{k}, it follows from the definition of X̆ that X = X̆∩R_{0}^{k}.

• The proof of Theorem 1.2 will show that the elementary extension of M_{1}

of domain R has the following property:

for any two integers k and m, if A is any subset of R_{0}^{k} defined by a formula φ(\overline{x}, \overline{a}) with free variables \overline{x} = (x_{1}, …, x_{k}) and parameters \overline{a} ∈ R_{0}^{m}, then the subset of R^{k} defined by φ(\overline{x}, \overline{a}) is Ă.

The first lemma is certainly well-known, but we could not find a reference for it.

Lemma 2.3.. – Let k be an integer, X a subset of R_{0}^{k}, and f : X → R_{0} be a uniformly continuous map. Then for each x ∈ \overline{X}, the following limit exists:

lim_{\begin{matrix}y∈X \\ y→x \end{matrix}} f(y)

Proof – Let ε ∈ R_{0}^{>0}. Since f is uniformly continuous on X, then we may associate with ε some δ(ε) ∈ R_{0}^{>0} such that whenever |v − u| < δ(ε) for u and v in X, we have |f(v) − f(u)| < ε. Then for each y ∈ X and z ∈ X such that |y − x| < δ(ε)/2 and |z − x| < δ(ε)/2, we have |z − y| < δ(ε) and |f(z) − f(y)| < ε.

Let C_{ε}^{−} (resp. C_{ε}^{+}) be the set of elements α ∈ R_{0} such that α < f(y) − ε (resp. α > f(y) + ε) for some y ∈ X satisfying |y − x| < δ(ε)/2. Let C^{−} (resp. C^{+}) be the union of subsets of the form C_{ε}^{−} (resp. C_{ε}^{+}) for ε ∈ R_{0}^{>0}.

Claim 1: for each a ∈ C^{−} and each b ∈ C^{+}, we have a < b. In particular, the set C^{−} ∩ C^{+} is empty.

There exist y_{1} and y_{2} in X, and ε_{1} and ε_{2} in R_{0}^{>0}, such that |y_{1} − x| < δ(ε_{1})/2, |y_{2} − x| < δ(ε_{2})/2, a < f(y_{1}) − ε_{1} and b > f(y_{2}) + ε_{2}. Let δ = min{δ(ε_{1}), δ(ε_{2})}. We fix y ∈ X such that |y −x| < δ/2. Then we have |f(y)−f(y_{1})| < ε_{1} < f(y_{1})−a and |f(y) − f(y_{2})| < ε_{2} < b − f(y_{2}), so we obtain a < f(y) < b.

Claim 2: for each r ∈ R_{0}^{>0}, there is a ∈ C^{−} and b ∈ C^{+} such that b − a < r.

Let y ∈ X such that |y − x| < δ(r/3)/2. Then for a = f(y) − r/2 and b = f(y) + r/2, we have a ∈ C_{r/3}^{−} and b ∈ C_{r/3}^{+}, and we obtain a ∈ C^{−}, b ∈ C^{+} and b − a < r.

Conclusion: By [3, §2, Lemma 2.8] (or [2]), there is a unique ω ∈ R satisfying a < ω < b for every a ∈ C^{−} and b ∈ C^{+}. Hence for each ε ∈ R_{0}^{>0}, there is δ(= δ(ε)) ∈ R_{0}^{>0} such that whenever |y − x| < δ/2 for y ∈ X, we have ω ≥ f(y) − ε and ω ≤ f(y) + ε, so |f(y) − ω| ≤ ε. Now ω = lim_{\begin{matrix}y∈X \\ y→x \end{matrix}} f(y) exists. □

Lemma 2.4.. – Let k be an integer. If E and F are two closed definable subsets of R_{0}^{k}, then \overline{E} ∩ \overline{F} = \overline{E ∩ F}. In particular, if \overline{E} ∩ \overline{F} is non-empty, then E ∩ F is non-empty too.

Proof – We have just to prove that \overline{E ∩ F} contains \overline{E} ∩ \overline{F}. Let x ∈ \overline{E} ∩ \overline{F}. For each ε ∈ R_{0}^{>0}, we fix u_{ε} ∈ E such that |u_{ε}−x| < ε. Let B_{ε} = {y ∈ R_{0}^{k} | |u_{ε}−y| ≤ ε}. Since x ∈ \overline{E} ∩\overline{F} and since |u_{ε} −x| < ε, we have x ∈ \overline{E_{ε}} ∩\overline{F_{ε}} where E_{ε} = B_{ε} ∩E and F_{ε} = B_{ε} ∩ F. Moreover, we note that E_{ε} and F_{ε} are closed and bounded definable subsets of R_{0}^{k}.

We show that E_{ε} ∩F_{ε} is non-empty. Let f_{ε} : E_{ε} ×F_{ε} → R_{0} defined by f_{ε}(z, z′) = |z − z′|. It is a definable continuous function, so its image is closed and bounded (see [4, Chapter 6 §1]). For each η ∈ R_{0}^{>0}, there exist u ∈ E_{ε} and v ∈ F_{ε} such that |u − x| < η/2 and |v − x| < η/2, so we have f(u, v) = |u − v| < η. Since the image of f_{ε} is closed and bounded, this implies that it contains zero. Hence there exist a ∈ E_{ε} and b ∈ F_{ε} such that f_{ε}(a, b) = 0. Now we have a = b ∈ E_{ε} ∩ F_{ε}, and E_{ε} ∩ F_{ε} is non-empty.

Since |u_{ε} − x| < ε, we have |y − x| < 2ε for any y ∈ B_{ε}. Thus, the previous paragraph proves that for each ε ∈ R_{0}^{>0}, there exists y ∈ E∩F such that |y−x| ≤ 2ε, so x ∈ \overline{E ∩ F}, as desired. □

Corollary 2.5.. – For each subset X of R_{0}^{k}, and each x ∈ X̆, there is a closed and bounded definable subset F of X such that x ∈ \overline{F}.

Proof – Since x ∈ X̆, there exists x_{0} ∈ X such that |x − x_{0}| < 1. We consider F_{1} = {y ∈ R_{0}^{k} | |y − x_{0}| ≤ 1}. Then F_{1} is a closed and bounded definable subset of R_{0}^{k}. By density of R_{0} in R and since |x − x_{0}| < 1, we have x ∈ \overline{F_{1}}.

Moreover, by the definition of X̆, there is a closed definable subset F_{2} of X such that x ∈ \overline{F_{2}}. Then F = F_{1} ∩ F_{2} is a closed and bounded definable subset of X, and Lemma 2.4 gives x ∈ \overline{F}. □

Proposition 2.6.. – Let k be an integer. If {A_{1}, …, A_{m}} is a partition of R_{0}^{k} into definable subsets, then {Ă_{1}, …, Ă_{m}} is a partition of R^{k}.

Proof – First we show that R^{k} is the union of Ă_{1}, …, Ă_{m}. Since each definable subset of R_{0}^{k} has a decomposition into cells (see [4, Chapter 3 §2] for more details), we may assume that A_{1}, …, A_{m} are cells.

Let x ∈ R^{k}. We show that x ∈ Ă_{j} for some j ∈ {1, …, m}. We may assume x ∉ R_{0}^{k}. By finiteness of the partition {A_{1}, …, A_{m}}, there exists r ∈ R^{>0} such that, for any s ∈]0, r], the set I = {i ∈ {1, …, m} | ∃a ∈ A_{i}, |x − a| < s} is constant. Since R_{0} is dense in R, the set I is non-empty, and by the definition of I, the point x is in the R^{k}-closure \overline{A_{i}} of A_{i} for each i ∈ I.

Let d be the smallest integer such that there is j ∈ {1, …, m} and a definable subset B of A_{j} of dimension d with x contained in \overline{B}. Let B_{i} = A_{i} ∩ ∂B for each i ∈ {1, …, m}. Then for each i ∈ {1, …, m}, the subset B_{i} of A_{i} is definable (see [4, Chapter 1 §3]) and we have dim B_{i} ≤ dim ∂B < dim B = d [4, Chapter 4 §1]. By the minimality of d, the point x is contained in the R^{k}-closure \overline{B_{i}} of B_{i} for no i ∈ {1, …, m}. Consequently, there is t ∈ R^{>0} such that |y − x| > t for any y ∈ ∪_{i=1}^{m}B_{i}, and we may choose t ∈ R_{0} as R_{0} is dense in R. Since ∪_{i=1}^{m}B_{i} = ∂B and since x ∈ \overline{B}, there exists b_{0} ∈ B such that |b_{0} − x| < t/2. Let B_{f} = \{b ∈ B | |b − b_{0}| ≤ t/2\}. By the choices of t and b_{0}, we have x ∈ \overline{B_{f}} and ∂B ∩ B_{f} = ∅. This implies that the set B_{f} is a closed definable subset of B and that x belongs to Ă_{j}, as desired.

We show that Ă_{i} ∩ Ă_{j} = ∅ for any distinct elements i and j of {1, …, m}. Otherwise there is a closed definable subset F_{i} (resp. F_{j}) of A_{i} (resp. A_{j}) such that \overline{F_{i}} ∩ \overline{F_{j}} ≠ ∅. By Lemma 2.4, the set F_{i} ∩ F_{j} is non-empty, contradicting A_{i} ∩ A_{j} = ∅. Thus {Ă_{1}, …, Ă_{m}} is a partition of R^{k}. □

3. Proof of the main theorem

We provide three preparatory results before the final argument.

Lemma 3.1.. – Let S_{0} = {(u, v) ∈ R_{0}^{2} | u < v} and S = {(u, v) ∈ R^{2} | u < v}. Then we have S = S̆_{0}.

Proof – First we show that S̆_{0} contains S. Let (u, v) ∈ S. Then there exists r ∈ R_{0}^{>0} such that v − u > r. Let (u_{0}, v_{0}) ∈ R_{0}^{2} such that |u − u_{0}| < r/4 and |v − v_{0}| < r/4, and let F = {(x, y) ∈ R_{0}^{2} | |x − u_{0}| ≤ r/4, |y − v_{0}| ≤ r/4}. In particular, we have (u, v) ∈ \overline{F}. Moreover we have v_{0} − u_{0} > r/2, so we obtain y−x > 0 for each (x, y) ∈ F, and F is a closed definable subset of S_{0}. Consequently (u, v) belongs to S̆_{0}, and S̆_{0} contains S.

Now we show that S contains S̆_{0}. Let (u, v) ∈ S̆_{0}. By Corollary 2.5, there is a closed and bounded definable subset F_{0} of S_{0} such that (u, v) ∈ \overline{F_{0}}. Let f : F_{0} → R_{0} defined by f(x, y) = y−x. Then f is a definable continuous function, so its image is closed and bounded. Let m = min{f(x, y) | (x, y) ∈ F_{0}}. Then we have b − a ≥ m for each (a, b) ∈ \overline{F_{0}}. Since F_{0} ⊆ S_{0}, we have m > 0 and we obtain (u, v) ∈ S. □

Corollary 3.2.. – Let T_{0} = {(u, v) ∈ R_{0}^{2} | u = v} and T = {(u, v) ∈ R^{2} | u = v}. Then we have T = T̆_{0}.

Proof – Let S_{0} = {(u, v) ∈ R_{0}^{2} | u < v} and S_{1} = {(u, v) ∈ R_{0}^{2} | u > v}. Then {S_{0}, S_{1}, T_{0}} is a partition of R_{0}^{2}, and Proposition 2.6 says that {S̆_{0}, S̆_{1}, T̆_{0}} is a partition of R^{2}. Now Lemma 3.1 gives T̆_{0} = R^{2} \ (S̆_{0} ∪ S̆_{1}) = T. □

Lemma 3.3.. – If G_{0} (resp. H_{0}, K_{0}) denotes the graph of · (resp. +, −) in R_{0}^{3} (resp. R_{0}^{3}, R_{0}^{2}), then the graph of · (resp. +, −) in R^{3} (resp. R^{3}, R^{2}) is Ğ_{0} (resp. H̆_{0}, K̆_{0}).

Proof – Since · is a continuous map over R_{0}, its graph G_{0} in R_{0}^{3} is closed, and we have Ğ_{0} = \overline{G_{0}}. Moreover, since · is a continuous map over R, its graph G in R^{3} is closed, and since G contains G_{0}, it contains \overline{G_{0}} too. But R_{0} is dense in R, hence for each (x, y) ∈ R^{2} we have (x, y, x · y) ∈ \overline{G_{0}}, and G is contained in \overline{G_{0}}. We conclude that G = \overline{G_{0}} = Ğ_{0}, as desired.

In the same way, we show that the graph of + (resp. −) in R^{3} (resp. R^{2}) is H̆_{0} (resp. K̆_{0}). □

Proof of Theorem 1.2 – We denote by L_{1} the language of M_{1} = (R_{0}, <, ⋯). For each function symbol f of L_{1} with arity k, we consider a relation symbol S_{f} such that S_{f}^{M_{1}} is the graph of f^{M_{1}}, and for each constant symbol c of L_{1}, we consider a relation symbol S_{c} such that S_{c}^{M_{1}} = c^{M_{1}}. We obtain a relational language L and a structure M_{0} = (R_{0}, <, ⋯) in L. We have just to prove that there is a unique L-structure M with domain R such that M is an elementary extension of M_{0}.

We note that, for any integer k, a subset X of R_{0}^{k} is definable (in M_{1}) if and only if it is definable in M_{0}.

Uniqueness:

First we assume that the structure M exists. Let S be any relation symbol of arity k of L. Let F be a closed and bounded definable subset of S^{M_{0}}, and let φ(\overline{x}, \overline{a}) be an L-formula with free variables \overline{x} = (x_{1}, …, x_{k}) and parameters \overline{a} ∈ R_{0}^{m} such that F is defined by φ(\overline{x}, \overline{a}). Let F̃ be the subset of R^{k} defined by φ(\overline{x}, \overline{a}). Since M is an elementary extension of M_{0}, then F̃ contains F and it is closed and bounded in R^{k}. Thus F̃ contains \overline{F}. By Corollary 2.5, this implies that S^{M} contains S̆^{M_{0}}.

In the same way, the complementary of S^{M} in R^{k} contains ˘ { \atop R_{0}^{k} \ S^{M_{0}}}. Now Proposition 2.6 gives S^{M} = S̆^{M_{0}}, so, if it exists such a structure M, then it is unique.

Existence:

We consider the L-structure M = (R, <, ⋯) where for each relation symbol S of arity k of L, we define S^{M} by S^{M} = S̆^{M_{0}}. By Lemmas 3.1 and 3.3 and Corollary 3.2, it is sufficient to show that M is an elementary extension of M_{0}.

First we note that for each relation symbol S of arity k of L, we have S^{M_{0}} = S̆^{M_{0}} ∩ R_{0}^{k} (Remark 2.2), so M_{0} is a substructure of M.

Claim 1: if A is a definable subset of R_{0}^{k} for an integer k, then we have Ă_{0} = R^{k} \ Ă where A_{0} = R_{0}^{k} \ A.

Since {A, A_{0}} is a partition of R_{0}^{k}, then Proposition 2.6 says that {Ă, Ă_{0}} is a partition of R^{k}.

Claim 2: if A and B are two definable subsets of R_{0}^{k} for an integer k, then we

have Ă ∩ B̆ = ˘ { \atop A ∩ B}.

It is sufficient to prove that ˘ { \atop A ∩ B} contains Ă ∩ B̆. Let x ∈ Ă ∩ B̆. Then there exist a closed definable subset E of A and a closed defnable subset F of B such that x ∈ \overline{E} ∩ \overline{F}. Now x belongs to \overline{E ∩ F} ⊆ ˘ { \atop A ∩ B} by Lemma 2.4, and we obtain

Ă ∩ B̆ = ˘ { \atop A ∩ B}.

Claim 3: for any two integers k and l, if A and B are definable subsets of R_{0}^{k}

and R_{0}^{l} respectively, then we have Ă × B̆ = ˘ { \atop A × B}.

Let x ∈ Ă × B̆. We show that x belongs to ˘ { \atop A × B}. We have x = (u, v) for u ∈ Ă and v ∈ B̆. Then there are two closed definable subsets F and G of A and B respectively such that u ∈ \overline{F} and v ∈ \overline{G}. For each ε ∈ R^{>0}, there f ∈ F and g ∈ G such that |u − f| < ε and |v − g| < ε, so |(u, v) − (f, g)| < ε. Consequently,

x = (u, v) belongs to \overline{F × G} ⊆ ˘ { \atop A × B}.

By Corollary 2.5, for any x ∈ ˘ { \atop A × B}, there is a closed and bounded definable subset H of A × B such that x ∈ \overline{H}. If H_{1} (resp. H_{2}) denotes the image of H by the projection π_{1} : A × B → A (resp. π_{2} : A × B → B), then by the continuity of the projections maps, the set H_{1} (resp. H_{2}) is closed and bounded, and H is contained in H_{1} × H_{2}. Now \overline{H} is contained in \overline{H_{1} × H_{2}}. But \overline{H_{1}} × \overline{H_{2}} is closed in R^{k+l} and it contains H_{1} × H_{2}, so it contains \overline{H_{1} × H_{2}}. Hence \overline{H} is contained in

\overline{H_{1}} × \overline{H_{2}}, and Ă × B̆ contains ˘ { \atop A × B}.

Claim 4: let π : R^{k+1} → R^{k} be the projection on the first k coordinates for an

integer k. If A is a definable subset of R_{0}^{k+1}, then we have π(Ă) = ˘ { \atop π(A)}.

First we note that the restriction π_{|R_{0}^{k+1}} : R_{0}^{k+1} → R_{0}^{k} of π to R_{0}^{k+1} is definable and continuous. In particular, the set π(A) is definable.

Let x ∈ Ă. By Corollary 2.5, there is a closed and bounded definable subset F of A such that x ∈ \overline{F}. Then, for each ε ∈ R^{>0}, there exists y ∈ F such that |y − x| < ε, so we have |π(y) − π(x)| < ε, and thus we obtain π(x) ∈ \overline{π(F)} But the restriction π_{|R_{0}^{k+1}} is definable and continuous, so π(F) is a closed and bounded

definable subset contained in π(A). Hence π(x) belongs to ˘ { \atop π(A)}, and ˘ { \atop π(A)} contains π(Ă).

Let x ∈ ˘ { \atop π(A)}. We show that x belongs to π(Ă). By definable choice [4, Chapter 6 §1], there is a definable map f : π(A) → R_{0} such that {(α, f(α)) | α ∈ π(A)} is contained in A. By cell decomposition [4, Chapter 3 §2], there are finitely many cells C_{1}, …, C_{s} of π(A) such that π(A) = ∪_{i=1}^{s}C_{i} and f is continuous on C_{i} for each i. By Proposition 2.6, we have x ∈ C̆_{r} for some r ∈ {1, …, s}. By Corollary 2.5, there is a closed and bounded definable subset G of C_{r} such that x ∈ \overline{G}. But f is continuous on G, so the graph Γ of its restriction f_{|G} : G → R_{0} to G is a closed and definable subset of A. Moreover, the continuity of f on G implies its uniform continuity on G [4, Chapter 6 §1], hence the following limit exists by Lemma 2.3:

u = lim_{\begin{matrix}y∈G \\ y→x \end{matrix}} f(y)

Now (x, u) belongs to \overline{Γ}, and since Γ is a closed and definable subset of A, we obtain (x, u) ∈ Ă and x ∈ π(Ă).

Claim 5: if φ(\overline{x}, \overline{a}) is an atomic formula with free variables \overline{x} = (x_{1}, …, x_{k}) and parameters \overline{a} = (a_{1}, …, a_{m}) in R_{0}^{m}, and if A is the definable subset of R_{0}^{k} defined by φ(\overline{x}, \overline{a}), then Ă is the M-definable subset of R^{k} defined by φ(\overline{x}, \overline{a}).

Let S be a relation symbol such that φ(\overline{x}, \overline{a}) = S(\overline{x}, \overline{a}). By the definition of S^{M}, we have S^{M} = S̆^{M_{0}}. Therefore, if B = S^{M_{0}} is the subset of R_{0}^{k+m} defined by φ(\overline{x}, \overline{y}), if C = R_{0}^{k} × \{\overline{a}\}, and if π : R^{k+m} → R^{k} is the projection on the first k coordinates, then we have A = π(B ∩ C). In the same way, the M-definable subset of R^{k} defined by φ(\overline{x}, \overline{a}) is π(S^{M} ∩ (R^{k} × \{\overline{a}\})). Since by Claim 3 we have

C̆ = R̆_{0}^{k} × ˘ ˆ\{\overline{a}\} = R^{k} × \{\overline{a}\}, and since by Claims 2 and 4 we have Ă = π(B̆ ∩ C̆), Claim 5 is proven.

Claim 6: let φ(\overline{x}, y, \overline{a}) be a formula with free variables \overline{x} = (x_{1}, …, x_{k}) and y, and parameters \overline{a} ∈ R_{0}^{m}. Let A be the subset of R_{0}^{k+1} defined by φ(\overline{x}, y, \overline{a}), and B be the subset of R_{0}^{k} defined by ∃y φ(\overline{x}, y, \overline{a}). If the subset of R^{k+1} defined by φ(\overline{x}, y, \overline{a}) is Ă, then the subset of R^{k} defined by ∃y φ(\overline{x}, y, \overline{a}) is B̆.

This follows from Claim 4.

Claim 7: let φ(\overline{x}, \overline{a}) be a formula with free variables \overline{x} = (x_{1}, …, x_{k}) and parameters \overline{a} ∈ R_{0}^{m}. Let A be the subset of R_{0}^{k} defined by φ(\overline{x}, \overline{a}), and B be the subset of R_{0}^{k} defined by ¬φ(\overline{x}, \overline{a}). If the subset of R^{k} defined by φ(\overline{x}, \overline{a}) is Ă, then the subset of R^{k} defined by ¬φ(\overline{x}, \overline{a}) is B̆.

This follows from Claim 1.

Claim 8: let φ(\overline{x}, \overline{a}) and ϕ(\overline{x}, \overline{a}) be formulas with free variables \overline{x} = (x_{1}, …, x_{k}) and parameters \overline{a} ∈ R_{0}^{m}. Let A be the subset of R_{0}^{k} defined by φ(\overline{x}, \overline{a}), and B be the subset of R_{0}^{k} defined by ϕ(\overline{x}, \overline{a}). If the subset of R^{k} defined by φ(\overline{x}, \overline{a}) is Ă and the one defined by ϕ(\overline{x}, \overline{a}) is B̆, then the subset of R^{k} defined by φ(\overline{x}, \overline{a}) ∧ ϕ(\overline{x}, \overline{a}) is

˘{ \atop A ∩ B}.

This follows from Claim 2.

Conclusion: it follows from Claims 5, 6, 7 and 8, and from Remark 2.2, that the structure M is an elementary extension of M_{0}. □

References

[1] Antongiulio  Fornasiero.  Tame  structures  and  open  cores.  2010.

http://www.logique.jussieu.fr/modnet/Publications/Preprint%20server/papers/237/237.pdf.

[2] Dana Scott. On completing ordered fields. In Applications of Model Theory to Algebra, Analysis, and Probability (Internat. Sympos., Pasadena, Calif., 1967), pages 274–278. Holt, Rinehart and Winston, New York, 1969.

[3] Saharon Shelah. Quite complete real closed fields. Israel J. Math., 142:261–272, 2004.

[4] Lou van den Dries. Tame topology and o-minimal structures, volume 248 of London Mathematical Society Lecture Note Series. Cambridge University Press, Cambridge, 1998.

Laboratoire de Mathématiques et Applications, Université de Poitiers

   E-mail address: olivier.frecon@math.univ-poitiers.fr