Topological properties of sets definable in weakly o-minimal structures^{1}

Roman Wencel^{2}

Department of Pure Mathematics, University of Leeds Mathematical Institute, Wrocław University

ABSTRACT

The paper is aimed at studying the topological dimension for sets definable in weakly o-minimal structures in order to prepare background for further investigation of groups, group actions and fields definable in the weakly o-minimal context. We prove that the topological dimension of a set definable in a weakly o-minimal structure is invariant under definable injective maps, strengthening an analogous result from [MMS] for sets and functions definable in models of weakly o-minimal theories. We pay special attention to large subsets of Cartesian products of definable sets, showing that if X, Y and S are non-empty definable sets and S is a large subset of X × Y, then for a large set of tuples ⟨\overline{a}_{1}, …, \overline{a}_{2^{k}}⟩ ∈ X^{2^{k}}, where k = dim(Y), the union of fibers S_{\overline{a}_{1}} ∪ … ∪ S_{\overline{a}_{2^{k}}} is large in Y. Finally, given a weakly o-minimal structure M, we find various conditions equivalent to the fact that the topological dimension in M enjoys the addition property.

0 Introduction

In the theory of o-minimal structures, cell decompositions have usually been an essential tool for introducing and effective investigation of various topological invariants of definable sets, the dimension and the Euler characteristic being classical examples (see for instance [vdD2, Chapter 4]).

In this paper we concentrate on studying the topological dimension for sets definable in weakly o-minimal structures. The dimension of an infinite set X ⊆ M^{m} definable in a weakly o-minimal structure M = (M, ≤, …) is defined as the biggest positive integer r for which there is a projection π : M^{m} ⟶ M^{r} such that π[X] has a non-empty interior in M^{r}. A finite set has dimension 0 if it is non-empty and −∞ otherwise (see [MMS, Definition 4.1]). If M is o-minimal, then this notion of dimension coincides with the usual one defined by cell decomposition. Some basic properties of the topological dimension in weakly o-minimal structures are collected in §1.

As illustrated by various examples from [MMS] and [V], the weak o-minimality of linearly ordered structures is not preserved under elementary equivalence. Therefore one cannot hope for a reasonable counterpart of the cell decomposition for arbitrary weakly o-minimal structures. A sensible way to avoid this kind of difficulty is to restrict one’s attention to the class of models of weakly o-minimal theories. In such a situation, D. Macpherson, D. Marker and C. Steinhorn established a version of cell decomposition (see [MMS, Theorem 4.6]). Naturally, one could ask how much of the cell decomposition survives if the hypothesis of weak o-minimality of the theory is relaxed to that of the structure. Our attempts towards answering this question are expressed by Lemma 2.4, where we find a decomposition of a definable set into finitely many subsets of ’simple nature’. However, these simple sets are rather remote from what has traditionally been understood under the name of a ’cell’.

Macpherson, Marker and Steinhorn show that the dimension of a set definable in a model of a weakly o-minimal theory is invariant under injective definable maps (see [MMS, Theorem 4.7]). Their proof uses cell decomposition and ω_{1}-saturatedness, and therefore cannot be easily generalized to sets and functions definable in general weakly o-minimal structures. Nevertheless, applying a completely different approach, in §2 we prove that the dimension of a set definable in a weakly o-minimal structure does not change under injective definable maps. The proof of this result easily reduces to showing that if X ⊆ M^{m} is a non-empty definable set and f : X ⟶ M is a definable function, then Γ(f) := \{⟨\overline{a}, f(\overline{a})⟩ : \overline{a} ∈ X\} ⊆ M^{m+1}, the graph of f, has dimension equal to the dimension of X. The difficulty with establishing the latter lies in showing that there is a projection witnessing the dimension of Γ(f) which drops the last coordinate.

12000 Mathematics Subject Classification. Primary 03C64.

2This research was supported by a Marie Curie Intra-European Fellowships within the 6th European Community Framework Programme. Contract number: MEIF-CT-2003-501326.

Imagine for example that there are some nasty one-dimensional definable set S ⊆ M^{2} and a definable function f : S ⟶ M such that dim(Γ(f)) = 2, i.e. some projection of Γ(f) contains an open box. Clearly, such a projection cannot drop the last coordinate. Suppose for instance that there are open intervals I, J ⊆ M for which

I × J ⊆ {⟨y, z⟩ ∈ M^{2} : (∃x ∈ M)(z = f(x, y))}.

For every a ∈ I, the set {x ∈ M : ⟨x, a⟩ ∈ S} is infinite. For the sake of simplicity, assume that {x ∈ M : ⟨x, a⟩ ∈ S} is convex and open whenever a ∈ I. Fix b ∈ J and let X = {⟨x, y⟩ ∈ S : y ∈ I, f(x, y) = b}. Note that {x ∈ M : ⟨x, a⟩ ∈ X} is a non-empty proper subset of {x ∈ M : ⟨x, a⟩ ∈ S} whenever a ∈ I. By the monotonicity theorem (see Theorem 1.2) we can find an open interval I′ ⊆ I such that each of the functions

x ⟼ inf{x ∈ M : ⟨x, a⟩ ∈ S}, x ⟼ sup{x ∈ M : ⟨x, a⟩ ∈ S}

is constant or strictly increasing on I′. As dim(S) = 1, the above functions must be both strictly increasing or both strictly decreasing. Moreover, for distinct a_{1}, a_{2} ∈ I′

{x ∈ M : ⟨x, a_{1}⟩ ∈ S} ∩ {x ∈ M : ⟨x, a_{1}⟩ ∈ S} = ∅.

Consequently, the set {x ∈ M : (∃y ∈ I′)(⟨x, y⟩ ∈ X)} is not a union of finitely many convex sets, which contradicts the weak o-minimality of M.

The general situation is much more complicated, however. Nevertheless, having in mind the above example and various special cases that may arise, we were able to state the list of inductive conditions of Theorem 2.11, from which the required result easily follows.

In §3 we study large subsets of Cartesian products of definable sets. The main result of §3 (i.e. Theorem 3.6) will constitute one of the crucial ingredients of our further study of groups, group actions and fields definable in weakly o-minimal structures. It will be used for example to show that a large subset of a group definable in a weakly o-minimal structure is generic. Theorem 3.6 says that if M = (M, ≤, …) is a weakly o-minimal structure, X ⊆ M^{m}, Y ⊆ M^{n}, S ⊆ X × Y, are nonempty definable sets, k = dim(Y) and S is large in X×Y, then the set of tuples ⟨\overline{a}_{1}, …, \overline{a}_{2^{k}}⟩ ∈ X^{2^{k}}, for which the union of fibers S_{\overline{a}_{1}} ∪ … ∪ S_{\overline{a}_{2^{k}}} is large in Y, is large in X^{2^{k}}.

We say that the topological dimension in a weakly o-minimal structure M has the addition property iff for every definable set S ⊆ M^{m+n} and a projection π : M^{m+n} ⟶ M^{m} dropping some n coordinates, if all fibers π^{−1}(\overline{a}) ∩ S, \overline{a} ∈ π[S] are of dimension k, then dim(S) = dim(π[S]) + k. The addition property holds in the o-minimal case but fails in general weakly o-minimal structures. We prove in §4 that it is closely related to the exchange property of the definable closure (Theorem 4.3) and equivalent to each of the following statements (Theorem 4.2).

• If I ⊆ M is an open interval and f : I ⟶ \overline{M}^{M} is a definable function (i.e. the set {⟨x, y⟩ ∈ I × M : y < f(x)} is definable), then there is an open interval I′ ⊆ I such that f ↾ I′ is continuous.

• If m ∈ ℕ_{+}, B ⊆ M^{m} is an open box and f : B ⟶ \overline{M}^{M} is a definable function (i.e. the set \{⟨\overline{x}, y⟩ ∈ B × M : y < f(\overline{x})\} is definable), then there is an open box B′ ⊆ B such that f ↾ B′ is continuous.

• If m ∈ ℕ_{+}, S ⊆ M^{m+1} is a non-empty definable set and π : M^{m+1} ⟶ M^{m} denotes a projection dropping one coordinate, then dim(S) = dim(π[S]) iff the set of tuples \overline{a} ∈ π[S] for which the fiber π^{−1}(\overline{a}) ∩ S is finite is large in π[S].

1 Notation and preliminaries

Let (M, ≤) be a dense linear ordering without endpoints. A set X ⊆ M is called convex in (M, ≤) iff for any a, b ∈ X and c ∈ M, if a ≤ c ≤ b, then c ∈ X. A non-empty convex set is called an interval in (M, ≤) iff it has both infimum and supremum in M ∪ {−∞, +∞}. The ordering of M determines a topology on M^{m}, m ∈ ℕ_{+}, whose basis consists of open boxes in M^{m}, i.e. sets of the form (a_{1}, b_{1}) × … × (a_{m}, b_{m}), where a_{i}, b_{i} ∈ M and a_{i} < b_{i}, 1 ≤ i ≤ m.

A first order structure M = (M, ≤, …) equipped with a dense linear ordering ≤ without endpoints is called weakly o-minimal iff every subset of M, definable in M, is a finite union of convex sets. Weak o-minimality, unlike o-minimality, is not preserved under elementary equivalence [MMS]. A first order complete theory is called weakly o-minimal iff all its models are weakly ominimal. If X ⊆ M is a non-empty set definable in a weakly o-minimal structure M, then any maximal convex subset of X is said to be a convex component of X. If Y is a convex component of X with inf Y = inf X [respectively: sup Y = sup X], then Y is called the first [the last] convex component of X.

Assume that M = (M, ≤, …) is a weakly o-minimal L-structure. A definable cut in M is an ordered pair ⟨C, D⟩ of non-empty definable subsets of M such that C < D and C ∪ D = M. In an obvious way the linear ordering (M, ≤) extends to the linear ordering of

\overline{M}^{M} := M ∪ \{⟨C, D⟩ : ⟨C, D⟩ is a definable cut in M and C, D are not intervals in (M, ≤)\}.

Note that M is dense in \overline{M}^{M} and every subset of M, definable in M, has infimum and supremum in \overline{M}^{M} ∪ \{−∞, +∞\}. If X ⊆ M^{m} is a non-empty definable [over A, A ⊆ M] set, then a function f : X ⟶ \overline{M}^{M} is called definable [over A] iff there is a formula φ(\overline{x}, y) ∈ L(M) [respectively: φ(\overline{x}, y) ∈ L(A)] such that |\overline{x}| = m, φ(M) ⊆ X × M and f(\overline{a}) = sup φ(\overline{a}, M) whenever \overline{a} ∈ X. If f, g : X ⟶ \overline{M}^{M} are definable functions such that f(\overline{a}) < g(\overline{a}) for \overline{a} ∈ X, then by (f, g)_{X} we will denote the set of tuples ⟨\overline{a}, b⟩ ∈ X × M for which f(\overline{a}) < b < g(\overline{a}). Throughout the paper we will also use the following convention. We will call a function f : X ⟶ M ∪ {−∞, +∞} [f : X ⟶ \overline{M}^{M} ∪ \{−∞, +∞\}] definable iff either f is a definable function from X to M [to \overline{M}^{M}], or (∀\overline{x} ∈ X)(f(\overline{x}) = −∞), or (∀\overline{x} ∈ X)(f(\overline{x}) = +∞).

If M = (M, …) is a first order structure, m ∈ ℕ_{+} and J is a proper subset of {1, …, m}, then by π_{J}^{m} we will denote the projection from M^{m} onto M^{m−|J|} dropping all the coordinates from J. In particular, π_{∅}^{m} is the identity map on M^{m}. If J is a non-empty subset of {1, …, m}, then by ϱ_{J}^{m} we will denote the projection from M^{m} onto M^{|J|} dropping all the coordinates from {1, …, m}\J. Usually, if J = {j_{1}, …, j_{k}}, we write π_{j_{1},...,j_{k}}^{m} instead of π_{{j_{1},...,j_{k}}}^{m} and ϱ_{j_{1},...,j_{k}}^{m} instead of ϱ_{{j_{1},...,j_{k}}}^{m}. For example we have π_{1,4}^{5}(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}) = ϱ_{2,3,5}^{5}(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}) = ⟨x_{2}, x_{3}, x_{5}⟩. A projection from M^{m} onto M^{r}, r ∈ {1, …, m}, is an arbitrary map of the form π_{J}^{m}, where J ⊊ {1, …, m} and |J| = m − r. Note that if m ≥ 3 and 1 ≤ i < j ≤ m, then π_{j−1}^{m−1} ∘ π_{i}^{m} = π_{i}^{m−1} ∘ π_{j}^{m} = π_{i,j}^{m}.

Throughout the rest of the paper, unless otherwise stated, we will work in an arbitrary weakly o-minimal structure M = (M, ≤, …). By a definable set (function) we will always mean a set (function) definable in the structure M. When talking about the interior or closure of a definable set X ⊆ M^{m} (notation: int(X), cl(X) respectively) we will always refer to the topology induced on M^{m} by the ordering (M, ≤).

Assume that m, n ∈ ℕ_{+}, S ⊆ M^{m+n} is a definable set, \overline{a} ∈ M^{m} and \overline{b} ∈ M^{n}. The fibers determined by \overline{a} and \overline{b} are defined as follows:

S\overline{a} = \{\overline{c} ∈ M^{n} : ⟨\overline{a}, \overline{c}⟩ ∈ S\}, S\overline{b} = \{\overline{c} ∈ M^{m} : ⟨\overline{c}, \overline{b}⟩ ∈ S\}.

Definition 1.1 Let I be a non-empty convex open subset of M. A function f : I ⟶ \overline{M}^{M} is called

(a) locally constant on I iff

(∀a ∈ I)(∃b, c ∈ I)(b < a < c and f ↾ (b, c) is constant);

(b) locally strictly increasing on I iff

(∀a ∈ I)(∃b, c ∈ I)(b < a < c and f ↾ (b, c) is strictly increasing);

(c) locally strictly decreasing on I iff

(∀a ∈ I)(∃b, c ∈ I)(b < a < c and f ↾ (b, c) is strictly decreasing);

(d) locally strictly monotone on I iff f is locally strictly increasing on I or locally strictly decreasing on I.

The following theorem, to be referred to as the monotonicity theorem, is a consequence of Theorem 3.3 from [MMS] and [Ar].

Theorem 1.2 Assume that A ⊆ M. If U ⊆ M is an infinite A-definable set and f : U ⟶ \overline{M}^{M} [respectively: f : U ⟶ M] is an A-definable function, then there is a partition of U into Adefinable sets X, I_{0}, …, I_{m} such that X is finite, I_{0}, …, I_{m} are non-empty convex open sets, and for every i ≤ m, f ↾ I_{i} if locally constant or locally strictly monotone [and continuous].

Lemma 1.3 Assume that I ⊆ M is an open interval and f, g : I ⟶ \overline{M}^{M} are definable functions such that f(a) < g(a) for a ∈ I and int((f, g)_{I}) = ∅. Then there is an open interval I′ ⊆ I such that

(a) the functions f ↾ I′ and g ↾ I′ are either both strictly increasing or both strictly decreasing;

(b) for any distinct a, b ∈ I′ we have that (f(a), g(a)) ∩ (f(b), g(b)) = ∅;

(c) for any a ∈ I′ and any open interval I_{1} with inf I_{1} < f(a) < g(a) < sup I_{1}, there is an open interval I′′ ⊆ I′, containing a, such that inf I_{1} < f(x) < g(x) < sup I_{1} whenever x ∈ I′′.

Proof. By the monotonicity theorem, there is an open interval I_{1} ⊆ I such that each of the functions f, g restricted to I_{1} is either strictly monotone or constant. As int((f, g)_{I}) = ∅, the functions f, g restricted to I_{1} are either both strictly decreasing or both strictly increasing. Below we only consider the first possibility.

Firstly, observe that f(a) > g(b) whenever inf I_{1} < a < b < sup I_{1}. For, if inf I_{1} < a < c < b < sup I_{1} and f(a) ≤ g(b), then (c, b) × (f(c), f(a)) ⊆ (f, g)_{I_{1}}, which contradicts our assumption. For a ∈ I_{1} define h_{1}(a) = lim_{x→a^{−}} f(x) = lim_{x→a^{−}} g(x) and h_{2}(a) = lim_{x→a^{+}} f(x) = lim_{x→a^{+}} g(x). By the weak

o-minimality of M, the set X := {a ∈ I_{1} : h_{1}(a) ≠ g(a) or h_{2}(a) ≠ f(a)} is finite, so there is an open interval I′ ⊆ I_{1} \ X. Clearly, I′ satisfies all our demands.

The proof of the following lemma can be easily derived from [Ar] and the proof of Theorem 4.8 from [MMS]. A similar technique will be used in the proof of Theorem 4.2.

Lemma 1.4 Assume that m ∈ ℕ_{+}, B ⊆ M^{m} is an open box and f : B ⟶ M is a definable function. Then there is an open box B′ ⊆ B such that f ↾ B′ is continuous.

The following lemma can be deduced from [Ar] and Theorem 4.3 from [MMS].

Lemma 1.5 Assume that m ∈ ℕ_{+}, B ⊆ M^{m} is an open box, f : B ⟶ M ∪ {−∞, +∞} and g : B ⟶ \overline{M}^{M} ∪ \{−∞, +∞\} are definable functions, f is continuous, and b ∈ M.

(a) If (∀\overline{a} ∈ B)(f(\overline{a}) < g(\overline{a})), then the set (f, g)_{B} contains an open box C. If additionally f is identically equal to b, then C may be chosen so that inf ϱ_{m+1}^{m+1}[C] = b.

(b) If (∀\overline{a} ∈ B)(f(\overline{a}) > g(\overline{a})), then the set (g, f)_{B} contains an open box C. If additionally f is identically equal to b, then C may be chosen so that sup ϱ_{m+1}^{m+1}[C] = b.

Assume that M is a weakly o-minimal structure and X ⊆ M^{m} is an infinite definable set. The dimension of X, denoted by dim(X), is the largest r for which there exists a projection π : M^{m} ⟶ M^{r} such that π[X] contains an open box. Non-empty finite sets are said to have dimension 0, while to an empty set we assign the dimension −∞. We shall use the convention that if d ∈ ℕ ∪ {−∞}, then d ≥ −∞ and d + (−∞) = −∞ + d = −∞.

Fact 1.6 Assume that m, n ∈ ℕ_{+} and X, Y ⊆ M^{m}, Z ⊆ M^{n} are definable sets.

(a) If X ⊆ Y, then dim(X) ≤ dim(Y).

(b) If k ∈ {1, …, m} and π : M^{m} ⟶ M^{k} is a projection, then dim(X)−(m−k) ≤ dim(π[X]) ≤ dim(X).

(c) If f : M^{m} ⟶ M^{m} is a permutation of variables, then dim(f[X]) = dim(X).

(d) dim(X × Z) = dim(X) + dim(Z).

(e) dim(X ∪ Y) = max{dim(X), dim(Y)}.

Proof. (a), (b), (c) and (d) are immediate. (e) follows from [Ar] and [MMS, Theorem 4.2].

Lemma 1.7 Assume that m ≥ 3, S ⊆ M^{m} is a definable set, i, j ∈ {1, …, m} and i ≠ j.

(a) If dim(S) = dim(π_{j}^{m}[S]) = dim(π_{i}^{m}[S])+1, then dim(π_{i}^{m}[S]) = dim(π_{i,j}^{m}[S]) = dim(π_{j}^{m}[S])− 1.

(b) If dim(S) = dim(π_{i}^{m}[S]) = dim(π_{i,j}^{m}[S]), then dim(S) = dim(π_{j}^{m}[S]).

Proof. (a) dim(π_{i,j}^{m}[S]) ≥ dim(π_{j}^{m}[S]) − 1 = dim(S) − 1 = dim(π_{i}^{m}[S]) ≥ dim(π_{i,j}^{m}[S]). (b) dim(S) ≥ dim(π_{j}^{m}[S]) ≥ dim(π_{i,j}^{m}[S]) = dim(S).

Definition 1.8 Assume that m ∈ ℕ_{+} and X, Y ⊆ M^{m} are non-empty definable sets. We say that X is large in Y iff dim(Y \ X) < dim(Y).

Fact 1.9 Assume that m ∈ ℕ_{+} and X, Y, Z ⊆ M^{m} are non-empty definable sets.

(a) If X, Y are finite, then X is large in Y iff Y ⊆ X.

(b) If X is large in Y, Y is large in Z and Y ⊆ Z, then X is large in Z.

(c) If X and Y are both large in Z, then X ∩ Y and X ∪ Y are large in Z.

Proof. (a) is obvious; (b) and (c) follow from Fact 1.6.

Lemma 1.10 Assume that m ≥ 2, B ⊆ M^{m} is an open box and f : B ⟶ \overline{M}^{M} is a definable function. For i ∈ \{1, …, m\}, \overline{a} ∈ π_{i}^{m}[B] and b ∈ ϱ_{i}^{m}[B] define f\overline{a} ; i(b) = f(\overline{c}), where \overline{c} ∈ B is the unique tuple such that ϱ_{i}^{m}(\overline{c}) = b and π_{i}^{m}(\overline{c}) = \overline{a}. Then there is an open box C ⊆ B such that for every i ∈ {1, …, m}, one of the following conditions holds.

(a) (∀\overline{a} ∈ π_{i}^{m}[C])(f\overline{a} ; i ↾ ϱ_{i}^{m}[C] is strictly increasing);

(b) (∀\overline{a} ∈ π_{i}^{m}[C])(f\overline{a} ; i ↾ ϱ_{i}^{m}[C] is strictly decreasing);

(c) (∀\overline{a} ∈ π_{i}^{m}[C])(f\overline{a} ; i ↾ ϱ_{i}^{m}[C] is constant).

Proof. Let X_{1} [respectively: X_{2}, X_{3}] be the set of all tuples \overline{a} ∈ π_{1}^{m}[B] for which there exists an open interval I(\overline{a}) ⊆ ϱ_{1}^{m}[B] such that inf I(\overline{a}) = inf ϱ_{1}^{m}[B] and the function f\overline{a} ; 1 ↾ I(\overline{a}) is locally strictly increasing [respectively: locally strictly decreasing, locally constant]. By the monotonicity theorem, π_{1}^{m}[B] = X_{1} ∪ X_{2} ∪ X_{3}, so at least one of the sets X_{1}, X_{2}, X_{3} has dimension m − 1. Suppose for example that dim(X_{1}) = m − 1, and fix an open box B_{0} ⊆ X_{1}. For \overline{a} ∈ B_{0} define

h_{1}(\overline{a}) = sup\{y ∈ ϱ_{1}^{m}[B] : f\overline{a} ; 1 ↾ (inf ϱ_{1}^{m}[B], y) is locally strictly increasing\}.

By Lemma 1.5, there are an open box B_{1} ⊆ B_{0} and an open interval I_{1} ⊆ ϱ_{1}^{m}[B] such that B_{1} × I_{1} ⊆ \{⟨\overline{x}, y⟩ ∈ B_{0} × M : inf ϱ_{1}^{m}[B] < y < h_{1}(\overline{x})\}. Now, fix b ∈ I_{1} and for \overline{a} ∈ B_{1} define

h_{2}(\overline{a}) = sup\{y ∈ (b, sup I_{1}) : f\overline{a} ; 1 ↾ (b, y) is strictly increasing\}.

Again, by Lemma 1.5, there are an open box B_{2} ⊆ B_{1} and an open interval I_{2} ⊆ I_{1} such that

B_{2} × I_{2} ⊆ \{⟨\overline{x}, y⟩ ∈ B_{1} × I_{1} : b < y < h_{2}(\overline{x})\}. Let B′ = I_{2} × B_{2}. The function f\overline{a} ; 1 ↾ ϱ_{1}^{m}[B′] is

strictly increasing whenever \overline{a} ∈ π_{1}^{m}[B′].

Repeating the above procedure for the remaining coordinates, one obtains an open box C ⊆ B′ as required by the assertion of the lemma.

2 Essential dimension theory and injective maps

In this section we start to develop the dimension theory for sets definable in weakly o-minimal structures. Among several other things we prove that the topological dimension of a set definable in a weakly o-minimal structure is invariant under injective definable maps. Before formulating the main theorem we prove a series of technical and preparatory lemmas.

Lemma 2.1 Assume that m ≥ 2, J is a non-empty proper subset of {1, …, m}, i ∈ {1, …, m}\J, S ⊆ M^{m} is a definable set and \overline{a} ∈ π_{i}^{m}[S]. Assume also that there are infinitely many tuples \overline{c} ∈ (π_{i}^{m})^{−1}(\overline{a}) ∩ S such that ϱ_{J}^{m}(\overline{c}) ∈ int[ϱ_{J}^{m}[(π_{J}^{m})^{−1}(π_{J}^{m}(\overline{c})) ∩ S]]. Then there is a definable set V ⊆ S such that ϱ_{J∪{i}}^{m}[V] ⊆ M^{|J|+1} is an open box and (in case |J| < m−1) π_{J∪{i}}^{m} is a singleton.

Proof. Assume that m, J, i, S and \overline{a} := ⟨a_{1}, …, a_{i−1}, a_{i+1}, …, a_{m}⟩ satisfy assumptions of the lemma. Permuting variables, without loss of generality we can assume that J = {1, …, |J|} and i = |J| + 1. In such a situation, let \overline{a}′ = ⟨a_{1}, …, a_{i−1}⟩ and

S_{1} = \{ \begin{matrix}S & if & i = m \\ \{⟨x_{1}, …, x_{i}⟩ ∈ M^{i} : ⟨x_{1}, …, x_{i}, a_{i+1}, …, a_{m}⟩ ∈ S\} & if & i < m. \end{matrix}

There is an open interval I ⊆ M such that \overline{a}′ ∈ int\{\overline{c} ∈ M^{i−1} : ⟨\overline{c}, d⟩ ∈ S_{1}\} whenever d ∈ I. Let

S′ = ⋃_{d∈I} int\{\overline{c} ∈ M^{i−1} : ⟨\overline{c}, d⟩ ∈ S_{1}\} × \{d\}.

We will be done if we demonstrate that int(S′) ≠ ∅. For this reason we will inductively find definable sets V_{1}, …, V_{i} ⊆ S′ such that

(a)_{s} (for s ∈ {1, …, i}) ϱ_{i+1−s,...,i}^{i}[V_{s}] ⊆ M^{s} is an open box;

(b)_{s} (for s ∈ {1, …, i − 1}) π_{i+1−s,...,i}^{i}[V_{s}] = {⟨a_{1}, …, a_{i−s}⟩}.

Note that for V_{1} := \{\overline{a}′\} × I, coditions (a)_{1} and (b)_{1} hold. Suppose that we have already found V_{s} ⊆ S′, 1 ≤ s < i, for which conditions (a)_{s} and (b)_{s} are satisfied. For \overline{x} ∈ ϱ_{i+1−s,...,i}^{i}[V_{s}],

define f(\overline{x}) ∈ \overline{M}^{M} ∪ \{+∞\}, f(\overline{x}) > a_{i−s}, as the supremum of the convex component of \{z ∈ M : ⟨a_{1}, …, a_{i−s−1}, z, \overline{x}⟩ ∈ S′\} containing a_{i−s}. By Lemma 1.5, the set \{⟨z, \overline{x}⟩ ∈ M^{s+1} : \overline{x} ∈ ϱ_{i+1−s,...,i}^{i}[V_{s}], a_{i−s} < z < f(\overline{x})\} ⊆ M^{s+1} contains an open box B. Clearly, the set V_{s+1} := {⟨a_{1}, …, a_{i−s−1}⟩} × B satisfies our demands.

Definition 2.2 Assume that m ≥ 2, J ⊆ {1, … m} and ∅ ≠ S ⊆ M^{m}. We say that the set S is J-open iff the following conditions are satisfied.

(a) (∀i ∈ \{1, …, m\} \ J)(∀\overline{a} ∈ π_{i}^{m}[S])((π_{i}^{m})^{−1}(\overline{a}) ∩ S is finite).

(b) If 0 < |J| < m and \overline{c} ∈ S, then there exists a definable set U ⊆ S containing \overline{c} such that ϱ_{J}^{m}[U] ⊆ M^{|J|} is an open box and π_{J}^{m}[U] = \{π_{J}^{m}(\overline{c})\}.

(c) If J = {1, …, m}, then S is open.

Fact 2.3 Assume that m ≥ 2, J, J_{1}, J_{2} ⊆ {1, …, m} and X, Y ⊆ M^{m}.

(a) If X, Y are J-open, then X ∪ Y is J-open.

(b) If X is J_{1}-open and Y is J_{2}-open, then X ∩ Y is J_{1} ∩ J_{2}-open or empty.

Lemma 2.4 Assume that m ≥ 2 and S ⊆ M^{m} is a non-empty definable set. Then there are pairwise disjoint definable sets X_{J}, J ⊆ {1, …, m}, such that S = ⋃_{J⊆{1,...,m}} X_{J} and for every

J ⊆ {1, …, m}, the set X_{J} is either J-open or empty.

Proof. Let J_{1}, …, J_{2^{m}} be an enumeration of all subsets of {1, …, m} such that |J_{i}| ≥ |J_{j}| whenever 1 ≤ i ≤ j ≤ 2^{m}. Clearly, J_{1} = {1, …, m} and J_{2^{m}} = ∅. We will define pairwise disjoint

sets X_{1}, …, X_{2^{m}} such that ⋃_{i=1}^{2^{m}} X_{i} = S and for every i ∈ {1, …, 2^{m}}, the set X_{i} is either J_{i}-open

or empty.

Let X_{1} = int(S). If X_{1} ≠ ∅, then X_{1} is J_{1}-open. For the inductive step, fix k ∈ {1, …, 2^{m} − 2} and suppose that the (pairwise disjoint) sets X_{1}, …, X_{k} have already been defined. Let Y = X \ (X_{1} ∪…∪X_{k}) and let X_{k+1} be the union of all sets of the form (ϱ_{J_{k+1}}^{m})^{−1}[B]∩(π_{J_{k+1}}^{m})^{−1}(π_{J_{k+1}}^{m}(\overline{c})), where \overline{c} ∈ Y and B ⊆ M^{|J_{k+1}|} is an open box such that \overline{c} ∈ (ϱ_{J_{k+1}}^{m})^{−1}[B]∩(π_{J_{k+1}}^{m})^{−1}(π_{J_{k+1}}^{m}(\overline{c})) ⊆ Y. In case X_{k+1} ≠ ∅, Lemma 2.1 and our enumeration of P({1, …, m}) guarantee that condition (a) of Definition 2.2 is satisfied. Condition (b) is obvious. Define also X_{2^{m}} as S \ (X_{1} ∪ … ∪ X_{2^{m}−1}). Certainly, X_{2^{m}} is ∅-open.

Lemma 2.5 Assume that m ∈ ℕ_{+}, i ∈ {1, …, m + 1}, B ⊆ M^{m} is an open box, and X ⊆ S ⊆ M^{m+1} are definable sets such that

• π_{i}^{m+1}[S] = π_{i}^{m+1}[X] = B;

• ϱ_{i}^{m+1}[(π_{i}^{m+1})^{−1}(\overline{c}) ∩ S] is an infinite convex set whenever \overline{c} ∈ B;

• ∅ ≠ (π_{i}^{m+1})^{−1}(\overline{c}) ∩ X ⊊ (π_{i}^{m+1})^{−1}(\overline{c}) ∩ S whenever \overline{c} ∈ B.

Then dim(S) = m + 1.

Proof. We use induction on m. For m = 1, the result is an easy consequence of Lemma 1.3. Suppose that it is true for dimensions smaller than m, where m ≥ 2, and fix B, S, X and i as in the statement of the lemma. Without loss of generality we can assume that i = m + 1. In such a situation, for \overline{c} ∈ B define f(\overline{c}) = inf\{d ∈ M : ⟨\overline{c}, d⟩ ∈ S\} and g(\overline{c}) = sup\{d ∈ M : ⟨\overline{c}, d⟩ ∈ S\}. Note that if one of the sets: \{\overline{c} ∈ B : f(\overline{c}) = −∞\}, \{\overline{c} ∈ B : g(\overline{c}) = +∞\} has non-empty interior, then by Lemma 1.5, S contains an open box. Hence we can assume that f(\overline{c}), g(\overline{c}) ∈ \overline{M}^{M} whenever \overline{c} ∈ B. Clearly, f and g are definable functions and f(\overline{c}) < g(\overline{c}) for \overline{c} ∈ B. Let B = B′ × I, where B′ is an open box and I is an open interval. By Lemma 1.10, without loss of generality we can assume that

• (∀\overline{a} ∈ B′)(f(\overline{a}, y) is strictly increasing) or

• (∀\overline{a} ∈ B′)(f(\overline{a}, y) is strictly decreasing), or

• (∀\overline{a} ∈ B′)(f(\overline{a}, y) is constant),

and similarly for g.

Suppose first that for \overline{a} ∈ B′, f(\overline{a}, y) is constant or strictly decreasing while g(\overline{a}, y) is constant or strictly increasing. Let b ∈ I. By the inductive hypothesis, dim(\{⟨\overline{x}, z⟩ : ⟨\overline{x}, b, z⟩ ∈ S\}) = m. Consequently, the set \{⟨\overline{x}, y, z⟩ : ⟨\overline{x}, b, z⟩ ∈ S, b < y < sup I\} ⊆ S has dimension m + 1. Similar argument works if for \overline{a} ∈ B′, we have that f(\overline{a}, y) is constant or strictly increasing while g(\overline{a}, y) is constant or strictly decreasing.

To finish the proof, we have to consider the case when for every \overline{a} ∈ B′, the functions f(\overline{a}, y), g(\overline{a}, y) are both strictly increasing or both strictly decreasing. Below we only deal with the first possibility.

For ⟨\overline{a}, b⟩ ∈ B′ × I, let h(\overline{a}, b) = inf\{f(\overline{a}, c) : c ∈ (b, sup I)\}. Below we consider two cases.

Case 1. There is \overline{a} ∈ B′ such that the set \{b ∈ I : h(\overline{a}, b) ≥ g(\overline{a}, b)\} is infinite, i.e. contains an open interval I′. Then the set \{z ∈ M : (∃y ∈ I′)(⟨\overline{a}, y, z⟩ ∈ X)\} is not a union of finitely many convex sets.

Case 2. For every \overline{a} ∈ B′, the set \{b ∈ I : h(\overline{a}, b) ≥ g(\overline{a}, b)\} is finite. For \overline{a} ∈ B′ define u(\overline{a}) = min(\{b ∈ I : h(\overline{a}, b) ≥ g(\overline{a}, b)\} ∪ \{sup I\}). By Lemma 1.5, there are an open box B_{1} ⊆ B′ and an open interval I_{1} ⊆ I such that for any \overline{a} ∈ B_{1} and b ∈ I_{1}, we have that h(\overline{a}, b) < g(\overline{a}, b). Again, using Lemma 1.5, without loss of generality we can assume that g(\overline{a}, b_{1}) > f(\overline{a}, b_{2}) for any \overline{a} ∈ B_{1} and b_{1} < b_{2} from I_{1}.

Now, fix b_{1} < b_{2} < b_{3} from I_{1} and define

X_{1} = \{⟨\overline{x}, z⟩ ∈ B_{1} × M : f(\overline{x}, b_{2}) < z < f(\overline{x}, b_{3})\};
S_{1} = \{⟨\overline{x}, z⟩ ∈ B_{1} × M : f(\overline{x}, b_{2}) < z < g(\overline{x}, b_{1})\}.

Note that π_{m}^{m}[X_{1}] = π_{m}^{m}[S_{1}] = B_{1}, and for every \overline{a} ∈ B_{1}, we have that (S_{1})\overline{a} is an infinite convex set such that ∅ ≠ (X_{1})\overline{a} ⊊ (S_{1})\overline{a}. By the inductive hypothesis, dim(S_{1}) = m. So the set \{⟨\overline{x}, y, z⟩ ∈ B_{1} × (b_{1}, b_{2}) × M : f(\overline{x}, b_{2}) < z < g(\overline{x}, b_{1})\} ⊆ S has dimension m + 1, which finishes the proof.

Definition 2.6 Assume that m ∈ ℕ_{+}, X ⊆ S ⊆ M^{m} are definable sets of dimension k ≥ 0 and \overline{a} ∈ X. We say that X is smooth at \overline{a} with respect to S iff k = 0, or k ≥ 1 and there are an open box B ⊆ M^{m} containing \overline{a} and a projection π : M^{m} ⟶ M^{k} such that B ∩ X = B ∩ S, π[B ∩ X] is an open box in M^{k}, and π ↾ B ∩ X is a homeomorphism from B ∩ X onto π[B ∩ X]. We say that X is locally smooth in S iff for every \overline{a} ∈ X, X is smooth at \overline{a} with respect to S.

Lemma 2.7 Assume that m ∈ ℕ_{+}, X ⊆ M^{m} is a definable set of dimension k ≥ 1, B′ ⊆ B ⊆ M^{m} are open boxes, \overline{a} ∈ B′ ∩ X, π : M^{m} ⟶ M^{k} is a projection, π[B ∩ X] is an open box in M^{k} and π ↾ B ∩ X is a homeomorphism from B ∩ X onto π[B ∩ X]. Then there is an open box B′′ ⊆ B′ containing \overline{a} such that π[B′′ ∩ X] is an open box in M^{k} and π ↾ B′′ ∩ X is a homeomorphism from B′′ ∩ X onto π[B′′ ∩ X].

Proof. Let g : π[B ∩ X] ⟶ B ∩ X be the map given by g(π(\overline{c})) = \overline{c} for \overline{c} ∈ B ∩ X. Since g is a homeomorphism from π[B ∩ X] onto B ∩ X, the preimage g^{−1}[B′ ∩ X] = π[B′ ∩ X] is open in M^{k}. Let B_{1} ⊆ g^{−1}[B′ ∩ X] be an open box containing π(\overline{a}). Then B′′ := B′ ∩ π^{−1}[B_{1}] is an open box in M^{m} satisfying our demands.

Lemma 2.8 Assume that m ∈ ℕ_{+}, U, V, Y are definable subsets of M^{m}, U, V ⊆ Y, U is locally smooth in Y, V is open in Y, U ∩ V ≠ ∅, and dim(U) = dim(Y) = k ≥ 1. Then dim(U ∩ V) = k and U ∩ V is locally smooth in Y.

Proof. Fix \overline{a} ∈ U ∩ V. We have to show that U ∩ V is smooth at \overline{a} with respect to Y. Since U is smooth at \overline{a} with respect to Y, there are an open box B_{1} ⊆ M^{m} containing \overline{a} and a projection π : M^{m} ⟶ M^{k} such that B_{1} ∩ Y ⊆ U, π[B_{1} ∩ Y] is an open box in M^{k}, and π ↾ B_{1} ∩ Y is a homeomorphism from B_{1} ∩Y onto π[B_{1} ∩Y]. Since V is open in Y, there is an open box B_{2} ⊆ M^{m} containing \overline{a} such that B_{2} ∩ Y ⊆ V. Hence B′ := B_{1} ∩ B_{2} is an open box in M^{m} containing \overline{a} such that B′ ∩ Y ⊆ U ∩ V. By Lemma 2.7, there is an open box B′′ ⊆ B′, containing \overline{a} such that π[B′′ ∩ Y] is an open box in M^{k} and π ↾ B′′ ∩ Y is a homeomorphism from B′′ ∩ Y onto π[B′′ ∩ Y]. This finishes the proof.

Lemma 2.9 Assume that m ∈ ℕ_{+}, i ∈ {1, …, m + 1} and S ⊆ M^{m+1} is a definable set with dim(π_{i}^{m+1}[S]) = m. Then dim(S) = m iff for every open interval I ⊆ M, there is a definable set X ⊆ π_{i}^{m+1}[S] such that dim(X) < m, and for any k ∈ ℕ_{+} and \overline{a}_{1}, …, \overline{a}_{k} ∈ π_{i}^{m+1}[S] \ X, we have

that I ⊈ ⋃_{l=1}^{k} ϱ_{i}^{m+1}[(π_{i}^{m+1})^{−1}(\overline{a}_{l}) ∩ S].

Proof. Without loss of generality we can assume that i = m + 1. For the left-to-right direction, suppose that there is an open interval I ⊆ M such that for every definable set X ⊆ π_{m+1}^{m+1}[S] with dim(X) < m, there are k ∈ ℕ_{+} and \overline{a}_{1}, …, \overline{a}_{k} ∈ π_{m+1}^{m+1}[S] \ X such that I ⊆ S_{\overline{a}_{1}} ∪ … ∪ S_{\overline{a}_{k}}. Denote by S_{1} the set of all tuples \overline{a} ∈ π_{m+1}^{m+1}[S] for which the set I ∩ S\overline{a} is non-empty and contains some open interval whose infimum is equal to inf I. Our assumptions guarantee that S_{1} contains an open box B. For \overline{a} ∈ B, let f(\overline{a}) ∈ \overline{M}^{M} be the the supremum of the first convex component of I ∩ S\overline{a}. The set \{⟨\overline{a}, b⟩ : inf(I) < b < f(\overline{a})\} is contained in S and (by Lemma 1.5) has dimension m + 1. Consequently, dim(S) = m + 1.

The right-to-left direction is trivial.

Lemma 2.10 Assume that m ∈ ℕ_{+}, S ⊆ M^{m} is a non-empty definable set, I ⊆ M is an open interval and X ⊆ S × I is a definable set such that dim((S × {b}) ∩ X) < dim(S) whenever b ∈ I. Then dim(X) ≤ dim(S).

Proof. We use induction on m. Let S ⊆ M be a non-empty definable set, I ⊆ M an open interval and X ⊆ S ×I a definable set such that dim((S ×{b})∩X) < dim(S) whenever b ∈ I. If S is finite, then (S × {b}) ∩ X = ∅ for b ∈ I, which means that X = ∅. If S is infinite, then (S × {b}) ∩ X is finite whenever b ∈ I. Hence X does not contain an open box, which means that dim(X) ≤ 1.

Now assume that S ⊆ M^{m+1} is a non-empty definable set, I ⊆ M is an open interval and X ⊆ S×I is a definable set such that dim((S×{b})∩X) < dim(S) whenever b ∈ I, and suppose that the Lemma holds for lower dimensions. If dim(S) = m + 1, then it is clear that X does not contain an open box. Assume that dim(S) ≤ m and suppose for a contradiction that dim(X) = dim(S)+1. Let π : M^{m+2} ⟶ M^{m+1} be a projection such that dim(π[X]) = dim(X) = dim(S) + 1. The projection π does not drop the last coordinate, so there is a unique projection π′ : M^{m+1} ⟶ M^{m} such that π(\overline{a}, b) = ⟨π′(\overline{a}), b⟩ for \overline{a} ∈ M^{m+1} and b ∈ M. Then

dim((π′[S] × {b}) ∩ π[X]) = dim(π[(S × {b}) ∩ X]) ≤ dim((S × {b}) ∩ X) < dim(S) = dim(π′[S]).

The first equality above holds because (π′[S] × {b}) ∩ π[X] = π[(S × {b}) ∩ X]. By the inductive assumption, dim(π[X]) ≤ dim(π′[S]). Hence dim(X) ≤ dim(S), a contradiction.

Theorem 2.11 Let m ∈ ℕ_{+}.

(a)_{m} If S ⊆ M^{m} is a non-empty definable set, then dim(cl(S) \ S) < dim(S).

(b)_{m} If X ⊆ S ⊆ M^{m} are non-empty definable sets and dim(X) = dim(S), then the set \{\overline{a} ∈ X : X is smooth at \overline{a} with respect to S\} is large in X.

(c)_{m} Assume that S ⊆ M^{m} is a non-empty definable set and f : S ⟶ M is a definable function. Then the set of continuity points of f is large in S.

(d)_{m} Assume that S ⊆ M^{m} is a non-empty definable set, f : S ⟶ M and g : S ⟶ \overline{M}^{M} are definable functions and f is continuous. If (∀\overline{a} ∈ S)(f(\overline{a}) < g(\overline{a})) [respectively: (∀\overline{a} ∈ S)(f(\overline{a}) > g(\overline{a}))], then there are an open interval I ⊆ M and a definable set X ⊆ S such that dim(X) = dim(S) and X × I ⊆ (f, g)_{S} [respectively: X × I ⊆ (g, f)_{S}].

(e)_{m} If S ⊆ M^{m+1} is a non-empty definable set and i ∈ {1, …, m + 1}, then dim(S) = dim(π_{i}^{m+1}[S]) iff for every open interval I ⊆ M, there is a definable set X ⊆ π_{i}^{m+1}[S] such that dim(X) < dim(π_{i}^{m+1}[S]), and for any k ∈ ℕ_{+} and \overline{a}_{1}, …, \overline{a}_{k} ∈ π_{i}^{m+1}[S] \ X,

I ⊈ ⋃_{l=1}^{k} ϱ_{i}^{m+1}[(π_{i}^{m+1})^{−1}(\overline{a}_{l}) ∩ S].

(f)_{m} Assume that S ⊆ M^{m+1} is a non-empty definable set, i ∈ {1, …, m + 1} and for every \overline{a} ∈ π_{i}^{m+1}[S], (π_{i}^{m+1})^{−1}(\overline{a}) ∩ S is finite. Then dim(S) = dim(π_{i}^{m+1}[S]).

(g)_{m} Assume that i ∈ {1, …, m+1} and X ⊆ S ⊆ M^{m+1} are non-empty definable sets. Assume also that π_{i}^{m+1}[X] = π_{i}^{m+1}[S], dim(π_{i}^{m+1}[S]) ≥ 1, and for every \overline{a} ∈ π_{i}^{m+1}[S], ϱ_{i}^{m+1}[(π_{i}^{m+1})^{−1}(\overline{a})∩ S] is an infinite convex set and ∅ ≠ (π_{i}^{m+1})^{−1}(\overline{a}) ∩ X ⊊ (π_{i}^{m+1})^{−1}(\overline{a}) ∩ S. Then dim(S) = dim(π_{i}^{m+1}[S]) + 1.

Proof. We use induction on m. Conditions (a)_{1} and (b)_{1} are obvious by the weak o-minimality of M. (c)_{1}, (d)_{1} and (e)_{1} are consequences of Theorem 1.2 and Lemmas 1.5 and 2.9 respectively. (e)_{1} implies (f)_{1}. Finally, (g)_{1} is consequence of Lemma 1.3. For the rest of the proof suppose that m ∈ ℕ_{+} and statements (a)_{m}–(g)_{m} are true.

Proof of (a)_{m+1}. Let S ⊆ M^{m+1} be a non-empty definable set. By Fact 1.6(e), it is enough to show that dim(cl(S) \ S) < dim(cl(S)). Suppose for a contradiction that dim(cl(S) \ S) = dim(cl(S)) = k. Clearly, this is not possible for k ∈ {0, m + 1}, so let 1 ≤ k ≤ m. There is i ∈ {1, …, m + 1} such that dim(π_{i}^{m+1}[cl(S) \ S]) = dim(π_{i}^{m+1}[cl(S)]) = k. To simplify notation, assume that i = m + 1 and let π = π_{m+1}^{m+1}, π′ = ϱ_{m+1}^{m+1}. Note that the set

X := \{\overline{a} ∈ π[cl(S) \ S] : π^{−1}(\overline{a}) ∩ S ≠ ∅\}

is large in π[cl(S) \ S]. Otherwise, by (b)_{m}, there is an open box B_{0} ⊆ M^{m} such that

B_{0} ∩ π[cl(S)] = B_{0} ∩ (π[cl(S) \ S] \ X)

and dim(B_{0} ∩ π[cl(S)]) = k. Consequently, for every open box B ⊆ B_{0} × M with B ∩ cl(S) ≠ ∅, we have that B ∩ S = ∅, which is impossible. So in particular dim(X) = k.

Let X_{1} be the set of all tuples \overline{a} ∈ X such that at least one of the convex components of the set π′[π^{−1}(\overline{a}) ∩ (cl(S) \ S)] precedes π′[π^{−1}(\overline{a}) ∩ S] and let X_{2} = X \ X_{1}. As X_{1} ∪ X_{2} = X, at least one of the sets X_{1}, X_{2} has dimension k. The proof is similar in both situations, therefore we only consider the case when dim(X_{1}) = k. For \overline{a} ∈ X_{1} denote by A(\overline{a}) the last convex component of π′[π^{−1}(\overline{a}) ∩ (cl(S) \ S)] preceding π′[π^{−1}(\overline{a}) ∩ S] and by B(\overline{a}) the first convex component of π′[π^{−1}(\overline{a}) ∩ S]. Define the following sets.

Y_{1} = \{\overline{a} ∈ X_{1} : sup A(\overline{a}) = inf B(\overline{a})\}
Y_{2} = \{\overline{a} ∈ X_{1} : sup A(\overline{a}) < inf B(\overline{a})\}

Again, at least one of the sets Y_{1}, Y_{2} has dimension k. In case dim(Y_{1}) = k, consider

X′ := ⋃_{\overline{a}∈Y_{1}}\{\overline{a}\} × A(\overline{a}) and S′ := ⋃_{\overline{a}∈Y_{1}}\{\overline{a}\} × (A(\overline{a}) ∪ B(\overline{a})).

By (g)_{m}, dim(S′) = k + 1, a contradiction. If dim(Y_{2}) = k, then define

X′′ := ⋃_{\overline{a}∈Y_{2}}\{\overline{a}\} × A(\overline{a}) and S′′ := ⋃_{\overline{a}∈Y_{2}}\{\overline{a}\} × \{b ∈ M : A(\overline{a}) < b < B(\overline{a})\}.

(g)_{m} implies that dim(X′′ ∪ S′′) = k + 1. But dim(X′′) = k, so dim(S′′) = k + 1. By lemma 2.10, there is b ∈ M such that dim(S′′ ∩ (M^{m} × {b})) = k. For such b, let Y_{3} = π[S′′ ∩ (M^{m} × {b})]. By (b)_{m}, there is an open box B_{1} ⊆ M^{m} such that B_{1} ∩ Y_{3} = B_{1} ∩ π[cl(S)] and dim(B_{1} ∩ Y_{3}) = k. There is b_{1} ∈ (−∞, b) such that (B_{1} × (b_{1}, b)) ∩ cl(S) ≠ ∅ and (B_{1} × (b_{1}, b)) ∩ S = ∅, which is impossible.

Proof of (b)_{m+1}. Assume that X ⊆ S ⊆ M^{m+1} are definable sets of dimension k ≥ 0. The assertion of (b)_{m+1} is obvious for k ∈ {0, m+1}. So let 1 ≤ k ≤ m and suppose for a contradiction that the set X′ := \{\overline{a} ∈ X : X is not smooth at \overline{a} with respect to S\} has dimension k. By Lemma 2.4 and Fact 1.6(e), there are J ⊊ {1, …, m + 1} and a k-dimensional J-open set X_{0} ⊆ X′. By (a)_{m+1}, the set X′_{0} := X_{0} \ cl(S \ X_{0}) is large in X_{0}. Fix i ∈ {1, …, m} \ J. Clearly, for every \overline{a} ∈ π_{i}^{m+1}[X′_{0}], the fiber (π_{i}^{m+1})^{−1}(\overline{a}) ∩ X′_{0} is finite. Hence, by (f)_{m}, dim(π_{i}^{m+1}[X′_{0}]) = k.

For \overline{a} ∈ π_{i}^{m+1}[X′_{0}] define

f(\overline{a}) = min ϱ_{i}^{m+1}[(π_{i}^{m+1})^{−1}(\overline{a}) ∩ X′_{0}];
g(\overline{a}) = min((ϱ_{i}^{m+1}[(π_{i}^{m+1})^{−1}(\overline{a}) ∩ X′_{0}] \ \{f(\overline{a})\}) ∪ \{+∞\}).

Let Y_{1} = \{\overline{a} ∈ π_{i}^{m+1}[X′_{0}] : g(\overline{a}) ∈ M\} and Y_{2} := \{\overline{a} ∈ π_{i}^{m+1}[X′_{0}] : g(\overline{a}) = +∞\}. Of course, Y_{1} ∪ Y_{2} = π_{i}^{m+1}[X′_{0}], so at least one of the sets Y_{1}, Y_{2} has dimension k. Below we only consider the case when dim(Y_{1}) = k.

By (b)_{m} and (c)_{m}, there are an open box B_{1} ⊆ M^{m} and a projection π : M^{m} ⟶ M^{k} such that B_{1} ∩ Y_{1} = B_{1} ∩ π_{i}^{m+1}[S], π[B_{1} ∩ Y_{1}] is an open box in M^{k}, π ↾ B_{1} ∩ Y_{1} is a homeomorphism from B_{1} ∩ Y_{1} onto π[B_{1} ∩ Y_{1}], and the functions f, g are continuous on B_{1} ∩ Y_{1}. Fix \overline{a} ∈ B_{1} ∩ Y_{1} and b, c ∈ M such that b < f(\overline{a}) < c < g(\overline{a}). There is an open box B_{2} ⊆ B_{1} containing \overline{a} such that b < f(\overline{x}) < c < g(\overline{x}) whenever \overline{x} ∈ B_{2} ∩ Y_{1}. Let \overline{c} ∈ X′_{0} be the unique tuple such that ϱ_{i}^{m+1}(\overline{c}) = f(\overline{a}) and π_{i}^{m+1}(\overline{c}) = \overline{a}. There is an open box B ⊆ M^{m+1} containing \overline{c} such that B ∩ cl(S \ X_{0}) = ∅, π_{i}^{m+1}[B] ⊆ B_{2} and b < inf ϱ_{i}^{m+1}[B] < f(\overline{x}) < sup ϱ_{i}^{m+1}[B] < g(\overline{x}) for \overline{x} ∈ π_{i}^{m+1}[B ∩ X′_{0}]. By Lemma 2.7, there is an open box B_{3} ⊆ π_{i}^{m+1}[B] containing \overline{a} such that π ↾ B_{3} ∩ Y_{1} is a homeomorphism from B_{3} ∩ Y_{1} onto π[B_{3} ∩ Y_{1}], an open box in M^{k}. Let B′ = B ∩(π_{i}^{m+1})^{−1}[B_{3}] and π′ = π ∘π_{i}^{m+1}. Then B′ ∩X = B′ ∩S, π′ ↾ B′ ∩X is a homeomorphism from B′ ∩ X onto π′[B′ ∩ X], an open box in M^{k}. This finishes the proof of (b)_{m+1}.

Proof of (c)_{m+1}. Assume that S ⊆ M^{m+1} is a non-empty definable set, f : S ⟶ M is a definable function and denote by X the set of discontinuity points of f. Suppose for a contradiction that dim(X) = dim(S) = k. By Lemma 1.4, without loss of generality we can assume that 1 ≤ k ≤ m. By (b)_{m+1}, there are an open box B ⊆ M^{m+1} and i ∈ {1, …, m + 1} such that B ∩ S = B ∩ X, dim(B ∩ X) = k and the projection π_{i}^{m+1} restricted to B ∩ X is a homeomorphism from B ∩ X onto π_{i}^{m+1}[B ∩ X]. Let g denote the inverse of this homeomorphism. By (c)_{m}, the function f ∘ g has a continuity point in π_{i}^{m+1}[B ∩ X]. Consequently, f has a continuity point in B ∩ X, which contradicts our choice of X.

Proof of (d)_{m+1}. Assume that S ⊆ M^{m+1} is a non-empty definable set, f : S ⟶ M and g : S ⟶ \overline{M}^{M} are definable functions, f is continuous, and f(\overline{a}) < g(\overline{a}) whenever \overline{a} ∈ S. In case dim(S) = m + 1, the assertion of (d)_{m+1} is a consequence of Lemma 1.5. The case dim(S) = 0 is trivial. So assume that 1 ≤ dim(S) = k ≤ m. By (b)_{m+1}, there are an open box B ⊆ M^{m+1} and i ∈ {1, …, m + 1} such that the projection π_{i}^{m+1} restricted to B ∩ S is a homeomorphism from B ∩ S onto π_{i}^{m+1}[B ∩ S] and dim(π_{i}^{m+1}[B ∩ S]) = k. Let h denote the inverse of this homeomorphism. By (d)_{m}, there are an open interval I and a definable set Z ⊆ π_{i}^{m+1}[B ∩ S] such that dim(Z) = k and Z × I ⊆ \{⟨\overline{x}, y⟩ : \overline{x} ∈ Z, f(h(\overline{x})) < y < g(h(\overline{x}))\}. Clearly, dim(h[Z]) = k. Moreover, h[Z] × I ⊆ \{⟨\overline{x}, y⟩ : \overline{x} ∈ B ∩ S and f(\overline{x}) < y < g(\overline{x})\}.

Proof of (e)_{m+1}. Assume that S ⊆ M^{m+2} is a non-empty definable set and i ∈ {1, …, m + 2}. If dim(S) = 0, then both sides of the equivalence in (e)_{m+1} are true. In case dim(π_{i}^{m+2}[S]) = dim(S) − 1 ∈ {0, m + 1}, they are both false. Also, by Lemma 2.9, if dim(S) = dim(π_{i}^{m+2}[S]) = m + 1, then the right side of the equivalence in (e)_{m+1} is valid. Having considered all the trivial cases, assume that 1 ≤ dim(π_{i}^{m+2}[S]) ≤ m.

For the left-to-right direction, assume that dim(S) = dim(π_{i}^{m+2}[S]) and I ⊆ M is an open interval. Denote by J the (necessarily non-empty) set of all j ∈ {1, …, m + 2} \ {i} for which dim(π_{i,j}^{m+2}[S]) = dim(π_{i}^{m+2}[S]) = dim(S). By Lemma 1.7(b), we have that dim(π_{i,j}^{m+2}[S]) = dim(π_{j}^{m+2}[S]) whenever j ∈ J. Without loss of generality we can assume that i < j for all j ∈ J. By (e)_{m}, there are definable sets X_{j} ⊆ π_{i,j}^{m+2}[S], j ∈ J, such that dim(X_{j}) < dim(π_{i,j}^{m+2}[S]),

and for any k ∈ ℕ_{+} and any \overline{a}_{1}, …, \overline{a}_{k} ∈ π_{i,j}^{m+2}[S] \ X_{j}, I ⊈ ⋃_{l=1}^{k} ϱ_{i}^{m+1}[(π_{i}^{m+1})^{−1}(\overline{a}_{l}) ∩ π_{j}^{m+2}[S]].

This implies that for any j ∈ J, k ∈ ℕ_{+} and \overline{b}_{1}, …, \overline{b}_{k} ∈ π_{i}^{m+2}[S] \ (π_{j−1}^{m+1})^{−1}[X_{j}], we have that

I ⊈ ⋃_{l=1}^{k} ϱ_{i}^{m+2}[(π_{i}^{m+2})^{−1}(\overline{b}_{l}) ∩ S]. Let X = ⋂_{j∈J}(π_{j−1}^{m+1})^{−1}[X_{j}] ∩ π_{i}^{m+2}[S]. Certainly, for any k ∈ ℕ_{+}

and \overline{b}_{1}, …, \overline{b}_{k} ∈ π_{i}^{m+2}[S] \ X we have that

I ⊈ ⋃_{l=1}^{k} ϱ_{i}^{m+2}[(π_{i}^{m+2})^{−1}(\overline{b}_{l}) ∩ S]

and dim(π_{j}^{m+1}[X]) < dim(π_{i,j}^{m+2}[S]) = dim(S) whenever j ∈ J. We claim that dim(X) < dim(π_{i}^{m+2}[S]). Suppose otherwise. Then there is j_{0} ∈ J such that

dim(S) = dim(π_{i,j_{0}}^{m+2}[S]) ≥ dim(π_{j_{0}−1}^{m+1}[X]) = dim(X) = dim(π_{i}^{m+2}[S]) = dim(S),

a contradiction.

For the right-to-left direction, assume that dim(S) = dim(π_{i}^{m+2}[S]) + 1 and fix j ∈ {1, …, m + 2} \ {i} such that dim(π_{j}^{m+2}[S]) = dim(S). Again, without loss of generality, we can assume that i < j. By Lemma 1.7, dim(π_{i}^{m+2}[S]) = dim(π_{i,j}^{m+2}[S]) = dim(π_{j}^{m+2}[S]) − 1. By (e)_{m}, there is an open interval I ⊆ M such that for any definable set X ⊆ π_{i,j}^{m+2}[S] with dim(X) < dim(π_{i,j}^{m+2}[S]), there are k ∈ ℕ_{+} and \overline{a}_{1}, …, \overline{a}_{k} ∈ π_{i,j}^{m+2}[S] \ X such that

I ⊆ ⋃_{l=1}^{k} ϱ_{i}^{m+1}[(π_{i}^{m+1})^{−1}(\overline{a}_{l}) ∩ π_{j}^{m+2}[S]].

Denote by S_{1} the set of all \overline{a}’s from π_{i,j}^{m+2}[S] for which the set ϱ_{i}^{m+1}[(π_{i}^{m+1})^{−1}(\overline{a}) ∩ π_{j}^{m+2}[S]] contains an interval of the form (inf I, y) with y ∈ I. Our assumptions guarantee that dim(S_{1}) = dim(π_{i,j}^{m+2}[S]) = dim(π_{i}^{m+2}[S]). For \overline{a} ∈ S_{1} denote by g(\overline{a}) the supremum of the first convex component of the set ϱ_{i}^{m+1}[(π_{i}^{m+1})^{−1}(\overline{a})∩π_{j}^{m+2}[S]]∩I. By (d)_{m}, there are an open interval I_{1} ⊆ I and a definable set S_{2} ⊆ S_{1} such that dim(S_{2}) = dim(S_{1}) = dim(π_{i}^{m+2}[S]) and S_{2} × I_{1} ⊆ \{⟨\overline{x}, y⟩ : \overline{x} ∈ S_{1}, inf I < y < g(\overline{x})\}. Hence, for every \overline{a} ∈ S_{2}, I_{1} ⊆ ϱ_{i,j}^{m+2}[(π_{i,j}^{m+2})^{−1}(\overline{a}) ∩ S].

Now, suppose that Z ⊆ π_{i}^{m+2}[S] is a definable set such that dim(Z) < dim(π_{i}^{m+2}[S]). For \overline{a} ∈ S_{2} denote by R(\overline{a}) the set of all elements b ∈ I_{1} for which there is a (necessarily unique) tuple \overline{c} ∈ π_{j}^{m+2}[S] such that π_{i}^{m+1}(\overline{c}) = \overline{a}, ϱ_{i}^{m+1}(\overline{c}) = b and ϱ_{j}^{m+2}[(π_{j}^{m+2})^{−1}(\overline{c}) ∩ S] (necessarily a non-empty set) is a union of some convex components of ϱ_{j−1}^{m+1}[(π_{j−1}^{m+1})^{−1}(\overline{a}) ∩ π_{i}^{m+2}[S]]. Let

S_{3} = \{\overline{a} ∈ S_{2} : R(\overline{a}) contains an open interval of the form (inf I_{1}, y) with y ∈ I_{1}\}.

Below we consider two cases.

Case 1. dim(S_{3}) = dim(S_{2}) = dim(π_{i}^{m+2}[S]). For \overline{a} ∈ S_{3} let h(\overline{a}) ∈ \overline{M}^{M} denote the supremum of the first convex component of R(\overline{a}). As previously, there are an open interval I_{2} ⊆ I_{1} and a definable set S_{4} ⊆ S_{3} such that dim(S_{4}) = dim(S_{3}) = dim(π_{i}^{m+2}[S]) and S_{4} × I_{2} ⊆ \{⟨\overline{x}, y⟩ : \overline{x} ∈ S_{3}, inf I_{1} < y < h(\overline{x})\}. Fix \overline{a} ∈ S_{4} \π_{j−1}^{m+1}[Z] and choose \overline{b}_{1}, …, \overline{b}_{k} ∈ (π_{j−1}^{m+1})^{−1}(\overline{a})∩π_{i}^{m+2}[S] so that ϱ_{j−1}^{m+1}(\overline{b}_{1}), …, ϱ_{j−1}^{m+1}(\overline{b}_{k}) are representatives of all the convex components of ϱ_{j−1}^{m+1}[(π_{j−1}^{m+1})^{−1}(\overline{a}) ∩

π_{i}^{m+2}[S]]. Clearly, I_{2} ⊆ ⋃_{l=1}^{k} ϱ_{i}^{m+2}[(π_{i}^{m+2})^{−1}(\overline{b}_{l}) ∩ S].

Case 2. dim(S_{3}) < dim(π_{i}^{m+2}[S]). Let S_{4} = S_{2} \ S_{3}. Of course, dim(S_{4}) = dim(π_{i}^{m+2}[S]). For \overline{a} ∈ S_{4} denote by T (\overline{a}) the set of all elements b ∈ I_{1} for which there is a (necessarily unique) tuple \overline{c} ∈ π_{j}^{m+2}[S] such that π_{i}^{m+1}(\overline{c}) = \overline{a}, ϱ_{i}^{m+1}(\overline{c}) = b and at least one of the convex components of ϱ_{j}^{m+2}[(π_{j}^{m+2})^{−1}(\overline{c})∩S] is a proper subset of some of the convex components of ϱ_{j−1}^{m+1}[(π_{j−1}^{m+1})^{−1}(\overline{a})∩ π_{i}^{m+2}[S]]. Note that for every \overline{a} ∈ S_{4}, T (\overline{a}) contains an open interval of the form (inf I, y), y ∈ I. As in Case 1, by (d)_{m}, there are an open interval I_{2} ⊆ I_{1} and a definable set S_{5} ⊆ S_{4} such that dim(S_{5}) = dim(S_{4}) = dim(π_{i}^{m+2}[S]) and for every \overline{c} ∈ π_{j}^{m+2}[S] with ϱ_{i}^{m+1}(\overline{c}) ∈ I_{2} and π_{i}^{m+1}(\overline{c}) ∈ S_{5}, at least one of the convex components of ϱ_{j}^{m+2}[(π_{j}^{m+2})^{−1}(\overline{c}) ∩ S] is a proper subset of some of the convex components of ϱ_{j−1}^{m+1}[(π_{j−1}^{m+1})^{−1}(π_{i}^{m+1}(\overline{c})) ∩ π_{i}^{m+2}[S]].

Fix b ∈ I_{2} and for \overline{a} ∈ S_{5} denote by \overline{c}(\overline{a}) the unique tuple from π_{j}^{m+2}[S] such that ϱ_{i}^{m+1}(\overline{c}(\overline{a})) = b and π_{i}^{m+1}(\overline{c}(\overline{a})) = \overline{a}. Let U(\overline{a}) be the first convex component of ϱ_{j}^{m+1}[(π_{j−1}^{m+1})^{−1}(\overline{a}) ∩ π_{i}^{m+2}[S]] such that ϱ_{j}^{m+2}[(π_{j}^{m+2})^{−1}(\overline{c}(\overline{a})) ∩ S] ∩ U(\overline{a}) is a non-empty proper subset of U(\overline{a}). Denote by V (\overline{a}) the first convex component of ϱ_{j}^{m+2}[(π_{j}^{m+2})^{−1}(\overline{c}(\overline{a})) ∩ S] ∩ U(\overline{a}). For every \overline{a} ∈ S_{5} denote by U′(\overline{a}) the unique definable subset of (π_{j−1}^{m+1})^{−1}(\overline{a}) ∩ π_{i}^{m+2}[S] such that ϱ_{j−1}^{m+1}[U′(\overline{a})] = U(\overline{a}) and by V′(\overline{a}) the unique definable subset of π_{i}^{m+2}[(π_{j}^{m+2})^{−1}(\overline{c}(\overline{a})) ∩ S] such that ϱ_{j−1}^{m+1}[V′(\overline{a})] = V (\overline{a}). Let S′ =

⋃_{\overline{a}∈S_{5}} U′(\overline{a}) and X′ = ⋃_{\overline{a}∈S_{5}} V′(\overline{a}). Note that dim(S′) = dim(π_{j−1}^{m+1}[S′]), π_{j−1}^{m+1}[S′] = π_{j−1}^{m+1}[X′] = S_{5},

and for every \overline{a} ∈ π_{j−1}^{m+1}[S′] we have that ϱ_{j−1}^{m+1}[(π_{j−1}^{m+1})^{−1}(\overline{a}) ∩ S′] is an infinite convex set and ∅ ≠ (π_{j−1}^{m+1})^{−1}(\overline{a}) ∩ X′ ⊊ (π_{j−1}^{m+1})^{−1}(\overline{a}) ∩ S′. This contradicts (g)_{m}.

Proof of (f)_{m+1}. Assume that S ⊆ M^{m+2} is a non-empty definable set, i ∈ {1, …, m + 2}, and for every \overline{a} ∈ π_{i}^{m+2}[S], the set (π_{i}^{m+2})^{−1}(\overline{a}) ∩ S is finite. Then for any open interval I ⊆ M,

k ∈ ℕ_{+} and \overline{a}_{1}, …, \overline{a}_{k} ∈ π_{i}^{m+2}[S], we have that I ⊈ ⋃_{l=1}^{k} ϱ_{i}^{m+2}[(π_{i}^{m+2})^{−1}(\overline{a}_{l}) ∩ S]. By (e)_{m+1},

dim(S) = dim(π_{i}^{m+2}[S]).

Proof of (g)_{m+1}. Of course, it is sufficient to prove (g)_{m+1} for i = m + 1. To make the notation simpler, let π = π_{m+2}^{m+2}. Assume that X ⊆ S ⊆ M^{m+2} are non-empty definable sets such that dim(π[S]) = dim(S) = k ≥ 1 and π[S] = π[X]. Assume also that for every \overline{a} ∈ π[S], the set B(\overline{a}) := \{b ∈ M : ⟨\overline{a}, b⟩ ∈ S\} is an infinite convex set of which \{b ∈ M : ⟨\overline{a}, b⟩ ∈ X\} is a non-empty proper subset. Denote by X′ the unique definable subset of X such that for every \overline{a} ∈ π[S], the set A(\overline{a}) := \{b ∈ M : ⟨\overline{a}, b⟩ ∈ X′\} is the first convex component of \{b ∈ M : ⟨\overline{a}, b⟩ ∈ X\}. Define the following sets.

Y_{1} = \{\overline{a} ∈ π[S] : inf B(\overline{a}) < inf A(\overline{a}) < sup B(\overline{a})\};
Y_{2} = \{\overline{a} ∈ π[S] : inf B(\overline{a}) < sup A(\overline{a}) < sup B(\overline{a})\};
Y_{3} = \{\overline{a} ∈ π[S] : inf A(\overline{a}) = inf B(\overline{a}), sup A(\overline{a}) = sup B(\overline{a}) and inf B(\overline{a}) ∈ B(\overline{a}) \ A(\overline{a})\};
Y_{4} = \{\overline{a} ∈ π[S] : inf A(\overline{a}) = inf B(\overline{a}), sup A(\overline{a}) = sup B(\overline{a}) and sup B(\overline{a}) ∈ B(\overline{a}) \ A(\overline{a})\};
Y_{5} = \{\overline{a} ∈ π[S] : inf A(\overline{a}) = sup A(\overline{a}) = inf B(\overline{a})\};
Y_{6} = \{\overline{a} ∈ π[S] : inf A(\overline{a}) = sup A(\overline{a}) = sup B(\overline{a})\}.

Clearly, π[S] = Y_{1} ∪ Y_{2} ∪ Y_{3} ∪ Y_{4} ∪ Y_{5} ∪ Y_{6}, so at least one of the sets Y_{1}, Y_{2}, Y_{3}, Y_{4}, Y_{5}, Y_{6} has dimension k. Below we consider 4 cases.

Case 1. dim(Y_{1}) = k. Lemma 2.5 implies that k < m + 1. By (b)_{m+1}, there are j ∈ {1, …, m + 1} and an open box B ⊆ M^{m+1} such that B ∩ π[S] = B ∩ Y_{1}, dim(B ∩ Y_{1}) = dim(π_{j}^{m+1}[B ∩ Y_{1}]) = k and the projection π_{j}^{m+1} restricted to B ∩ Y_{1} is a homeomorphism. By (g)_{m}, the set π_{j}^{m+2}[π^{−1}[B∩Y_{1}]∩S] ⊆ M^{m+1} has dimension k+1, which implies that dim(S) = k+1, a contradiction.

Case 2. dim(Y_{2}) = k. This case is similar to Case 1.

Case 3. dim(Y_{3}) = k. Define functions f : Y_{3} ⟶ M and g : Y_{3} ⟶ \overline{M}^{M} as follows:

f(\overline{a}) = inf A(\overline{a}), g(\overline{a}) = sup A(\overline{a}).

Our assumptions guarantee that f(\overline{a}) < g(\overline{a}) whenever \overline{a} ∈ Y_{1}. By (c)_{m+1}, there is a definable set Z ⊆ Y_{3}, large in Y_{3}, such that the function f restricted to Z is continuous. By (d)_{m+1}, there are an open interval I and a definable set Z′ ⊆ Z such that dim(Z′) = k and Z′ × I ⊆ \{⟨\overline{x}, y⟩ : \overline{x} ∈ Z and f(\overline{x}) < y < g(\overline{x})\}. Hence dim(S) = k + 1.

Case 4. One of the sets Y_{4}, Y_{5}, Y_{6} has dimension k. This case is similar to Case 3.

The following corollary is a direct consequence of condition (f)_{m} from Theorem 2.11.

Corollary 2.12 Assume that m ∈ ℕ_{+}, S ⊆ M^{m} is a non-empty definable set and f : S ⟶ M is a definable function. Then dim(Γ(f)) = dim(S).

Theorem 2.13 Assume that M = (M, ≤, …) is a weakly o-minimal structure, m ∈ ℕ_{+}, S ⊆ M^{m} is a non-empty definable set and f : S ⟶ M^{n} is an injective definable map. Then dim(S) = dim(Γ(f)) = dim(f[S]).

Proof. Let f = ⟨f_{1}, …, f_{n}⟩, where f_{1}, …, f_{n} are definable maps from S into M. Let g_{1} = f_{1} and for 1 ≤ k < n define a map g_{k+1} : Γ(g_{k}) ⟶ M as follows: g_{k+1}(\overline{a}, g_{k}(\overline{a})) = f_{k+1}(\overline{a}). It is clear that Γ(f) = Γ(g_{n}). By Corollary 2.12, dim(Γ(f)) = dim(S). The same argument with f replaced by f^{−1} shows that dim(Γ(f)) = dim(f[S]).

Corollary 2.14 Assume that m ∈ ℕ_{+}, J_{1}, J_{2} are distinct subsets of {1, …, m + 1} and X_{1}, X_{2} ⊆ M^{m+1} are definable sets such that X_{1} is J_{1}-open and X_{2} is J_{2}-open. Then

dim(X_{1} ∩ X_{2}) < max{dim(X_{1}), dim(X_{2})}.

Proof. Assume that m, J_{1}, J_{2}, X_{1}, X_{2} satisfy assumptions of the lemma. In case dim(X_{1}) ≠ dim(X_{2}) or X_{1} ∩ X_{2} = ∅, the assertion of the lemma is trivial. So let dim(X_{1}) = dim(X_{2}) = k and let X_{1} ∩ X_{2} ≠ ∅. Without loss of generality we can assume that J_{2} \ J_{1} ≠ ∅. Below, for a fixed i ∈ J_{2} \ J_{1} we consider two cases.

Case 1. dim(π_{i}^{m+1}[X_{2}]) < k. Then for every \overline{a} ∈ π_{i}^{m+1}[X_{1}∩X_{2}], the fiber (π_{i}^{m+1})^{−1}(\overline{a})∩X_{1}∩X_{2} is finite, and using condition (f)_{m} from Theorem 2.11 we conclude that

dim(X_{1} ∩ X_{2}) = dim(π_{i}^{m+1}[X_{1} ∩ X_{2}]) ≤ dim(π_{i}^{m+1}[X_{2}]) < k.

Case 2. dim(π_{i}^{m+1}[X_{2}]) = k. Suppose for a contradiction that dim(X_{1} ∩ X_{2}) = k and let Z = π_{i}^{m+1}[X_{1} ∩ X_{2}]. Again, by condition (f)_{m} from Theorem 2.11, dim(Z) = k. There is a unique definable set X ⊆ X_{1} ∩ X_{2} such that π_{i}^{m+1}[X] = Z and for every \overline{a} ∈ Z, we have that

ϱ_{i}^{m+1}[(π_{i}^{m+1})^{−1}(\overline{a}) ∩ X] = \{min ϱ_{i}^{m+1}[(π_{i}^{m+1})^{−1}(\overline{a}) ∩ X_{1} ∩ X_{2}]\}.

Clearly, dim(X) = k. Moreover, there is a unique definable set S ⊆ X_{2} such that π_{i}^{m+1}[S] = Z and for every \overline{a} ∈ Z, ϱ_{i}^{m+1}[(π_{i}^{m+1})^{−1}(\overline{a}) ∩ S] is the convex component of ϱ_{i}^{m+1}[(π_{i}^{m+1})^{−1}(\overline{a}) ∩ X_{2}] containing ϱ_{i}^{m+1}[(π_{i}^{m+1})^{−1}(\overline{a}) ∩ X]. Now, condition (g)_{m} from Theorem 2.11 implies that dim(X_{2}) = k + 1, a contradiction.

3 Large subsets of Cartesian products

This section has been motivated in two ways. Firstly, if M is an o-minimal structure, X, Y are non-empty definable subsets of M^{m}, M^{n} respectively, and S is a large subset of X × Y, then the set of all \overline{a}’s from X for which the fiber S\overline{a} is large in Y, is large in X. Of course, such a statement fails in general weakly o-minimal structures. Nevertheless, I was eager to know if there exists a number k such that for ”almost all” tuples ⟨\overline{a}_{1}, …, \overline{a}_{k}⟩ ∈ X^{k}, the union S_{\overline{a}_{1}} ∪ … ∪ S_{\overline{a}_{k}} is large in Y. Another motivation comes from the area of groups and fields definable in the o-minimal context. One could easily rewrite all proofs from [Pi] so that the use of generic types is replaced by the addition property of the dimension. Such proofs works for weakly o-minimal structures with the addition property (for the definition see §4). A natural question to ask was how important is the assumption of the addition property. It turned out that a weaker version of topologization of groups and fields in weakly o-minimal context is possible modulo some technical fact concerning large definable subsets of cartesian products of definable sets. The aim of this section is to provide a proof of that technical fact. Below in a series of lemmas we will show that if X ⊆ M^{m}, Y ⊆ M^{n} and S ⊆ X × Y are definable sets and S is large in X × Y, then we can find \overline{a}_{1}, …, \overline{a}_{k} ∈ X, where k = 2^{dim(Y)} such that the set S_{\overline{a}_{1}} ∪ … ∪ S_{\overline{a}_{k}} is large in Y.

Lemma 3.1 Let m, n ∈ ℕ_{+} and assume that X ⊆ M^{m}, Y ⊆ M^{n} and S ⊆ X × Y are non-empty definable sets. The following conditions are equivalent.

(a) S is large in X × Y.

(b) For any definable sets U ⊆ X and V ⊆ Y with dim(U) = dim(X) and dim(V) = dim(Y), we have that (U × V) ∩ S ≠ ∅.

Proof. (a)⟹(b). If there are definable sets U ⊆ X and V ⊆ Y such that dim(U) = dim(X), dim(V) = dim(Y) and (U × V) ∩ S = ∅, then dim((X × Y) \ S) = dim(U × V) = dim(X × Y), which means that S is not large in X × Y.

(b)⟹(a). Assume that S is not large in X × Y and let W = (X × Y) \ S. By condition (b)_{m+n} from Theorem 2.11, there are open boxes B ⊆ M^{m} and C ⊆ M^{n} such that (B × C) ∩ W = (B × C) ∩ (X × Y) = (B ∩ X) × (C ∩ Y) and dim((B × C) ∩ W) = dim(X × Y). Clearly, dim(B ∩ X) = dim(X), dim(C ∩ Y) = dim(Y) and ((B ∩ X) × (C ∩ Y)) ∩ S = ∅.

Lemma 3.2 Let m, n ∈ ℕ_{+} and assume that X ⊆ M^{m}, Y ⊆ M^{n} and S ⊆ X × Y are non-empty definable sets.

(a) If the set \{\overline{a} ∈ X : S\overline{a} is large in Y \} is large in X, then S is large in X × Y.

(b) If the set \{\overline{b} ∈ Y : S\overline{b} is large in X\} is large in Y, then S is large in X × Y.

Proof. As both cases are similar, we only prove (a). Suppose that S is not large in X × Y. Then by Lemma 3.1, there are definable sets U ⊆ X and V ⊆ Y such that dim(U) = dim(X), dim(V) = dim(Y) and (U × V) ∩ S = ∅. Hence dim(\{\overline{a} ∈ X : S\overline{a} is not large in Y \} = dim(X).

Lemma 3.3 Let m ∈ ℕ_{+} and assume that X ⊆ M^{m}, U ⊆ M, S ⊆ X × U are non-empty definable sets, U is open and S is large in X ×U. Then the set X′ := \{⟨\overline{a}, \overline{b}⟩ ∈ X ×X : S\overline{a} ∪S\overline{b} is large in U\} is large in X × X.

Proof. The assertion of the lemma is obvious for X finite. So assume that dim(X) = k ≥ 1 and suppose for a contradiction that X′ is not large in X × X. Then by Lemma 3.1, there are definable sets X_{1}, X_{2} ⊆ X such that dim(X_{1}) = dim(X_{2}) = k and (∀\overline{a} ∈ X_{1})(∀\overline{b} ∈ X_{2})(S\overline{a} ∪ S\overline{b} is not large in U). Let Y_{1} = ⋃_{\overline{a}∈X_{1}}\{\overline{a}\} × int(U \ S\overline{a}) and Y_{2} = ⋃_{\overline{b}∈X_{2}}\{\overline{b}\} × int(U \ S\overline{b}). Clearly,

(Y_{1} ∪ Y_{2}) ∩ S = ∅ and

(∀\overline{a} ∈ X_{1})(∀\overline{b} ∈ X_{2})((Y_{1})\overline{a} ∩ (Y_{2})\overline{b} is a non-empty open subset of U).

Below we consider two cases.

Case 1. There is \overline{a} ∈ X_{1} such that

(∗) dim(\{\overline{b} ∈ X_{2} : (Y_{1})\overline{a} ∩ (Y_{2})\overline{b} is a union of some convex components of (Y_{1})\overline{a}\}) = k.

Fix \overline{a} ∈ X_{1} for which (∗) holds. There are a definable set X′_{2} ⊆ X_{2} and an open interval I ⊆ U such that dim(X′_{2}) = k and (∀\overline{b} ∈ X′_{2})(I ⊆ (Y_{1})\overline{a} ∩ (Y_{2})\overline{b}). Consequently, (X′_{2} × I) ∩ S = ∅, which means that S is not large in X × U.

Case 2. For every \overline{a} ∈ X_{1}, the set

\{\overline{b} ∈ X_{2} : (Y_{1})\overline{a} ∩ (Y_{2})\overline{b} is not a union of some convex components of (Y_{1})\overline{a}\}

is large in X_{2}. By Lemma 3.2, the set

\{⟨\overline{a}, \overline{b}⟩ ∈ X_{1} × X_{2} : (Y_{1})\overline{a} ∩ (Y_{2})\overline{b} is not a union of some convex components of (Y_{1})\overline{a}\}

is large in X_{1} × X_{2}. Hence, by Lemma 3.1, there are definable sets X′_{1} ⊆ X_{1} and X′_{2} ⊆ X_{2} such that dim(X′_{1}) = dim(X′_{2}) = k and for any \overline{a} ∈ X′_{1} and \overline{b} ∈ X′_{2}, (Y_{1})\overline{a} ∩ (Y_{2})\overline{b} is not a union of some convex components of (Y_{1})\overline{a}. Fix \overline{b}_{0} ∈ X′_{2} and for \overline{a} ∈ X′_{1} denote by A(\overline{a}) the first convex component of (Y_{1})\overline{a} which is not contained in (Y_{2})_{\overline{b}_{0}} and has a non-empty intersection with (Y_{2})_{\overline{b}_{0}}. Denote also by B(\overline{a}) the first convex component of A(\overline{a}) ∩ (Y_{2})_{\overline{b}_{0}}. Let P = ⋃_{\overline{a}∈X′_{1}}\{\overline{a}\} × A(\overline{a})

and R = ⋃_{\overline{a}∈X′_{1}}\{\overline{a}\} × B(\overline{a}). Our construction guarantees that π_{m+1}^{m+1}[P] = π_{m+1}^{m+1}[R] = X′_{1} and

dim(P) = dim(R) = k. Moreover, for every \overline{a} ∈ X′_{1}, P\overline{a} is an infinite convex set and R\overline{a} is a non-empty proper subset of P\overline{a}. By condition (g)_{m} from Theorem 2.11, this is impossible.

Lemma 3.4 Let m, n ∈ ℕ_{+} and assume that X ⊆ M^{m}, U ⊆ M^{n}, S ⊆ X × U are non-empty definable sets, U is open, and S is large in X × U. Then the set

\{⟨\overline{a}_{1}, …, \overline{a}_{2^{n}}⟩ ∈ X^{2^{n}} : S_{\overline{a}_{1}} ∪ … ∪ S_{\overline{a}_{2^{n}}} is large in U\}

is large in X^{2^{n}}.

Proof. We use induction on n. The case n = 1 is a consequence of Lemma 3.3, so suppose that the result holds for dimension n. Assume that X ⊆ M^{m} is a nonempty definable set of dimension k, U ⊆ M^{n+1} is a non-empty open definable set and S ⊆ X × U is a definable set, large in X × U. We will show that the set

\{⟨\overline{a}_{1}, …, \overline{a}_{2^{n+1}}⟩ ∈ X^{2^{n+1}} : S_{\overline{a}_{1}} ∪ … ∪ S_{\overline{a}_{2^{n+1}}} is large in U\}

is large in X^{2^{n+1}}. Let Y = ⋃_{\overline{a}∈X}\{\overline{a}\} × int(U \ S\overline{a}). We will be done if we prove that the set

\{⟨\overline{a}_{1}, …, \overline{a}_{2^{n+1}}⟩ ∈ X^{2^{n+1}} : Y_{\overline{a}_{1}} ∩ … ∩ Y_{\overline{a}_{2^{n+1}}} = ∅\}

is large in X^{2^{n+1}}.

Clearly, dim(Y) ≤ k + n. For every \overline{a} ∈ X, the fiber Y\overline{a} is an open (possibly empty) subset of U. This implies that for every \overline{d} ∈ Y, ϱ_{m+1}^{m+n+1}[(π_{m+1}^{m+n+1})^{−1}(π_{m+1}^{m+n+1}(\overline{d})) ∩ Y] ⊆ M is an open set containing ϱ_{m+1}^{m+n+1}(\overline{d}).

Claim 1. For every b ∈ M, dim((ϱ_{m+1}^{m+n+1})^{−1}(b) ∩ Y) < k + n.

Proof of Claim 1. Suppose for a contradiction that for some b ∈ M, dim((ϱ_{m+1}^{m+n+1})^{−1}(b)∩Y) =

k+n. For ⟨\overline{a}, \overline{c}⟩ ∈ π_{m+1}^{m+n+1}[(ϱ_{m+1}^{m+n+1})^{−1}(b)∩Y], where |\overline{a}| = m and |\overline{c}| = n, let f(\overline{a}, \overline{c}) ∈ \overline{M}^{M}∪\{+∞\} be the supremum of the set

\{b_{1} > b : (b, b_{1}) ⊆ ϱ_{m+1}^{m+n+1}[(π_{m+1}^{m+n+1})^{−1}(\overline{a}\overline{c}) ∩ Y]\}.

By condition (d)_{m+n} from Theorem 2.11, there are a definable set X_{1} ⊆ π_{m+1}^{m+n+1}[(ϱ_{m+1}^{m+n+1})^{−1}(b) ∩ Y] and an open interval I ⊆ M such that dim(X_{1}) = k + n and

X_{1} × I ⊆ \{⟨\overline{a}, \overline{c}, b_{1}⟩ : ⟨\overline{a}, \overline{c}⟩ ∈ π_{m+1}^{m+n+1}[(ϱ_{m+1}^{m+n+1})^{−1}(b) ∩ Y] and b < b_{1} < f(\overline{a}, \overline{c})\}.

This means that dim(Y) = k + n + 1, a contradiction.

Claim 1 implies that for every b ∈ M, the set Y (b) := π_{m+1}^{m+n+1}[(ϱ_{m+1}^{m+n+1})^{−1}(b) ∩ Y] ⊆ X × M^{n} has dimension lower than k + n. Moreover, for any \overline{a} ∈ X and b ∈ M, Y (b)\overline{a} is an open subset of M^{n}. By the inductive hypothesis, the set

V (b) := \{⟨\overline{a}_{1}, …, \overline{a}_{2^{n}}⟩ ∈ X^{2^{n}} : Y (b)_{\overline{a}_{1}} ∩ … ∩ Y (b)_{\overline{a}_{2^{n}}} = ∅\}

is large in X^{2^{n}}. Consequently, by Lemma 3.2, the set Z := ⋃_{b∈M} V (b) × {b} ⊆ X^{2^{n}} × M is large in

X^{2^{n}} × M.

Claim 2. For any \overline{a}_{1}, …, \overline{a}_{2^{n}} ∈ X, the fiber ((X^{2^{n}} × M) \ Z)_{⟨\overline{a}_{1},...,\overline{a}_{2^{n}} ⟩} ⊆ M is open.

Proof of Claim 2. Let \overline{a}_{1}, …, \overline{a}_{2^{n}} ∈ X. By our choice of Y, the set Y_{\overline{a}_{1}} ∩ … ∩ Y_{\overline{a}_{2^{n}}} ⊆ M^{m+1} is open. Moreover, for every b ∈ M^{m+1} we have that

b ∈ ((X^{2^{n}} × M) \ Z)_{⟨\overline{a}_{1},...,\overline{a}_{2^{n}} ⟩} ⟺
⟨\overline{a}_{1}, …, \overline{a}_{2^{n}}⟩ ∉ V (b) ⟺
Y (b)_{\overline{a}_{1}} ∩ … ∩ Y (b)_{\overline{a}_{2^{n}}} ≠ ∅ ⟺
(∃\overline{c} ∈ M^{n})(⟨b, \overline{c}⟩ ∈ Y_{\overline{a}_{1}} ∩ … ∩ Y_{\overline{a}_{2^{n}}})  .

Thus, if b ∈ ((X^{2^{n}} × M) \ Z)_{⟨\overline{a}_{1},...,\overline{a}_{2^{n}} ⟩}, then there are an open interval I containing b and an open box B ⊆ M^{n} such that I × B ⊆ Y_{\overline{a}_{1}} ∩ … ∩ Y_{\overline{a}_{2^{n}}}, which implies that I ⊆ ((X^{2^{n}} × M) \ Z)_{⟨\overline{a}_{1},...,\overline{a}_{2^{n}} ⟩}.

By Lemma 3.3, the set

Z′ := \{⟨\overline{a}_{1}, …, \overline{a}_{2^{n+1}}⟩ ∈ X^{2^{n+1}} : Z_{⟨\overline{a}_{1},...,\overline{a}_{2^{n}} ⟩} ∪ Z_{⟨\overline{a}_{2^{n}+1},...,\overline{a}_{2^{n+1}} ⟩} = M\}

is large in X^{2^{n+1}}. Clearly, if ⟨\overline{a}_{1}, …, \overline{a}_{2^{n+1}}⟩ ∈ Z′, then for every b ∈ M, ⟨\overline{a}_{1}, …, \overline{a}_{2^{n}}⟩ ∈ V (b) or ⟨\overline{a}_{2^{n}+1}, …, \overline{a}_{2^{n+1}}⟩ ∈ V (b). Consequently, Y_{\overline{a}_{1}} ∩ … ∩ Y_{\overline{a}_{2^{n+1}}} = ∅. This finishes the proof.

The following lemma is obvious.

Lemma 3.5 Assume that m, n ∈ ℕ_{+}, X ⊆ M^{m} is a non-empty definable set, Y ⊆ M^{n} is a nonempty finite set and S ⊆ X × Y is a definable set, large in X × Y. Then the set \{\overline{a} ∈ X : S\overline{a} = Y \} is large in X.

Theorem 3.6 Let m, n ∈ ℕ_{+} and assume that X ⊆ M^{m}, Y ⊆ M^{n}, S ⊆ X × Y are non-empty definable sets, S is large in X × Y, and dim(Y) = k. Then the set

\{⟨\overline{a}_{1}, …, \overline{a}_{2^{k}}⟩ ∈ X^{2^{k}} : S_{\overline{a}_{1}} ∪ … ∪ S_{\overline{a}_{2^{k}}} is large in Y \}

is large in X^{2^{k}}.

Proof. We proceed inductively on n. For n = 1 the result easily follows from Lemmas 3.3 and 3.5. Suppose that it holds for dimension n and assume that X ⊆ M^{m} and Y ⊆ M^{n+1} are non-empty definable sets, k = dim(Y) and S ⊆ X × Y is a definable set, large in X × Y. For k = 0 the assertion of the theorem holds by Lemma 3.5. In case k = n + 1, it is true by Lemma 3.4 and the fact that dim(Y \ int(Y)) ≤ n. To complete the proof, assume that k ∈ {1, …, n}. By Lemma 2.4, there are J_{0}, …, J_{l}, distinct proper subsets of {1, …, n + 1} and pairwise disjoint definable sets Y_{0}, …, Y_{l} such that Y_{0} ∪ … ∪ Y_{l} = Y and for every i ≤ l, the set Y_{i} is J_{i}-open. Clearly, without loss of generality we can assume that dim(Y_{i}) = k whenever i ≤ l. Then for any i ≤ l, S is large in X × Y_{i}. Since the intersection of finitely many subsets of X^{2^{k}} all of which are large in X^{2^{k}} is large in X^{2^{k}}, without loss of generality we can assume that Y = Y_{i} for some i ≤ l. As Y is J_{i}-open, there is a projection π : M^{n+1} ⟶ M^{n} such that for every \overline{a} ∈ π[Y], the fiber π^{−1}(\overline{a}) ∩ Y is finite. By condition (f)_{m} of Theorem 2.11, dim(Y) = dim(π[Y]). Consequently, dim(X × Y) = dim(X × π[Y]). For \overline{a} ∈ M^{m} and \overline{b} ∈ M^{n+1} define π′(\overline{a}, \overline{b}) = ⟨\overline{a}, π(\overline{b})⟩. The set S′ := (X × π[Y]) \ π′[(X × Y) \ S] is large in X × π[Y] because

dim(π′[(X × Y) \ S]) ≤ dim((X × Y) \ S) < dim(X × Y) = dim(X × π[Y]).

By the inductive assumption, the set

X′ := \{⟨\overline{a}_{1}, …, \overline{a}_{2^{k}}⟩ ∈ X^{2^{k}} : S′_{\overline{a}_{1}} ∪ … ∪ S′_{\overline{a}_{2^{k}}} is large in π[Y]\}

is large in X^{2^{k}}. Note that for any \overline{a}_{1}, …, \overline{a}_{2^{k}} ∈ X, π^{−1}[S′_{\overline{a}_{1}} ∪ … ∪ S′_{\overline{a}_{2^{k}}}] ∩ Y ⊆ S_{\overline{a}_{1}} ∪ … ∪ S_{\overline{a}_{2^{k}}}. Consequently, if ⟨\overline{a}_{1}, …, \overline{a}_{2^{k}}⟩ ∈ X′, then S_{\overline{a}_{1}} ∪ … ∪ S_{\overline{a}_{2^{k}}} is large in Y. This finishes the proof.

A natural question that appears in mind after having completed the proof of Theorem 3.6 is whether one could replace the number 2^{k} by a smaller one. To be more precise, given a weakly o-minimal structure M, define a function f_{M} : ℕ ⟶ ℕ as follows: f_{M}(k) is the smallest number l such that if m, n ∈ ℕ_{+}, X ⊆ M^{m}, Y ⊆ M^{n} and S ⊆ X × Y are non-empty definable sets, S is large in X × Y, and dim(Y) = k, then the set \{⟨\overline{a}_{1}, …, \overline{a}_{l}⟩ ∈ X^{l} : S_{\overline{a}_{1}} ∪ … ∪ S_{\overline{a}_{l}} is large in Y \} is large in X^{l}. It is well known that if M is o-minimal, then f_{M}(k) = 1. According to Theorem 3.6, f_{M}(k) ≤ 2^{k} for any weakly o-minimal structure M. Below we give an example of a weakly o-minimal structure M with f_{M}(k) ≥ k + 1 for k ∈ ℕ.

Example. Let M = (M, ≤, …) be a weakly o-minimal structure in which there are: a convex open definable set U ⊆ M and a definable function f : U ⟶ M which is locally constant but not piecewise constant (see [MMS, Example 2.6.1]). For k ≥ 2 and 1 ≤ j < k define

X_{j,k} = {⟨x_{1}, …, x_{k}⟩ ∈ M^{k} : x_{1} = f(x_{1+j})}; S_{k} = M^{k} \ (X_{1,k} ∪ … ∪ X_{k−1,k}).  (1)

It is easy to see that S_{k+1} is large in f[M] × M^{k} for k ∈ ℕ_{+}. Moreover, if a_{1}, …, a_{k} ∈ f[U], then the set (S_{k+1})_{a_{1}} ∪ … ∪ (S_{k+1})_{a_{k}} is not large in M^{k}.

4 The addition property and the exchange property

For a definable set S ⊆ M^{m+n} and numbers d ∈ {−∞, 0, 1, …, n} and d′ ∈ {−∞, 0, 1, …, m} we define

X(S, m, d) = \{\overline{a} ∈ M^{m} : dim(S\overline{a}) = d\},
Y (S, n, d′) = \{\overline{b} ∈ M^{n} : dim(S\overline{b}) = d′\}.

Clearly, the sets X(S, m, d) and Y (S, n, d′) are definable.

Definition 4.1 We say that dim has the addition property in M iff one [equivalently: both] of the following conditions is [are] true.

(a) If m, n ∈ ℕ_{+}, S ⊆ M^{m+n} is a definable set, and d ∈ {−∞, 0, 1, …, n}, then

dim ( ⋃_{\overline{a}∈X(S,m,d)}\{\overline{a}\} × S\overline{a}) = dim(X(S, m, d)) + d.

(b) If m, n ∈ ℕ_{+}, S ⊆ M^{m+n} is a definable set, and d′ ∈ {−∞, 0, 1, …, m}, then

dim ( ⋃_{\overline{a}∈Y (S,n,d′)} S\overline{b} × \{\overline{b}\}) = dim(Y (S, n, d′)) + d′.

The following theorem relates the addition property of M to two statements concerning definable functions with values in \overline{M}^{M}. The addition property (condition (d)) is also shown to be equivalent to a seemingly weaker statement (c). In fact the proof of equivalence of (c) and (d) does not depend on the assumption of weak o-minimality of M. It goes through in every first order structure with a sufficiently good dimension function (for details we refer the reader to [vdD1], section 1).

Theorem 4.2 The following conditions are equivalent.

(a) For any open interval I ⊆ M and any definable function f : I ⟶ \overline{M}^{M}, there is an open interval I′ ⊆ I such that f ↾ I′ is continuous.

(b) If m ∈ ℕ_{+}, B ⊆ M^{m} is an open box and f : B ⟶ \overline{M}^{M} is a definable function, then there is an open box B′ ⊆ B such that f ↾ B′ is continuous.

(c) If m ∈ ℕ_{+}, i ∈ {1, …, m} and S ⊆ M^{m+1} is a non-empty definable set, then dim(S) = dim(π_{i}^{m+1}[S]) iff the set of tuples \overline{a} ∈ π_{i}^{m+1}[S] for which the fiber (π_{i}^{m+1})^{−1}(\overline{a}) ∩ S is finite, is large in π_{i}^{m+1}[S].

(d) dim has the addition property in M.

Proof. (a)⟹(b). Assume that (a) holds. In order to prove (b) we use induction on m. There is nothing to do in dimension 1, so fix a positive integer m, and suppose that for any open box B ⊆ M^{m} and any definable function f : B ⟶ \overline{M}^{M}, there is an open box B′ ⊆ B such that f ↾ B′ is continuous.

Let C ⊆ M^{m} be an open box, I ⊆ M an open interval and f : C × I ⟶ \overline{M}^{M} a definable function. By Lemma 1.10, without loss of generality we can assume that for every \overline{a} ∈ C, the function f(\overline{a}, y) is either constant or strictly monotone. For b ∈ I, denote by S(b) the set of continuity points of f(\overline{x}, b). By the inductive hypothesis, S(b) is large in C whenever b ∈ I. By Lemma 3.2, the set ⋃_{b∈I} S(b) × {b} is large in C × I, so it contains an open box C_{1} × I_{1}, where I_{1} is

an open interval. Clearly, for every b ∈ I_{1}, the function f(\overline{x}, b) restricted to C_{1} is continuous. By (a), for every \overline{a} ∈ C_{1}, the set of discontinuity points of f(\overline{a}, y) restricted to I_{1} is finite. For \overline{a} ∈ C_{1}, let

u(\overline{a}) = min(\{b ∈ I_{1} : b is a discontinuity point of f(\overline{a}, y)\} ∪ \{sup I_{1}\}).

By Lemma 1.5, there are an open box C_{2} ⊆ C_{1} and an open interval I_{2} ⊆ I_{1} such that for every \overline{a} ∈ C_{2}, the function f(\overline{a}, y) is continuous on I_{2}. Now, it is easy to check that f ↾ C_{2} × I_{2} is continuous.

(b)⟹(c) Assume that (b) holds. Clearly, we will be done if we show that for any m ∈ ℕ_{+}, the following condition (∗)_{m} is true.

(∗)_{m} If S ⊆ M^{m+1} is a non-empty definable set and π : M^{m+1} ⟶ M^{m} denotes the projection dropping the last coordinate, then dim(S) = dim(π[S]) iff the set \{\overline{a} ∈ π[S] : π^{−1}(\overline{a}) ∩ S is finite} is large in π[S].

In order to prove (∗)_{1}, consider a non-empty definable set S ⊆ M^{2} and let π : M^{2} ⟶ M be be the projection dropping the second coordinate. If dim(S) = 2, then dim(π[S]) = 1 and the set {a ∈ π[S] : π^{−1}(a) ∩ S is finite} is not large in π[S]. If dim(S) = 0, then dim(π[S]) = 0 and π^{−1}(a) ∩ S is finite for every a ∈ π[S]. Thus in both cases the equivalence from (∗)_{1} holds. To complete the proof of (∗)_{1}, assume that dim(S) = 1. Then dim(π[S]) ∈ {0, 1}.

Case 1. dim(π[S]) = 1.

Suppose for a contradiction that the set {a ∈ π[S] : π^{−1}(a) ∩ S is finite} is not large in π[S]. Then the set {a ∈ π[S] : π^{−1}(a) ∩ S is infinite} contains an open interval I. For a ∈ I define f(a), g(a) ∈ \overline{M}^{M} ∪ \{−∞, +∞\} as infimum and supremum (respectively) of the first convex component of int({b ∈ M : ⟨a, b⟩ ∈ S}). By (b), there is an open interval J ⊆ I such that the functions f, g are both continuous on J. By Lemma 1.5, the set (f, g)_{J} contains an open box, which means that dim(S) = 2, a contradiction.

Case 2. dim(π[S]) = 0, i.e. π[S] is finite.

Since dim(S) = 1, there is a ∈ π[S] such that π^{−1}(a) ∩ S is infinite, which means that the set {a ∈ π[S] : π^{−1}(a) ∩ S is finite} is not large in π[S]. This finishes the proof of (∗)_{1}.

Assume that m > 1, π : M^{m+1} ⟶ M^{m} is the projection dropping the last coordinate, S ⊆ M^{m+1} is a non-empty definable set and suppose that (∗)_{k} holds if 1 ≤ k < m. For the left-to-right direction, assume that dim(S) = dim(π_{i}^{m+1}[S]).

Case 1’. dim(π_{i}^{m+1}[S]) = m.

Suppose for a contradiction that the set \{\overline{a} ∈ π[S] : π^{−1}(\overline{a}) ∩ S is finite\} is not large in π[S]. Then the set \{\overline{a} ∈ π[S] : π^{−1}(\overline{a}) ∩ S is infinite\} contains an open box B. For \overline{a} ∈ B define f(\overline{a}), g(\overline{a}) ∈ \overline{M}^{M} ∪ \{−∞, +∞\} as infimum and supremum (respectively) of the first convex component of int(\{b ∈ M : ⟨\overline{a}, b⟩ ∈ S\}). By (b), there is an open box B′ ⊆ B such that the functions f and g restricted to B′ are both continuous. Hence (by Lemma 1.5), the set (f, g)_{B′} contains an open box, which implies that dim(S) = m + 1 > dim(π[S]), a contradiction.

Case 2’. dim(π[S]) < m.

Denote by J the set of all j’s from {1, …, m} for which dim(π_{j,m+1}^{m+1}[S]) = dim(π[S]) = dim(S). The assumption dim(π[S]) < m guarantees that J ≠ ∅. Let j ∈ J. By (∗)_{m−1}, the set

X_{j} := \{\overline{b} ∈ π_{j,m+1}^{m+1}[S] : (π_{j}^{m})^{−1}(\overline{b}) ∩ π[S] is finite\}

is large in π_{j,m+1}^{m+1}[S]. By Lemma 1.7(b), dim(π_{j}^{m+1}[S]) = dim(π_{j,m+1}^{m+1}[S]). Again, by (∗)_{m−1}, the set

X′_{j} := \{\overline{b} ∈ π_{j,m+1}^{m+1}[S] : (π_{m}^{m})^{−1}(\overline{b}) ∩ π_{j}^{m+1}[S] is finite\}

is large in π_{j,m+1}^{m+1}[S]. So Y_{j} := X_{j} ∩ X′_{j} is also large in π_{j,m+1}^{m+1}[S]. Let

Z = π[S] ∩ ⋃_{j∈J}(π_{j}^{m})^{−1}[Y_{j}].

The definition of Z guarantees that

(∗) dim(π_{j}^{m}[π[S] \ Z]) < dim(π_{j,m+1}^{m+1}[S]) for j ∈ J,

and π^{−1}(\overline{a}) ∩ S is finite whenever \overline{a} ∈ Z. We claim that Z is large in π[S]. Suppose not. Then

(∗∗) dim(π[S] \ Z) = dim(π[S]) = dim(S).

Since dim(π[S]) < m, there is j_{0} ∈ {1, …, m} such that

(∗ ∗ ∗) dim(π_{j_{0}}^{m}[π[S] \ Z]) = dim(π[S] \ Z).

(∗), (∗∗) and (∗ ∗ ∗) imply that j_{0} ∈ J and dim(π_{j_{0},m+1}^{m+1}[S]) > dim(S), which is impossible.

For the right-to-left direction, assume that the set \{\overline{a} ∈ π[S] : π^{−1}(\overline{a}) ∩ S is finite\} is large in π[S] and dim(S) = dim(π[S]) + 1. We consider three cases.

Case 1”. dim(π[S]) = m.

In this situation, dim(S) = m + 1, i.e. S contains an open box B. This implies that the set \{\overline{a} ∈ π[S] : π^{−1}(\overline{a}) ∩ S is finite\} is not large in π[S], a contradiction.

Case 2”. 0 < dim(π[S]) < m.

Let T = \{\overline{a} ∈ π[S] : π^{−1}(\overline{a}) ∩ S is infinite\}. Since dim(S) ≤ m, there are distinct j_{1}, …, j_{k} ∈ {1, … m} such that dim(π_{j_{1},...,j_{k}}^{m+1}[S]) = dim(S) = m + 1 − k. Then dim(π_{j_{1},...,j_{k},m+1}^{m+1}[S]) ≥ dim(π_{j_{1},...,j_{k}}^{m+1}[S]) − 1 = dim(S) − 1 = dim(π[S]) ≥ dim(π_{j_{1},...,j_{k},m+1}^{m+1}[S]). Consequently, dim(π_{j_{1},...,j_{k},m+1}^{m+1}[S]) = dim(π[S]). Let

X = \{\overline{a} ∈ π_{j_{1},...,j_{k},m+1}^{m+1}[S] : (π_{j_{1},...,j_{k}}^{m})^{−1}(\overline{a}) ∩ π[S] is finite\} ∩ (π_{j_{1},...,j_{k},m+1}^{m+1}[S] \ π_{j_{1},...,j_{k}}^{m}[T]). Since

dim(π_{j_{1},...,j_{k}}^{m}[T]) ≤ dim(T) < dim(π[S]),

by the inductive hypothesis and Fact 1.9, X is large in π_{j_{1},...,j_{k},m+1}^{m+1}[S]. Note that (π_{m+1−k}^{m+1−k})^{−1}(\overline{a})∩ π_{j_{1},...,j_{k}}^{m+1}[S] is a finite subset of M^{m+1−k} whenever \overline{a} ∈ X. Again, by the inductive hypothesis, dim(π_{j_{1},...,j_{k},m+1}^{m+1}[S]) = dim(π_{j_{1},...,j_{k}}^{m+1}[S]). Summing up, dim(π[S]) = dim(S), a contradiction.

Case 3”. dim(π[S]) = 0.

The assumptions guarantee that π[S] is finite and π^{−1}(\overline{a}) ∩ S is finite for all \overline{a} ∈ π[S]. Hence S is finite and dim(S) = 0.

(c)⟹(d) Suppose that (c) holds. Using induction on n, we will prove condition (a) from Definition 4.1. Let S ⊆ M^{m+1} be a definable set. Applying (c) to the sets ⋃_{\overline{a}∈X(S,m,0)}\{\overline{a}\} × S\overline{a} and

⋃_{\overline{a}∈X(S,m,1)}\{\overline{a}\} × S\overline{a}, we easily get

dim ( ⋃_{\overline{a}∈X(S,m,0)}\{\overline{a}\} × S\overline{a}) = dim(X(S, m, 0)) and
dim ( ⋃_{\overline{a}∈X(S,m,1)}\{\overline{a}\} × S\overline{a}) = dim(X(S, m, 1)) + 1.

Now, assume that S is a definable subset of M^{m} × M^{n+1}, d ∈ {−∞, 0, …, n + 1}, and suppose that condition (d) holds for definable subsets of M^{m} × M^{n}. By (c), without loss of generality we can assume that d ≥ 1. For

⟨i, j⟩ ∈ Q_{d} := ({d} × {−∞, 0, …, d − 1}) ∪ ({−∞, 0, …, d − 1} × {d − 1})

define

X_{i,j}(S, m, d) = \{\overline{a} ∈ M^{m} : dim(S\overline{a}) = d, dim(X(S\overline{a}, n, 0)) = i and dim(X(S\overline{a}, n, 1)) = j\}.

It is clear that X(S, m, d) is a disjoint union of the sets X_{i,j}(S, m, d) as ⟨i, j⟩ ranges over Q_{d}. Let

Y_{i,j}^{0} = \{\overline{a}\overline{b} ∈ (X_{i,j}(S, m, d) × M^{n}) ∩ π_{m+n+1}^{m+n+1}[S] : dim(S_{\overline{a}\overline{b}}) = 0\} and
Y_{i,j}^{1} = \{\overline{a}\overline{b} ∈ (X_{i,j}(S, m, d) × M^{n}) ∩ π_{m+n+1}^{m+n+1}[S] : dim(S_{\overline{a}\overline{b}}) = 1\}.

One easily sees that Y_{i,j}^{0} ⊆ X(S, m + n, 0) and Y_{i,j}^{1} ⊆ X(S, m + n, 1) whenever ⟨i, j⟩ ∈ Q_{d}. For every \overline{a} ∈ X_{i,j}(S, m, d), dim((Y_{i,j}^{0})\overline{a}) = i and dim((Y_{i,j}^{1})\overline{a}) = j. Using Fact 1.6(e), condition (c) and the inductive hypothesis, we obtain

dim (  ⋃_{\overline{a}∈X_{i,j}(S,m,d)}\{\overline{a}\} × S\overline{a}) = dim (  ⋃_{\overline{a}\overline{b}∈(X_{i,j}(S,m,d)×M^{n})∩π_{m+n+1}^{m+n+1}[S]}\{\overline{a}\overline{b}\} × S_{\overline{a}\overline{b}}) =

max \{dim ( ⋃_{\overline{a}\overline{b}∈Y_{i,j}^{0}}\{\overline{a}\overline{b}\} × S_{\overline{a}\overline{b}}) , dim ( ⋃_{\overline{a}\overline{b}∈Y_{i,j}^{1}}\{\overline{a}\overline{b}\} × S_{\overline{a}\overline{b}})\} =

max{dim(Y_{i,j}^{0}), dim(Y_{i,j}^{1}) + 1} = max{dim(X_{i,j}(S, m, d)) + i, dim(X_{i,j}(S, m, d)) + j + 1} = dim(X_{i,j}(S, m, d)) + max(i, j + 1) = dim(X_{i,j}(S, m, d)) + d.

Now, by Fact 1.6(e),

dim ( ⋃_{\overline{a}∈X(S,m,d)}\{\overline{a}\} × S\overline{a}) = dim ( ⋃_{⟨i,j⟩∈Q_{d}} ⋃_{\overline{a}∈X_{i,j}(S,m,d)}\{\overline{a}\} × S\overline{a}) =

max \{dim ( ⋃_{\overline{a}∈X_{i,j}(S,m,d)}\{\overline{a}\} × S\overline{a}) : ⟨i, j⟩ ∈ Q_{d}\} =

max{dim(X_{i,j}(S, m, d)) : ⟨i, j⟩ ∈ Q_{d}} + d = dim(X(S, m, d)) + d.

(d)⟹(a) Assume that I ⊆ M is an open interval and f : I ⟶ \overline{M}^{M} is a definable function such that for every a ∈ I, f is not continuous at a. By the monotonicity theorem, there is an open interval I′ ⊆ I such that f ↾ I′ is strictly monotone. Suppose for instance that f ↾ I′ is strictly increasing. For a ∈ I′, define

f_{1}(a) = sup_{x∈I′∩(−∞,a)} f(x) and f_{2}(a) = inf_{x∈I′∩(a,+∞)} f(x).

As I′ is contained in the set of discontinuity points of f, we have that f_{1}(a) < f_{2}(a) for a ∈ I′. Moreover, (f_{1}(a), f_{2}(a)) ∩ (f_{1}(b), f_{2}(b)) = ∅ whenever a, b are distinct elements from I′. Let S = ⋃_{a∈I}{a} × (f_{1}(a), f_{2}(a)). Clearly, S witnesses the fact that the addition property of dim

fails in M.

Lemma 4.3 (a) If for every N ≻ M, dcl has the exchange property in N, then dim has the addition property in M.

(b) If dim has the addition property in M, then dcl has the exchange property in M.

Proof. (a) Suppose that dim does not have the addition property in M. By Theorem 4.2, there are an open interval I ⊆ M and a definable function f : I ⟶ \overline{M}^{M} such that each element a ∈ I is a discontinuity point of f. By Theorem 1.2, there is an open interval I_{1} ⊆ I such that f ↾ I_{1} is strictly monotone. Suppose for example that f ↾ I_{1} is strictly increasing. For a ∈ I_{1} define g(a) = lim_{x⟶a^{−}} f(x) and h(a) = lim_{x⟶a^{+}} f(x). Our assumptions guarantee that g(a) ≤ f(a) ≤ h(a)

and g(a) < h(a) for a ∈ I_{1}. Hence at least one of the sets

X_{1} := {a ∈ I_{1} : g(a) < f(a)}, X_{2} := {a ∈ I_{1} : f(a) < h(a)}

contains an open interval I_{2}. Suppose for instance that I_{2} ⊆ X_{1} and define an L(M)-formula φ(x, y) as follows:

φ(x, y) ≡ (x ∈ I_{2}) ∧ (g(x) < y < f(x)).

Fix a ∈ I_{2} and N ≻ M such that φ(a, N) is not contained in dcl(a). Clearly, dcl does not have the exchange property in N.

(b) Suppose that dcl does not have the exchange property in M. There are A ⊆ M and a, b ∈ M such that a ∈ dcl(Ab) \ dcl(A) and b ∉ dcl(Aa). There is a formula φ(x, y) ∈ L(A) such that φ(M, b) = {a} and |φ(M, d)| ≤ 1 whenever d ∈ M. Let X = {c ∈ M : φ(c, M) ≠ ∅}. Note that a ∈ int(X), because a ∉ dcl(A). Denote by I convex component of int(X) containing a. Again, b ∈ int(φ(a, M)), because b ∉ acl(Aa). Assume that b belongs to the k-th convex component of int(φ(a, M)). As a ∉ dcl(A), there is an A-definable convex open set I′ ⊆ I containing a such that for every c ∈ I′, the set int(φ(c, M)) has at least k convex components. There is a formula ψ(x, y) ∈ L(A) such that ψ(M) ⊆ I′ × M and for every c ∈ I′, ψ(c, M) is the k-th convex component of int(φ(c, M)). Now, for c ∈ I′ let f(c) = sup(ψ(c, M)). Clearly, there is an open interval I′′ ⊆ I′ on which f is strictly monotone. But then f is not continuous on any subinterval of I′′. This finishes the proof.

References

[Ar] R. Arefiev, On monotonicity for weakly o-minimal structures, preprint.

[MMS] D. Macpherson, D. Marker, and C. Steinhorn, Weakly o-minimal structures and real closed fields, Trans. Amer. Math. Soc. 352 (2000), 5435-5483.

[Pi] A. Pillay, On groups and fields definable in o-minimal structures, J. Pure Appl. Algebra 53 (1988), 239-255.

[V] V. Verbovskiy, Non-uniformly weakly o-minimal group. Algebra and model theory, 3 (Erlogol, 2001), 136–145, 161, Novosibirsk State Tech. Univ., Novosibirsk, 2001.

[vdD1] L. van den Dries, Dimension of definable sets, algebraic boundedness and henselian fields, Ann. Pure Appl. Logic 45 (1989), 189-209.

[vdD2] L. van den Dries, Tame Topology and o-minimal Structures, London Mathematical Society Lecture Notes Series, vol. 248, Cambridge: Cambridge University Press 1998.

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