# Bi-Colored Expansions of Geometric Theories

S. Jalili***Department of Mathematics and Computer Science, Amirkabir
University of Technology (Tehran Polytechnic), Hafez Avenue 15194, P.O.Box
15875-4413, Tehran, Iran. E-mail<EMAIL_ADDRESS>M.
Pourmahdian†††Department of Mathematics and Computer Science, Amirkabir
University of Technology (Tehran Polytechnic), Hafez Avenue 15194, P.O.Box
15875-4413, Tehran, Iran; School of Mathematics, Institute for Research in
Fundamental Sciences (IPM), P.O. Box 19395-5746, Tehran, Iran. E-mail:
<EMAIL_ADDRESS>M. Khani‡‡‡ Department of Mathematical Sciences, Isfahan
University of Technology, Isfahan, 84156-83111, Iran<EMAIL_ADDRESS>

###### Abstract

This paper concerns the study of Bi-colored expansions of geometric theories
in the light of the Fraïssé-Hrushovski construction method. Substructures of
models of a geometric theory $T$ are expanded by a color predicate $p$, and
the dimension function associated with the pre-geometry of the $T$-algebraic
closure operator together with a real number $0<\alpha\leqslant 1$ is used to
define a pre-dimension function $\delta_{\alpha}$. The pair
$(\mathcal{K}_{\alpha}^{+},\leqslant_{\alpha})$ consisting of all such
expansions with a hereditary positive pre-dimension along with the notion of
substructure $\leqslant_{\alpha}$ associated to $\delta_{\alpha}$ is then used
as a natural setting for the study of generic bi-colored expansions in the
style of Fraïssé-Hrushovski construction. Imposing certain natural conditions
on $T$, enables us to introduce a complete axiomatization
$\mathbb{T}_{\alpha}$ for the class of rich structures in this class. We will
show that if $T$ is a dependent theory (NIP) then so is $\mathbb{T}_{\alpha}$.
We further prove that whenever $\alpha$ is rational the strong dependence
transfers to $\mathbb{T}_{\alpha}$. We conclude by showing that if $T$ defines
a linear order and $\alpha$ is irrational then $\mathbb{T}_{\alpha}$ is not
strongly dependent.
Keywords: Geometric Theories; Fraïssé-Hrushovski construction; Bi-Colored
Expansions; Rich Structures; Dependence (NIP); Strong Dependence.
AMS Subject Classification 2020 03C45.

##  1 Introduction

Extending methods of stability and applying them to a larger class of theories
is a dominating theme in current research of pure model theory. In addition to
showing the prevalence of these methods, this line of research also presents
potential applications beyond model theory. To instantiate general concepts in
stability-hierarchy and perhaps examine some related open
questions/conjectures one would need to look for some new examples,
conceivably through adapting known model-theoretic methods. Our main aim in
this paper is to study the Fraïssé-Hrushovski method beyond the realm of
stability/simplicity. We aim to take the first step of the more general plan
of modifying the Fraïssé-Hrushovski construction to study Bi-colored
expansions of geometric theories without the independence property (NIP).

There are two major (independent) research themes with which our proposed plan
is naturally connected, as explained below.

Our first source of contribution is to the Bi-colored fields, that is the
comprehensive studies concerning the expansion of algebraically closed fields
with a unary predicate $p$—often called a color predicate— interpreted either
by an arbitrary set (Black fields) [1, 2, 19], an additive subgroup (Red
fields) [5, 20], or a multiplicative subgroup (Green fields) [6, 20]. All
examples obtained by this construction are $\omega$-stable, either with Morley
rank $\omega$ (non-collapsed constructions)[2, 19, 20] or finite Morley rank
(collapsed constructions)[1, 6].

The other theme that our results are naturally connected to is the study of
the generic expansions of models of geometric theories by a unary predicate
which can be interpreted either by a submodel (Lovely pairs)[7, 8, 21, 26] or
more generally by submodels of reducts [9, 10, 11, 13].

While our methodology is inspired by the first subject, in our study of the
bi-colored expansions of theories, we do not restrict ourselves, a priori, to
any particular class of theories in stability hierarchy. This would give a
clearer understanding of the axiomatization of the resulting expansion in the
style of the second subject. Our presentation somehow shows that both subjects
could be put in a common framework where the Fraïssé-Hrushovski construction
is seen as a sort of generalization of the theme of the generic expansions.

In more technical terms, we assume that $T$ is a complete geometric theory
without finite models in a countable language $\mathcal{L}$. As a convenient
routine, we also assume that $T$ admits elimination of quantifiers in
$\mathcal{L}$. As $T$ is geometric, the algebraic closure operator gives rise
to a pre-geometry and subsequently a notion of dimension function $\dim$,
where $\dim(\bar{a})$ for any finite tuple $\bar{a}$ is the size of a basis of
$\bar{a}$. We further require that $T$ satisfies the free-amalgamation
property and its corresponding pre-geometry is indecomposable. Subsequently,
we expand $\mathcal{L}$ to $\mathcal{L}_{p}=\mathcal{L}\cup\\{p\\}$ by adding
a unary predicate $p$ called the coloring predicate. We consider the class of
all $\mathcal{L}_{p}$-structures whose universe $M$ is a model of
$T^{\forall}$, fix a real number $0<\alpha\leqslant 1$, and for $M\models
T^{\forall}$ and a finite subset $A$ of $M$, define the pre-dimension function

$\displaystyle\delta_{\alpha}(A)=\dim(A)-\alpha|p(A)|.$

Now, as $T$ is geometric, the dimension function enjoys certain definability
conditions which makes $\mathcal{K}_{\alpha}^{+}$, the class of
$\mathcal{L}_{p}$-structures $M\in\mathcal{K}_{\alpha}^{+}$ such that
$\delta_{\alpha}(A)\geqslant 0$ for all $A\subseteq_{\text{fin}}M$,
$\mathcal{L}_{p}$-axiomatizable.

There is a notion $\leqslant_{\alpha}$ of closed substructures in
$\mathcal{K}_{\alpha}^{+}$ associated to the pre-dimension function
$\delta_{\alpha}$. The free-amalgamation of $T$ implies that
$(\mathcal{K}_{\alpha}^{+},\leqslant_{\alpha})$ has the amalgamation property,
and this guaranties that $(\mathcal{K}_{\alpha}^{+},\leqslant_{\alpha})$ has
$\lambda$-rich models, for each $\lambda\geqslant\aleph_{0}$.

Using ideas from [18], we give a complete axiomatization $\mathbb{T}_{\alpha}$
for the class all $\omega_{1}$-rich structures. This axiomatization together
with a description of types enables us to characterize coheirs in models of
$\mathbb{T}_{\alpha}$ and prove that if $T$ is dependent then so is
$\mathbb{T}_{\alpha}$.

Yet we will show that our construction separates between rational and
irrational $\alpha$ via the notion of strong dependence. The strong dependence
is transferred to $\mathbb{T}_{\alpha}$ for rational $\alpha$, but this is not
the case when $\alpha$ is irrational. Indeed, assuming that $T$ defines a
linear order and $\alpha$ is irrational we show that $\mathbb{T}_{\alpha}$ is
not strongly dependent. This fact mimics the well-known result in ab-initio
Fraïssé-Hrushovski construction that differentiate between rational/irrational
$\alpha$ via $\omega$-stability/strict-stability.

The structure of the paper is as follows. In section 2 after fixing the
setting for our underlying theory $T$, we introduce the class
$(\mathcal{K}^{+}_{\alpha},\leqslant_{\alpha})$. In Section 3 we prove certain
properties of this class, present an axiomatization $\mathbb{T}_{\alpha}$ for
its rich structures, and prove it to be a complete theory. Finally in Section
4 we have given the mentioned results on dependence and strong dependence of
$\mathbb{T}_{\alpha}$.

##  2 Bi-Colored Structures

###  2.1 Preliminaries, Setting and Conventions

Let $T$ be as in the introduction. It is clear, by the quantifier-elimination
assumption, that
$\operatorname{acl}_{M_{1}}(M_{0})\cong\operatorname{acl}_{M_{2}}(M_{0})$ for
all $M_{0}\subseteq M_{1},M_{2}\models T$.

We use capital letters $M,N,P,K$ to denote the $\mathcal{L}$-structures and
$A,B,C,D$ and $X,Y,Z$ respectively for finite and infinite sets. Instead of
$X\cup Y$ we would write $XY$. For tuples $\bar{a},\bar{b}$ in a model of $M$
of $T$, by the quantifier-free type of $\bar{a}$ over $\bar{b}$, denoted by
$\operatorname{qftp}_{\mathcal{L}}(\bar{a}/\bar{b})$, we mean the set of all
quantifier-free formulas $\varphi(\bar{x},\bar{b})$ such that
$M\models\varphi(\bar{a},\bar{b})$.

The theory $T$ is assumed to be geometric, so it eliminates the quantifier
$\exists^{\infty}$ and the algebraic closure, $\operatorname{acl}$, satisfies
the exchange property in its models. We denote the dimension obtained by
$\operatorname{acl}$ in $T$ by $\dim$. So for tuples
$\bar{a}=(a_{1},...,a_{n})$ and $\bar{b}=(b_{1},\dots,b_{m})$ of $M$ and a set
$Y\subseteq M$, we let $\dim(\bar{a}/Y)$ be the size of the
$\operatorname{acl}$-base of $\\{a_{1},\ldots,a_{n}\\}$ over $Y$, and
$\dim(\bar{a}/\bar{b})=\dim(\bar{a}/\\{b_{1},\dots,b_{m}\\})$. If
$\varphi(\bar{x},\bar{y})$ is an $\mathcal{L}$-formula and $\bar{b}\in
M^{|\bar{y}|}$, then
$\dim(\varphi(M,\bar{b}))=\max\\{\dim(\bar{a}/\bar{b}):\hskip 8.53581pt\
M\models\varphi(\bar{a},\bar{b})\\}$.

We say that $Y$ is $\dim$-independent from $Z$ over $X$, and write
$Y\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}_{X}^{\dim}Z$,
if $\dim(\bar{a}/X)=\dim(\bar{a}/XZ)$ for each tuple $\bar{a}\in
Y^{|\bar{a}|}$. If in addition, $Y\cap Z=X$, we say that $Y$ and $Z$ are free
over $X$. To emphasize the freeness of $Y$ and $Z$ over $X$ we would write the
set $YZ$ as $Y\oplus_{X}Z$ (we also use the same notation for the free-amalgam
of structures).

Some properties of the $\dim$ function and $\dim$-independence is recalled in
the following for later reference, assuming that all subsets and tuples are
from (a sufficiently saturated) model $M$ of $T$.

######  Fact 2.1.

The $\dim$ function has the following properties.

  * •

Finite character. $\dim(\bar{a}/Y)=\min\\{\dim(\bar{a}/B):\hskip
8.53581ptB\subseteq_{\text{fin}}Y\\}$.

  * •

Additivity. $\dim(\bar{a}\bar{b}/Z)=\dim(\bar{a}/Z)+\dim(\bar{b}/\bar{a}Z)$.

  * •

Monotonicity. $\dim(\bar{a}/Y)\geqslant\dim(\bar{a}/Z)$ for $Y\subseteq Z$.

  * •

Definability. For each formula $\varphi(\bar{x},\bar{y})$ and
$k\leqslant|\bar{x}|$ the set of tuples $\bar{b}\in M^{|\bar{y}|}$ with
$\dim(\varphi(M,\bar{b}))=k$ is definable by a formula
$d_{\varphi,k}(\bar{y})$.

  * •

$\bigvee$-Definability. If $\dim(\bar{a}/\bar{b})\leqslant n$, then there is a
formula $\psi(\bar{x},\bar{y})$ such that
$\psi(\bar{x},\bar{b})\in\operatorname{qftp}_{\mathcal{L}}(\bar{a}/\bar{b})$
and if $M\models\psi(\bar{a}^{{}^{\prime}},\bar{b}^{{}^{\prime}})$ then
$\dim(\bar{a}^{{}^{\prime}}/\bar{b}^{{}^{\prime}})\leqslant n$, for any
$\bar{a}^{{}^{\prime}},\bar{b}^{{}^{\prime}}$.

We need a stronger version of Definability, as stated below. The first part of
the following statement appears in [10], Lemma 2.3.

######  Lemma 2.2.

Let $M$ be an $\omega$-saturated model of $T$, $\varphi(\bar{x};\bar{y})$ an
$\mathcal{L}$-formula and $k$ a natural number. Then

  1. 1.

There is a formula $H_{\varphi}(\bar{y})$ that defines the set of tuples
$\bar{b}\in M^{|\bar{y}|}$ such that there exists $\bar{a}$ satisfying
$\varphi(\bar{x};\bar{b})$ with
$\bar{a}\cap\operatorname{acl}(\bar{b})=\emptyset$.

  2. 2.

There exists an $\mathcal{L}$-formula $D_{\varphi,k}(\bar{y})$ which defines
the set of tuples $\bar{b}\in M^{|\bar{y}|}$ for which there exists $\bar{a}$
satisfying $\varphi(\bar{x};\bar{b})$ with $\dim(\bar{a}/\bar{b})\geqslant k$
and $\bar{a}\cap\operatorname{acl}(\bar{b})=\emptyset$.

###### Proof.

We use 1 to prove 2. Certainly, we may assume that $k\leqslant|\bar{x}|=n$.
For $\sigma\in S_{n}$ and
$\psi(x_{\sigma(1)};x_{\sigma(2)},\dots,x_{\sigma(n)},\bar{y})=\varphi(\bar{x},\bar{y})$
the assertion

$\exists x_{\sigma(1)}\big{[}\varphi(\bar{x},\bar{y})\wedge
x_{\sigma(1)}\notin\operatorname{acl}(x_{\sigma(2)},\dots,x_{\sigma(n)},\bar{y})\big{]}$

is definable by a formula
$H_{\psi}(x_{\sigma(2)};\dots,x_{\sigma(n)},\bar{y}).$ Now by applying $1$ to
$H_{\psi}$ and iterating this procedure the following statement

$\displaystyle\exists x_{\sigma(k)}\dots\exists x_{\sigma(1)}$
$\displaystyle\big{[}\varphi(\bar{x},\bar{y})\wedge\bigwedge_{i=1}^{k}x_{\sigma(i)}\notin\operatorname{acl}(x_{\sigma(i+1)},\dots,x_{\sigma(k+1)},\dots,x_{\sigma(n)},\bar{y})\big{]}$

is definable by an $\mathcal{L}$-formula
$\rho(x_{\sigma(k+1)},\dots,x_{\sigma(n)};\bar{y}).$ Informally $\rho$ states
that there exists a set
$\\{x_{\sigma(1)},\dots,x_{\sigma(k)}\\}\subseteq\\{x_{1},\dots,x_{n}\\}$
which has dimension at least $k$ over $\bar{y}$. Now by applying $1$ to $\rho$
we get that the statement

$\exists x_{\sigma(k+1)}\dots\exists
x_{\sigma(n)}\big{(}\rho(x_{\sigma(k+1)},\dots,x_{\sigma(n)};\bar{y})\wedge\bigwedge_{i=k+1}^{n}x_{\sigma(i)}\notin\operatorname{acl}(\bar{y})\big{)}$

is definable by an $\mathcal{L}$-formula $D_{\varphi,k,\sigma}(\bar{y})$. Let
$D_{\varphi,k}(\bar{y})=\bigvee_{\sigma\in
S_{n}}D_{\varphi,k,\sigma}(\bar{y})$. Note that any tuple $\bar{b}$ in $M$
satisfies $D_{\varphi,k}(\bar{y})$ if and only if there exists a tuple
$\bar{a}$ which satisfies $\varphi(\bar{x},\bar{b})$, avoids
$\operatorname{acl}(\bar{b})$ and includes a subtuple
$(a_{i_{1}},\dots,a_{i_{k}})$ with
$\dim((a_{1},\dots,a_{n})/\bar{b})\geqslant\dim((a_{i_{1}},\dots,a_{i_{k}})/\bar{b})=k$.
Therefore $D_{\varphi,k}(\bar{y})$ is the desired formula. ∎

######  Fact 2.3.

  1. 1.

Symmetry. $Y\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}_{X}^{\dim}Z$
if and only if $Z\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}_{X}^{\dim}Y$.

  2. 2.

Transitivity. $Y\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}_{X}^{\dim}Z_{1}Z_{2}$
if and only if $Y\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}_{X}^{\dim}Z_{1}$
and $Y\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}_{XZ_{1}}^{\dim}Z_{2}$.

  3. 3.

$\operatorname{acl}$-Preservation. $Y\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}_{X}^{\dim}Z$
if and only if $\operatorname{acl}(Y)\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}_{\operatorname{acl}(X)}^{\dim}\operatorname{acl}(Z)$.

  4. 4.

$\dim$-Morley Sequences. Any order indiscernible sequence $\\{a_{i}:\hskip
8.53581pti\in I\\}$ over $X$ is a $\dim$-Morley sequence, i.e. for any two
disjoint subsequences $J_{1}$ and $J_{2}$ of $I$ we have $J_{1}\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}_{X}^{\dim}J_{2}$.

We will need formulas that are “strong” in the sense that they carry enough
information about $\dim$. These are characterized by the following Definition.

######  Definition 2.4.

We say that an $\mathcal{L}$-formula $\varphi(\bar{x},\bar{y})$ is strong with
respect to the distinct tuples $\bar{d},\bar{a}$ if
$\varphi(\bar{x},\bar{a})\in\operatorname{qftp}_{\mathcal{L}}(\bar{d}/\bar{a})$
and for each $\bar{d}^{{}^{\prime}}\bar{a}^{{}^{\prime}}$ in a sufficiently
saturated model $M\models T$,

  1. 1.

if $M\models\varphi(\bar{d}^{{}^{\prime}},\bar{a}^{{}^{\prime}})$ then
$\bar{d}^{{}^{\prime}}\bar{a}^{{}^{\prime}}$ is distinct, and

  2. 2.

if $M\models\varphi(\bar{d}^{{}^{\prime}},\bar{a}^{{}^{\prime}})$ and
$\dim(\bar{d}^{{}^{\prime}}/\bar{a}^{{}^{\prime}})=\dim(\bar{d}/\bar{a})$ then
for each partition $P=(\bar{x}_{1},\bar{x}_{2})$ of $\bar{x}$,
$\dim(\bar{d}^{{}^{\prime}}_{2}/\bar{d}^{{}^{\prime}}_{1}\bar{a}^{{}^{\prime}})\leqslant\dim(\bar{d}_{2}/\bar{d}_{1}\bar{a})$.

######  Remark 2.5.

The second item above implies, in particular, that
$\dim(\bar{d}^{{}^{\prime}}_{1}/\bar{a}^{{}^{\prime}})\geqslant\dim(\bar{d}_{1}/\bar{a})$.

The $\bigvee$-Definability (Fact 2.1) implies that a strong formula for any
pair of distinct sequence always exists. We shortly give a proof of this in
the following, where for a sequence of distinct variables $\bar{x}$ we denote
by $\mathcal{P}$ the set of all partitions $P=(\bar{x}_{1},\bar{x}_{2})$ of
$\bar{x}$.

######  Lemma 2.6.

  1. 1.

Suppose that $\bar{d},\bar{a}$ is a pair of distinct sequences. Then there is
a strong formula $\varphi(\bar{x},\bar{y})$ with respect to $\bar{d},\bar{a}$.

  2. 2.

If $\varphi(\bar{x},\bar{y})$ is strong with respect to $\bar{d},\bar{a}$,
$T\models(\theta(\bar{x},\bar{y})\to\varphi(\bar{x},\bar{y}))$ and
$\theta(\bar{x},\bar{a})\in\operatorname{qftp}_{\mathcal{L}}(\bar{d}/\bar{a})$
then $\theta(\bar{x},\bar{y})$ is also strong.

###### Proof.

We prove only item 1, and 2 would be clear. For each partition
$P=(\bar{x}_{1},\bar{x}_{2})$ of $\bar{x}$, by the $\bigvee$-Definability
(Fact 2.1) there is a formula $\varphi_{P}(\bar{x}_{2},\bar{x}_{1},\bar{y})$
such that
$\varphi_{P}(\bar{x}_{2},\bar{d}_{1},\bar{a})\in\operatorname{qftp}_{\mathcal{L}}(\bar{d}_{2}/\bar{d}_{1}\bar{a})$
and if
$M\models\varphi_{P}(\bar{d}^{{}^{\prime}}_{2},\bar{d}^{{}^{\prime}}_{1},\bar{a}^{{}^{\prime}})$
then
$\dim(\bar{d}^{{}^{\prime}}_{2}/\bar{d}^{{}^{\prime}}_{1}\bar{a}^{{}^{\prime}})\leqslant\dim(\bar{d}_{2}/\bar{d}_{1}\bar{a})$.
Now we simply let $\varphi(\bar{x},\bar{y})$ be the conjunction of all
formulas $\varphi_{P}$ with the formula that states $\bar{x}$ and $\bar{y}$
are distinct. ∎

In the next sections, we deal with a class of bi-colored expansions of models
of $T$. For this class to have certain desired features, we need $T$ to
satisfy yet two other properties, that is the free-amalgamation, and the
indecomposability, as introduced in the following. These features will later
lead us to an axiomatizations for the “rich” structures.

Recall that an $\mathcal{L}$-embedding $f:M\to N$ is algebraically closed if
$\operatorname{acl}(f(M))=f(M)$.

######  Definition 2.7.

We say that $T$ has the free-amalgamation property over algebraically closed
substructures if for each $M_{0},M_{1},M_{2}\models T^{\forall}$ and each pair
of algebraically closed embeddings $f_{1}:M_{0}\to M_{1}$, $f_{2}:M_{0}\to
M_{2}$, there are $M\models T$ and embeddings $g_{1}:M_{1}\to M$ ,
$g_{2}:M_{2}\to M$ such that

  1. 1.

$g_{1}\circ f_{1}=g_{2}\circ f_{2}$, and

  2. 2.

$g_{1}(M_{1})$ and $g_{2}(M_{2})$ are free over $g_{1}\circ f_{1}(M_{0})$ in
$M$.

We call $M$ a free amalgam of $M_{1}$ and $M_{2}$ over $M_{0}$, and although
$M$ is not unique up to isomorphism and by abuse of notation, denote it by
$M_{1}\oplus_{M_{0}}M_{2}$.

######  Observation 2.8.

Let $T$ have the free-amalgamation property and $M$ be a $\lambda$-saturated
model of $T$. Suppose that $\Sigma(\bar{x})$ is a partial type over $X$ which
has a solution with no intersection with $\operatorname{acl}(X)$. Then for
each $Y\supseteq X$ with $|Y|<\lambda$ there is a solution
$\bar{d}^{{}^{\prime}}$ to $\Sigma$ in $M$ such that
$\bar{d}^{{}^{\prime}}\cap Y=\emptyset$. Furthermore $\bar{d}^{{}^{\prime}}$
can be chosen in such that $X{\bar{d}^{{}^{\prime}}}$ and $Y$ are free over
$X$.

###### Proof.

There is a solution $\bar{d}$ in $M\oplus_{\operatorname{acl}(X)}M$ which
satisfies the conditions required. By saturation we can find a solution
$\bar{d}^{{}^{\prime}}$ in $M$ such that $X\cup{\bar{d}^{{}^{\prime}}}$ and
$Y$ are free over $X$. ∎

######  Definition 2.9.

We call $T$ indecomposable if no finite-dimensional algebraically closed set
$X$ can be written as a finite union $X=Y_{1}\ldots Y_{n}$ with
$\dim(Y_{i})<\dim(X)$ for $i\leqslant n$.

The assumption of indecomposability is meant to provide the following
desirable property for bases.

######  Lemma 2.10.

Assume that $T$ is indecomposable and $M\models T$. Let
$B=\\{d_{1},\dots,d_{m}\\}$ be an independent set over $A\subseteq M$. Then
for each natural number $n$ there is a subset
$D\subseteq\operatorname{acl}(Ad_{1},\dots,d_{m})$ with $|D|=n$ such that each
$m$-element subset of $BD$ is a base for
$\operatorname{acl}(Ad_{1},\dots,d_{m})$.

###### Proof.

By assumption $\dim(d_{1},\dots,d_{m}/A)=m$. Let $S_{k}$ be the group of
permutations of $\\{1,\dots,k\\}$. Since $T$ is indecomposable, the set

$\displaystyle
D_{m+1}:=\operatorname{acl}(Ad_{1},\dots,d_{m})\setminus\bigcup_{i=1}^{m}\operatorname{acl}(Ad_{1},\dots,\hat{d_{i}},\dots,d_{m})$

is non-empty. Let $d_{m+1}\in D_{m+1}$ be arbitrary. By the exchange property,
the set $\\{d_{\sigma(1)},\dots,d_{\sigma(m)}\\}$ is a base for
$\\{d_{1},\dots,d_{m},d_{m+1}\\}$ over $A$ for any $\sigma\in S_{m+1}$.

Again by indecomposability of $T$, the set

$\displaystyle
D_{m+2}:=\operatorname{acl}(Ad_{1},\dots,d_{m+1})\setminus\bigcup_{1\leqslant
i<j\leqslant
m+1}\operatorname{acl}(Ad_{1},\dots,\hat{d_{i}},\dots,\hat{d_{j}},\dots,d_{m+1})$

is non-empty. Now let $d_{m+2}$ be an arbitrary element of $D_{m+2}$. Again by
the exchange property the set $\\{d_{\sigma(1)},\dots,d_{\sigma(m)}\\}$ is
base for $\\{d_{1},\dots,d_{m+2}\\}$ over $A$ for any $\sigma\in S_{m+2}$.

Iterating this process we obtain the set $D=\\{d_{m+1},\dots,d_{m+n}\\}$ as
desired. ∎

Examples of indecomposable geometric theories with free-amalgamation include
strongly minimal expansions of divisible abelian groups, o-minimal expansions
of ordered divisible abelian groups (ODAG), any completion of the theory of
algebraically closed valued fields, and the theory of $p$-adic fields. The
indecomposability of the mentioned theories indeed stems in the fact that no
vector space over an infinite field can be written as a finite union of its
proper subvector spaces.

###  2.2 Bi-colored Expansions

From here onward we assume that $T$ is an indecomposable geometric theory
satisfying the free-amalgamation property. Before proceeding, we need to fix
some more notation and conventions.

##### Notation and Conventions.

We expand $\mathcal{L}$ to $\mathcal{L}_{p}=\mathcal{L}\cup\\{p\\}$ by adding
a unary predicate $p$ called the coloring predicate. We denote by
$\mathcal{K}$ the class of all $\mathcal{L}_{p}$-structures whose universe $M$
is a model of $T^{\forall}$. For $X\subseteq M$, by $\langle X\rangle_{M}$, or
$\langle X\rangle$ we denote the $\mathcal{L}_{p}$-structure generated by $X$
in $M$. In this sense we call $M$ finitely generated if there is a finite set
$A\subseteq M$ such that $M=\langle A\rangle$. For technical convenience, we
take $\emptyset$ also as a finitely generated structure in $\mathcal{K}$. For
the ease of reference, we say $x$ is colored if $M\models p(x)$. We would
simply write $p(x)$ instead of $M\models p(x)$ when the meaning is clear from
the context. We write $p(\bar{x})$ instead of $\bigwedge_{i=1}^{n}p(x_{i})$,
when $\bar{x}=(x_{1},\ldots,x_{n})$. Also by $p(X/Y)$ we mean the set of
colored elements of $X\setminus Y$. Finally we fix a real number
$0<\alpha\leqslant 1$ for the rest of the paper. We will occasionally add the
assumption that $\alpha$ is/is not rational. The $\mathcal{L}$ and
$\mathcal{L}_{p}$-algebraic closures are respectively denoted by
$\operatorname{acl}$ and $\operatorname{Acl}$.

######  Definition 2.11.

For $M\in\mathcal{K}$ and a finite subset $A$ of $M$, define

$\displaystyle\delta_{\alpha}(A)=\dim(A)-\alpha|p(A)|.$

More generally for $A\subseteq_{\text{fin}}M$ and $X\subseteq M$ we define
$\delta_{\alpha}(A/X)=\dim(A/X)-\alpha|p(A/X)|$. The function
$\delta_{\alpha}$ is called a pre-dimension map.

The assumption of quantifier elimination makes $\delta_{\alpha}(A)$
independent from the ambient $M$. It is also routine to check that

$\displaystyle\delta_{\alpha}(AB/C)=\delta_{\alpha}(A/BC)+\delta_{\alpha}(B/C),$

and that $\delta_{a}$ is submodular, that is

$\delta_{\alpha}(AB)+\delta_{\alpha}(A\cap
B)\leqslant\delta_{\alpha}(A)+\delta_{\alpha}(B).$

##### More on Notation.

Denote by $\mathcal{K}_{\alpha}^{+}$ the subclass of $\mathcal{K}$ consisting
of the $\mathcal{L}_{p}$-structures $M\in\mathcal{K}_{\alpha}^{+}$ such that
$\delta_{\alpha}(A)\geqslant 0$ for all $A\subseteq_{\text{fin}}M$. When $A$
is a subset of a structure $M$ in the $\mathcal{K}^{+}_{\alpha}$, we simply
say that $A$ is in $\mathcal{K}^{+}_{\alpha}$. Furthermore in this situation
for $N\in\mathcal{K}_{\alpha}^{+}$ we say that $f:A\to N$ is an
$\mathcal{L}_{p}$-embedding if
$\operatorname{qftp}_{\mathcal{L}}(f(A))=\operatorname{qftp}_{\mathcal{L}}(A)$
and $f(A)$ has the same color in $N$ as $A$ in $M$. Finally, to avoid
repeating the assumption that a structure is in $\mathcal{K}^{+}_{\alpha}$ in
each statement, we simply assume that each structure and set in the following
belongs to $\mathcal{K}^{+}_{\alpha}$ unless we emphasize otherwise.

It is clear by our previous conventions that
$\emptyset\in\mathcal{K}^{+}_{\alpha}$. Also note that by
$\bigvee$-Definability (Fact 2.1), $\mathcal{K}_{\alpha}^{+}$ is axiomatizable
in $\mathcal{L}_{p}$ by the theory consisting of the
$\mathcal{L}_{p}$-sentences

$\displaystyle\neg\exists
x_{1},\ldots,x_{n}\big{(}\psi_{l}(x_{1},\dots,x_{n})\wedge\bigvee_{\begin{subarray}{c}Y\subseteq\\{x_{1},\dots,x_{n}\\}\\\
|Y|\geq k\end{subarray}}p(Y)\big{)}$

where $\psi_{l}$ asserts that the dimension of $(x_{1},\ldots,x_{n})$ is
bounded by $l$, and $l<\alpha k$.

Notice that since $\alpha>0$ all algebraic points over $\emptyset$ are non-
colored.

We need to review some basic concepts concerning the properties of the pre-
dimension function $\delta_{\alpha}$.

######  Definition 2.12.

  1. 1.

For $A\subseteq_{\text{fin}}M$ and $X\subseteq M$ we say that $A$ is closed in
$X$ and write $A\leqslant_{\alpha}X$ if $A\subseteq X$ and
$\delta_{\alpha}(B/A)\geqslant 0$ for all $A\subseteq
B\subseteq_{\text{fin}}X$.

  2. 2.

For $X,Y$ arbitrary subsets of $M$, we say $X$ is closed in $Y$ and write
$X\leqslant_{\alpha}Y$ if $X\subseteq Y$ and $\delta_{\alpha}(A/X)\geqslant 0$
for all $A\subseteq_{\text{fin}}Y$.

  3. 3.

For structures $M,N$ we say that $M$ is closed in $N$ and write
$M\leqslant_{\alpha}N$ if $M$ is a substructure of $N$ and
$M\leqslant_{\alpha}N$ as sets in the sense above.

For simplicity, we omit the subscript $\alpha$ from $\leqslant_{\alpha}$ and
$\delta_{\alpha}$. It is easy to see that if $X\leqslant M$, then $\neg p(x)$
for all $x\in\operatorname{acl}_{M}(X)\setminus X$. Also in this case, both
$\langle X\rangle_{M}$ and $\operatorname{acl}_{M}(X)$ are closed in $M$.

######  Remark 2.13.

$(\mathcal{K}_{\alpha}^{+},\leqslant)$ is a so-called smooth class, that is
for all $M,M_{1},M_{2},X$,

  1. 1.

$\emptyset,M\leqslant M$.

  2. 2.

If $M\leqslant M_{1}$ and $M_{1}\leqslant M_{2}$, then $M\leqslant M_{2}$.

  3. 3.

If $M\leqslant M_{2}$ then $M\leqslant M_{1}$ for all $M\subseteq
M_{1}\subseteq M_{2}$.

  4. 4.

if $M\leqslant M_{1}$ then $M\cap X\leqslant X$ for all $X\subseteq M_{1}$.

Note also that items 2 and 4 above easily imply that if $M_{1},M_{2}\leqslant
M$ then $M_{1}\cap M_{2}\leqslant M$.

######  Definition 2.14.

Assuming that $X,A,B\subseteq M$,

  1. 1.

The closure of $X$ in $M$, denoted by $\operatorname{cl}_{M}(X)$, is the
smallest subset $Y$ of $M$ containing $X$ such that $Y\leqslant M$.

  2. 2.

We say that $B$ is an intrinsic extension of $A$ and write $A\leqslant_{i}B$
if $A\subseteq B$ but there is no $A^{{}^{\prime}}\neq B$ with $A\subseteq
A^{{}^{\prime}}\leqslant B$, equivalently, $\delta(B)\lneqq\delta(A^{\prime})$
for all $A\subseteq A^{\prime}\varsubsetneqq B$.

  3. 3.

A pair $(A,B)$ is called minimal if $A\subseteq B$, $A\nleqslant B$ but
$A\leqslant C$ for all $A\subseteq C\subsetneqq B$.

It is clear that if $(A,B)$ is a minimal pair, then $A\leqslant_{i}B$.
Moreover if $B$ is an intrinsic extension of $A$ then it is possible to find a
tower $B_{0}=A\subseteq B_{1}\subseteq\ldots\subseteq B_{n}=B$ where each
$(B_{i},B_{i+1})$ is minimal.

The following statements are well-known facts about basic properties of
$\operatorname{cl}_{M}$.

######  Fact 2.15.

  1. 1.

$\operatorname{cl}_{M}(X)$ is the intersection of all closed subsets of $M$
that contain $X$.

  2. 2.

$\operatorname{cl}_{M}(X)=\bigcup_{A\subseteq X}\operatorname{cl}_{M}(A)$.

  3. 3.

$\operatorname{cl}_{M}(A)=\bigcup\\{B\subseteq M:A\leqslant_{i}B\\}$.

When $M$ is clear from the context, we write $\operatorname{cl}(X)$ instead of
$\operatorname{cl}_{M}(X)$.

######  Definition 2.16.

For $A\subseteq M$ and natural number $n$, we define
$\operatorname{cl}^{n}_{M}(A)$ as the union of all $B\subseteq M$ such that
$A\leqslant_{i}B$ and $|B-A|<n$.

######  Remark 2.17.

Let $B,M\in\mathcal{K}_{\alpha}^{+}$ and $A\subseteq M$ . Then

  1. 1.

There is $f:\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ such that for any
embedding $g:B\to M$ the size of $\operatorname{cl}^{n}_{M}(g(B))$ is bounded
by $f(|B|,n)$.

  2. 2.

$\operatorname{cl}_{M}(A)\subseteq\operatorname{Acl}_{M}(A)$.

  3. 3.

When $\alpha$ is rational, then $\operatorname{cl}_{M}(A)$ is finite; hence
$\operatorname{cl}_{M}(A)=\operatorname{cl}_{M}^{n}(A)$ for some $n$.

  4. 4.

By item 1, $\operatorname{cl}_{M}(A)$ is countable when $\alpha$ is
irrational.

The following definition singles out two different types of closed extensions.
This distinction will be highlighted in the axiomatization of rich structures
given in section $3.1$.

######  Lemma 2.18.

Suppose that $A\subseteq M$ is finite. Then there is a finite set $B$
containing $A$ such that $\langle A\rangle=\langle B\rangle$ and and
$B\leqslant\langle B\rangle=\langle A\rangle$.

###### Proof.

If $c_{1},\ldots,c_{k}\in\langle A\rangle\setminus A$ are colored elements
then

$\displaystyle 0\leqslant\delta(Ac_{1},\ldots,c_{k})=\delta(A)-k\alpha.$

Hence if $C\subseteq\langle A\rangle$ and each elements of $C$ is colored then

$\displaystyle|C\setminus A|\leqslant\delta(A)/\alpha\leqslant|A|/\alpha.$

So let $B=A\cup p(\langle A\rangle)$. ∎

##### Remark and Convention.

For $B$ as in the above lemma, each $x\in\langle B\rangle-B$ is non-colored.
Hence we make the convention that whenever we write $\langle A\rangle$ we
assume that $A\leqslant\langle A\rangle$. Notice that by this convention for
$A\subseteq B$ we have that $\langle A\rangle\leqslant\langle B\rangle$ if and
only if $A\leqslant B$.

######  Definition 2.19.

Assume that $\langle A\rangle\subseteq\langle B\rangle$ and $A\leqslant B$ .
Call $B$

  1. 1.

algebraic over $A$ if $\dim(b/A)=0$ for all $b\in B\setminus A$, and

  2. 2.

transcendental over $A$ if $\dim(b/A)=1$ for all $b\in B\setminus A$.

######  Remark 2.20.

It can be easily seen that if $A\leqslant B\subseteq\langle B\rangle$ then
there exists $B_{1}$ such that $A\leqslant B_{1}\leqslant B$ with $B_{1}$
algebraic over $A$ and $B$ transcendental over $B_{1}$. Furthermore, if $B$ is
algebraic over $A$ and $f:\langle B\rangle\to M$ is an
$\mathcal{L}_{p}$-embedding in some $M\in\mathcal{K}_{\alpha}^{+}$ then
$\operatorname{cl}_{M}(f(B))=\operatorname{cl}_{M}(f(A))\oplus_{f(A)}f(B)$.
So, in particular, for each natural $n$ we have
$\operatorname{cl}^{n}_{M}(f(B))=\operatorname{cl}^{n}_{M}(f(A))\oplus_{f(A)}f(B)$.

##  3  Rich Structures and their Axiomatization

###  3.1 Rich Structures

We have already defined what we mean by $T$ having the free-amalgamation
property (see Definition 2.7). Along the same lines, we say that
$(\mathcal{K}_{\alpha}^{+},\leqslant)$ has the free-amalgamation property
(defined more precisely after the following definition) if the conditions in
Definition 2.7 hold with $M_{i},M$ structures in the class
$\mathcal{K}^{+}_{\alpha}$, and $f_{i},g_{i}$ strong embeddings in the sense
of the following definition.

######  Definition 3.1.

By a strong embedding $f:M\to N$ we mean an $\mathcal{L}_{p}$-embedding $f$
such that $f(M)\leqslant N$.

So by the class $(\mathcal{K}^{+}_{\alpha},\leqslant)$ having the amalgamation
property we mean that for each $M_{0},M_{1},M_{2}$ and each pair of strong
embeddings $f_{1}:M_{0}\to M_{1}$, $f_{2}:M_{0}\to M_{2}$, there are $M$ and
strong embeddings $g_{1}:M_{1}\to M$, $g_{2}:M_{2}\to M$ such that $g_{1}\circ
f_{1}=g_{2}\circ f_{2}$.

######  Lemma 3.2.

The class $(\mathcal{K}^{+}_{\alpha},\leqslant)$ has the amalgamation
property. Moreover, if both $f_{1},f_{2}$ are algebraically-closed then we can
choose $M\in\mathcal{K}^{+}_{\alpha}$ such that $g_{1}(M_{1})$ and
$g_{2}(M_{2})$ are free over $g_{1}\circ f_{1}(M_{0})$ in $M$.

###### Proof.

Assume that $\mathcal{L}_{p}$-structures $M_{0},M_{1},M_{2}$ in
$\mathcal{K}_{\alpha}^{+}$ and the strong embeddings $f_{1}$, $f_{2}$ are as
above.

We can assume, without loss of generality that $M_{1},M_{2}$ are models of
$T$. This can be obtained by $\mathcal{L}$-embedding of each $M_{1}$ and
$M_{2}$ into a model of $T$ and a suitable extension of the coloring.

We may also assume that $f_{1}$, $f_{2}$ are algebraically-closed strong
embeddings. This is because $\operatorname{acl}_{M_{1}}(M_{0})$ and
$\operatorname{acl}_{M_{2}}(M_{0})$ are $\mathcal{L}_{p}$-isomorphic over
$M_{0}$, as both $f_{1}$ and $f_{2}$ are strong (see the explanation after
Definition 2.12).

Now since $T$ has the free-amalgamation property, there are a model $M$ and
$\mathcal{L}$-embeddings $g_{i}:M_{i}\to M$, such that $g_{1}\circ
f_{1}=g_{2}\circ f_{2}$. Defining $p(M)=g_{1}(p(M_{1}))\cup g_{2}(p(M_{2}))$,
it is clear that $M\in\mathcal{K}^{+}_{\alpha}$.

It remains only to prove that $g_{1},g_{2}$ are strong embeddings. For
notational simplicity we denote again by $M_{1},M_{2}$ and $M_{0}$ their
respective images in $M$, that is $g_{1}(M_{1})$, $g_{2}(M_{2})$ and
$g_{1}\circ f_{1}(M_{0})$. In the following we show that $M_{1}\leqslant M$
(and $M_{2}\leqslant M$ follows similarly).

Let $A\subseteq M\setminus M_{1}$ be arbitrary. Writing $A=BC$ where
$B\subseteq{M}_{2}$, $C\subseteq M\setminus{M}_{2}$ we have

$\displaystyle\delta(A/{M}_{1})=\delta(C/{M}_{1})+\delta(B/{M}_{1}C).$

Now $\delta(C/{M}_{1})\geqslant 0$ because all elements of $C$ are non-
colored. Also $\delta(B/{M}_{1}C)=\delta(B/{M_{0}}C)\geqslant 0$, because
${M}_{1}\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}^{\dim}_{{M}_{0}}{M}_{2}$.

∎

As we have let $\emptyset$ in $\mathcal{K}^{+}_{\alpha}$, the amalgamation
property implies the joint embedding property. It is also clear that if $M$ is
in $\mathcal{K}_{\alpha}^{+}$ then so are all its substructures. In this
sense, we call $(\mathcal{K}_{\alpha}^{+},\leqslant)$ a Fraïssé class, and it
is natural to look for its Fraïssé limit, or the rich structures in the sense
below.

######  Definition 3.3.

An $\mathcal{L}_{p}$-structure $M$ in $\mathcal{K}_{\alpha}^{+}$ is called
$\lambda$-rich, for $\lambda\geqslant\aleph_{0}$ a cardinal, if

  1. 1.

$M\models T$,

  2. 2.

If $M_{1}\leqslant M_{2}$ are generated by a set of cardinality $<\lambda$,
then each strong embedding $f:M_{1}\to M$ extends to a strong embedding
$g:M_{2}\to M$.

The Amalgamation and Joint Embedding properties together with the fact that
$\mathcal{K}^{+}_{\alpha}$ is closed under the union of $\leqslant$-chains,
imply that the $\lambda$-rich structures exist. Indeed a union of
$\leqslant$-chains of models of $T$, and the quantifier-elimination (for this
union to be a model) give such a structure. We prefer to state this also as
the following fact.

######  Fact 3.4.

$(\mathcal{K}_{\alpha}^{+},\leqslant)$ contains $\lambda$-rich structures, for
all $\lambda\geqslant\aleph_{0}$.

######  Lemma 3.5.

Each $\lambda$-rich structure is $\lambda$-saturated as a model of $T$.

###### Proof.

Let $M$ be $\lambda$-rich. Assume that $\Sigma(x)$ is a partial $1$-type over
a set $X\subseteq{M}$, where $|X|<\lambda$ and without loss of generality
assume that $X$ is closed in ${M}$. Let $d\not\in{M}$ be a solution of
$\Sigma(x)$ in an $\mathcal{L}$-elementary extension ${N}$ of ${M}$. Extend
the coloring of ${M}$ to ${N}$ by letting $\neg p(x)$ for each $x\in{N}-{M}$,
so that now ${N}\in\mathcal{K}_{\alpha}^{+}$. Observe that
$\operatorname{cl}_{M}(X)\leqslant{M},{N}$ and hence we keep the notation
$\operatorname{cl}(X)$. It is clear that
$\langle\operatorname{cl}(X)d\rangle\in\mathcal{K}^{+}_{\alpha}$,
$\langle\operatorname{cl}(X)\rangle\leqslant\langle\operatorname{cl}(X)d\rangle$
and $\langle\operatorname{cl}(X)\rangle\leqslant{M}$. Now since $M$ is
$\lambda$-rich, there is a strong embedding
$f:\langle\operatorname{cl}(X)d\rangle\to{M}$, and hence $f(d)$ is the
solution of $\Sigma(x)$ in ${M}$. ∎

It is worth noting that in the literature, a $\lambda$-rich structure is
sometimes called $\lambda$-ultra-homogeneous. Moreover, it is clear from the
definition (letting $M_{1}=\emptyset$) that for each $M_{2}$ as above, there
is a strong embedding $g:M_{2}\to M$. This property of $M$ is called
$\lambda$-universality.

We are now set to provide an axiomatization for the rich structures.

###  3.2 Axiomatization

Below for the $\mathcal{L}$-formula $\varphi(\bar{x},\bar{y})$ and a natural
number $k$, we let $D_{\varphi,k}(\bar{y})$ and $d_{\varphi,k}(\bar{y})$ be
the formulas introduced in Lemma 2.2 and Fact 2.1 respectively.

######  Definition 3.6.

Let $\mathbb{T}_{\alpha}$ be an $\mathcal{L}_{p}$-theory whose models
$\mathbb{M}$ satisfy the following.

  1. 1.

$\mathbb{M}\models T$

  2. 2.

$\mathbb{M}\in\mathcal{K}_{\alpha}^{+}$ (that is $\delta(A)\geqslant 0$ for
all $A\subseteq_{\text{fin}}\mathbb{M}$).

  3. 3.

For each $\bar{a}\leqslant\bar{a}\bar{d}$, where $\bar{d}=(d_{1},...,d_{n})$
is transcendental over $\bar{a}$, $\dim(\bar{d}/\bar{a})=k$, and
$\varphi(\bar{x},\bar{y})$ is strong with respect to $\bar{d},\bar{a}$,

$\displaystyle\mathbb{M}\models$ $\displaystyle\forall\bar{y}\hskip
8.53581pt\Big{[}D_{\varphi,k}(\bar{y})\wedge
d_{\varphi,k}(\bar{y})\to\exists\bar{z}\hskip
5.69054pt\big{(}\varphi(\bar{z},\bar{y})\wedge\bigwedge_{p(d_{i})}p(z_{i})\wedge\bigwedge_{\neg
p(d_{i})}\neg p(z_{i})\hskip 5.69054pt\big{)}\Big{]}\hskip 5.69054pt$

The items above can be expressed as first-order axiom-schemes (for the second
item this is pointed out before Definition 2.12). In the rest of this section
we aim to establish the following theorem.

######  Theorem.

$\mathbb{T}_{\alpha}$ is a complete theory that axiomatizes $\omega_{1}$-rich
structures of $\mathcal{K}_{\alpha}^{+}$.

Towards a proof of the theorem above, what we can prove in the rest of this
subsection with the information so far, is that $\omega_{1}$-rich structures
in general, and $\omega$-rich structures when $\alpha$ is rational, are a
model of $\mathbb{T}_{\alpha}$. Proving that models of $\mathbb{T}_{\alpha}$
are rich and therefore giving the full proof of the theorem will be possible
only in Section 3.3, after some more auxiliary lemmas in the next section.

######  Lemma 3.7.

Any $\omega_{1}$-rich structure $\mathbb{M}$ is a model of
$\mathbb{T}_{\alpha}$ (and hence $\mathbb{T}_{\alpha}$ is consistent).

###### Proof.

By Lemma 3.5, $\mathbb{M}$ is $\omega_{1}$-saturated as a model of $T$. Now we
only need to prove that item 3 in Definition 3.6 holds. Let $\bar{a},\bar{d}$
and $\varphi$ be as in there, and $\bar{a}^{{}^{\prime}}$ in $\mathbb{M}$ be
such that $\mathbb{M}\models D_{\varphi,k}(\bar{a}^{{}^{\prime}})\wedge
d_{\varphi,k}(\bar{a}^{{}^{\prime}})$. So there is
$\bar{d}^{{}^{\prime}}\in\mathbb{M}^{|\bar{d}|}$ such that
$\mathbb{M}\models\varphi(\bar{d}^{{}^{\prime}},\bar{a}^{{}^{\prime}})$,
$\bar{d}^{{}^{\prime}}\cap\operatorname{acl}(\bar{a}^{{}^{\prime}})=\emptyset$,
and $\dim(\bar{d}^{{}^{\prime}}/\bar{a}^{{}^{\prime}})=k$. Since
$\langle\operatorname{cl}(\bar{a}^{{}^{\prime}})\rangle$ is countable, by
$\omega_{1}$-saturation of $\mathbb{M}$ and Observation 2.8 we may assume that
$\langle\operatorname{cl}(\bar{a}^{{}^{\prime}})\rangle\cap\bar{d}^{{}^{\prime}}=\emptyset$
and, $\bar{d}^{\prime}$ and
$\langle\operatorname{cl}(\bar{a}^{\prime})\rangle$ are free over
$\bar{a}^{\prime}$. Consider the $\mathcal{L}$-structure generated by
$\operatorname{cl}(\bar{a}^{\prime})\bar{d}^{\prime}$ in $\mathbb{M}$ and make
it into the $\mathcal{L}_{p}$-structure $N$ by coloring
$\operatorname{cl}(\bar{a}^{\prime})\bar{d}^{\prime}$ with the same colors as
$\operatorname{cl}(\bar{a})\bar{d}$ and leaving the rest of the elements non-
colored. It is clear that
$\langle\operatorname{cl}(\bar{a}^{\prime})\rangle\leqslant\mathbb{M}$. We
claim that also
$\langle\operatorname{cl}(\bar{a}^{\prime})\rangle\leqslant{N}$. If this claim
is proved then since $\mathbb{M}$ is $\omega_{1}$-rich, there is a strong
embedding $f:N\to\mathbb{M}$ which fixes
$\langle\operatorname{cl}(\bar{a}^{{}^{\prime}})\rangle$ pointwise. If we put
$f(\bar{d}^{{}^{\prime}})=\bar{e}$, then

$\displaystyle\mathbb{M}\models\big{(}\varphi(\bar{e},\bar{a}^{{}^{\prime}})\wedge\bigwedge_{p(d_{i})}p(e_{i})\wedge\bigwedge_{\neg
p(d_{i})}\neg p(e_{i})\big{)}.$

Now to prove the claim, let $\bar{d}^{\prime}_{1}\subseteq\bar{d}^{\prime}$
and $\bar{e}\subseteq
N\setminus\operatorname{cl}(\bar{a}^{{}^{\prime}})\bar{d}^{\prime}$. Notice
that
$\dim(\bar{d}^{\prime}_{1}/\langle\operatorname{cl}(\bar{a}^{{}^{\prime}})\rangle)=\dim(\bar{d}^{{}^{\prime}}_{1}/\bar{a}^{{}^{\prime}})$
since $\bar{d}\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}^{\dim}_{\operatorname{cl}(\bar{a}^{{}^{\prime}})}\bar{a}^{{}^{\prime}}$.
Moreover as $\varphi$ is strong, by 2.5 we have
$\dim(\bar{d}^{{}^{\prime}}_{1}/\bar{a}^{{}^{\prime}})\geqslant\dim(\bar{d}_{1}/\bar{a})$
. Therefore

$\displaystyle\delta(\bar{d}^{\prime}_{1}\bar{e}/\langle\operatorname{cl}(\bar{a}^{{}^{\prime}})\rangle)$
$\displaystyle=\dim(\bar{d}^{\prime}_{1}\bar{e}/\langle\operatorname{cl}(\bar{a}^{{}^{\prime}})\rangle)-\alpha|p(\bar{d}^{\prime}_{1}\bar{e})|$
$\displaystyle=\dim(\bar{d}^{\prime}_{1}/\langle\operatorname{cl}(\bar{a}^{{}^{\prime}})\rangle)-\alpha|p(\bar{d}^{\prime}_{1})|$
$\displaystyle=\dim(\bar{d}^{{}^{\prime}}_{1}/\bar{a}^{{}^{\prime}})-\alpha|p(\bar{d}^{\prime}_{1})|$
$\displaystyle\geqslant\dim(\bar{d}_{1}/\bar{a})-\alpha|p(\bar{d}_{1})|\geqslant
0.$

∎

It follows from the proof above that in case $\alpha$ is rational,
$\omega$-rich structures are a model of $\mathbb{T}_{\alpha}$.

###  3.3 Auxiliary Lemmas

The following lemmas and their proofs are inspired by [18], Lemmas $3.1$ and
$4.1$. When $A\leqslant B$ and $B$ is transcendental over $A$, these lemmas
provide us with a way to find a set $D$ containing $B$ (in both cases of
rational and irrational $\alpha$) whose pre-dimension is as much close from
below to $\delta(B)$ as we wish, yet $A\leqslant D$ and $D$ is transcendental
over $A$.

We will in particular utilize Lemmas 3.9, 3.10, and 3.11 to show that any
$\omega_{1}$-saturated model of $\mathbb{T}_{\alpha}$, for an irrational
$\alpha$ is in fact semi-generic (Theorem 3.15). Similarly we will make use of
Lemmas 3.12 and 3.13 to prove that any $\omega$-saturated model of
$\mathbb{T}_{\alpha}$, for a rational $\alpha$ is $\omega$-rich (Theorem
3.21).

######  Lemma 3.8.

For each $n\in\omega$ there exists an $\epsilon_{n}>0$ such that for each
$A\subseteq B$, if $|B\setminus A|<n$ and $\delta(B/A)<0$ then
$\delta(B/A)<-\epsilon_{n}$.

###### Proof.

The set $V_{n}=\\{\delta(B/A):\ A\subseteq B,\ |B\setminus A|<n\\}$ is finite.
Therefore there exists $\epsilon_{n}>0$, depending on $n$, which keeps every
negative value in $V_{n}$ bounded away from $0$, as required by the lemma. ∎

Recall that by the finite version of the $\Delta$-system lemma [17], if
$\\{B_{i}\\}_{i\in\omega}$ is a countably-infinite family of finite sets of
bounded size, then there is an arbitrarily-large finite subfamily of $B_{i}$’s
with the same mutual intersection. The following lemma shows that when the
$B_{i}$ are subsets of a structure in $\mathcal{K}^{+}_{\alpha}$, then it is
possible that the intersection in question is closed in each $B_{i}$.

######  Lemma 3.9.

For $k\geqslant 1$, let $\\{B_{i}:\hskip 5.69054pti\in\omega\\}$ be a family
of $k$-element subsets of a structure $M$. Then for each natural number $n$
there are a subset $\Omega$ of $\omega$ with $|\Omega|\geqslant n$ and
$A\subseteq_{\text{fin}}M$ such that

  1. 1.

$A\leqslant B_{i}$, for each $i\in\Omega$ and

  2. 2.

$B_{i}\cap B_{j}=A$, for $i\neq j\in\Omega$.

###### Proof.

Let $\epsilon_{k}$ be as in Lemma 3.8. By the finite version of
$\Delta$-system lemma, for each natural number $n$ there are a finite subset
$\Omega^{\prime}\subset\omega$ with $|\Omega^{\prime}|\geqslant n+\lfloor
k/\epsilon_{k}\rfloor$ and a finite subset $A$ of $M$ such that $B_{i}\cap
B_{j}=A$ for each $i\neq j\in\Omega^{\prime}$. But we need a further step to
show that there are at most $\lfloor{\frac{k}{\epsilon_{k}}}\rfloor$-many
$B_{i}$ in which $A$ is not closed.

Put $Z=\\{i\in\Omega^{\prime}:\hskip 5.69054ptA\nleqslant
B_{i}\\}=\\{{i_{1}},\ldots,{i_{m}}\\}$. We show that
$m\leqslant\lfloor\frac{k}{\epsilon_{k}}\rfloor$. For $i\in Z$ since
$A\nleqslant B_{i}$, there is $C_{i}$ with $A\subsetneq C_{i}\subseteq B_{i}$
and $\delta(C_{i}/A)<0$. Put $C=\bigcup_{i\in Z}C_{i}$. Then

$\displaystyle\delta(C/A)=\delta(C_{i_{m}}/AC_{i_{1}},...,C_{i_{m-1}})+...+\delta(C_{i_{2}}/AC_{i_{1}})+\delta(C_{i_{1}}/A)$
$\displaystyle\leqslant-m\epsilon_{k}.$

So
$0\leqslant\delta(C)\leqslant-m\epsilon_{k}+\delta(A)\leqslant-m\epsilon_{k}+k$.
Therefore $m\leqslant\lfloor k/\epsilon_{k}\rfloor$. ∎

######  Lemma 3.10.

($\alpha$ irrational) Assume that $A\leqslant B$ and $B$ is transcendental
over $A$. Then for each $\epsilon>0$, there is a finite set $D$ containing $B$
and transcendental over $A$ such that

  1. 1.

$-\epsilon<\delta(D/B)\leqslant 0$,

  2. 2.

$\delta(B)\leqslant\delta(D^{\prime})$ for all $B\subseteq
D^{{}^{\prime}}\subsetneqq D$,

  3. 3.

$A\leqslant D$.

###### Proof.

Without loss of generality we may assume that $0<\epsilon<\alpha$ and
$\delta(B/A)>\epsilon$. Note that the set $\\{m-\alpha n:\
m,n\in\mathbb{N}\\}$ is dense in $\mathbb{R}$ (by Dirichlet’s rational
approximation theorem). Hence there are natural numbers $1\leqslant s<k$ such
that $s/k<\alpha<(s+\epsilon)/k$.

By Lemma 2.10 and richness there is a set $D=Bd_{1},...,d_{k}$ such that
$p(d_{i})$ for all $i\leqslant k$ and each $s$-element subset of
$\\{d_{1},\ldots,d_{k}\\}$ together with $B$ is a base for $D$. We claim such
$D$ satisfies the conditions required by the lemma.

For the first condition note that $-\epsilon<\delta(D/B)=s-k\alpha<0$. For the
second condition, let $B\subseteq D^{\prime}\subseteq D$ and
$|D^{\prime}-B|=l$. If $l\leqslant s$ then $\delta(D^{\prime}/B)=l-l\alpha>0$
because $\alpha<1$. Also if $s<l<k$ then
$\delta(D^{\prime}/B)=s-l\alpha>s-\alpha k>0$. For the third condition, note
that $A\leqslant E$ for any proper subset $E$ of $D$ containing $A$. It only
remains to show that $\delta(D/A)>0$. But this is also easy to see, since
$\delta(D/A)=\delta(D/B)+\delta(B/A)$ where $\epsilon<\delta(B/A)$ and
$\delta(D/B)>-\epsilon$. Finally each $d_{i}$ is transcendental over $B$, and
hence over $A$. ∎

######  Lemma 3.11.

($\alpha$ irrational) Assume that $A\leqslant B$ and $B$ is transcendental
over $A$. Then for each $\mu>0$ and natural number $n$ there is $D^{*}$
containing $B$ such that

  1. 1.

$A\leqslant D^{*}$ and $D^{*}$ is transcendental over $A$,

  2. 2.

$\delta(D^{*}/A)<\mu$,

  3. 3.

$B\leqslant C$, for each $C$ with $B\subseteq C\subseteq D^{*}$ and
$|C\setminus B|<n$.

###### Proof.

Consider a real number
$0<\lambda<\min\\{\delta(B/A),\frac{1}{n}\epsilon_{n}\\}$, where
$\epsilon_{n}$ is as in 3.8. By 3.10 there exists $D$ such that
$-\lambda<\delta(D/B)<0$, $A\leqslant D$ and $D$ is transcendental over $A$.
Let $\gamma=-\delta(D/B)$, so $0<\gamma<\lambda$. Also let $k\geqslant 1$ be
the natural number such that

$\displaystyle k\gamma\leqslant\delta(B/A)<(k+1)\gamma.$

Let $D^{*}=D_{1}\oplus_{B}\ldots\oplus_{B}D_{k}$ be obtained as the union of
$k$ copies of $D$ (with the same colors) free from one another over $B$. We
first show that $A\leqslant D^{*}$ (hence $D^{*}\in\mathcal{K}_{\alpha}^{+}$)
and $\delta(D^{*}/A)<\mu$.

To show that $A\leqslant D^{*}$, we need to show that $\delta(C/A)\geqslant 0$
for all $C$ with $A\subseteq C\subseteq D^{*}$. So let $B_{0}=C\cap B$ and
$C_{i}=C\cap D_{i}$ for each $i<k$. There are two cases to consider:

Case 1. $B_{0}=B$.

By 3.10, $\delta(D^{{}^{\prime}}/B)>-\gamma$ for all $D^{{}^{\prime}}$ that
$B\subseteq D^{{}^{\prime}}\subseteq D$. So $\delta(C_{i}/B)>-\gamma$ for each
$i<k$. Therefore

$\displaystyle\delta(C)=\delta(B)+\sum_{i<k}\delta(C_{i}/B)>\delta(B)-k\gamma\geqslant\delta(A).$

Case 2. $B_{0}\neq B$.

Since $A\leqslant B$, it follows that $\delta(B_{0}/A)\geqslant 0$.
Furthermore each $C_{i}$ is proper subset of $D_{i}$ which means
$\delta(C_{i}/B_{0})\geqslant 0$ for all $i<k$ by 3.10. Hence

$\displaystyle\delta(C/A)=\sum_{i<k}\delta(C_{i}/B_{0})+\delta(B_{0}/A)\geqslant
0.$

So in general $A\leqslant D^{*}$ and hence $D^{*}\in\mathcal{K}_{\alpha}^{+}$.

Now it is easy to see that $\delta(D^{*}/A)<\mu$, because

$\displaystyle\delta(D^{*}/A)=\delta(D^{*}/B)+\delta(B/A)\leqslant\delta(B/A)-k\gamma<\gamma<\lambda<\mu.$

Finally we prove that item 3 in the statement of the lemma is fulfilled. Let
$C$ be subset of $D^{*}$ containing $B$ with $|C\setminus B|<n$. Let
$C_{i}=C\cap D_{i}$ for each $i<k$. Since $|C\setminus B|<n$ it follows that
$|\\{i;\hskip 5.69054ptC_{i}\neq B\\}|<n$. Hence

$\displaystyle\delta(C)=\delta(B)+\sum_{i<k}\delta(C_{i}/B)\geqslant\delta(B)-n\lambda.$

So $\delta(C/B)\geqslant-n\lambda\geqslant-\epsilon_{n}$. But 3.8 implies that
$\delta(C/B)\geqslant 0$. ∎

######  Lemma 3.12.

($\alpha$ rational) Assume that $\alpha=\frac{m}{n}$ with $m,n$ co-prime and
$0<m<n$. Let $A\leqslant B$, $B$ be transcendental over $A$, $\delta(B/A)>0$
and $t$ be a natural number. Then there is $D$ containing $B$ such that

  1. 1.

$A\leqslant D$ and $D$ is transcendental over $A$.

  2. 2.

$(B,D)$ is a minimal pair, $|D\setminus B|>t$, and $\delta(D/B)=-\frac{1}{n}$.

###### Proof.

Similar to the proof of 3.10 we introduce suitable $s<k$ and
$D=Bd_{1},...,d_{k}$ such that $p(d_{i})$ for all $i\leqslant k$, and each of
the $s$-element subsets of $\\{d_{1},\ldots,d_{k}\\}$ together with $B$ forms
a basis for $D$.

Let $1\leqslant k^{{}^{\prime}},s^{{}^{\prime}}<n^{t+1}$ such that
$m^{t+1}k^{{}^{\prime}}-1=s^{{}^{\prime}}n^{t+1}$ witness the fact that
$(m^{t+1},n^{t+1})=1$. Put $s^{{}^{\prime}}n^{t}=s$ and
$k^{{}^{\prime}}m^{t}=k$. Then $\delta(D/B)=s-\frac{m}{n}k=-\frac{1}{n}$.

The other requirements on $D$ can be verified in a similar way to the proof of
3.10. ∎

######  Lemma 3.13.

($\alpha$ rational) Under the assumptions of the previous lemma, there is a
transcendental extension $D^{*}$ of $A$ such that $\delta(D^{*}/A)=0$ and for
each $D^{{}^{\prime}}$ containing $B$ with $|D^{{}^{\prime}}\setminus B|<t$ we
have $B\leqslant D^{{}^{\prime}}$.

###### Proof.

Suppose that $\delta(B/A)=\frac{p}{n}$. Similar to the proof of 3.11 we
consider $D^{*}$ to be a disjoint free union of $p$-copies of $D$ over $A$ and
show that $D^{*}$ has the desired properties. ∎

###  3.4 Completeness

We now turn to the theory $\mathbb{T}_{\alpha}$ and showing that it is
complete.

####  3.4.1 $\alpha$ irrational

In this situation to prove the completeness of $\mathbb{T}_{\alpha}$ we adopt
the notion of semi-genericity as appears in [4].

Recall the notation from subsection $2.2$ that for a set
$A\in\mathcal{K}_{\alpha}^{+}$ an embedding $f:A\to\mathbb{M}$ is called
$\mathcal{L}_{p}$-embedding if
$\operatorname{qftp}_{\mathcal{L}}(f(A))=\operatorname{qftp}_{\mathcal{L}}(A)$
and for each $x\in A$, $x$ and $f(x)$ have the same color.

######  Definition 3.14.

An $\mathcal{L}_{p}$-structure $M\in\mathcal{K}_{\alpha}^{+}$ is called semi-
generic, if

  1. 1.

$M\models T$, and

  2. 2.

whenever $A\leqslant B$, $B$ is transcendental over $A$ and $f:A\to M$ is an
$\mathcal{L}_{p}$-embedding then for each natural number $n$ there exists an
$\mathcal{L}_{p}$-embedding $\hat{f}_{n}:B\to M$ extending $f$ and such that

$\displaystyle\operatorname{cl}^{n}_{M}(\hat{f}_{n}(B))=\hat{f}_{n}(B)\oplus_{f(A)}\operatorname{cl}^{n}_{M}(f(A)).\quad(*)$

The proof of the following lemma is inspired by [18], Proposition 4.4. .

######  Theorem 3.15.

Each $\omega_{1}$-saturated model of $\mathbb{T}_{\alpha}$ is semi-generic.

###### Proof.

We find it beneficial to provide a sketch of the proof beforehand. Assuming
that $A,B$ are as in item 2 in 3.14, we use the axioms and saturation to find
infinitely-many mutually-free copies over $A$ of a set $D^{*}$ as in 3.11,
whose pre-dimension is close enough to that of $A$. We will then show that
there are only finitely many of these copies in which the corresponding image
of $B$ violates $(*)$ in 3.14, and therefore there is a copy with the desired
property.

Now, assume that $\mathbb{M}$ is an $\omega_{1}$-saturated model of
$\mathbb{T}_{\alpha}$, $A\leqslant B$, $B$ is transcendental over $A$ and
$f:A\to\mathbb{M}$ is an $\mathcal{L}_{p}$-embedding. Let $n$ be a given
natural number, $\epsilon_{n}$ be as in 3.8 and $0<\epsilon<\epsilon_{n}$. By
LABEL:{lem:3.3.4} there exists $D^{*}\in\mathcal{K}_{\alpha}^{+}$ extending
$B$ such that $A\leqslant D^{*}$, $D^{*}$ is transcendental over $A$ and
$\delta(D^{*}/A)<\epsilon$. Moreover as in the mentioned lemma,
$B\leqslant_{i}D^{*}$ and for every $B\subseteq C\subseteq D^{*}$ with
$|C\setminus B|<n$ we have $B\leqslant C$. Now by applying 2.8 to $X=f(A)$,
$Y=\operatorname{cl}(f(A))$, and
$\Sigma(\bar{x})=\\{\phi(\bar{x},f(\bar{a})):\hskip
8.53581pt\phi(\bar{x},\bar{a})\in\operatorname{qftp}_{\mathcal{L}}(D^{*}/A)\\}$
regarding the fact that $D^{*}$ is transcendental over $A$, there exists an
$\mathcal{L}$-embedding $h:D^{*}\to\mathbb{M}$ such that
$h(D^{*})\cap\operatorname{cl}(f(A))=f(A)$ and $h(D^{*})$ and
$\operatorname{cl}(f(A))$ are free over $f(A)$.

Now for a finite subset $A^{{}^{\prime}}$ of $\operatorname{cl}(f(A))$
including $f(A)$ we let
$D^{{}^{\prime}}=A^{{}^{\prime}}\oplus_{f(A)}h(D^{*})$. We color the
$\mathcal{L}$-structure $\langle D^{{}^{\prime}}\rangle_{\mathbb{M}}$ by
letting $h(D^{*})$ have the same color as $D^{*}$, while keeping the color of
$A^{{}^{\prime}}$ and leaving the rest non-colored. Denote the resulting
$\mathcal{L}_{p}$-structure by $N$. Then
$D^{{}^{\prime}}\in\mathcal{K}_{\alpha}^{+}$, since
$N\in\mathcal{K}_{\alpha}^{+}$. Also $A^{{}^{\prime}}\leqslant
D^{{}^{\prime}}$ and $D^{{}^{\prime}}$ is transcendental over
$A^{{}^{\prime}}$.

Let $\\{D^{\prime}_{i}\\}$, indexed by the natural numbers, be disjoint copies
of $D$ mutually free over $A$. For each natural number $m$ let
$\mathcal{D}^{{}^{\prime}}_{m}=D^{{}^{\prime}}_{1}\oplus_{A}\ldots\oplus_{A}D^{{}^{\prime}}_{m}$
be the free union of $m$ disjoint copies of $D^{{}^{\prime}}$ over
$A^{{}^{\prime}}$ (so
$D^{{}^{\prime}}_{i}\cong_{A^{{}^{\prime}}}D^{{}^{\prime}}$). Axioms of
$\mathbb{T}_{\alpha}$ (Axiom 3) and $\omega_{1}$-saturation of $\mathbb{M}$
ensure that there exists a copy of $\mathcal{D}^{{}^{\prime}}_{m}$ over
$A^{{}^{\prime}}$, since $\mathcal{D}^{{}^{\prime}}_{m}$ is transcendental
over $A^{{}^{\prime}}$. Thus for each $m$ there are $m$ copies of
$D^{{}^{\prime}}$ disjoint over $f(A)$ satisfying
$\operatorname{qftp}_{\mathcal{L}}(h(D^{{}^{\prime}})/A^{{}^{\prime}})$. Hence
by $\omega_{1}$-saturation there are infinity many copies of $D^{*}$ disjoint
over $f(A)$ satisfying
$\operatorname{qftp}_{\mathcal{L}}(h(D^{*})/\operatorname{cl}(f(A)))$.

Let $g_{i}:D^{{}^{\prime}}\to\mathbb{M}$ be the embedding that represents the
$i$’th copy of $D^{{}^{\prime}}$ over $f(A)$, $B_{i}=g_{i}(B)$ and
$D^{*}_{i}=g_{i}(D^{{}^{\prime}})$. By the construction,
$D^{*}_{i}\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}^{\dim}_{f(A)}\operatorname{cl}(f(A))$
and consequently $B_{i}\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}^{\dim}_{f(A)}\operatorname{cl}(f(A))$.

Let $Z=\\{i\in\omega:\hskip
8.53581pt\operatorname{cl}^{n}(B_{i})\nsubseteq\operatorname{cl}(f(A))\oplus_{f(A)}B_{i}\\}$.
Under the assumption that $Z$ is finite, the rest of the proof goes as
follows. Pick some $i\not\in Z$. We claim that
$\operatorname{cl}^{n}(B_{i})=\operatorname{cl}^{n}(f(A))\oplus_{f(A)}B_{i}$,
that is the condition of semi-genericity is satisfied by $\hat{f}_{n}=g_{i}$.
Suppose that $E\subseteq\mathbb{M}$ such that $B_{i}\leqslant_{i}E$ and
$|E-B_{i}|<n$. Then $E\subseteq\operatorname{cl}(f(A))\oplus_{f(A)}B_{i}$. If
$E_{1}=E\setminus B_{i}$, then $E_{1}\subseteq\operatorname{cl}(f(A))$ and
$E_{1}\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}^{\dim}_{f(A)}B_{i}$.
Therefore $f(A)\leqslant_{i}f(A)\cup E_{1}$ and
$E_{1}\subseteq\operatorname{cl}^{n}(f(A))$. Hence
$E\subseteq\operatorname{cl}^{n}(f(A))\oplus_{f(A)}B_{i}$ and
$\operatorname{cl}^{n}(B_{i})=\operatorname{cl}^{n}(f(A))\oplus_{f(A)}B_{i}$.

Now it remains to show that the assumption that $Z$ is infinite leads to a
contradiction. For $i\in Z$, denote by $C_{i}\subseteq\mathbb{M}$ an intrinsic
extension of $B_{i}$ not included in
$\operatorname{cl}(f(A))\oplus_{f(A)}B_{i}$ with $|C_{i}\setminus B_{i}|<n$.
Put $H_{i}=D^{*}_{i}\cup C_{i}$ and choose a natural number $s$ such that
$|H_{i}|<s$ for all $i\in Z$. As every subset of $D^{*}_{i}$ that includes
$B_{i}$ and has less than $n$ more elements than $B_{i}$ is closed in
$D^{*}_{i}$, we have $C_{i}\nsubseteq D^{*}_{i}$.

By Lemma 3.9 there are an arbitrary large finite set $\Omega\subseteq Z$ and a
set $F$ such that $H_{i}\cap H_{j}=F$ and $F\leqslant H_{i}$ for all $i\neq
j\in\Omega$. If $A^{{}^{\prime}}=F\cap\operatorname{cl}(f(A))$ then
$A^{{}^{\prime}}\leqslant F$ and by transitivity $A^{{}^{\prime}}\leqslant
H_{i}$ and $\delta(H_{i}/A^{{}^{\prime}})\geqslant 0$.

On the other hand we have that $C_{i}\setminus D^{*}_{i}\neq\emptyset$. So

$\displaystyle\delta(H_{i}/A^{{}^{\prime}})=\delta(D^{*}_{i}\cup
C_{i}/A^{{}^{\prime}})$
$\displaystyle=\delta(D^{*}_{i}/A^{{}^{\prime}})+\delta(C_{i}/D^{*}_{i}).$

Notice that $\delta(C_{i}/D^{*}_{i})<0$. Hence as we have $|C_{i}\setminus
D^{*}_{i}|<n$, it follows that
$\delta(C_{i}/D^{*}_{i})<-\epsilon_{n}<-\epsilon$. Furthermore as
$D^{*}_{i}\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}^{\dim}_{f(A)}\operatorname{cl}(f(A))$,
we have that $\delta(D^{*}_{i}/A^{{}^{\prime}})=\delta(D^{*}_{i}/A)<\epsilon$.
Hence $\delta(H_{i}/A^{{}^{\prime}})<0$, a contradiction. ∎

######  Definition 3.16.

By $X\stackrel{{\scriptstyle w}}{{\cong}}Y$, read as $X$ is weakly isomorphic
to $Y$, we mean that there is an $\mathcal{L}$-isomorphism $f:\langle
X\rangle\to\langle Y\rangle$ mapping $X$ to $Y$ and such that
$p(x)\leftrightarrow p(f(x))$ for all $x\in X$.

######  Remark 3.17.

Notice that the above notion of isomorphism is weaker than $\langle
X\rangle\cong_{\mathcal{L}_{p}}\langle Y\rangle$, since the colours are
preserved only for elements of $X$. Weak isomorphism of $X$ and $Y$, as in the
definition above, also implies that for every $A\subseteq X$ and natural
number $i$, if $\operatorname{cl}^{i}_{\langle X\rangle}(A)\subseteq X$ and
$\operatorname{cl}^{i}_{\langle Y\rangle}(f(A))\subseteq Y$ then
$f(\operatorname{cl}^{i}_{\langle X\rangle}(A))=\operatorname{cl}^{i}_{\langle
Y\rangle}(f(A))$ and therefore $\operatorname{cl}^{i}_{\langle
X\rangle}(A)\stackrel{{\scriptstyle w}}{{\cong}}\operatorname{cl}^{i}_{\langle
Y\rangle}(f(A))$.

######  Theorem 3.18.

Let $\mathbb{M}_{1}$ and $\mathbb{M}_{2}$ be two $\omega_{1}$-saturated models
of $\mathbb{T}_{\alpha}$ and $\phi(\bar{x})$ an $\mathcal{L}_{p}$-formula.
Then there exists $n=n_{\varphi}$ such that for all
$\bar{a}_{1}\in\mathbb{M}_{1}^{|\bar{a}_{1}|}$ and
$\bar{a}_{2}\in\mathbb{M}_{2}^{|\bar{a}_{2}|}$,

$\displaystyle\operatorname{cl}^{n}_{\mathbb{M}_{1}}(\bar{a}_{1})\stackrel{{\scriptstyle
w}}{{\cong}}\operatorname{cl}^{n}_{\mathbb{M}_{2}}(\bar{a}_{2})\Rightarrow\big{(}\mathbb{M}_{1}\models\varphi(\bar{a}_{1})\Leftrightarrow\mathbb{M}_{2}\models\varphi(\bar{a}_{2})\big{)}.$

Note that the idea of the following proof is borrowed from [4], Lemma 1.30,
but there are essential alterations that make the rewrite necessary.

###### Proof.

Assume that $\mathbb{M}_{1},\mathbb{M}_{2}$,
$\bar{a}_{1}\in\mathbb{M}_{1}^{|\bar{x}|}$ and
$\bar{a}_{2}\in\mathbb{M}_{2}^{|\bar{x}|}$ are as in the statement of the
theorem. We proceed by induction on the complexity of
$\mathcal{L}_{p}$-formulas. As it appears, the induction step for the
$\mathcal{L}$-atomic formulas as well as the boolean connectives are
straightforward. For an $\mathcal{L}_{p}$-atomic formula of the form
$\varphi:=p(t(x_{1},\dots,x_{n}))$ where $t(x_{1},\dots,x_{n})$ is an
$\mathcal{L}$-term, we may take $n_{\varphi}=1$. Then notice that
$\operatorname{acl}_{\mathbb{M}_{i}}(\bar{a}_{i})\cap
p(\mathbb{M}_{i})\subseteq\operatorname{cl}^{1}_{\mathbb{M}_{i}}(\bar{a}_{i})$
for $i=1,2$. Hence
$\operatorname{cl}^{1}_{\mathbb{M}_{1}}(\bar{a}_{1})\stackrel{{\scriptstyle
w}}{{\cong}}\operatorname{cl}^{1}_{\mathbb{M}_{2}}(\bar{a}_{2})$ implies that
$\operatorname{acl}_{\mathbb{M}_{1}}(\bar{a}_{1})\stackrel{{\scriptstyle
w}}{{\cong}}\operatorname{acl}_{\mathbb{M}_{2}}(\bar{a}_{2})$. Therefore
$t^{\mathbb{M}_{1}}(\bar{a}_{1})\in p(\mathbb{M}_{1})$ if and only if
$t^{\mathbb{M}_{2}}(\bar{a}_{2})\in p(\mathbb{M}_{2})$.

Now assuming that the statement holds for $\psi(y,\bar{x})$ we need to
introduce a natural number $n_{\varphi}$ so that it holds also for the formula
$\varphi(\bar{x})=\exists y\psi(y,\bar{x})$.

By Remark 2.17 let the natural number $p_{1}$ be sufficiently large so that
$|\operatorname{cl}^{n_{\psi}}_{N}(\bar{a}d)|<p_{1}$ for all
$N\in\mathcal{K}_{\alpha}^{+}$ and $\bar{a},d\in N$ with
$|\bar{a}|=|\bar{x}|$. Put $p=\max\\{p_{1},n_{\psi}\\}$. For
$N\in\mathcal{K}_{\alpha}^{+}$ and $\bar{a}\in N^{|\bar{x}|}$ and $i<p$, we
define inductively $A_{i}^{N}(\bar{a})$ by setting
$A_{0}^{N}(\bar{a})=\bar{a}$ and
$A^{N}_{i+1}(\bar{a})=\operatorname{cl}^{p}_{N}(A_{i}^{N}(\bar{a}))$. Now let
$n_{\varphi}\geqslant 1$ be such that
$A^{N}_{p}(\bar{a})\subseteq\operatorname{cl}^{n_{\varphi}}_{N}(\bar{a})$ for
all $N\in\mathcal{K}^{+}_{\alpha}$ and $\bar{a}\in N^{|\bar{x}|}$. We will
show that $n_{\varphi}$ satisfies the requirement of the theorem.

Let $\bar{a}_{1}\in\mathbb{M}_{1}^{|\bar{x}|}$ and
$\bar{a}_{2}\in\mathbb{M}_{2}^{|\bar{x}|}$ be such that
$\operatorname{cl}^{n_{\varphi}}_{\mathbb{M}_{1}}(\bar{a}_{1})\stackrel{{\scriptstyle
w}}{{\cong}}\operatorname{cl}^{n_{\varphi}}_{\mathbb{M}_{2}}(\bar{a}_{2})$.
For $b_{1}\in\mathbb{M}_{1}$ we show that there is $b_{2}\in\mathbb{M}_{2}$
with
$\operatorname{cl}^{n_{\psi}}_{\mathbb{M}_{1}}(\bar{a}_{1}b_{1})\stackrel{{\scriptstyle
w}}{{\cong}}\operatorname{cl}^{n_{\psi}}_{\mathbb{M}_{2}}(\bar{a}_{2}b_{2})$.

Denote by $H_{0}$ the set with elements $\bar{a}_{1}b_{1}$ and let
$H_{1}=\operatorname{cl}^{n_{\psi}}_{\mathbb{M}_{1}}(H_{0})$. Since
$\operatorname{cl}^{n_{\varphi}}_{\mathbb{M}_{1}}(\bar{a}_{1})\stackrel{{\scriptstyle
w}}{{\cong}}\operatorname{cl}^{n_{\varphi}}_{\mathbb{M}_{2}}(\bar{a}_{2})$,
for each $i\leqslant p$,
$A_{i}^{\mathbb{M}_{1}}:=A_{i}^{\mathbb{M}_{1}}(\bar{a}_{1})\stackrel{{\scriptstyle
w}}{{\cong}}A_{i}^{\mathbb{M}_{2}}:=A_{i}^{\mathbb{M}_{2}}(\bar{a}_{2})$.
Since $|H_{1}\setminus H_{0}|<p$, there is $j<p$ such that

$\displaystyle(A^{\mathbb{M}_{1}}_{j+1}\setminus
A^{\mathbb{M}_{1}}_{j})\cap(H_{1}\setminus H_{0})=\emptyset.$

Moreover the fact that $p>|H_{1}\setminus A^{\mathbb{M}_{1}}_{j}|$ implies
$A^{\mathbb{M}_{1}}_{j}\leqslant A^{\mathbb{M}_{1}}_{j}H_{1}$.

Now by Remark 2.20 we may find $G_{1}\subseteq H_{1}$ such that
$A^{\mathbb{M}_{1}}_{j}G_{1}$ is algebraic over $A^{\mathbb{M}_{1}}_{j}$ and
$A^{\mathbb{M}_{1}}_{j}H_{1}$ is transcendental over
$A^{\mathbb{M}_{1}}_{j}G_{1}$. Note that since $n_{\varphi}\geqslant 1$, we
have $\operatorname{acl}_{\mathbb{M}_{1}}(\bar{a}_{1})\stackrel{{\scriptstyle
w}}{{\cong}}\operatorname{acl}_{\mathbb{M}_{2}}(\bar{a}_{2})$. Hence there is
$G_{2}\subseteq\mathbb{M}_{2}$ such that
$A_{j}^{\mathbb{M}_{1}}G_{1}\stackrel{{\scriptstyle
w}}{{\cong}}A_{j}^{\mathbb{M}_{2}}G_{2}$. Furthermore, again by Remark 2.20,

$\displaystyle\operatorname{cl}^{p}_{\mathbb{M}_{2}}(A_{j}^{\mathbb{M}_{2}}G_{2})=\operatorname{cl}^{p}_{\mathbb{M}_{2}}(A_{j}^{\mathbb{M}_{2}})\oplus_{A_{j}^{\mathbb{M}_{2}}}A_{j}^{\mathbb{M}_{2}}G_{2}.\hskip
8.53581pt(\ast)$

Let $g:A^{\mathbb{M}_{1}}_{j}G_{1}\to\mathbb{M}_{2}$ be the
$\mathcal{L}_{p}$-embedding witnessing
$A_{j}^{\mathbb{M}_{1}}G_{1}\stackrel{{\scriptstyle
w}}{{\cong}}A_{j}^{\mathbb{M}_{2}}G_{2}$. Now since
$A^{\mathbb{M}_{1}}_{j}H_{1}$ is transcendental over
$A^{\mathbb{M}_{1}}_{j}G_{1}$, by semi-genericity of $\mathbb{M}_{2}$, we find
an $\mathcal{L}_{p}$-embedding
$\hat{g}:A_{j}^{\mathbb{M}_{1}}H_{1}\to\mathbb{M}_{2}$ extending $g$ such that

$\displaystyle\operatorname{cl}^{p}_{\mathbb{M}_{2}}(A_{j}^{\mathbb{M}_{2}}\hat{g}(H_{1}))=\operatorname{cl}^{p}_{\mathbb{M}_{2}}(A_{j}^{\mathbb{M}_{2}}G_{2})\oplus_{A_{j}^{\mathbb{M}_{2}}G_{2}}A_{j}^{\mathbb{M}_{2}}\hat{g}(H_{1}).$

Now by $(\ast)$ we have

$\displaystyle\operatorname{cl}^{p}_{\mathbb{M}_{2}}(A_{j}^{\mathbb{M}_{2}}\hat{g}(H_{1}))=\operatorname{cl}^{p}_{\mathbb{M}_{2}}(A_{j}^{\mathbb{M}_{2}})\oplus_{A_{j}^{\mathbb{M}_{2}}}A_{j}^{\mathbb{M}_{2}}\hat{g}(H_{1}),$

that is

$\operatorname{cl}^{p}_{\mathbb{M}_{2}}(A_{j}^{\mathbb{M}_{2}}\hat{g}(H_{1}))=A_{j+1}^{\mathbb{M}_{2}}\oplus_{A_{j}^{\mathbb{M}_{2}}}A_{j}^{\mathbb{M}_{2}}\hat{g}(H_{1}).$

Now letting $b_{2}=\hat{g}(b_{1})$, we will show that
$\operatorname{cl}^{n_{\psi}}_{\mathbb{M}_{1}}(\bar{a}_{1}b_{1})\stackrel{{\scriptstyle
w}}{{\cong}}\operatorname{cl}^{n_{\psi}}_{\mathbb{M}_{2}}(\bar{a}_{2}b_{2})$.

By the definition of $\hat{g}$ we have
$A_{j}^{\mathbb{M}_{2}}\hat{g}(H_{1})\stackrel{{\scriptstyle
w}}{{\cong}}A_{j}^{\mathbb{M}_{1}}H_{1}$ and
$\operatorname{cl}_{\mathbb{M}_{1}}^{n_{\psi}}(\bar{a}_{1},b_{1})=H_{1}\subseteq
A_{j}^{\mathbb{M}_{1}}H_{1}$. Note that

$\displaystyle\operatorname{cl}_{\mathbb{M}_{2}}^{n_{\psi}}(\bar{a}_{2},b_{2})\subseteq\operatorname{cl}_{\mathbb{M}_{2}}^{p}(\bar{a}_{2},b_{2})\subseteq\operatorname{cl}_{\mathbb{M}_{2}}^{p}(A_{j}^{\mathbb{M}_{2}}\hat{g}(H_{1}))=A_{j+1}^{\mathbb{M}_{2}}\hat{g}(H_{1}).$

Moreover, since $A_{j+1}^{\mathbb{M}_{2}}$ and $\hat{g}(H_{1})$ are free over
$A_{j}^{\mathbb{M}_{2}}$, we have
$\operatorname{cl}_{\mathbb{M}_{2}}^{n_{\psi}}(\bar{a}_{2},b_{2})\subseteq
A_{j}^{\mathbb{M}_{2}}\hat{g}(H_{1})$. So it follows that
$\operatorname{cl}^{n_{\psi}}_{\mathbb{M}_{1}}(\bar{a}_{1},b_{1})=\operatorname{cl}^{n_{\psi}}_{\langle
A^{\mathbb{M}_{1}}_{j}H_{1}\rangle}(\bar{a}_{1},b_{1})$ and
$\operatorname{cl}^{n_{\psi}}_{\mathbb{M}_{2}}(\bar{a}_{2},b_{2})=\operatorname{cl}^{n_{\psi}}_{\langle
A^{\mathbb{M}_{2}}_{j}\hat{g}(H_{1})\rangle}(\bar{a}_{2},b_{2})$. On the other
hand by 3.17 we have that

$\displaystyle\operatorname{cl}^{n_{\psi}}_{\langle
A^{\mathbb{M}_{1}}_{j}H_{1}\rangle}(\bar{a}_{1},b_{1})\stackrel{{\scriptstyle
w}}{{\cong}}\operatorname{cl}^{n_{\psi}}_{\langle
A^{\mathbb{M}_{2}}_{j}\hat{g}(H_{1})\rangle}(\bar{a}_{2},b_{2}).$

Therefore
$\operatorname{cl}^{n_{\psi}}_{\mathbb{M}_{1}}(\bar{a}_{1},b_{1})\stackrel{{\scriptstyle
w}}{{\cong}}\operatorname{cl}^{n_{\psi}}_{\mathbb{M}_{2}}(\bar{a}_{2},b_{2})$.
∎

The completeness of $\mathbb{T}_{\alpha}$ is finally established as follows.

######  Theorem 3.19.

$\mathbb{T}_{\alpha}$ is complete.

###### Proof.

If $\mathbb{M}_{1}$ and $\mathbb{M}_{2}$ are two $\omega_{1}$-saturated models
of $\mathbb{T}_{\alpha}$ then by Theorem 3.18,
$\mathbb{M}_{1}\equiv\mathbb{M}_{2}$. Therefore $\mathbb{T}_{\alpha}$ is
complete. ∎

######  Corollary 3.20.

Let $\mathfrak{C}$ be a monster model of $\mathbb{T}_{\alpha}$.

  1. 1.

Assume that $\bar{a}_{1},\bar{a}_{2}$ in $\mathfrak{C}$ are small tuples and
$X$ is a closed small subset of $\mathfrak{C}$. Then,

$\displaystyle\operatorname{tp}(\bar{a}_{1}/X)=\operatorname{tp}(\bar{a}_{2}/X)\Leftrightarrow\langle\operatorname{cl}(X\bar{a}_{1})\rangle\cong_{\langle
X\rangle}\langle\operatorname{cl}(X\bar{a}_{2})\rangle$

  2. 2.

$\mathfrak{C}$ is $\lambda$-rich, for all $\lambda<|\mathfrak{C}|$.

####  3.4.2 $\alpha$ rational

When $\alpha$ is rational, we show the completeness of $\mathbb{T}_{\alpha}$
by proving that any $\omega$-saturated model of $\mathbb{T}_{\alpha}$ is
$\omega$-rich and that any two $\omega$-rich structures of
$\mathcal{K}^{+}_{\alpha}$ are back and forth equivalent.

######  Theorem 3.21.

Suppose that $\mathbb{M}$ is an $\omega$-saturated model of
$\mathbb{T}_{\alpha}$ then $\mathbb{M}$ is $\omega$-rich.

###### Proof.

Let $M\leqslant N$ be two finitely generated structures in
$\mathcal{K}_{\alpha}^{+}$ and $f:M\to\mathbb{M}$ be a strong
$\mathcal{L}_{p}$-embedding. We prove that there is a strong
$\mathcal{L}_{p}$-embedding $g:N\to\mathbb{M}$ extending $f$.

Since $M,N$ are finitely generated, there is $A\subseteq_{\text{fin}}M$ and
$B\subseteq_{\text{fin}}N$ such that $M=\langle A\rangle$, $N=\langle
B\rangle$ and $A\leqslant B\leqslant N$. By 2.20 we have two specific cases to
consider.

##### Case 1.

$B$ is algebraic over $A$.

Since $A\leqslant N\in\mathcal{K}_{\alpha}^{+}$, any $x\in B\setminus A$ is
non-colored. So by 2.20 it is clear that there is a strong
$\mathcal{L}_{p}$-embedding $g:B\to\mathbb{M}$ extending $f$.

##### Case 2.

$B$ is transcendental over $A$.

There are two subcases to consider:

Case 2.1 $\delta(B/A)>0$. By 3.13 there is $D\in\mathcal{K}_{\alpha}^{+}$
extending $B$ with $A\leqslant D$, $D$ transcendental over $A$ and
$\delta(D/A)=0$. Using the same method as in the proof of 3.15 for each
natural number $n$ we can find an $\mathcal{L}_{p}$-embedding
$g:D\to\mathbb{M}$ extending $f$ such that $g(B)$ is $n$-strong, i.e. for each
$C\subseteq\mathbb{M}$ containing $g(B)$ with $|C\setminus g(B)|<n$ we have
$B\leqslant C$. Hence by $\omega$-saturation there is indeed a strong
$\mathcal{L}_{p}$-embedding $g:B\to\mathbb{M}$ extending $f$.

Case 2.2. $\delta(B/A)=0$. This can be dealt with as Case $2.1$ by taking $D$
above in the place of $B$.

Now by 2.20 there is $B_{1}$ with $A\leqslant B_{1}\leqslant B$ such that
$B_{1}$ is algebraic over $A$ and $B$ is transcendental over $B_{1}$. By Case
$1$ there is an $\mathcal{L}_{p}$-embedding $g_{1}:B_{1}\to\mathbb{M}$
extending $f$ and by Case 2, there is an $\mathcal{L}_{p}$-embedding
$g:B\to\mathbb{M}$ extending $g_{1}$. ∎ It is easy to see that $\omega$-rich
structures are back and forth equivalent and hence so are any two
$\omega$-saturated models of $\mathbb{T}_{\alpha}$. This yields the following
sought-for theorem.

######  Theorem 3.22.

$\mathbb{T}_{\alpha}$ is complete.

######  Remark 3.23.

3.20 also holds for rational $\alpha$.

######  Remark 3.24.

If $T$ is an o-minimal expansion of a densely ordered group and $\alpha=1$
then $\mathbb{T}_{\alpha}$ happens to be the theory $T^{\operatorname{indep}}$
as introduced in [12].

##  4 Dependence and Strong Dependence

In this section, we will prove that if $T$ is dependent, that is when it has
Not-the-Independence Property, then so is $\mathbb{T}_{\alpha}$. We will show
that $\mathbb{T}_{\alpha}$ inherits the strong dependence from $T$ as well,
when $\alpha$ is rational. Moreover we will show that when $T$ defines a
linear order, $\mathbb{T}_{\alpha}$ is not strongly dependent.

Recall that an $\mathcal{L^{\prime}}$-theory $T^{\prime}$ is called dependent
if each $\mathcal{L^{\prime}}$-formulas $\varphi(\bar{x},\bar{y})$ is
dependent, that is there are no model $M^{\prime}\models T^{\prime}$, and
sequences $\\{\bar{a}_{i}:\hskip 8.53581pti\in\omega\\}$ and
$\\{\bar{b}_{J}:\hskip 8.53581ptJ\subseteq\omega\\}$ such that
$M\models\varphi(\bar{a}_{i},\bar{b}_{J})$ if and only if $i\in J$. It is
known that in this definition suffices it to assume $|\bar{x}|=1$.

From now on we add the further assumption that $T$ is dependent, and let
$\mathfrak{C}$ be a monster model of $\mathbb{T}_{\alpha}$. In this section we
use boldface letters $\mathbb{M},\mathbb{N}$ for models of
$\mathbb{T}_{\alpha}$.

By standard facts on dependent theories, to show that $\mathbb{T}_{\alpha}$ is
dependent, we verify that for a given model $\mathbb{M}$ of
$\mathbb{T}_{\alpha}$, the number of coheir extensions of any $1$-type $p(x)$
over $\mathbb{M}$ is at most $2^{|\mathbb{M}|}$. The following notion of
$D$-independence would be of great help ([23]) for our counting of the
coheirs.

######  Definition 4.1.

Let $\mathbb{M}$ be an arbitrary model of $\mathbb{T}_{\alpha}$,
$A,B\subseteq_{\text{fin}}\mathbb{M}$ and $X\subseteq\mathbb{M}$. Define

  1. 1.

$D(A)=\inf\\{\delta(C):\hskip 8.53581ptA\subseteq
C\subseteq_{\text{fin}}\mathbb{M}\\}$,

  2. 2.

$D(B/A)=D(BA)-D(A)$,

  3. 3.

$\operatorname{CL}(A)=\\{x\in\mathbb{M}:\hskip 8.53581ptD(xA)=D(A)\\}$,

  4. 4.

$D(A/X)=\inf\\{D(A/X_{0}):\hskip 8.53581ptX_{0}\subseteq_{\text{fin}}X\\}$.

Note that the function $D$ is automorphism invariant, i.e. for any
automorphism $f:\mathfrak{C}\to\mathfrak{C}$ we have $D(A/X)=D(f(A)/f(X))$ for
any $A,X\subseteq\mathfrak{C}$.

######  Fact 4.2.

When $\alpha$ is rational it can be easily seen that
$D(A)=\delta(\operatorname{cl}(A))$ and
$D(B/A)=\delta(\operatorname{cl}(BA))-\delta(\operatorname{cl}(A))$. Therefore
the set containing all positive values $D(B/A)$ is discrete, and it does not
contain any infinite decreasing sequence.

######  Definition 4.3.

Let $\mathbb{M}\models\mathbb{T}_{\alpha}$,
$A,B\subseteq_{\text{fin}}\mathbb{M}$ and $Z\subseteq\mathbb{M}$. We say that
$A,B$ are $D$-independent over $Z$ and write $A\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}_{Z}^{D}B$
whenever $D(A/Z)=D(A/ZB)$ and
$\operatorname{cl}(AZ)\cap\operatorname{cl}(BZ)=\operatorname{cl}(Z)$.
Moreover for arbitrary subsets $X,Y$ of $\mathbb{M}$ we say that $X,Y$ are
$D$-independent over $Z$ and write $X\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}_{Z}^{D}Y$ if
$C\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}_{Z}^{D}E$ for
any $C\subseteq_{\text{fin}}X$ and $E\subseteq_{\text{fin}}Y$. In this case we
also say that $\operatorname{tp}(X/YZ)$ is a $D$-independent extension of
$\operatorname{tp}(X/Z)$.

The following facts, and the technique to prove the following lemma, are
available in [3], Section 3.

######  Fact 4.4.

The relation $\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}^{D}$ has the
following properties.

  1. 1.

$D$-symmetry. If $A\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}_{B}^{D}C$,
then $C\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}_{B}^{D}A$.

  2. 2.

$D$-transitivity. $A\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}_{B}^{D}C$ and
$A\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}_{BC}^{D}E$ if
and only if $A\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}_{B}^{D}CE$.

  3. 3.

$D$-local character. For any $A$ and $X$ there exists a countable set
$X_{0}\subseteq X$ such that $A\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}_{X_{0}}^{D}X$.
In the case of a rational $\alpha$, $X_{0}$ is finite.

  4. 4.

$D$-existence. For all sets $X,Y,Z$ with $Y\subseteq Z$ there exists
$X^{{}^{\prime}}$ such that
$\operatorname{tp}(X/Y)=\operatorname{tp}(X^{{}^{\prime}}/Y)$ and
$X^{{}^{\prime}}\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}_{Y}^{D}Z$.

######  Fact 4.5.

By $D$-existence, it is possible to find a $D$-Morley sequence for a given
type $p(\bar{x})\in S_{|\bar{x}|}(M)$; that is a sequence
$(\bar{b}_{i})_{i<\omega}$ of realizations of $p(\bar{x})$ such that
$\bar{b}_{0},\ldots,\bar{b}_{n}\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}_{M}^{D}\bar{b}_{n+1}$
for each $n<\omega$.

######  Lemma 4.6.

Let $Z\leqslant\mathbb{M}$ and $X,Y\subseteq\mathbb{M}$. Then
$X\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}_{Z}^{D}Y$ if
and only if

  1. 1.

$\operatorname{cl}(XZ)\cap\operatorname{cl}(YZ)=Z$,

  2. 2.

$\operatorname{cl}(XY)=\operatorname{cl}(XZ)\cup\operatorname{cl}(YZ)$,

  3. 3.

$\operatorname{cl}(XZ)\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}_{Z}^{\dim}\operatorname{cl}(YZ)$.

The above three conditions imply that
$\operatorname{cl}(XY)=\operatorname{cl}(XZ)\oplus_{Z}\operatorname{cl}(YZ)$.

######  Lemma 4.7.

Suppose that $q(\bar{x})\in S(\mathfrak{C})$ is a coheir extension of
$p(\bar{x})\in S(\mathbb{M})$. Then $q(\bar{x})$ is a $D$-independent
extension of $p(\bar{x})$.

###### Proof.

For $\bar{a}\models q(\bar{x})$, we need to show that $\bar{a}\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}_{\mathbb{M}}^{D}\mathfrak{C}$.
Take a small set $\mathbb{N}$ with
$\mathbb{M}\leqslant\mathbb{N}\leqslant\mathfrak{C}$ and
$D(\bar{a}/\mathbb{N})=D(\bar{a}/\mathfrak{C})$. By $D$-existence and
$D$-transitivity (Fact 4.4) we can also find $\mathbb{N}^{{}^{\prime}}$ such
that $\mathbb{N}^{{}^{\prime}}\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}_{\mathbb{M}}^{D}\mathbb{N}$
and $\mathbb{N}^{{}^{\prime}}\equiv_{\mathbb{M}}\mathbb{N}$.

##### Claim.

$D(\bar{a}/\mathbb{N}^{{}^{\prime}})=D(\bar{a}/\mathbb{N})$.
Proof of claim. Otherwise, assume that
$D(\bar{a}/\mathbb{N}^{{}^{\prime}})=\gamma>D(\bar{a}/\mathbb{N})$. So we can
find $\bar{b}\subseteq_{\text{fin}}\operatorname{cl}(\bar{a}\mathbb{N})$ and
$\bar{e}\subseteq_{\text{fin}}\mathbb{N}$ such that
$\delta(\bar{b}/\bar{e})<\gamma$. By Fact 2.1 there is
$\varphi(\bar{z},\bar{x},\bar{e})\in\operatorname{tp}(\bar{b}\bar{a}/\bar{e})$
such that for any $\bar{b}^{{}^{\prime}}$ and $\bar{e}^{{}^{\prime}}$ if
$\mathfrak{C}\models\varphi(\bar{b}^{{}^{\prime}},\bar{a},\bar{e}^{{}^{\prime}})$
then
$\bar{b}^{{}^{\prime}}\subseteq\operatorname{cl}(\bar{a}\bar{e}^{{}^{\prime}})$
and $\delta(\bar{b}^{{}^{\prime}}/\bar{a}\bar{e}^{{}^{\prime}})<\gamma$. So
the formula $\exists\bar{z}\hskip 5.69054pt\varphi(\bar{z},\bar{x},\bar{e})$
is in $\operatorname{tp}(\bar{a}/\mathfrak{C})$. But since
$\operatorname{tp}_{\mathcal{L}}(\bar{a}/\mathfrak{C})$ is a coheir extension
of $\operatorname{tp}_{\mathcal{L}}(\bar{a}/\mathbb{M})$, it is also
$\mathbb{M}$-invariant. So there is
$\bar{e}^{{}^{\prime}}\subseteq\mathbb{N}^{{}^{\prime}}$ such that
$\exists\bar{z}\hskip
5.69054pt\varphi(\bar{z},\bar{x},\bar{e}^{{}^{\prime}})\in\operatorname{tp}_{\mathcal{L}}(\bar{a}/\mathfrak{C})$.
Therefore $\delta(\bar{b}^{{}^{\prime}}/\bar{a}\bar{e}^{{}^{\prime}})<\gamma$
for some $\bar{b}^{{}^{\prime}}\subseteq\operatorname{cl}(\bar{a}\mathbb{N})$.
This implies that $D(\bar{a}/\mathbb{N}^{{}^{\prime}})<\gamma$, a
contradiction.
∎(Proof of claim)
Writing the above claim as $\bar{a}\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}^{D}_{\mathbb{N}^{\prime}}\mathbb{N}$
together with our assumption that $\mathbb{N}^{\prime}\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}^{D}_{\mathbb{M}}\mathbb{N}$
and $D$-transitivity gives $\mathbb{N}^{\prime}\bar{a}\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}^{D}_{\mathbb{M}}\mathbb{N}$.
Therefore again by $D$-transitivity we have $\bar{a}\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}^{D}_{\mathbb{M}}\mathfrak{\mathbb{N}}$,
and hence $\bar{a}\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}^{D}_{\mathbb{M}}\mathfrak{C}$
because $D(\bar{a}/\mathfrak{C})=D(\bar{a}/\mathbb{N})$.

Finally we have to show that
$\operatorname{cl}(\bar{a}\mathbb{M})\cap\mathfrak{C}=\mathbb{M}$. Let
$e\in\operatorname{cl}(\bar{a}\mathbb{M})\cap\mathfrak{C}$. By Remark 2.17
there exists an algebraic $\mathcal{L}_{p}$-formula
$\varphi(\bar{x},e,\bar{m})\in q(\bar{x})$ witnessing
$e\in\operatorname{cl}(\bar{a}\mathbb{M})\cap\mathfrak{C}$. If
$\operatorname{tp}(e/\mathbb{M})$ has infinitely many realizations
$(e_{i})_{i<\omega}$ in $\mathfrak{C}$ then as $q(\bar{x})$ is a coheir
extension of $p(\bar{x})$, the formula $\varphi(\bar{a},e_{i},\bar{m})$ holds
for each $i<\omega$, which contradicts the fact that
$e\in\operatorname{acl}(\bar{a}\mathbb{M})$. Therefore
$\operatorname{tp}(e/\mathbb{M})$ is algebraic and hence $e\in\mathbb{M}$. ∎
Recall that coheir extensions are in particular non-forking. In the following
proposition using a similar idea as in [14] Lemma 4.3, we address the fact
that when $\alpha$ is rational, the above lemma holds even for non-dividing
extensions. However we would not require this in the rest.

######  Proposition 4.8.

($\alpha$ rational) Assume that $\mathbb{M},\mathbb{N}$ are models of
$\mathbb{T}_{\alpha}$ such that
$\mathbb{M}\leqslant\mathbb{N}\leqslant\mathfrak{C}$. If
$\operatorname{tp}(\bar{a}/\mathbb{N})$ is a non-dividing extension of
$\operatorname{tp}(\bar{a}/\mathbb{M})$, then $\bar{a}\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}_{\mathbb{M}}^{D}\mathbb{N}$.

###### Proof.

We first show that $D(\bar{a}/\mathbb{M})=D(\bar{a}/\mathbb{N})$. Assume on
the contrary that $D(\bar{a}/\mathbb{M})>D(\bar{a}/\mathbb{N})$. So there is
$\bar{b}\subseteq\mathbb{N}\setminus\mathbb{M}$ such that
$D(\bar{a}/\mathbb{M})>D(\bar{a}/\mathbb{M}\bar{b})$. Assume that
$(\bar{b}_{i})_{i<\omega}$ is a Morley sequence of
$\operatorname{tp}(\bar{b}/\mathbb{M})$ with respect to $\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}_{\mathbb{M}}^{D}$,
as in Fact 4.5. Clearly
$D(\bar{a}/\mathbb{M})>D(\bar{a}/\mathbb{M}\bar{b_{i}})$ for each $i<\omega$.
Now by the $D$-symmetry
$D(\bar{b}_{i}/\mathbb{M})>D(\bar{b}_{i}/\mathbb{M}\bar{a})$. On the other
hand,

$\displaystyle
D(\bar{b}_{i}/\mathbb{M}\bar{b}_{0},\ldots,\bar{b}_{i-1})\geqslant
D(\bar{b}_{i}/\mathbb{M}\bar{a}\bar{b}_{0},\ldots,\bar{b}_{i-1}).$ (1)

Now if equality occurred in (1) then we would have

$\displaystyle D(\bar{b}_{i}/\mathbb{M})$
$\displaystyle=D(\bar{b}_{i}/\mathbb{M}\bar{b}_{0},\ldots,\bar{b}_{i-1})$
$\displaystyle=D(\bar{b}_{i}/\mathbb{M}\bar{a}\bar{b}_{0},\ldots,\bar{b}_{i-1})$
$\displaystyle\leqslant D(\bar{b}_{i}/\mathbb{M}\bar{a})$

which is a contradiction. So we have

$\displaystyle
D(\bar{b}_{i}/\mathbb{M}\bar{b}_{0},\ldots,\bar{b}_{i-1})>D(\bar{b}_{i}/\mathbb{M}\bar{a}\bar{b}_{0},\ldots,\bar{b}_{i-1}).$

Writing the above as $\bar{b}_{i}\not\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}_{\mathbb{M}\bar{b}_{0},\ldots,\bar{b}_{i-1}}^{D}\bar{a}$,
again by the $D$-symmetry we have $\bar{a}\not\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}_{\mathbb{M}\bar{b}_{0},\ldots,\bar{b}_{i-1}}^{D}\bar{b}_{i}$.
That is

$\displaystyle
D(\bar{a}/\mathbb{M}\bar{b}_{0},\ldots,\bar{b}_{i-1})>D(\bar{a}/\mathbb{M}\bar{b}_{0},\ldots,\bar{b}_{i}),$

for each $i<\omega$, which is a contradiction, because in the case of a
rational $\alpha$ we cannot have a decreasing sequence of $D$-dimensions (see
Fact 4.2).

Now the fact that
$\operatorname{cl}(\bar{a}\mathbb{M})\cap\mathbb{N}=\mathbb{M}$ follows from a
similar argument to the proof of Lemma 4.7. ∎

We now get to the point to prove that the dependence is transferred from $T$
to $\mathbb{T}_{\alpha}$.

######  Theorem 4.9.

If $T$ is dependent, then so is $\mathbb{T}_{\alpha}$.

###### Proof.

Let $\mathbb{M}$ and $\mathbb{N}$ be models of $\mathbb{T}_{\alpha}$ such that
$\mathbb{M}\leqslant\mathbb{N}\leqslant\mathfrak{C}$. Assume that
$q_{1}(x),q_{2}(x)\in S_{1}(\mathbb{N})$ are two coheir extensions of a type
$p(x)\in S_{1}(\mathbb{M})$. If $a_{1}\models q_{1}(x)$ and $a_{2}\models
q_{2}(x)$ then by Corollary 3.20 there is an $\mathcal{L}_{p}$-isomorphism
between $\operatorname{cl}(a_{1}\mathbb{M})$ and
$\operatorname{cl}(a_{2}\mathbb{M})$ mapping $a_{1}$ to $a_{2}$ and fixing
$\mathbb{M}$ pointwise, and therefore
$\operatorname{tp}(\operatorname{cl}(a_{1}\mathbb{M})/\mathbb{M})=\operatorname{tp}(\operatorname{cl}(a_{2}\mathbb{M})/\mathbb{M})$.
Since $\operatorname{cl}(a_{i}\mathbb{M})$ is contained in the
$\mathcal{L}_{p}$-algebraic closure of $a_{i}\mathbb{M}$, it follows that
$\operatorname{tp}(\operatorname{cl}(a_{i}\mathbb{M})/\mathbb{N})$ is a coheir
of $\operatorname{tp}(\operatorname{cl}(a_{i}\mathbb{M})/\mathbb{M})$. Thus by
Lemma 4.6 and Lemma 4.7 we have
$\operatorname{cl}(a_{i}\mathbb{N})=\operatorname{cl}(a_{i}\mathbb{M})\oplus_{\mathbb{M}}\mathbb{N}$.
This means that $q_{1}(x)=q_{2}(x)$ if and only if there is an
$\mathcal{L}$-isomorphism between $\operatorname{cl}(a_{1}\mathbb{M})$ and
$\operatorname{cl}(a_{2}\mathbb{M})$ mapping $a_{1}$ to $a_{2}$ and fixing
$\mathbb{N}$ pointwise. In other words, $q_{1}(x)=q_{2}(x)$ if and only if
$\operatorname{tp}_{\mathcal{L}}(\operatorname{cl}(a_{1}\mathbb{M})/\mathbb{N})=\operatorname{tp}_{\mathcal{L}}(\operatorname{cl}(a_{2}\mathbb{M})/\mathbb{N})$.

Now since $T$ is dependent, there are at most $2^{|\mathbb{M}|}$
$\mathcal{L}$-coheirs of
$\operatorname{tp}_{\mathcal{L}}(\operatorname{cl}(a_{1}\mathbb{M})/\mathbb{M})$
over $\mathbb{N}$. On the other hand by the above argument any coheir
extension of $p(x)$ is uniquely determined by
$\operatorname{tp}_{\mathcal{L}}(\operatorname{cl}(a\mathbb{M})/\mathbb{N})$
for any realization $a$ of $p(x)$. Therefore there are at most
$2^{|\mathbb{M}|}$ coheirs of $p(x)$. ∎

By the proof of the theorem above, the number of coheirs is the same in $T$
and $\mathbb{T}_{\alpha}$. This gives also the following corollary on the
stability.

######  Corollary 4.10.

If $T$ is stable, then so is $\mathbb{T}_{\alpha}$.

###### Proof.

If $T$ is stable, then for any
$\mathbb{M}\leqslant\mathbb{N}\models\mathbb{T}_{\alpha}$ every $1$-type over
$\mathbb{M}$ has a unique coheir extension to $\mathbb{N}$, and hence
$\mathbb{T}_{\alpha}$ is also stable. ∎

Our next result is that when $\alpha$ is rational, then the strong dependence
also transfers from $T$ to $\mathbb{T}_{\alpha}$. Before stating the result,
let us recall some basic definitions concerning the strong dependence.

Let $T^{{}^{\prime}}$ be an $\mathcal{L}^{{}^{\prime}}$-theory and $X$ an
index set. Let $I_{t}=\\{\bar{a}_{ti}:\hskip 5.69054pti<\omega\\}$, for $t\in
X$, be sequences of tuples in a monster model of $T^{{}^{\prime}}$ such that
$|\bar{a}_{ti}|=|\bar{a}_{sj}|$, for each $i,j\in\omega$ and $t,s\in X$. For a
small set of parameters $A$, we say that sequences $\\{I_{t}:\hskip
5.69054ptt\in X\\}$ are  mutually indiscernible over $A$ if each sequence
$I_{t}$ is indiscernible over $A\cup\bigcup\\{I_{s}:\hskip 5.69054pts\neq
t\\}$.

Let $\Sigma(\bar{x})$ be a partial type over $A$ and $\kappa$ be a (finite or
infinite) cardinal. We say that dp-rk$(\Sigma,A)<\kappa$ if for each family
$\\{I_{t}:\hskip 5.69054ptt<\kappa\\}$ of mutually indiscernible sequences
over $A$ and for any realization $\bar{b}$ of $\Sigma$ , there is $t<\kappa$
such that $I_{t}$ is indiscernible over $A\bar{b}$.

A dependent theory $T^{{}^{\prime}}$ is strongly dependent if for $\bar{x}$ of
any length, dp-rk$(\bar{x}=\bar{x},\emptyset)<\aleph_{0}$ (for an account on
strong-dependence see [23]).

The following fact states some equivalent forms of strong dependence which are
use for 4.13.

######  Fact 4.11 ( [12, 22]).

The following are equivalent for a complete theory $T^{{}^{\prime}}$.

  1. 1.

$T^{{}^{\prime}}$ is strongly dependent.

  2. 2.

For every indiscernible sequence $I=\\{\bar{b}_{i}:\hskip
8.53581pti<\kappa\\}$ with tuples $\bar{b}_{i}$ at most countable, and every
finite set $C$, there is a convex equivalence relation $\sim$ on $\kappa$ with
finitely many classes and such that $\operatorname{tp}(\bar{b}_{i}/C)$ depends
only on the $\sim$-class of $i$. In other words, the set
$\\{\operatorname{tp}(\bar{b}_{i}/C):\hskip 8.53581pti\in\kappa\\}$ is finite
and for each $i\in\kappa$ the set $\kappa(i)=\\{j:\hskip
8.53581pt\operatorname{tp}(\bar{b}_{j}/C)=\operatorname{tp}(\bar{b}_{i}/C)\\}$
is a convex subset of $\kappa$.

  3. 3.

For every indiscernible sequence $\\{\bar{b}_{i}:\hskip 8.53581pti\in I\\}$
with tuples $\bar{b}_{i}$ at most countable, and every finite set $C$, there
is a convex equivalence relation $\sim$ on $I$ with finitely many classes and
such that $\\{\bar{b}_{i}:\hskip 8.53581pti\in j/\sim\\}$ is indiscernible
over $C$. In other words, in the above item the sequence
$I_{i}=\\{\bar{b}_{j}:\hskip 8.53581ptj\in\kappa(i)\\}$ is indiscernible over
$C$.

######  Lemma 4.12.

Let $I=\\{\bar{a}_{i}:\hskip 8.53581pti<\kappa\\}$ be an
$\mathcal{L}_{p}$-indiscernible sequence over $A$. Then

  1. 1.

$I$ is $\mathcal{L}_{p}$-indiscernible also over $\operatorname{cl}(A)$.

  2. 2.

There is a suitable enumeration $\bar{b}_{i}=\bar{a}_{i}\bar{c}_{i}$ of each
$\langle\operatorname{cl}(A\bar{a}_{i})\rangle$ such that
$I^{{}^{\prime}}=\\{\bar{b}_{i}:\hskip 8.53581pti<\kappa\\}$ is an
$\mathcal{L}_{p}$-indiscernible sequence over $A$.

###### Proof.

1\. Let $\\{\bar{c}_{i}:\hskip 8.53581pti<\kappa\\}$ be an
$\mathcal{L}_{p}$-indiscernible sequence over $\operatorname{cl}(A)$ that
realizes the $EM$-type of $I$. This sequence has the same type over $A$ as the
type of $I$, and therefore, there is an $\mathcal{L}_{p}$-automorphism
$\sigma:\mathfrak{C}\to\mathfrak{C}$ fixing $A$ pointwise such that
$\sigma(\bar{c}_{i})=\bar{a}_{i}$, for each $i<\kappa$. So,
$\\{\bar{a}_{i}:\hskip 8.53581pti<\kappa\\}$ is
$\mathcal{L}_{p}$-indiscernible over
$\sigma(\operatorname{cl}(A))=\operatorname{cl}(A)$.

2\. As
$\langle\operatorname{cl}(A\bar{a}_{i})\rangle\subseteq\operatorname{dcl}(\operatorname{cl}(A\bar{a}_{i}))$,
it is enough to prove 2 for the sequence
$I^{\prime}=\\{\operatorname{cl}(A\bar{a}_{i}):\hskip 8.53581pti<\kappa\\}$.
Since $\bar{a}_{i}$ and $\bar{a}_{0}$ have the same type over $A$, enumerate
each $\operatorname{cl}(A\bar{a}_{i})$ in accordance with
$\operatorname{cl}(A\bar{a}_{0})$ using the $\mathcal{L}_{p}$-automorphism
$\sigma_{i}$ that fixes $A$ pointwise and sends $\bar{a}_{0}$ to
$\bar{a}_{i}$. Denote this enumeration by $\bar{d}_{i}$ and put
$J=\\{\bar{d}_{i}:\hskip 8.53581pti<\kappa\\}$. By standard application of
Ramsey’s theorem, let $I^{\prime}=\\{\bar{b}_{i}:\hskip 8.53581pti<\kappa\\}$
be the $A$-indiscernible sequence satisfying the $EM$-type of $J$ over $A$.
Since each $\bar{b}_{i}$ contains an automorphic image of $\bar{a}_{i}$, by
strong homogeneity of $\mathfrak{C}$, we may assume that each $\bar{b}_{i}$
includes $\bar{a}_{i}$.

We now need to show that $\bar{b}_{i}$ is indeed an enumeration of
$\operatorname{cl}(A\bar{a}_{i})$. Recall that $\bar{d}_{i}$ is already an
enumeration of $\operatorname{cl}(A\bar{a}_{i})$. Let $A_{0}$ be a finite
subset of $A$ and $(s_{1},\dots,s_{l})$ be a finite subtuple of $\bar{a}_{0}$.
Then $\operatorname{cl}^{n}(A_{0}s_{1},\dots,s_{l})=\\{e_{1},\dots,e_{k}\\}$
is a finite subtuple of $\bar{d}_{0}$. Thus,
$|\operatorname{cl}^{n}(A_{0},\sigma_{i}(s_{1}),\dots,\sigma_{i}(s_{l}))|=k$,
for each $i<\kappa$. Now the following two facts are covered by the $EM$-type
of $J$. First that for each $i<\kappa$, $\bar{b}_{i}$ includes
$\operatorname{cl}(A\bar{a}_{i})$. Second, that for any finite subset $A_{0}$
of $A$ and finite subtuple $(t_{1},\dots,t_{l})$ of $\bar{a}_{i}$, there is a
finite subtuple $(f_{1},\dots,f_{k})$ of $\bar{b}_{i}$ such that
$\operatorname{cl}^{n}(A_{0}t_{1},\dots,t_{l})=\\{f_{1},\dots,f_{k}\\}$. Hence
$\bar{b}_{i}$ is an enumeration of $\operatorname{cl}(A\bar{a}_{i})$, as
required. ∎

######  Theorem 4.13.

($\alpha$ rational) If $T$ is strongly dependent, then so is
$\mathbb{T}_{\alpha}$.

###### Proof.

Let $I=\\{\bar{b}_{i}:\hskip 8.53581pti<\kappa\\}$ be an indiscernible
sequence and $C$ a finite set, as in 4.11. By $D$-local character and since
$\alpha$ is rational, there exists a finite set $H\subseteq I$ such that
$C\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}^{D}_{H}I$,
and in particular $C\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}^{D}_{H}\bar{b}_{i}$
for each $i$. So by 4.6 for every $i<\kappa$,
$\langle\operatorname{cl}(\bar{b}_{i}CH)\rangle=\langle\operatorname{cl}(\bar{b}_{i}H)\rangle\oplus_{\langle\operatorname{cl}(H)\rangle}\langle\operatorname{cl}(CH)\rangle$.
Hence by 3.20, $\operatorname{tp}(\bar{b}_{i}/\operatorname{cl}(CH))$ depends
only on $\operatorname{tp}(\bar{b}_{i}/\operatorname{cl}(H))$ and
$\operatorname{tp}_{\mathcal{L}}(\langle\operatorname{cl}(\bar{b}_{i}H)\rangle/\operatorname{cl}(HC))$.
We will find a convex partition of $\kappa$ in each part of which these two
are fixed, and the statement follows from 4.11.

Since $H$ is a finite subset of $I$ and $I$ is
$\mathcal{L}_{p}$-indiscernible, there is a finite partition
$J_{1},\dots,J_{k}$ of $\kappa$ into convex sets such that each
$I_{j}=\\{\bar{b}_{i}:\hskip 8.53581pti\in J_{j}\\}$ is indiscernible over
$H$. So in particular $\operatorname{tp}(\bar{b}_{i}/H)$–determined by
$\langle\operatorname{cl}(\bar{b}_{i}H)\rangle$– is fixed for each $i\in
J_{j}$ and $1\leqslant j\leqslant k$. We may assume that each $J_{j}$ is
infinite, as otherwise, the set $I_{j}$ can be added to $H$.

Since $I_{j}$ is indiscernible over $H$, by 4.12 it is also indiscernible over
$\operatorname{cl}(H)$. Moreover again by 4.12 we may assume that
$I^{{}^{\prime}}_{j}=\\{\langle\operatorname{cl}(\bar{b}_{i}H)\rangle:\hskip
8.53581pti\in J_{j}\\}$ is indiscernible over $H$.

Now since $T$ is strongly dependent, by 4.11, there exists a convex
equivalence relation $\sim_{j}$ over $J_{j}$ with finitely many equivalence
classes such that $\\{\langle\operatorname{cl}(\bar{b}_{i}H)\rangle:\hskip
8.53581pti\in\alpha/\sim_{j}\\}$ is $\mathcal{L}$-indiscernible over
$\operatorname{cl}(CH)$. So, in particular,
$\operatorname{tp}_{\mathcal{L}}(\langle\operatorname{cl}(\bar{b}_{i}H)\rangle/\operatorname{cl}(CH))$
is fixed on any equivalence class of $\sim_{j}$. Hence
$\sim=\bigcup_{j=1}^{k}\sim_{j}$ forms a convex equivalence relation on
$\kappa$ with finitely many equivalence classes such that
$\operatorname{tp}(\bar{b}_{i}/\operatorname{cl}(HC))$ (and hence
$\operatorname{tp}(\bar{b}_{i}/C)$) depends only on the $\sim$-class of $i$. ∎

We will show in the following that in the particular case that $T$ defines a
linear order, $\mathbb{T}_{\alpha}$ is not strongly dependent for any
irrational $\alpha$.

######  Theorem 4.14.

Assume that $T$ defines a linear order and $\alpha$ is irrational. Then
$\mathbb{T}_{\alpha}$ is not strongly dependent.

###### Proof.

Let $<^{\mathfrak{C}}$ be the interpretation of the order defined by $T$. We
construct $\mathcal{L}_{p}$-mutually indiscernible sequences $\\{I_{i}:\hskip
5.69054pti<\omega\\}$ and find $b_{0}\in\mathfrak{C}$ such that
$I_{i}=\\{a_{ij}:\ j<\omega\\}$ is not $\mathcal{L}_{p}$-indiscernible over
$b_{0}$ for each $i<\omega$.

Let $J=\\{c_{ij}:\ (i,j)\in\omega\times\omega\\}$ be an
$\mathcal{L}$-indiscernible sequence over $\emptyset$, ordered by
$\omega\times\omega$ with horizontal lexicographic order. For convenience, we
may assume the sequence $J$ is increasing with respect to $<^{\mathfrak{C}}$.
We now inductively define a sequence $(D_{n},E_{n},F_{n})_{n<\omega}$ of
subsets of $\mathfrak{C}$ and a sequence $(s_{n+1},k_{n+1})_{n<\omega}$ of
tuples of natural numbers which, in particular, have the following properties:

  1. 1.

$E_{n}\subseteq J_{n}=\\{c_{nj}:j\in\omega\\}$ and $|E_{n+1}|=s_{n+1}$.

  2. 2.

$|F_{n+1}|=k_{n+1}-s_{n+1}$ and
$F_{n+1}\subseteq\operatorname{acl}(D_{n}E_{n+1})$. Furthermore each
$s_{n+1}$-element subset of $E_{n+1}F_{n+1}$ is a basis over $D_{n}$.

  3. 3.

$D_{0}\subseteq D_{1}\subseteq\dots\subseteq D_{n}$,
$D_{n}\subseteq\operatorname{acl}(\bigcup_{i=0}^{n}E_{i})$ and $D_{n}\cap
J=\bigcup_{i=0}^{n}E_{i}$.

For $n=0$, let $E_{0}=D_{0}=\\{c_{01}\\}$ and $F_{0}=\emptyset$. Suppose by
induction that the sets $(D_{n},E_{n},F_{n})$ are constructed and $D_{n}$
satisfies condition $3$ above. By Dirichlet’s rational approximation theorem
there is a pair of natural numbers $s_{n+1},k_{n+1}$ with
$-(1-\alpha)/2^{n+1}<s_{n+1}-\alpha k_{n+1}<0$. Let $E_{n+1}\subseteq J_{n+1}$
be the set $\\{c_{(n+1)1},c_{(n+1)2},\ldots,c_{(n+1)s_{n}}\\}$. Since $J$ is
an indiscernible sequence over $\emptyset$, by Fact 2.3, $E_{n+1}$ is
$\dim$-independent from $D_{n}$ over $\emptyset$. Hence
$\dim(E_{n+1}/D_{n})=s_{n+1}$. By indecomposablity of $T$ let $F_{n+1}$ be the
set consisting of $k_{n+1}-s_{n+1}$ elements in
$\operatorname{acl}(E_{n+1}D_{n})$, with the property that each
$s_{n+1}$-element subsets of $E_{n+1}F_{n+1}$ forms a basis over $D_{n}$.
Finally set $D_{n+1}=D_{n}\cup E_{n+1}\cup F_{n+1}$. Note that
$\dim(D_{n+1}/D_{n})=s_{n}$ and $|D_{n+1}\setminus D_{n}|=k_{n}$. Furthermore,
$D_{n+1}\subseteq\operatorname{acl}(\bigcup_{i=0}^{n+1}E_{i})$.

Now set $D=\bigcup_{i=0}^{\infty}D_{i}$ and $I=J\setminus D$. Notice that
$I=J\setminus\bigcup_{i=0}^{\infty}E_{i}$. So $I$ is also an
$\mathcal{L}$-indiscernible sequence over $\emptyset$. So we can re-enumerate
$I$ and represent it by $\\{d_{ij}:\hskip
5.69054pt(i,j)\in\omega\times\omega\\}$. By the construction,
$D\subseteq\operatorname{acl}(\bigcup_{i=0}^{\infty}E_{i})$. On the other hand
since $J$ is $\mathcal{L}$-indiscernible, it follows from Fact 2.3 that
$D\mathrel{\raise
0.86108pt\hbox{\ooalign{$|$\cr\raise-3.87495pt\hbox{$\smile$}}}}^{dim}_{\emptyset}I$
and
$\operatorname{acl}(I)\cap\operatorname{acl}(D)=\operatorname{acl}(\emptyset)$.

Let $H=\langle D\cup I\rangle$. Then $H=\langle
D\oplus_{\emptyset}I\rangle=\langle D\rangle\oplus_{\emptyset}\langle
I\rangle$. Now we recolor $H$ by letting only the elements in $D$ colored and
the rest non-colored. Hence the following conditions are fulfilled.

  * •

For each $n$, $D_{n}\in\mathcal{K}_{\alpha}^{+}$, and hence
$D\in\mathcal{K}_{\alpha}^{+}$.

  * •

$I,H\in\mathcal{K}_{\alpha}^{+}$.

  * •

Every finite subset of $I$ as well as $I$ itself are closed in $H$.

  * •

$(D_{n},D_{n+1})$ is a minimal pair, for each $n\in\omega$. Hence
$\operatorname{cl}_{H}(c_{01})=D$.

Suppose that $f:H\to\mathfrak{C}$ is a strong embedding and $b_{n}=f(c_{n1})$
and $a_{ij}=f(d_{ij})$ for each $i,j,n\in\omega$. Then it is clear that
$I_{i}=\\{a_{ij}:\hskip 5.69054ptj\in\omega\\}$ are $\mathcal{L}_{p}$-mutually
indiscernible sequences over $\emptyset$. Furthermore
$a_{n0}<^{\mathfrak{C}}b_{n}<^{\mathfrak{C}}a_{n1}$ for each $n\in\omega$. We
claim that non of the $I_{i}$ is $\mathcal{L}_{p}$-indiscernible over $b_{0}$
for each $i\in\omega$.

Assume for a contradiction, that $I_{i}$ is indiscernible over $b_{0}$, for
some $i$ . Note that since
$b_{i}\in\operatorname{cl}(b_{0})\subseteq\operatorname{Acl}(b_{0})$, there
exists an algebraic $\mathcal{L}_{p}$-formula $\psi(x,b_{0})$ which is
satisfied by $b_{i}$. Let $\varphi(x,y,b_{0})$ be the formula

$\exists t\quad(x<^{\mathfrak{C}}t<^{\mathfrak{C}}y\wedge\psi(t,b_{0})).$

Then $\varphi(a_{i0},a_{i1},b_{0})$ holds. But since $I_{i}$ is indiscernible
over $b_{0}$, $\varphi(a_{ij},a_{ij+1},b_{0})$ holds for each $j\in\omega$.

But this gives infinitely many elements satisfying $\psi(x,b_{0})$ which
contradicts the algebraicity of $\psi(x,b_{0})$.

∎

######  Corollary 4.15.

If $\alpha$ is irrational then $\mathbb{ODAG}_{\alpha}$ and
$\mathbb{RCF}_{\alpha}$ are not strongly dependent.

Finally bring the paper to an end with a curious counterexample on the
distality of $\mathbb{T}_{\alpha}$ assuming $T$ itself is distal.

Recall from [24] that a dependent theory $T^{{}^{\prime}}$ is distal if for
any indiscernible sequence $I$, every set $A$, tuple $\bar{b}$ and
$A$-indiscernible sequence $I^{{}^{\prime}}=I_{1}+I_{2}$ with $I_{1}$ and
$I_{2}$ without endpoints and
EM-$\operatorname{tp}(I^{{}^{\prime}})$=EM-$\operatorname{tp}(I)$, if
$I_{1}+\bar{b}+I_{2}$ is indiscernible, then it is $A$-indiscernible.

Unlike the strong dependence we will see that distality does not transfer to
$\mathbb{T}_{\alpha}$, both for rational and irrational $\alpha$, even when we
start with as distal a theory as the o-minimal theories.

######  Example 4.16.

Let $T$ be any o-minimal expansion of ODAG. Then $\mathbb{T}_{\alpha}$ is not
distal for any $0<\alpha\leqslant 1$.
Let $A=\\{a\\}$ be a one-element set, $I^{{}^{\prime}}=I_{1}+I_{2}$ with both
$I_{1}$ and $I_{2}$ sequences of elements in $\mathfrak{C}$ ordered by
rational numbers and an element $b\in\mathfrak{C}$ such that $I_{1}+b+I_{2}$
is $\mathcal{L}$-indiscernible over $A$. Put $I=I_{1}$. It follows from the
Fact 4.5 that $I_{1}+b+I_{2}$ is a $\dim$-Morley sequence over $A$. In
particular $a+b\notin\langle I^{{}^{\prime}}A\rangle$. Now to find an
$\mathcal{L}_{p}$-indiscernible sequence which violates the distality we color
the elements of $\mathcal{P}=\langle I^{{}^{\prime}}\\{b\\}A\rangle$ by
letting $a+b$ colored and the rest non-colored. Then it follows that
$\mathcal{P}\in\mathcal{K}_{\alpha}^{+}$ and if $f:\mathcal{P}\to\mathfrak{C}$
is a strong embedding then we have that $f(I^{{}^{\prime}})$ is an
$\mathcal{L}_{p}$-indiscernible over $f(A)$ and $f(I_{1})+f(b)+f(I_{2})$ is an
$\mathcal{L}_{p}$-indiscernible but it is not the case that
$f(I_{1})+f(b)+f(I_{2})$ is an $\mathcal{L}_{p}$-indiscernible over $f(A)$.
Hence this shows that $\mathbb{T}_{\alpha}$ is not distal.

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