# Spectral spaces in o-minimal and other NIP theories

Elías Baro, José F. Fernando, and Daniel Palacín

###### Abstract.

We study some model-theoretic notions in NIP by means of spectral topology. In
the o-minimal setting we relate the o-minimal spectrum with other topological
spaces such as the real spectrum and the space of infinitesimal types of
Peterzil and Starchenko. In particular, we prove for definably compact groups
that the space of closed points is homeomorphic to the space of infinitesimal
types. We also prove that with the spectral topology the set of invariant
types concentrated in a definably compact set is a normal spectral space whose
closed points are the finitely satisfiable types.

On the other hand, for arbitrary NIP structures we equip the set of invariant
types with a new topology, called the honest topology. With this topology the
set of invariant types is a normal spectral space whose closed points are the
finitely satisfiable ones, and the natural retraction from invariant types
onto finitely satisfiable types coincides with Simon’s $F_{M}$ retraction.

###### 2020 Mathematics Subject Classification:

03C45, 03C64, 14P10

All authors are supported by Spanish STRANO PID2021-122752NB-I00 and Grupos
UCM 910444. First and second authors were also supported by STRANO
MTM2017-82105-P. The third author was also supported by the contract
2020-T1/TIC-20313 from Community of Madrid, as well as by the Deutsche
Forschungsgemeinschaft (DFG, German Research Foundation) - Project number
2100310301, part of the ANR-DFG program GeoMod.

##  1\. Introduction

Spectral spaces (see Definition 2.2) constitute a topological framework
related with several areas of mathematics, specially with algebraic geometry
via the Zariski spectrum of a ring (and afterwards, following the abstract
development of Grothendieck). In the late 1970s, the real spectra of a ring
was introduced by Coste and Roy in their foundational paper [5] as a
substitute in real algebraic geometry for the Zariski spectra of a ring in
ordinary algebraic geometry. Recall that if $R$ is a real closed field,
$V\subset R^{n}$ is a real algebraic set and $\mathcal{R}(V)$ denotes the ring
of regular functions on $V$, then the real spectra
$\text{Sper}(\mathcal{R}(V))$ is a spectral space that reflects the
semialgebraic properties of $V$. For example, there is a natural injection of
$V$ into $\text{Sper}(\mathcal{R}(V))$, which is continuous with respect to
the Euclidean topology of $V$. Moreover, we can identify canonically every
semialgebraic subset $S$ of $V$ with a constructible subset $\widetilde{S}$ of
$\text{Sper}(\mathcal{R}(V))$ such that $\widetilde{S}\cap V=S$ (see [3, Prop.
7.2.2]). The space $\text{Sper}(\mathcal{R}(V))$ is an object whose (standard)
topological properties yield information about (semialgebraic) topological
properties of $V$. For example, a semialgebraic subset $S$ of $V$ is
semialgebraically connected if and only if $\widetilde{S}$ is connected. Or
more notably, it is possible to construct a (standard) sheaf theory of rings
on $\text{Sper}(\mathcal{R}(V))$ whose sections are the semialgebraic
continuous functions. In general, the space $\text{Sper}(\mathcal{R}(V))$ is
quasi-compact but it is not Hausdorff. Nonetheless, since it is normal, the
subset of its closed points constitutes a quasi-compact Hausdorff subspace
(see Fact 2.6).

Spectral spaces within model theory were first considered by Pillay [22], who
introduced the so-called o-minimal spectra of definable sets. Before
proceeding, we first recall some basic model-theoretic concepts and fix some
notation. Given a first order structure $\mathcal{M}$ in a language
$\mathcal{L}$ and an $\mathcal{L}$-formula $\phi(x)$, possibly with
parameters, we denote by $\phi(M)=\\{a\in M^{n}:\phi(a)\text{ holds}\\}$ the
definable subset of $M^{n}$ given by $\phi(x)$, where $x$ is an $n$-tuple of
variables. We will say that $\phi(x)$ is an $\mathcal{L}_{M}$-formula, denoted
by $\phi(x)\in\mathcal{L}_{M}$, when the parameters are from $M$. We shall not
distinguish between a definable set and a formula defining it. Notice that the
family of definable subsets of $M^{n}$ constitutes a Boolean algebra and thus
it has an associated set of ultrafilters. This set is called the set of
complete $n$-types over $M$ and it is denoted by $S_{n}(\mathcal{M})$. One can
turn $S_{n}(\mathcal{M})$ into a topological space by considering the Stone
topology, where a base of open sets is given by the sets of the form
$[Z]:=\\{p\in S_{n}(\mathcal{M}):Z\in p\\}$ for $Z\subset M^{n}$ definable.
Given a definable set $X\subset M^{n}$, we denote the set of $n$-types
concentrated in $X$ by $S_{X}(\mathcal{M}):=[X]$, which with the induced Stone
topology is a quasi-compact Hausdorff and totally disconnected space that
contains a copy of $X$. This compactification is the natural one from a model-
theoretic point of view. However, in o-minimal expansions of real closed
fields Pillay [22] consider a thicker topology. Recall that an
$\mathcal{L}$-structure $\mathcal{M}$ expanding a linear order is _o-minimal_
if every definable subset of $M$ is a finite union of points and intervals
with end points in $M\cup\\{\pm\infty\\}$.

The versatility of o-minimality has recently led to spectacular applications
in several areas ranging from diophantine geometry, non-archimedean geometry,
number theory and additive combinatorics. Real algebraic geometry has
(partially) served as a guide in the development of o-minimal geometry.
Influenced by the real spectra, in [22] the set $S_{X}(\mathcal{M})$ is
equipped with a spectral topology whose basic open sets are the sets $[U]$
with $U\subset X$ open definable. If the o-minimal structure under
consideration is an expansion $\mathcal{R}$ of a real closed field then
$S_{X}(\mathcal{R})$ equipped with the spectral topology is a normal spectral
space that we denote by $S^{\mathrm{t}}_{X}(\mathcal{R})$ and called it _o-
minimal spectrum_. In particular, the set of closed points
$\beta_{X}(\mathcal{R})$ is a quasi-compact and Hausdorff space and there is a
natural continuous retraction

$r:S^{\mathrm{t}}_{X}(\mathcal{R})\rightarrow\beta_{X}(\mathcal{R}).$

The space $S^{\mathrm{t}}_{X}(\mathcal{R})$ has been considered by several
authors mainly as a tool to develop sheaf theories in the o-minimal setting
[22, 10, 2, 11]. In [28], Tressl also considers
$S^{\mathrm{t}}_{X}(\mathcal{R})$ and relates it with the real spectra of the
ring of continuous definable functions. On the other hand, the main theme of
the unpublished paper [15] is the relationship between the specialization
order and the Rudin-Keisler order.

The purpose of this paper is two-fold. First, we relate the o-minimal spectrum
with some relevant model-theoretic notions. The main results concerning this
part are the following. Let $\mathcal{R}$ be an o-minimal structure expanding
a real closed field, and let $G$ be a definable group. In [19], to study types
modulo infinitesimals, Peterzil and Starchenko introduce a compactification of
$G$ called $S^{\mu}_{G}(\mathcal{R})$ which is related with the Samuel
compactification of a topological group. Their motivation to consider this
compactification is to obtain definable subgroups of $G$ as stabilizers of
certains elements of $S^{\mu}_{G}(\mathcal{R})$. In the paper we show:

######  Theorem (Corollary 3.14).

Let $G$ be a definably compact group. Then $\beta_{G}(\mathcal{R})$ and
$S^{\mu}_{G}(\mathcal{R})$ are canonically homeomorphic.

On the other hand, the set of invariant types has played a fundamental role in
the development of model theory. Invariant types have been intensively studied
in o-minimality [9, 25], in the dp-minimal context [26] and also in the more
general framework of NIP [17, 23]. If $\mathcal{R}$ is an o-minimal expansion
of a real closed field and $\bar{\mathcal{R}}$ is a sufficiently saturated
elementary extension of $\mathcal{R}$, given an $R$-definable subset
$X\subset\bar{R}^{n}$, we say that a type $p\in S_{X}(\bar{\mathcal{R}})$ is
$R$-invariant if for every automorphism $\sigma$ of $\bar{\mathcal{R}}$ fixing
$R$ pointwise we have that $p=\sigma(p)$, where $\sigma(p):=\\{\sigma(Y):Y\in
p\\}$. We will denote the set of $R$-invariant types as
$S_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}$. An important subset of $R$-invariant
types is the one of finitely satisfiable ones, which we will denote by
$S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}$ (see 4.1). Recall that a type $p\in
S_{X}(\bar{\mathcal{R}})$ is _finitely satisfiable_ in $R$ if for every $Y\in
p$ we have that $Y\cap R^{n}\neq\emptyset$. Both
$S_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}$ and $S_{X}(\bar{\mathcal{R}})^{\rm
fs}_{R}$ are closed subsets of $S_{X}(\mathcal{R})$ with respect to the Stone
topology. In particular, both are spectral spaces with the topology inherited
from $S_{X}^{\rm t}(\bar{\mathcal{R}})$. In Corollary 4.4 and 4.5 we show that
$S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}$ is closed in $S_{X}^{\rm
t}(\bar{\mathcal{R}})$ and the inherited topology is in fact the Stone
topology. Additionally, we define $S_{X}(\bar{\mathcal{R}})^{\rm bdd}$ as the
set of types that contains a set $\\{x\in\bar{R}^{n}:\|x\|<r\\}$ for some
positive $r\in\bar{R}$. In this framework we have:

######  Theorem (Theorem 4.8 and Corollary 4.10).

Let $X\subset\bar{R}^{n}$ be a closed definable set. Then

$S_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}\cap\beta_{X}(\bar{\mathcal{R}})\cap
S_{X}(\bar{\mathcal{R}})^{\rm bdd}=S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}.$

In particular, if $X$ is definably compact then $S_{X}(\bar{\mathcal{R}})^{\rm
inv}_{R}\cap\beta_{X}(\bar{\mathcal{R}})=S_{X}(\bar{\mathcal{R}})^{\rm
fs}_{R}$ and we have a canonical continuous retraction $r|_{S^{\rm
t}_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}}:S^{\rm t}_{X}(\bar{\mathcal{R}})^{\rm
inv}_{R}\rightarrow S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}$.

Thus, the spectral topology identifies finitely satisfiable types inside the
invariant types. It is worth noticing that the above results are based on the
following result that is interesting by itself: _given a definable subset
$Z\subset\bar{R}^{n}$ such that $Z\cap R^{n}\neq\emptyset$, there is an open
definable subset $U\subset\bar{R}^{n}$ such that $Z\subset U$ and $Z\cap
R^{n}=U\cap R^{n}$_ (_i.e._ external definable subsets of $R^{n}$ can be
defined via open subsets of $\bar{R}^{n}$). Another key ingredient in the
proof of the above result is the characterization of forking in o-minimal
structures provided by Dolich [9].

The second objective of the paper is to use the intuition gained from the
above theorem to analyse certain objects in abstract NIP contexts. Recall that
a complete first order $\mathcal{L}$-theory has the _non independence
property_ (NIP) if there is no $\mathcal{L}$-formula $\phi(x,y)$ and there is
no model $\mathcal{M}$ of the theory for which there are sequences
$(a_{i})_{i\in\mathbb{N}}$ and $(b_{I})_{I\subset\mathbb{N}}$ such that
$\phi(a_{i},b_{I})$ holds if and only if $i\in I$. NIP theories include stable
theories (as the theory of algebraically closed fields), o-minimal theories
(as the theory of real closed fields) or $P$-minimal theories (as the theory
of the $p$-adic field).

Let $T$ be a complete NIP $\mathcal{L}$-theory, let $\bar{\mathcal{M}}$ be a
saturated elementary extension of a model $\mathcal{M}$ of $T$ and let
$X\subset M^{n}$ be an $M$-definable subset. In [23], Simon constructs a
retraction

$F_{M}:S_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}\rightarrow
S_{X}(\bar{\mathcal{M}})^{\rm fs}_{M},$

which is continuous with respect to the Stone topology. As the author claims:

> _This map remains rather puzzling. It would be nice to have a better
> understanding of it, for instance a different construction leading to it._

Despite of the theorem above, the retractions $r|_{S^{\rm
t}_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}}$ and $F_{M}$ are completely unrelated
(see Example 5.2). Nevertheless, since a canonical retraction exist under the
presence of a normal spectral topology, it seems natural to ask:

_Can one endow $S_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$ with a topology so
that it becomes a normal spectral space such that the closed points of this
topology are the finitely satisfiable types and the natural retraction from
$S_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$ to the closed points
$S_{X}(\bar{\mathcal{M}})^{\rm fs}_{M}$ is exactly $F_{M}$?_

We give a positive answer to this question by considering a new topology on
$S_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$, that we call the _honest topology_.
Recall that an $\bar{M}$-definable subset $Z\subset\bar{M}^{n}$ is _honest_ if
for every $M$-definable subset $X\subset M^{n}$ with $Z\cap M^{n}\subset X$ we
have $Z\subset X(\bar{M})$, where $X(\bar{M})$ denotes the set defined in
$\bar{\mathcal{M}}$ by the formula that defines $X$. Honest sets were
introduced in [6], where it is proven assuming NIP that for every
$\bar{M}$-definable subset $Y\subset\bar{M}^{n}$, there is an
$\bar{M}$-definable honest subset $Z\subset\bar{M}^{n}$ such that $Z\cap
M^{n}=Y\cap M^{n}$.

The set of invariant types with honest topology is denoted by $S^{{\rm
h}_{M}}_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$ and a base of closed sets is
given by the sets $[Z]$ for $Z$ an honest set. We show that:

######  Theorem (Theorem 5.13).

Let $T$ be a complete NIP theory. The set $S^{{\rm
h}_{M}}_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$ is a normal spectral topological
space whose closed points are exactly the finitely satisfiable types, and the
natural retraction onto the closed points

$r_{M}^{\rm h}:S^{{\rm h}_{M}}_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}\rightarrow
S_{X}(\bar{\mathcal{M}})^{\rm fs}_{M}$

coincides with $F_{M}$.

We finish this introduction with the organization of the paper. Section 2
serves mainly as a preliminary section. In Subsection 2.1 we summarize all the
basic definitions and results concerning spectral spaces regarding $S^{\rm
t}_{X}(\mathcal{M})$. As far as possible, we work in a topological first order
structure $\mathcal{M}$, which includes for example the theory of the $p$-adic
field (and not only the o-minimal setting). All results are well-known (as
well as the results of Subsection 2.3), but in some cases we provide
alternative proofs with a model-theoretic flavor instead of using the language
of spectral spaces. In Subsection 2.2 we give a model-theoretic
characterization of the closed points $\beta_{X}(\mathcal{M})$ of the spectral
space $S^{\rm t}_{X}(\mathcal{M})$ via a universal property. In Subsection 2.3
we compute a bound of the cardinal of the closure $\textrm{cl}^{\rm t}(p)$ of
a type $p\in S^{\rm t}_{X}(\mathcal{M})$ in terms of $\dim(p)$.

In Section 3 we relate the o-minimal spectrum with two natural objects. Even
though we already know that $S_{X}(\mathcal{R})$ is not homeomorphic to the
real spectrum of the ring of continuos functions, in Subsection 3.1 we show
that their closed points are naturally homeomorphic. In Subsection 3.2 we show
the already explained relation between $S_{G}^{\rm t}(\mathcal{R})$ and
$S^{\mu}_{G}(\mathcal{R})$ (the first theorem of this introduction).

In Section 4 we handle the study of invariant and coheir types. In Subsection
4.1 we show that $S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}$ is closed in
$S_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}$ and that the induced spectral
topology in $S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}$ coincides with the Stone
one. In Subsection 4.2 we prove that if $X$ is closed then the invariant
bounded types that are closed with respect to the spectral topology are
exactly the finitely satisfiable types (the second theorem of this
introduction).

In Section 5 we analyze Simon’s retraction $F_{M}$. We introduce the honest
topology in the set of invariant types and we show that the space
$S_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$ is a normal spectral space whose set
of closed points is $S_{X}(\bar{\mathcal{M}})^{\rm fs}_{M}$ (the last theorem
of this introduction).

##  2\. The spectral topology

In this section we equip the space of types of an o-minimal expansion of a
real closed field with the so-called spectral topology as it is done in [22].
We isolate those topological properties of expansions of real closed fields
that are needed and work with topological structures with rather general
considerations. We recall the notion of topological structure introduced in
[20].

######  Definition 2.1.

A first-order structure $\mathcal{M}$ in a language $\mathcal{L}$ is a
topological structure if there is a formula $\theta(x,\bar{y})$ in
$\mathcal{L}$, with $x$ a single variable, such that the collection
$\\{\theta(M,\bar{b})\\}_{\bar{b}\in M^{|y|}}$ forms a basis of a
$\mathrm{T}_{1}$ topology without isolated points.

Let $\mathcal{M}$ denote a topological first-order $\mathcal{L}$-structure and
note that if $X$ is a definable set, then so is the closure
$\mathrm{cl}_{X}(V)$ of $V$ in $X$ for every definable subset $V$ of $X$.
Assume further that $\mathcal{M}$ satisfies the following two mild conditions.
We will justify in Subsection 2 our choice.

  1. (A1)

For every $n$, every definable subset of $M^{n}$ is a Boolean combination of
closed definable subsets of $M^{n}$.

  2. (A2)

It is completely definably normal, that is, for any disjoint definable closed
subsets $X_{1}$ and $X_{2}$ of a definable subset $X\subset M^{n}$, there are
disjoint open definable neighborhoods $U_{1}$ and $U_{2}$ in $X$ of $X_{1}$
and $X_{2}$ respectively.

In [24, Cor. 4.2] the authors prove that certain structures of dp-minimal
theories which are topological structures satisfy (A1). These include for
example dp-minimal expansions of divisible ordered abelian groups and dp-
minimal expansions of valued fields. For example, o-minimal and weakly
o-minimal theories expanding an ordered group, or P-minimal theories expanding
a ($p$-adically closed) field.

On the other hand, it is well-known that any o-minimal expansion $\mathcal{M}$
of a real closed field satisfies (A2). Indeed, note that, given a definable
set $Z\subset M^{n}$ in an o-minimal expansion $\mathcal{M}$ of a real closed
field, the distance map $\mathrm{dist}(x,Z)$ is definable, see [12, Lem.
6.3.5]. If $\mathcal{M}$ is a valued field, then we can use the valuation in a
similar way in order to prove (A2). Hence, if $\mathcal{M}$ is the $p$-adic
field or it is an o-minimal expansion of a real closed field, then
$\mathcal{M}$ satisfies both (A1) and (A2).

Regarding weakly o-minimal expansions of ordered groups, to the best of our
knowledge it is not known if they satisfy (A2):

######  Question.

_Is every weakly o-minimal expansion of an ordered group definably normal?_

We point out that if $\mathcal{M}$ is a weakly o-minimal structure expanding
an ordered group then two disjoint closed subsets of $M^{1}$ can be easily
separated by definable open sets.

###  2.1. Spectral topology

Given a definable set $X\subset M^{n}$, we denote as usual
$S_{X}(\mathcal{M})=S_{n}(\mathcal{M})\cap[X]$ the set of $n$-types over $M$
which concentrate on $X$. This set of types is naturally equipped with the
Stone topology, a quasi-compact Hausdorff totally disconnected topology.
However, observe that the topology on $\mathcal{M}$ does not play any role in
the definition of the Stone topology. Henceforth we will identify a definable
set with the formula that defines it and therefore an element of
$S_{X}(\mathcal{M})$ we will be written as an ultrafilter of sets, or as a
maximal consistent collection of formulas.

We now consider a topology on the set $S_{X}(M)$ coarser than the Stone
topology, which was introduced in [22]. A basic open subset in this new
topology is a set $[\phi]$ where $\phi(\mathcal{M})$ is an open subset of $X$.
The set $S_{X}(\mathcal{M})$ endowed with this topology will be denoted by
$S^{\mathrm{t}}_{X}(\mathcal{M})$ and the closure of a set $C$ by
$\mathrm{cl}^{\rm t}(C)$. We point out that $S^{\mathrm{t}}_{X}(\mathcal{M})$
is a _spectral space_ , a notion that we recall next (see [8]).

######  Definition 2.2.

A spectral space is a topological space $S$ that satisfies the following
conditions:

  1. (S1)

It is quasi-compact and $\mathrm{T}_{0}$.

  2. (S2)

The set of all quasi-compact open subsets forms a basis of open sets.

  3. (S3)

The intersection of two quasi-compact open subsets of $S$ is again quasi-
compact.

  4. (S4)

It is _sober_ , that is, every nonempty closed and irreducible subset of $S$
is the closure of a unique point.

The topology on $S$ is called the _spectral topology_.

We include a short proof for the sake of completeness of the following result.

######  Fact 2.3.

[22, Lem. 1.1] The topological space $S^{\mathrm{t}}_{X}(\mathcal{M})$ is
spectral.

###### Proof.

It is clear that the map $S_{X}(\mathcal{M})\to
S^{\mathrm{t}}_{X}(\mathcal{M})$ given by $p\mapsto p$ is continuous and thus
$S^{\mathrm{t}}_{X}(\mathcal{M})$ is quasi-compact. Also, it is
$\mathrm{T}_{0}$ by (A1) because any two distinct types differ in a closed
set. Moreover, the continuity of the map above yields that every basic open
set is quasi-compact. Hence, the set of all quasi-compact open sets is
precisely the set of all basic open sets, by model theoretic compactness. Thus
(S2) and (S3) follow. Finally, to show that $S^{\mathrm{t}}_{X}(\mathcal{M})$
is sober, fix a nonempty closed and irreducible subset $C$ of
$S^{\mathrm{t}}_{X}(\mathcal{M})$. The set

$\Sigma(x)=\\{\phi(x)\in\mathcal{L}_{M}\ |\ C\subset[\phi],\ [\phi]\text{
closed}\\}\cup\\{\neg\phi(x)\in\mathcal{L}_{M}\ |\ C\not\subset[\phi],\
[\phi]\text{ closed}\\}$

is consistent and in fact it determines a unique complete type by (A1). If $p$
denotes this unique completion, we conclude that $C=\mathrm{cl}^{\rm t}(p)$.
Consequently (S4) also holds because any type $q$ satisfying
$C=\mathrm{cl}^{\rm t}(q)$ coincides with the completion $p$ of $\Sigma$. ∎

######  Remark 2.4.

1) Condition (A1) is equivalent to the fact that
$S^{\mathrm{t}}_{X}(\mathcal{M})$ is $\mathrm{T}_{0}$.

2) If a subset $\widetilde{V}$ of $S^{\rm t}_{X}(\mathcal{M})$ is open and
quasi-compact, then $\widetilde{V}=[V]$ for some open definable subset $V$ of
$X$. In particular, if $\widetilde{C}\subset S^{\rm t}_{X}(\mathcal{M})$
satisfies that $S^{\rm t}_{X}(\mathcal{M})\setminus\widetilde{C}$ is open and
quasi-compact, then $\widetilde{C}=[C]$ for a closed definable subset
$C\subset X$. Thus, the collection of sets of the form $[U]$ and $[C]$ with
$U\subset X$ definable open and $C\subset X$ definable closed constitutes a
subbasis of the constructible topology of $S^{\rm t}_{X}(\mathcal{M})$ (see
[8, Def. 1.3.11]). So by (A1), the sets of the form $[Y]$ for $Y$ a definable
subset of $X$ is a basis of the constructible topology. Consequently, the
constructible topology of $S^{\rm t}_{X}(\mathcal{M})$ is the Stone topology
of $S_{X}(\mathcal{M})$.

As we already noticed in the Introduction, spectral spaces have played an
important role in real algebraic geometry. In [5] the authors introduce the
real spectra of a ring and prove that it is a normal spectral space with good
properties. Assumption (A2) implies the same properties in our context:

######  Lemma 2.5.

The space $S^{\mathrm{t}}_{X}(\mathcal{M})$ is normal. Moreover, for all
$p,p_{1},p_{2}\in S^{\mathrm{t}}_{X}(\mathcal{M})$ with
$p_{1},p_{2}\in\mathrm{cl}^{\rm t}(p)$, either $p_{1}\in\mathrm{cl}^{\rm
t}(p_{2})$ or $p_{2}\in\mathrm{cl}^{\rm t}(p_{1})$.

###### Proof.

Indeed, let $C_{1}$ and $C_{2}$ be two disjoint closed subsets of
$S^{\mathrm{t}}_{X}(\mathcal{M})$. Then there are families of closed definable
subsets $\\{X_{i}\\}_{i\in I}$ and {$Y_{j}\\}_{j\in J}$ such that
$C_{1}=\bigcap_{i}[X_{i}]$ and $C_{2}=\bigcap_{j}[Y_{j}]$. By quasi-
compactness there are two disjoint closed definable subsets $X_{i_{0}}$ and
$Y_{j_{0}}$ such that $C_{1}\subset[X_{i_{0}}]$ and $C_{2}\subset[Y_{j_{0}}]$.
As $\mathcal{M}$ is definably normal (condition (A2)), we conclude that there
are disjoint definable open subsets $U$ and $V$ with $X_{i_{0}}\subset U$ and
$Y_{j_{0}}\subset V$, so $C_{1}\subset[U]$ and $C_{2}\subset[V]$ with
$[U]\cap[V]=\emptyset$, as required.

For the second part, let $p_{1},p_{2}\in\mathrm{cl}^{\rm t}(p)$ and suppose
that $p_{1}\notin\mathrm{cl}^{\rm t}(p_{2})$ and $p_{2}\notin\mathrm{cl}^{\rm
t}(p_{1})$. Then there are two closed definable subsets $C_{1}$ and $C_{2}$ of
$X$ such that $C^{\prime}_{1}:=C_{1}\setminus C_{2}\in p_{1}$ and
$C^{\prime}_{2}:=C_{2}\setminus C_{1}\in p_{2}$. Consider the open definable
set $U:=X\setminus(C_{1}\cap C_{2})$ of $X$. Since $U$ is definably normal and
both $C^{\prime}_{1}$ and $C^{\prime}_{2}$ are closed in $U$, there are open
definable sets $V_{1}$ and $V_{2}$ of $U$ such that $V_{1}\cap
V_{2}=\emptyset$ with $C^{\prime}_{1}\subset V_{1}$ and $C^{\prime}_{2}\subset
V_{2}$. In particular, since $p_{i}\in[V_{i}]$ for each $i=1,2$, we obtain
$p\in[V_{1}]\cap[V_{2}]$, so $V_{1}\cap V_{2}\neq\emptyset$, a contradiction.∎

Another milestone in real algebraic geometry is the paper of Carral and Coste
[4], where they compare the cohomological and Krull dimensions of a normal
spectral space. In their analysis the set of closed points plays an important
role, because the cohomological dimension of a normal spectral space coincides
with the cohomological dimension of its closed points. Recall that in a
topological space $S$, given $x,y\in S$, we say that $x$ is a specialization
of $y$ if $y\in\overline{\\{x\\}}$. Usually, this is denoted by
$x\rightsquigarrow y$. Note that the arrow $\rightsquigarrow$ induces a
partial order, and that with this notation Lemma 2.5 says that for every $p\in
S^{t}_{X}(\mathcal{M})$ the set $\mathrm{cl}^{\rm t}(p)$ is totally ordered
under specialization. We denote by $\text{Max}(S)$ the set of _maximal points_
of $S$ with respect to the specialization order. Clearly, the maximal points
are the closed points of $S$. In the context of normal spectral spaces, the
maximal points have the following crucial property:

######  Fact 2.6.

Let $S$ be a normal spectral space. Then $\text{Max}(S)$ is a quasi-compact
Hausdorff space. Moreover, for each $x\in S$ there is a unique
$r(x)\in\text{Max}(S)$ with $r(x)\in\text{cl}(x)$, and the retraction map
$S\rightarrow\text{Max}(S):x\mapsto r(x)$ is continuous, closed and proper.

###### Proof.

See [4, Prop. 3] and [8, Prop. 4.1.2]. ∎

If $X$ is a semialgebraic set, then the set of maximal points of the spectrum
of the ring of semialgebraic functions on $X$ (see Subsection 3.1) is
traditionally denoted with the letter $\beta$ because it is related with the
Stone-Čech compactification of $X$, see [13]. Since our context is similar to
the latter, given a definable set $X$ of $\mathcal{M}$ we will denote by
$\beta_{X}(\mathcal{M})$ the set of closed points of
$S^{\mathrm{t}}_{X}(\mathcal{M})$, that is,

$\beta_{X}(\mathcal{M})=\\{p\in S^{\mathrm{t}}_{X}(\mathcal{M})\ |\
\\{p\\}\text{ is closed}\\}.$

Note that $X$ is included in $\beta_{X}(\mathcal{M})$ by considering the map
$a\mapsto\operatorname{tp}(a/M)$ for $a$ in $X$. Moreover, the induced
topology from $M^{n}$ on $X$ coincides with the one induced from
$\beta_{X}(\mathcal{M})$ on $X$. The following is a straightforward
consequence of Lemma 2.5 and Fact 2.6.

######  Corollary 2.7.

(1) The space $\beta_{X}(\mathcal{M})$ is quasi-compact and Haussdorf.

(2) There is a continuous closed proper retraction
$r:S^{\mathrm{t}}_{X}(\mathcal{M})\rightarrow\beta_{X}(\mathcal{M})$ that maps
$p\in S^{\mathrm{t}}_{X}(\mathcal{M})$ to the unique
$m\in\beta_{X}(\mathcal{M})$ such that $m\in\mathrm{cl}^{\rm t}(p)$.

###### Proof.

We give a proof specialized to our context for the sake of the reader. Let us
show first that for every $p\in S^{\mathrm{t}}_{X}(\mathcal{M})$ there is
$m\in\beta_{X}(\mathcal{M})$ such that $m\in\text{cl}^{\rm t}(p)$. Since
$S^{\mathrm{t}}_{X}(\mathcal{M})$ is quasi-compact, by Zorn’s lemma the set
$\\{\text{cl}^{\rm t}(y):y\in\text{cl}^{\rm t}(p)\\}$ has a minimal element
$\text{cl}^{t}(m)$ with respect to the inclusion partial order. The space
$S^{\mathrm{t}}_{X}(\mathcal{M})$ is $\mathrm{T}_{0}$ and therefore
$m\in\beta_{X}(\mathcal{M})$, as required. Moreover, by Lemma 2.5 such a type
$m\in\beta_{X}(\mathcal{M})\cap\text{cl}^{t}(p)$ is the unique closed point in
$\text{cl}^{\rm t}(p)$. Indeed, if $m,m^{\prime}\in\mathrm{cl}^{\rm
t}(p)\cap\beta_{X}(\mathcal{M})$ were distinct, then there would exist two
disjoint respective open sets $\widetilde{U}$ and $\widetilde{V}$ of $m$ and
$m^{\prime}$ with $p\in\widetilde{U}\cap\widetilde{V}$, which is a
contradiction.

Once we have established that for every $p\in S^{\mathrm{t}}_{X}(\mathcal{M})$
there is a unique specialization $m\in\beta_{X}(\mathcal{M})$ such that
$m\in\mathrm{cl}^{\rm t}(p)$, we define

$\displaystyle r:S^{\mathrm{t}}_{X}(\mathcal{M})$
$\displaystyle\rightarrow\beta_{X}(\mathcal{M})$ $\displaystyle p$
$\displaystyle\mapsto r(p)=m.$

It remains to check that the retraction $r$ is continuous and closed. To show
that $r$ is continuous, we fix a definable closed subset $C$ of $X$. We first
claim that:

(1) $r^{-1}([C])=\bigcap\left\\{[V]\ |\ C\subset V\subset X\text{ open and
definable}\right\\}.$

Fix $p\in r^{-1}([C])$ and suppose there is an open definable subset $V$ of
$X$ containing $C$ such that $p\notin[V]$. Since $r(p)\in\mathrm{cl}^{\rm
t}(p)$, we deduce $r(p)\notin[V]$, so $r(p)\notin[C]$, which is a
contradiction. To prove the other inclusion, suppose that $p\in[V]$ for every
open definable subset $V$ of $X$ containing $C$, and assume that $p\notin
r^{-1}([C])$. Then $r(p)\notin[C]$, so $r(p)\in[X\setminus C]$. Since
$X\setminus C$ is open and $r(p)\in\mathrm{cl}^{\rm t}(p)$, we deduce that
$p\in[X\setminus C]$, so $p\notin[C]$. Thus, we have

(2) $[C]\subset\bigcup\left\\{[X\setminus F]\ |\ p\in[F]\text{ and }F\subset
X\text{ definable and closed}\right\\}.$

Indeed, suppose there is $q\in[C]$ such that $q\in[F]$ for every closed
definable subset $F$ of $X$ with $p\in[F]$. Thus $q\in\mathrm{cl}^{\rm t}(p)$,
so $r(q)=r(p)$. Since $q\in[C]$ and $C$ is closed, $r(q)\in[C]$ and hence
$r(p)\in[C]$, which is a contradiction. Therefore, we obtain (2). Now, since
$[C]$ is quasi-compact in the Stone topology we deduce that $[C]$ is contained
in $[X\setminus F]$ for some closed definable subset $F$ of $X$ with
$p\in[F]$. Thus, since $p\in[V]$ for every open definable subset $V$ of $X$
containing $C$, and $X\setminus F$ is an open definable subset with $C\subset
X\setminus F$, the choice of $p$ yields that $p\in[X\setminus F]$, which is a
contradiction. Therefore, we obtain (1).

Finally, since $\mathcal{M}$ is definably normal, we deduce that

(3) $r^{-1}([C])=\bigcap\left\\{[\mathrm{cl}_{X}(V)]\ |\ C\subset V\subset
X\text{ open and definable}\right\\},$

where $\mathrm{cl}_{X}(V)$ denotes the topological closure of $V\subset X$ in
$X$. Indeed, by (1) it is enough to show that if $p\in[\mathrm{cl}_{X}(V)]$
for every open definable subset $V$ of $X$ containing $C$, then $p\in[V]$ for
any such $V$. Fix an open definable subset $V$ of $X$ containing $C$ and note
that there exists an open definable subset $U$ such that $C\subset
U\subset\mathrm{cl}_{X}(U)\subset V$. It is enough to take two disjoint open
definable sets $U$ and $W$ with $C\subset U$ and $V^{c}\subset W$, which exist
as $\mathcal{M}$ is definably normal (by the (A2) assumption). Hence, we have
that $p\in[\mathrm{cl}_{X}(U)]$ as $U$ is open and definable, so $p\in[V]$, as
desired. Therefore, we obtain (3) and $r$ is continuous.

Once we have showed that
$r:S^{\mathrm{t}}_{X}(\mathcal{M})\rightarrow\beta_{X}(\mathcal{M})$ is
continuous, let us remark that it is also closed and proper because
$S^{\mathrm{t}}_{X}(\mathcal{M})$ is quasi-compact and
$\beta_{X}(\mathcal{M})$ is Hausdorff. ∎

The set of closed points contains a set which is important from a model-
theoretic point if view.

######  Lemma 2.8.

Let $X\subset M^{n}$ be definable. The set $\beta_{X}(\mathcal{M})$ contains
the set of all realized types, which is dense in
$S^{\mathrm{t}}_{X}(\mathcal{M})$.

###### Proof.

Since $\mathcal{M}$ is equipped with a $\mathrm{T}_{1}$ topology, the
definable set $\\{m\\}$ is closed for each $m\in X$. Thus, the set
$[x=m]\subset S^{\mathrm{t}}_{X}(\mathcal{M})$ is also closed. As $[x=m]$ only
contains the type $\operatorname{tp}(m/M)$, realized types belong to
$\beta_{X}(\mathcal{M})$. Furthermore, the set of realized types is dense
since any consistent $\mathcal{L}_{M}$-formula is realized in $\mathcal{M}$. ∎

We point out that in general $\beta_{X}(\mathcal{M})$ does not have a good
behavior under elementary extensions nor intersections. To illustrate this we
consider the following toy example.

######  Remark 2.9.

Let $\mathcal{R}$ be the field structure of the real numbers and let
$X=\mathbb{R}$. It is immediate to verify that

$\beta_{X}(\mathcal{R})=\left\\{\operatorname{tp}(a/\mathbb{R})\right\\}_{a\in\mathbb{R}}\cup\\{p_{-},p_{+}\\},$

where $p_{-}:=\\{x<a\ |\ a\in\mathbb{R}\\}$ and $p_{+}:=\\{x>a\ |\
a\in\mathbb{R}\\}$. We remark:

  1. (i)

The space $\beta_{X}(\mathcal{R})$ does not behave well under elementary
extensions nor restrictions. Indeed, if $\mathcal{R}_{1}$ is an elementary
extension of $\mathcal{R}$ and $\varepsilon\in R_{1}$ is a positive
infinitesimal, then
$\operatorname{tp}(\varepsilon/R_{1})\in\beta_{X}(\mathcal{R}_{1})$ whereas
$\operatorname{tp}(\varepsilon/\mathbb{R})\notin\beta_{X}(\mathcal{R})$.

  2. (ii)

Let $Y=(0,\infty)\subset X$. Then
$\beta_{Y}(\mathcal{R})\neq\beta_{X}(\mathcal{R})\cap[Y]$. Indeed, using the
notation from (i), we have

$\beta_{Y}(\mathcal{R})=\left\\{\operatorname{tp}(a/\mathbb{R})\right\\}_{a\in\mathbb{R}_{>0}}\cup\\{\operatorname{tp}(\varepsilon/\mathbb{R}),p_{+}\\}.$

Hence, the type
$\operatorname{tp}(\varepsilon/\mathbb{R})\in\beta_{Y}(\mathcal{R})$ but it
does not belong to $\beta_{X}(\mathcal{R})\cap[Y]$.

  3. (iii)

It also follows that $\beta_{X}(\mathcal{R})$ is not closed with respect to
the spectral topology. For example, the type
$\operatorname{tp}(\varepsilon/\mathbb{R})\in\mathrm{cl}^{\rm
t}(\beta_{X}(\mathcal{R}))$. In this case, the subset $\beta_{X}(\mathcal{R})$
is closed if and only if $X$ is finite.

###  2.2. A universal property for closed points

Let $\mathcal{M}$ be a topological structure as above satisfying (A1) and (A2)
and let $X\subset M^{n}$ be a definable set. We consider the subspace topology
on $X$ and we establish a universal property for $\beta_{X}(\mathcal{M})$ in
the vein of [16, Lem. 3.7].

######  Definition 2.10.

Let $C$ be a topological space. A continuous map $f:X\rightarrow C$ is
definably separated by closed sets if for every two disjoint closed subsets
$C_{1}$ and $C_{2}$ of $C$ there are two disjoint closed definable subsets
$X_{1}$ and $X_{2}$ of $X$ such that $f^{-1}(C_{1})\subset X_{1}$ and
$f^{-1}(C_{2})\subset X_{2}$.

######  Remark 2.11.

The map $i:X\hookrightarrow\beta_{X}(\mathcal{M})$ is clearly definably
separated by closed sets. Indeed, given two disjoint closed subsets $C_{1}$
and $C_{2}$ of $\beta_{X}(\mathcal{M})$, by quasi-compactness there are two
closed definable subsets $V_{1}$ and $V_{2}$ of $X$ such that each
$C_{j}\subset\beta_{X}(\mathcal{M})\cap[V_{j}]$ and

$\left(\beta_{X}(\mathcal{M})\cap[V_{1}]\right)\cap\left(\beta_{X}(\mathcal{M})\cap[V_{2}]\right)=\emptyset.$

Since each $i^{-1}\left(\beta_{X}(\mathcal{M})\cap[V_{j}]\right))=V_{j}$, it
follows immediately that $V_{1}\cap V_{2}=\emptyset$ and $i^{-1}(C_{j})\subset
V_{j}$, as required.

It is easy to show that given a continuous map $f:X\rightarrow C$ from $X$ to
a topological space $C$, if there exist a continuous extension
$\hat{f}:\beta_{X}(\mathcal{M})\rightarrow C$ of $f$, then $f$ is definably
separated by closed sets, by the remark above. We show the converse:

######  Proposition 2.12.

Let $f:X\rightarrow C$ be a continuous map, where $C$ is quasi-compact and
Hausdorff. If $f$ is a definably separated by closed sets, then there is a
continuous extension $\hat{f}:\beta_{X}(\mathcal{M})\rightarrow C$ of $f$ to
$\beta_{X}(\mathcal{M})$.

###### Proof.

We first define $\hat{f}$ as follows: Given a type
$p\in\beta_{X}(\mathcal{M})$, set

$\hat{f}(p):=\bigcap_{U\in p\text{ open}}\overline{f(U)}.$

The intersection above is not empty, (by quasi-compactness) because it has the
finite intersection property. Thus, it suffices to show that the above
intersection contains a single point. Suppose otherwise there are $a$ and $b$
in $\hat{f}(p)$ with $a\neq b$. Since $C$ is regular, there are open
respective neighborhoods $U_{a}$ and $U_{b}$ of $a$ and $b$ such that
$\overline{U}_{a}\cap\overline{U}_{b}=\emptyset$. As $f:X\to C$ is a definably
separated by closed sets, there exist two disjoint closed definable subsets
$X_{a}$ and $X_{b}$ of $X$ such that $f^{-1}(\overline{U}_{a})\subset X_{a}$
and $f^{-1}(\overline{U}_{b})\subset X_{b}$. We can assume that
$p\notin[X_{a}]$, so that $p\in[X\setminus X_{a}]$. Since $X\setminus X_{a}$
is open, $a\in\hat{f}(p)\subset\overline{f(X\setminus X_{a})}$. Hence, as
$a\in U_{a}$ and $a\in\overline{f(X\setminus X_{a})}$, it follows that
$U_{a}\cap f(X\setminus X_{a})\neq\emptyset$. In particular,
$f^{-1}(\overline{U}_{a})\cap(X\setminus X_{a})\neq\emptyset$, which is a
contradiction as $f^{-1}(\overline{U}_{a})\subset X_{a}$.

Once we have proved that $\hat{f}$ is a function, we prove that it is
continuous. Let $Z$ be a closed subset of $C$ and suppose that there is some
$p\in\overline{\hat{f}^{-1}(Z)}\setminus\hat{f}^{-1}(Z)$. Since
$\hat{f}(p)\notin Z$, we have

$Z\cap\bigcap_{U\in p\text{ open}}\overline{f(U)}=\emptyset,$

so $Z\cap\overline{f(U)}=\emptyset$ for some definable open set $U$ with
$p\in[U]$ (by quasi-compactness). Since $[U]$ is open and
$p\in\overline{\hat{f}^{-1}(Z)}$, it follows that
$[U]\cap\hat{f}^{-1}(Z)\neq\emptyset$, so there is some $q\in[U]$ such that
$\hat{f}(q)\in Z$. By definition we also have $\hat{f}(q)\in\overline{f(U)}$,
hence $\hat{f}(q)\in Z\cap\overline{f(U)}$, which is a contradiction. ∎

######  Remark 2.13.

Let $X$ be definable in $\mathcal{M}$ and let $\mathcal{M}^{\prime}$ be an
expansion of $\mathcal{M}$ satisfying (A1) and (A2). Then we can consider also
$\beta_{X}(\mathcal{M}^{\prime})$. It turns out that there is a closed
continuous surjective map

$j:\beta_{X}(\mathcal{M}^{\prime})\rightarrow\beta_{X}(\mathcal{M}).$

Indeed, the inclusion continuous map $X\hookrightarrow\beta_{X}(\mathcal{M})$
is definably separated by closed sets with respect to $\mathcal{M}^{\prime}$
and therefore there exists a continuous extension $j$. The map $j$ is closed
and therefore it is surjective because $X$ is dense in both
$\beta_{X}(\mathcal{M})$ and $\beta_{X}(\mathcal{M}^{\prime})$.

Let us describe the map $j$. Given $p\in S_{X}(\mathcal{M}^{\prime})$, denote
by $p^{\mathcal{M}}$ the restriction of $p$ to the language of $\mathcal{M}$,
and let $r:S^{\mathrm{t}}_{X}(\mathcal{M})\rightarrow\beta_{X}(\mathcal{M})$
be the natural retraction provided in Corollary 2.7. The map

$\beta_{X}(\mathcal{M}^{\prime})\rightarrow\beta_{X}(\mathcal{M}),\ p\mapsto
r(p^{\mathcal{M}})$

is continuous and coincide with $j$ on realized types, so they are equal.

The following example shows that the map $j$ from Remark 2.13 is not a
homeomorphism in general.

######  Example 2.14.

Let $\mathcal{R}$ be the field of real numbers and $\mathcal{R}_{\rm exp}$ be
the field of real numbers with the exponentiation. Set $X=\mathbb{R}^{2}$ and
consider in $\beta_{X}(\mathcal{R}_{\rm exp})$ the types

$\displaystyle q_{1}$ $\displaystyle:=\\{Y\subset\mathbb{R}^{2}\ |\ Y\text{
definable in }\mathcal{R}_{\rm exp}\ \&\ \exists
t_{0}>0,\forall\,t>t_{0},(t,0)\in Y\\},$ $\displaystyle q_{2}$
$\displaystyle:=\\{Y\subset\mathbb{R}^{2}\ |\ Y\text{ definable in
}\mathcal{R}_{\rm exp}\ \&\ \exists t_{0}>0,\forall\,t>t_{0},(t,e^{-t})\in
Y\\}.$

In this case $q_{1}^{\mathcal{R}}\in\beta_{X}(\mathcal{R})$ but
$q_{2}^{\mathcal{R}}\notin\beta_{X}(\mathcal{R})$. Moreover,
$q_{1}^{\mathcal{R}}\in\mathrm{cl}^{\rm t}(q_{2}^{\mathcal{R}})$. Indeed,
given an open semialgebraic set $U$ of $\mathbb{R}^{2}$ such that $(t,0)\in U$
for every $t$ large enough, we have $(t,e^{-t})\in U$ for $t$ large enough.
Thus,

$j(q_{1})=r(q_{1}^{\mathcal{R}})=r(q_{2}^{\mathcal{R}})=j(q_{2}).$

###  2.3. Closed types in topological structures with a dimension

Let $\mathcal{M}$ be a topological structure as above satisfying (A1) and
(A2). Assume further that $\mathcal{M}$ is equipped with a dimension
satisfying that for all definable sets $V\subset X\subset M^{n}$ we have that
$\dim(\mathrm{cl}_{X}(V)\setminus V)<\dim(V)$. This is the case for example if
$\mathcal{M}$ is an o-minimal expansion of a real closed field or the $p$-adic
field.

######  Remark 2.15.

The existence of such a well-behaved dimension map implies assumption (A1).
Indeed, note first that assumption (A1) is equivalent to the fact that for any
two different types there is an open (equivalently, closed) set which contains
one of these points and not the other. This follows by a standard topological
argument [27, Ex. 3.1.1]. Now, let $p$ and $q$ be two distinct types in
$S_{X}(\mathcal{M})$. Let $V$ be a definable set which distinguish them, that
is to say that $V$ is in one type but not the other. Clearly, we can take $V$
so that $\dim(V)$ is minimal with this property. Assume that $V$ is not closed
(as otherwise we are done) and suppose that $V\in p$ but $V\notin q$. Since
$\dim(\mathrm{cl}_{X}(V)\setminus V)<\dim(V)$ and
$(\mathrm{cl}_{X}(V)\setminus V)\notin p$, we have that
$(\mathrm{cl}_{X}(V)\setminus V)\notin q$ as well due to the minimality of the
dimension of $V$. Thus, $\mathrm{cl}_{X}(V)\notin q$ and
$\mathrm{cl}_{X}(V)\in p$, so $p$ and $q$ are distinguished by the closed set
$\mathrm{cl}_{X}(V)$.

Let $X\subset M^{n}$ be a definable set in $\mathcal{M}$. By Corollary 2.7 we
know that the space of closed types $\beta_{X}(\mathcal{M})$ is quasi-compact
and Hausdorff and there is a natural retraction
$r:S^{\mathrm{t}}_{X}(M)\rightarrow\beta_{X}(\mathcal{M})$. The following
lemma exhibits the relation between dimension and the spectral topology, where
the dimension $\dim(p)$ of the type $p$ is defined as $\min\\{\dim(X)\ |\ X\in
p\\}$.

######  Lemma 2.16.

Let $p,q\in S_{X}(\mathcal{M})$. The following holds:

  1. (1)

If $q\in\mathrm{cl}^{\rm t}(p)$ and $q\neq p$, then $\dim(q)<\dim(p)$.

  2. (2)

$\dim(r(p))\leq\dim(p)$, with equality if and only if
$p\in\beta_{X}(\mathcal{M})$.

In particular, for every $p\in S_{X}(\mathcal{M})$ the number of elements in
$\mathrm{cl}^{\rm t}(p)$ is less than or equal to $1+\dim(p)$.

###### Proof.

(1) Let $W\subset X$ be a definable set in $p$ of minimal dimension. Since
$q\in\mathrm{cl}^{\rm t}(p)$, we have $\mathrm{cl}_{X}(W)\in p$ and
$\mathrm{cl}_{X}(W)\in q$, so we can assume $\dim(p)=\dim(X)$ after
substituting $X$ by $\mathrm{cl}_{X}(W)$.

Since the spectral topology is $\mathrm{T}_{0}$ and $p\neq q$, it follows
$p\notin\mathrm{cl}^{\rm t}(q)$. Therefore, there is a definable open subset
$U$ of $X$ such that $p\in[U]$ but $q\notin[U]$. Since
$p\in[\mathrm{cl}_{X}(U)]$, we deduce $q\in[\mathrm{cl}_{X}(U)\setminus U]$,
so

$\dim(q)\leq\dim(\mathrm{cl}_{X}(U)\setminus
U)<\dim(\mathrm{cl}_{X}(U))=\dim(U)=\dim(X)=\dim(p),$

as required.

(2) follows immediately from (1). The bound of the cardinality of
$\mathrm{cl}^{\rm t}(p)$ follows from (1) and Lemma 2.5. ∎

As we pointed out in Lemma 2.8 the space $\beta_{X}(\mathcal{M})$ contains all
realized types which concentrate on $X$, but it also may contain types of
maximal dimension as happens in the example of Remark 2.9. Nonetheless, these
two kinds of closed types may have distinct fibers with respect to the
retraction $r$. In fact, the size of the fiber characterizes types of maximal
dimension in the o-minimal setting.

######  Proposition 2.17.

Let $\mathcal{R}$ be an o-minimal expansion of a real closed field. Let
$p\in\beta_{R^{n}}(\mathcal{R})$ be any type. Then $r^{-1}(\\{p\\})=\\{p\\}$
if and only if $\dim(p)=n$.

###### Proof.

Suppose first that there is some $q\neq p$ with $r(q)=p$. Note that
$q\not\in\beta_{R^{n}}(\mathcal{R})$ and thus $\dim(p)<\dim(q)\leq n$ by Lemma
2.16(2).

To prove the other direction, assume that $\dim(p)<n$. We can assume there is
an elementary extension $\mathcal{N}$ of $\mathcal{R}$ and a realization
$(\bar{a},a)\in N^{n-1}\times N$ of $p$ with
$a\in\operatorname{dcl}(R,\bar{a})$. Consider the realized type
$p^{\prime}:=\operatorname{tp}(a/N)$, and note there exists some non-realized
type $q^{\prime}=\operatorname{tp}(b/N)$ with
$r_{\mathcal{N}}(q^{\prime})=p^{\prime}$, where
$r_{\mathcal{N}}:S^{\mathrm{t}}_{N^{n}}(\mathcal{N})\rightarrow\beta_{N^{n}}(\mathcal{N})$.
For example, take $b:=a+\varepsilon$ for some infinitesimal $\varepsilon>0$
with respect to $N$. Set $q=\operatorname{tp}(\bar{a},b/R)$, which is not $p$
because $b\notin\operatorname{dcl}(R,\bar{a})$.

We claim that $r(q)=p$. To see this, let $U\subset R^{n}$ be an open
$R$-definable set such that $p\in[U]$. We denote by $U(N)\subset N^{n}$ its
realization in $\mathcal{N}$. Consider now the $N$-definable open set

$W=\\{y\in N\ |\ (\bar{a},y)\in U(N)\\}$

and note that $a\in W$. Thus, there is an open interval $V=(c,d)\subset W$
with $c,d\in N$ such that $c<a<d$. It follows that $p^{\prime}\in[V]$, so
$q^{\prime}\in[V]$ because $r_{\mathcal{N}}(q^{\prime})=p^{\prime}$. Hence,
$b\in W$ and so $(\bar{a},b)\in U(N)$, which yields that $q\in[U]$. Therefore,
as $q\neq p$, we obtain |$r^{-1}(\\{p\\})|>1$, as required. ∎

##  3\. The relation with other topological constructions

In this section we fix an o-minimal expansion $\mathcal{R}$ of a real closed
field and we study its connections with its definable real sprectrum, as well
as with its space of $\mu$-types introduced by Peterzil and Starchenko [19].

###  3.1. The relation with the real spectrum

Let $\mathcal{R}$ be an o-minimal expansion of a real closed field. Let
$\mathcal{C}^{\mathcal{R}}(X)$ be the ring of continuous functions from $X$ to
$R$ that are definable in $\mathcal{R}$, that is,

$\mathcal{C}^{\mathcal{R}}(X)=\left\\{f:X\to R\ |\ f\text{ is continuous and
$R$-definable}\right\\}.$

The aim of this section is to prove that $\beta_{X}(\mathcal{R})$ is
homeomorphic to the set of closed points of the real spectrum
$\mathrm{Sper}(\mathcal{C}^{\mathcal{R}}(X))$ of
$\mathcal{C}^{\mathcal{R}}(X)$. Recall that the set
$\mathrm{Sper}(\mathcal{C}^{\mathcal{R}}(X))$ is the collection of _prime
cones_ of $\mathcal{C}^{\mathcal{R}}(X)$, that is, subsets $\alpha$ of
$\mathcal{C}^{\mathcal{R}}(X)$ such that
$\mathfrak{p}_{\alpha}:=\alpha\cap-\alpha$ is a prime ideal of
$\mathcal{C}^{\mathcal{R}}(X)$ and $\alpha/\mathfrak{p}_{\alpha}$ is the
positive cone of a total order of
$\mathcal{C}^{\mathcal{R}}(X)/\mathfrak{p}_{\alpha}$ (we refer the reader to
[3, Ch. 7] for the basic definitions concerning the real spectrum). Let
$\rho_{\alpha}:\mathcal{C}^{\mathcal{R}}(X)\to\mathcal{C}^{\mathcal{R}}(X)/\mathfrak{p}_{\alpha}$
be the natural homomorphism. The subsets

$D_{r}(f):=\\{\alpha\in\mathrm{Sper}(\mathcal{C}^{\mathcal{R}}(X))\ |\
\rho_{\alpha}(f)>0\\}$

for $f\in\mathcal{C}^{\mathcal{R}}(X)$, which are called basic open subsets,
constitute a basis of the _spectral topology_ of
$\mathrm{Sper}(\mathcal{C}^{\mathcal{R}}(X))$.

As we see in the following well-known fact, the ring
$\mathcal{C}^{\mathcal{R}}(X)$ has the special feature that the real spectrum
is homeomorphic to the (Zariski) spectrum, and therefore we will work with the
latter. Before proceeding, let us recall some basic definitions on the
(Zariski) spectrum. Let $\mathrm{Spec}(\mathcal{C}^{\mathcal{R}}(X))$ denote
the spectrum of $\mathcal{C}^{\mathcal{R}}(X)$, that is, the set of prime
ideals of $\mathcal{C}^{\mathcal{R}}(X)$, and we regard it as a topological
space with the usual Zariski topology. Recall that given
$f\in\mathcal{C}^{\mathcal{R}}(X)$, a basic open set is

$D(f)=\left\\{\mathfrak{p}\in\mathrm{Spec}(\mathcal{C}^{\mathcal{R}}(X))\ |\
f\notin\mathfrak{p}\right\\}.$

######  Fact 3.1.

The support map

$\mathrm{supp}:\mathrm{Sper}(\mathcal{C}^{\mathcal{R}}(X))\rightarrow\mathrm{Spec}(\mathcal{C}^{\mathcal{R}}(X)),\
\alpha\mapsto\mathfrak{p}_{\alpha}$

is a homeomorphism.

###### Proof.

The support map is continuous and its image is the set of real prime ideals of
$\mathcal{C}^{\mathcal{R}}(X)$ (see [3, Prop. 7.1.8]). Each
$\mathfrak{p}\in\mathrm{Spec}(\mathcal{C}^{\mathcal{R}}(X))$ is real prime
because the field of fractions of $\mathcal{C}^{\mathcal{R}}(X)/\mathfrak{p}$
is a real field, so $\mathrm{supp}$ is surjective. Now, the field of fractions
of $\mathcal{C}^{\mathcal{R}}(X)/\mathfrak{p}$ admits a unique ordering and
therefore the map $\mathrm{supp}$ is injective. Indeed, given
$f\in\mathcal{C}^{\mathcal{R}}(X)$ we have $(f-|f|)(f+|f|)=0$, so either
$f-|f|\in\mathfrak{p}$ or $f+|f|\in\mathfrak{p}$, where $|f|$ denotes the
absolute value of $f$. Since $|f|=(\sqrt{|f|})^{2}$, the ordering of
$\mathcal{C}^{\mathcal{R}}(X)/\mathfrak{p}$ is completely determined by
$\mathfrak{p}$.

Finally, we note that $\text{supp}(D_{r}(f))=D(f+|f|)$ for every
$f\in\mathcal{C}^{\mathcal{R}}(X)$, which yields that the inverse of the
support map is continuous. ∎

The real spectrum of a ring is a normal spectral space by [3, Rmk. 7.1.17,
Prop. 7.1.25], and therefore we deduce the following.

######  Fact 3.2.

The space $\mathrm{Spec}(\mathcal{C}^{\mathcal{R}}(X))$ is a normal spectral
space.

###### Proof.

For the sake of the reader, we include the proof of normality without using
the machinery of the real spectrum. Let $C_{1}$ and $C_{2}$ be two disjoint
closed subsets of $\mathrm{Spec}(\mathcal{C}^{\mathcal{R}}(X))$, which we may
assume to be basic closed subsets, by quasi-compactness of
$\mathrm{Spec}(\mathcal{C}^{\mathcal{R}}(X))$. Hence, there are two continuous
definable functions $f_{1},f_{2}\in\mathcal{C}^{\mathcal{R}}(X)$ such that
$C_{i}=D(f_{i})^{c}$ for $i=1,2$. Consider the closed definable sets
$Z(f_{1})=f_{1}^{-1}(\\{0\\})$ and $Z(f_{2})=f_{2}^{-1}(\\{0\\})$.

As $X$ is definably normal, there are two open definable subsets $U_{1}\supset
Z(f_{1})$ and $U_{2}\supset Z(f_{2})$ such that $U_{1}\cap U_{2}=\emptyset$.
Let $g_{1},g_{2}\in\mathcal{C}^{\mathcal{R}}(X)$ be the functions defined by
$g_{i}(x)=\text{dist}(x,X\setminus U_{i})$ for each $x\in X$ and $i=1,2$. Note
that $Z(g_{1})=X\setminus U_{1}$ and $Z(g_{2})=X\setminus U_{2}$. We claim
that $C_{i}\subset D(g_{i})$. Otherwise, for $\mathfrak{p}\in C_{i}\setminus
D(g_{i})$ we have that $g_{i},f_{i}\in\mathfrak{p}$. As

$Z(f_{i}^{2}+g_{i}^{2})=Z(f_{i})\cap Z(g_{i})\subset U_{i}\cap(X\setminus
U_{i})=\emptyset,$

the function $f_{i}^{2}+g_{i}^{2}\in\mathfrak{p}$ is a unit, a contradiction.
Hence $C_{i}\subset D(g_{i})$ for $i=1,2$. It remains to show that
$D(g_{1})\cap D(g_{2})=\emptyset$. Indeed, since

$Z(g_{1}\cdot g_{2})=Z(g_{1})\cup Z(g_{2})=(X\setminus U_{1})\cup(X\setminus
U_{2})=X\setminus(U_{1}\cap U_{2})=X,$

the function $g_{1}\cdot g_{2}$ is the constant function $x\mapsto 0$ and
therefore $g_{1}\cdot g_{2}\in\mathfrak{p}$. Hence, as $\mathfrak{p}$ is
prime, either $\mathfrak{p}\notin D(g_{1})$ or $\mathfrak{p}\notin D(g_{2})$.
∎

Now, consider the map

$\iota:S^{\mathrm{t}}_{X}(\mathcal{R})\to\mathrm{Spec}(\mathcal{C}^{\mathcal{R}}(X)),\
p\mapsto\\{f\in\mathcal{C}^{\mathcal{R}}(X)\ |\ Z(f)\in p\\},$

where $Z(f):=f^{-1}(\\{0\\})$ is a definable closed set because $f$ is
continuous. In fact, any definable closed set $Z$ of $X$ is of this form by
considering the map $f:X\to\mathbb{R}$ given by $f(x)=\mathrm{dist}(x,Z)$.

######  Proposition 3.3.

The map
$\iota:S^{\mathrm{t}}_{X}(\mathcal{R})\to\mathrm{Spec}(\mathcal{C}^{\mathcal{R}}(X))$
is injective and spectral, that is, it is continuous and the inverse image of
a quasi-compact open is quasi-compact.

###### Proof.

To prove that it is continuous, consider an arbitrary basic open set $D(f)$
given by some $f\in\mathcal{C}^{\mathcal{R}}(X)$. Then

$\iota^{-1}(D(f))=\\{p\in S^{\mathrm{t}}_{X}(\mathcal{R})\ |\ \iota(p)\in
D(f)\\}=\\{p\in S^{\mathrm{t}}_{X}(\mathcal{R})\ |\ Z(f)\notin
p\\}=[X\setminus Z(f)]$

and $\iota$ is continuous because the set $Z(f)$ is a closed subset of $X$. To
verify that the inverse image of a quasi-compact open set
$U\subset\mathrm{Spec}(\mathcal{C}^{\mathcal{R}}(X))$ is quasi-compact, note
that $U$ is a finite union of sets of the form $D(f_{i})$ for $i\in I$, where
$I$ finite and each $f_{i}\in\mathcal{C}^{\mathcal{R}}(X)$. Thus,

$\iota^{-1}(U)=\bigcup_{i\in I}\iota^{-1}(D(f_{i}))=\bigcup_{i\in
I}[X\setminus Z(f_{i})]$

is quasi-compact, as so is each set of the form $[V]$ for $V\subset X$
definable since $S_{X}(\mathcal{R})\to S^{\mathrm{t}}_{X}(\mathcal{R})$ is
continuous.

Finally, by [28, Cor. 2.6 and Prop. 3.1] the map $\iota$ is injective. We
provide also a direct proof. Consider $p,q\in S^{\mathrm{t}}_{X}(\mathcal{R})$
with $p\neq q$ such that $\iota(p)=\iota(q)$. By assumption (A1), there exists
a definable closed subset $Z$ of $X$ such that $Z\in p$ and $Z\notin q$. Since
$Z$ is closed, there is some $f\in\mathcal{C}^{\mathcal{R}}(X)$ such that
$Z(f)=Z$, so $f\in\iota(p)$ but $f\notin\iota(q)$, a contradiction. ∎

In general, the map $\iota$ is not surjective even when $\mathcal{R}$ has only
the field structure.

######  Example 3.4.

The following two examples correspond to [14, Rmk. 1.2] (cf. [3, Rmk. 2.6.5])
and [28, pp. 1], respectively. In both cases, one finds two functions
$f,g\in\mathcal{C}^{\mathcal{R}}(X)$ with $Z(f)=Z(g)$ and a prime ideal
$\mathfrak{p}\in\mathrm{Spec}(\mathcal{C}^{\mathcal{R}}(X))$ containing one
but not the other, showing that $\iota$ cannot be surjective.

  1. (i)

Let $\mathcal{R}$ be the field of real numbers and
$X=\\{(x,y)\in\mathbb{R}\times\mathbb{R}\ |\ y>0\\}\cup\\{(0,0)\\}$. Consider
the semialgebraic functions $f(x,y)=y$ and $g(x,y)=x^{2}+y^{2}$. Their zero
set agree on $X$ but for each $k\in\mathbb{N}$ the limit at $(0,0)$ of the
semialgebraic function $h_{k}=g^{k}/f$ does not exist. Therefore the prime
ideal

$\mathfrak{p}=\left\\{h\in\mathcal{C}^{\mathcal{R}}(X)\ \big{|}\
\exists\varepsilon>0,\forall t\in[0,\varepsilon)\,\lim_{x\to
t}h(x,0)=0\right\\}$

contains $f$ but not $g$.

  2. (ii)

Let $\mathcal{R}$ be the field of real numbers with the exponential map $\exp$
and $X=\mathbb{R}$. Consider the definable functions $f(x)=x$ and
$g(x)=\exp(-1/x^{2})$ for $x>0$ and $g(0)=0$. Their zero set agree on $X$ but
for each $k\in\mathbb{N}$ the limit at $0$ of the definable function
$h_{k}=g^{k}/f$ does not exist. Therefore the prime ideal $\mathfrak{p}$ given
by

$\left\\{h\in\mathcal{C}^{\mathcal{R}}(X)\ \big{|}\ \forall k\geq
1,\exists\varepsilon_{k}>0,\exists c_{k}>1,\forall x\in[0,\varepsilon_{k})\
|h(x)|<c_{k}\cdot x^{k}\right\\}$

contains $g$ but not $f$.

However, for locally closed semialgebraic sets Lojasiewicz’s inequality yields
the surjectivity of $\iota$. The following result already appears in [28, Thm.
8.1]. We include a (different) proof for the sake of completeness.

######  Fact 3.5.

Suppose that $\mathcal{R}$ is a polynomially bounded expansion of a real
closed field. Assume that $X$ is a locally closed and definable. Then the
spectral map
$\iota:S^{\mathrm{t}}_{X}(\mathcal{R})\to\mathrm{Spec}(\mathcal{C}^{\mathcal{R}}(X))$
is surjective and in particular a homeomorphism.

###### Proof.

Let $\mathfrak{p}\in\mathrm{Spec}(\mathcal{C}^{\mathcal{R}}(X))$, and consider
the following collection of definable sets

$\Sigma=\\{Z(f)\ |\ f\in\mathfrak{p}\\}\cup\\{X\setminus Z(f)\ |\
f\notin\mathfrak{p}\\}.$

Let us show that $\Sigma$ is consistent. Assume, towards a contradiction, that

$Z(f_{1})\cap\cdots\cap Z(f_{n})\cap\big{(}X\setminus
Z(g_{1})\big{)}\cap\cdots\cap\big{(}X\setminus Z(g_{m})\big{)}=\emptyset$

for some $f_{1},\ldots,f_{n}\in\mathfrak{p}$ and some
$g_{1}\ldots,g_{m}\in\mathcal{C}^{\mathcal{R}}(X)$. Define
$f:=f^{2}_{1}+\cdots+f^{2}_{n}$ and $g:=g_{1}\cdots g_{m}$. Thus,
$f\in\mathfrak{p}$ and $g\notin\mathfrak{p}$, because $\mathfrak{p}$ is prime.
Note that

$Z(f)=\bigcap_{i=1}^{n}Z(f_{i})\subset\bigcup_{j=1}^{m}Z(g_{j})=Z(g).$

Therefore, by [28, Thm. 7.3] (cf. [3, Thm. 2.6.6], Lojasiewicz’s inequality)
there is some $k\in\mathbb{N}$ and some $h\in\mathcal{C}^{\mathcal{R}}(X)$
such that $g^{k}=hf$. In particular, $g\in\mathfrak{p}$, which is a
contradiction.

Hence, take a complete type $p$ extending $\Sigma$. It is obvious that
$\mathfrak{p}\subset\iota(p)$, since $\Sigma\subset p$. On the other hand,
given $f\in\iota(p)$ we have by definition that $Z(f)\in p$, so $X\setminus
Z(f)\notin\Sigma\subset p$. Hence, $f\in\mathfrak{p}$, yielding that
$\iota(p)=\mathfrak{p}$. ∎

As shown in [4, Props. 2 and 3], the normality of
$\mathrm{Spec}(\mathcal{C}^{\mathcal{R}}(X))$ implies that the set of maximal
ideals $\mathrm{Max}(\mathcal{C}^{\mathcal{R}}(X))$ is quasi-compact and
Hausdorff. Furthermore, there is a continuous retraction map
$r_{\mathrm{Spec}}:\mathrm{Spec}(\mathcal{C}^{\mathcal{R}}(X))\rightarrow\mathrm{Max}(\mathcal{C}^{\mathcal{R}}(X))$.

######  Theorem 3.6.

The map
$\iota|_{\beta_{X}(\mathcal{R})}:\beta_{X}(\mathcal{R})\rightarrow\mathrm{Max}(\mathcal{C}^{\mathcal{R}}(X))$
is a homeomorphism and the following diagram is commutative:

(4)
${S^{\mathrm{t}}_{X}(\mathcal{R})}$${\mathrm{Spec}(\mathcal{C}^{\mathcal{R}}(X))}$${\beta_{X}(\mathcal{R})}$${\mathrm{Max}(\mathcal{C}^{\mathcal{R}}(X))}$$\scriptstyle{\iota}$$\scriptstyle{r}$$\scriptstyle{r_{\mathrm{Spec}}}$$\scriptstyle{\iota|_{\beta_{X}(\mathcal{R})}}$

###### Proof.

We first show that for every
$\mathfrak{m}\in\mathrm{Max}(\mathcal{C}^{\mathcal{R}}(X))$ there exists some
$p\in\beta_{X}(\mathcal{R})$ such that $\iota(p)=\mathfrak{m}$. Indeed, let

$\Sigma=\\{Z(f)\ |\ f\in\mathfrak{m}\\}.$

Similarly as we argued above $\Sigma$ is consistent. Otherwise, let
$f_{1},\ldots,f_{n}\in\mathfrak{m}$ be such that

$Z(f_{1})\cap\ldots\cap Z(f_{n})=\emptyset$

and set $f=f_{1}^{2}+\dots+f_{n}^{2}$. It follows that $f\in\mathfrak{m}$ but
$Z(f)=\emptyset$, which is a contradiction. Now, let $p\in S_{X}(\mathcal{R})$
be a complete type extending $\Sigma$. We claim that $\iota(p)=\mathfrak{m}$.
Indeed, by definition $\iota(p)$ is a prime ideal that contains
$\mathfrak{m}$, so $\iota(p)=\mathfrak{m}$ by maximality of $\mathfrak{m}$.
Now, since the map $\iota$ is injective and continuous by Proposition 3.3, the
set $\iota^{-1}(\\{\mathfrak{m}\\})=\\{p\\}$ is closed, hence
$p\in\beta_{X}(\mathcal{R})$, as required.

Now we claim that for every $q\in\beta_{X}(\mathcal{R})$ we have
$\iota(q)\in\mathrm{Max}(\mathcal{C}^{\mathcal{R}}(X))$. Indeed, the prime
ideal $\iota(q)$ is contained in a maximal ideal $\mathfrak{m}$. Moreover, the
previous paragraph and Proposition 3.3 yield that there is a unique
$p\in\beta_{X}(\mathcal{R})$ such that $\iota(p)=\mathfrak{m}$. We claim that
$p\in\mathrm{cl}^{\rm t}(q)$. Otherwise, suppose that there is some definable
open set $U$ such that $p\in[U]$ and $q\notin[U]$. Thus, there is some
$f\in\mathcal{C}^{\mathcal{R}}(X)$ with $Z(f)=X\setminus U$, so that
$Z(f)\notin p$. In particular, by definition $f\notin\iota(p)$, so
$f\notin\iota(q)$, because $\iota(q)\subset\mathfrak{m}=\iota(p)$. It follows
that $Z(f)=X\setminus U\notin q$, so $U\in q$, which is a contradiction.
Therefore, we have proved that $p\in\mathrm{cl}^{\rm t}(q)$ and since $q$ is
closed point, we get that $p=q$ and hence $\iota(q)=\iota(p)=\mathfrak{m}$ is
maximal, as required.

Altogether, we have proved that

$\iota|_{\beta_{X}(\mathcal{R})}:\beta_{X}(\mathcal{R})\rightarrow\mathrm{Max}(\mathcal{C}^{\mathcal{R}}(X))$

is a continuous bijection and so a homeomorphism, because as both spaces are
quasi-compact and Hausdorff the map $\iota|_{\beta_{X}(\mathcal{R})}$ is
closed.

Finally, observe that given a type $q\in S^{\mathrm{t}}_{X}(\mathcal{R})$, the
ideal $r_{\mathrm{Spec}}(\iota(q))$ is the unique maximal ideal which contains
$\iota(q)$. Since $r(q)\in\mathrm{cl}^{\rm t}(q)$, it follows that
$\iota(q)\subset\iota(r(q))$, so

$\iota(r(q))=r_{\mathrm{Spec}}(\iota(q))$

and diagram (4) is commutative as required. ∎

We finish this section noting that $S^{\mathrm{t}}_{X}(\mathcal{R})$ is
locally connected, similarly as it happens with the spectrum of the ring of
continuous semialgebraic functions (see [1, Cor. 3.10]).

######  Lemma 3.7.

The topological space $S^{\mathrm{t}}_{X}(\mathcal{R})$ is locally connected.

###### Proof.

We first note that if $U$ is a definable open subset of $X$ which is definably
connected, then $[U]$ is connected in $S^{\mathrm{t}}_{X}(\mathcal{R})$.
Otherwise, there are open disjoint non-empty subsets $C_{1}$ and $C_{2}$ of
$[U]$ such that $[U]=C_{1}\cup C_{2}$. Since $C_{1}$ and $C_{2}$ are open in
$S^{\mathrm{t}}_{X}(\mathcal{R})$, we have $C_{1}=\bigcup_{i\in I}[V_{i}]$ and
$C_{2}=\bigcup_{j\in J}[W_{j}]$ for $V_{i}$ and $W_{j}$ open definable subsets
of $X$. The space $S_{X}(\mathcal{R})$ is quasi-compact with respect to the
Stone topology, so we can assume that $I$ and $J$ are finite. Thus, the sets
$V:=\bigcup_{i\in I}V_{i}$ and $W:=\bigcup_{j\in J}W_{j}$ are definable, open
and closed disjoint non-empty subsets of $U$ such that $U=V\cup W$, which is a
contradiction.

Let $p\in S^{\mathrm{t}}_{X}(\mathcal{R})$ and let $\widetilde{U}$ be an open
neighborhood of $p$ in $S^{\mathrm{t}}_{X}(\mathcal{R})$. Then, there is an
open definable subset $U$ of $X$ such that $p\in[U]\subset\widetilde{U}$. By
[12, Prop. 3.2.18] the definable set $U$ has finitely many definably connected
components which are open. Therefore, we can assume that $U$ is definably
connected. In particular, the set $[U]$ is connected for the spectral
topology, as required. ∎

###  3.2. The relation with infinitesimal types

Fix a definable group $G\subset R^{n}$ in the o-minimal expansion
$\mathcal{R}$ of a real closed field. As usual, by [21] we can regard $G$ as a
topological group. Moreover, by Robson’s embedding theorem [12, Thm. 10.1.8]
we can assume that the group topology coincides with the topology induced by
the ambient space $R^{n}$. We say that a set $X$ is a $G$-set if $G$ acts on
$X$ and is a $G$-space if in addition $X$ is a topological space and $G$ acts
continuously on $X$ (i.e. the group action $(g,x)\mapsto g\cdot x$ is a
continuous map).

######  Lemma 3.8.

There is an action from $G$ on $\beta_{G}(\mathcal{R})$ given by the map
$(g,p)\mapsto g\cdot p$. In particular, the set $\beta_{G}(\mathcal{R})$ is a
$G$-set.

###### Proof.

Given $g\in G$, the map $\pi_{g}:S^{\mathrm{t}}_{G}(\mathcal{R})\to
S^{\mathrm{t}}_{G}(\mathcal{R})$ given by $p\mapsto g\cdot p$ is a
homeomorphism. Hence, it maps closed points to closed points. ∎

Once we have an action from $G$ on $\beta_{G}(\mathcal{R})$, it is natural to
ask whether this action is continuous. We will prove that the action is
continuous if and only if $\beta_{G}(\mathcal{R})$ is canonically isomorphic
to the compactification $S^{\mu}_{X}(\mathcal{R})$ introduced in [19].

We recall the definition of $S^{\mu}_{X}(\mathcal{R})$ and we refer the reader
to [19, A.1] for further details and its basic properties.

######  Definition 3.9.

The infinitesimal type $\mu$ of $G$ is the partial type over $R$ consisting of
all $\mathcal{L}_{R}$-formulas defining open neighborhoods of the identity.

Fix a definable $G$-set $X\subset R^{n}$ and assume that $G$ acts definably on
$X$. Given two $\mathcal{L}_{R}$-formulas $\varphi(x)$ and $\psi(x)$ such that
$\varphi(R)\subset G$ and $\psi(R)\subset X$, we write $\varphi\cdot\psi$ to
denote the $\mathcal{L}_{R}$-formula

$(\varphi\cdot\psi)(x)=\exists u\exists v(\varphi(u)\wedge\psi(v)\wedge
x=uv).$

For $p\in S_{X}(\mathcal{R})$ we define $\mu\cdot p$ as the partial type

$(\mu\cdot p)(x)=\bigcup\left\\{(\theta\cdot\psi)(x)\ |\ \theta\in\mu\text{
and }\psi\in p\right\\}.$

For $p,q\in S_{X}(\mathcal{R})$ we say that $p\sim_{\mu}q$ if $\mu\cdot p$ and
$\mu\cdot q$ are equivalent partial types (i.e. every formula in $\mu\cdot q$
is implied by $\mu\cdot p$, and vice versa). We denote the equivalence class
of $p$ by $[p]_{\mu}$ and the quotient $S_{X}(\mathcal{R})/\sim_{\mu}$ by
$S^{\mu}_{X}(\mathcal{R})$. We consider the quotient topology on it. It is a
quasi-compact Hausdorff space, and a basis for its topology is given by open
sets of the form

$U_{\varphi}^{\mu}=\\{[p]_{\mu}\in S^{\mu}_{X}(\mathcal{R})\ |\ \mu\cdot
p\vdash\varphi\\}$

for $\varphi(x)$ an $\mathcal{L}_{R}$-formula with $\varphi(R)\subset X$, see
[19, Claim A.3]. Moreover, the natural projection
$S_{X}(\mathcal{R})\rightarrow S^{\mu}_{X}(\mathcal{R})$ is continuous and
closed [19, Claim A.2].

Henceforth we will work with the natural definable continuous action of $G$ on
itself (even though it is possible to extend our results to other $G$-spaces,
see Remark 3.16). Our purpose is to compare $\beta_{G}(\mathcal{R})$ with
$S^{\mu}_{G}(\mathcal{R})$. As first step, we have the following:

######  Lemma 3.10.

The natural projection $\pi:S^{\mathrm{t}}_{G}(\mathcal{R})\rightarrow
S^{\mu}_{G}(\mathcal{R})$ is continuous and preserves the action of $G$. In
particular, the map $\pi|_{\beta_{G}(\mathcal{R})}$ is also a continuous
surjection.

###### Proof.

Let $V$ be the preimage of $U_{\varphi}^{\mu}$ under $\pi$. Given $q\in V$,
since $\mu\cdot q\vdash\varphi$ and $\mu\cdot\mu=\mu$, we can find by
compactness some $\theta\in\mu$ and $\psi\in q$ such that
$(\theta\cdot\theta\cdot\psi)\vdash\varphi$. By assumption, the definable set
$(\theta\cdot\psi)(R)$ is open, as so is $\theta(R)$. Hence, $V$ is also open
since $q\in[\theta\cdot\psi]\subset V$. We have used that
$[\theta\cdot\psi]\subset V$ because
$(\theta\cdot\theta\cdot\psi)\vdash\varphi$. ∎

One of the main properties of the space $S_{G}^{\mu}(\mathcal{R})$ is that $G$
acts continuoulsy on it [19, Claim A.5]. We now give an equivalent condition
for the continuity of the action of $G$ on $\beta_{G}(\mathcal{R})$.

######  Proposition 3.11.

The action of $G$ on $\beta_{G}(\mathcal{R})$ is continuous if and only if the
map $\pi|_{\beta_{G}(\mathcal{R})}$ is a homeomorphism.

###### Proof.

Assume that the action of $G$ on $\beta_{G}(\mathcal{R})$ is continuous and
consider the map $i:G\rightarrow\beta_{G}(\mathcal{R})$ given by
$g\mapsto\text{tp}(g/R)$. Note that $i$ is by Remark 2.11 definably separated
(by closed sets). Thus, by [19, Claim A.12] there is a continuous map
$i_{*}:S^{\mu}_{G}(\mathcal{R})\rightarrow\beta_{G}(\mathcal{R})$ that extends
$i$. Notice now that
$i_{*}\circ\pi|_{\beta_{G}(\mathcal{R})}:\beta_{G}(\mathcal{R})\to\beta_{G}(\mathcal{R})$
is the identity on realized types. Hence, it is the identity, so
$\beta_{G}(\mathcal{R})$ and $S^{\mu}_{G}(\mathcal{R})$ are homeomorphic.
Reciprocally, if $\pi|_{\beta_{G}(\mathcal{R})}$ is a homeomorphism, then we
deduce since it preserves the action of $G$ that the action of $G$ over
${\beta_{G}(\mathcal{R})}$ is continuous from the fact the action of $G$ over
$S_{G}^{\mu}(\mathcal{R})$ is continuous [19, Claim A.5]. ∎

Next, we prove that $\beta_{G}(\mathcal{R})$ and $S_{G}^{\mu}(\mathcal{R})$
are homeomorphic whenever $G$ is definably compact. We will need the following
topological lemma:

######  Lemma 3.12.

Let $Z$ be a definably compact subset of $G$, and let $V$ be an open definable
subset of $G$ with $Z\subset V$. Then there is a definable open neighborhood
$W$ of the identity such that $WZ\subset V$.

###### Proof.

Let $f:G\times G\rightarrow G$ be the continuous group operation. Note that
$f^{-1}(V)$ is an open definable subset of $G\times G$, and $\\{e\\}\times
Z\subset f^{-1}(V)$. Given $\varepsilon>0$ and $z\in G$ we denote
$\widetilde{B}_{\varepsilon}(z):=B_{\varepsilon}(z)\cap G$. For every $z\in Z$
we can consider

$f(z):=\sup\\{\varepsilon>0:\exists\delta>0,\,\widetilde{B}_{\varepsilon}(e)\cdot\widetilde{B}_{\delta}(z)\subset
V\\}>0.$

Suppose there is no $\varepsilon>0$ such that for every $z\in Z$ we have
$f(z)>\varepsilon$. Then by definable choice there is a definable curve
$\alpha:(0,\infty)\rightarrow Z$ such that $f(\alpha(t))<t$ for every $t>0$.
By o-minimality there is some $\varepsilon_{1}>0$ such that
$\alpha:(0,\varepsilon_{1})\rightarrow Z$ is continuous. As $Z$ is definably
compact, we have $z_{0}:=\lim_{t\to 0}\alpha(t)\in Z$, so there are
$\varepsilon_{0}>0$ and $\delta_{0}>0$ such that
$\widetilde{B}_{\varepsilon_{0}}(e)\cdot\widetilde{B}_{\delta_{0}}(z_{0})\subset
V$. Since $\alpha$ is continuous, the open definable set
$\alpha^{-1}(\widetilde{B}_{\delta_{0}}(z_{0}))$ contains an interval of the
form $(0,\varepsilon_{2})$. Thus,
$\alpha(t)\in\widetilde{B}_{\delta_{0}}(z_{0})$ for every $t<\varepsilon_{2}$.
In particular, for each $t<\varepsilon_{2}$ there is $\delta_{t}>0$ with
$\widetilde{B}_{\delta_{t}}(\alpha(t))\subset\widetilde{B}_{\delta_{0}}(z_{0})$,
so

$\widetilde{B}_{\varepsilon_{0}}(e)\cdot\widetilde{B}_{\delta_{t}}(\alpha(t))\subset\widetilde{B}_{\varepsilon_{0}}(e)\cdot\widetilde{B}_{\delta_{0}}(z_{0})\subset
V.$

It follows that $f(\alpha(t))>\varepsilon_{0}$ for every $t<\varepsilon_{2}$,
which contradicts the fact that $\varepsilon_{0}<f(\alpha(t))<t$ for every
$t<\min\\{\varepsilon_{0},\varepsilon_{2}\\}$. Consequently, there is an
$\varepsilon>0$ such that $W:=\widetilde{B}_{\varepsilon}(e)$ satisfies
$WZ\subset V$, as required. ∎

We can now state and prove the main result of this section.

######  Theorem 3.13.

For each type $p\in S_{G}(\mathcal{R})$ we have that $r(p)\sim_{\mu}p$.
Furthermore, if $p$ is bounded, then $r(p)$ is the unique type in
$\beta_{G}(\mathcal{R})$ satisfying this, i.e.,

$[p]_{\mu}\cap\beta_{G}(\mathcal{R})=\\{r(p)\\}.$

###### Proof.

Let $p\in S_{G}(\mathcal{R})$ and note first that $q\sim_{\mu}p$ if and only
if $q\vdash\mu\cdot p$, see [19, Claim 2.7]. Thus, the partial type $\mu\cdot
p$ determines a closed subset

$[\mu\cdot p]=\left\\{q\in S^{\mathrm{t}}_{G}(\mathcal{R})\ |\ q\vdash\mu\cdot
p\right\\}$

in $S^{\mathrm{t}}_{G}(\mathcal{R})$, because $[\mu\cdot p]$ is the inverse
image of $\\{[p]_{\mu}\\}$ under $\pi:S^{\mathrm{t}}_{G}(\mathcal{R})\to
S_{G}^{\mu}(\mathcal{R})$, which is continuous by Lemma 3.10. Hence, we deduce
that $\mathrm{cl}^{\rm t}(p)\subset[\mu\cdot p]$, so $r(p)\in[\mu\cdot p]$,
which yields that $r(p)\sim_{\mu}p$.

Assume now that $p$ is bounded, and let us show:
$[p]_{\mu}\cap\beta_{G}(\mathcal{R})=\\{r(p)\\}$. We can clearly assume
$r(p)=p$ and suppose, to get a contradiction, that there exists some
$q\in\beta_{G}(\mathcal{R})\setminus\\{p\\}$ such that $q\sim_{\mu}p$. Since
$\beta_{G}(\mathcal{R})$ is quasi-compact and Hausdorff, there are two
disjoint open definable subsets $U_{1}\subset G$ and $U_{2}\subset G$ with
$p\in[U_{1}],q\in[U_{2}]$ and
$\overline{U_{1}}\cap\overline{U_{2}}=\emptyset$, where $\overline{U_{i}}$
denotes the topological closure of $U_{i}$ in $G$ for $i=1,2$. In particular,
since $p$ is bounded, we can assume that $U_{1}$ is also bounded, so that
$\overline{U_{1}}$ is definably compact and contained in the open definable
set $G\setminus\overline{U_{2}}$. By Lemma 3.12 there is an open definable
neighborhood $W$ of $e$ such that $W\cdot\overline{U_{1}}\subset
G\setminus\overline{U_{2}}$. On the other hand, note that $W\cdot
U_{1}\in\mu\cdot p$. Since $q\sim_{\mu}p$, we have $q\vdash\mu\cdot p$ and
hence $q\in[W\cdot U_{1}]$, which is a contradiction, because
$U_{2}\cap(W\cdot U_{1})=\emptyset$.∎

######  Corollary 3.14.

Let $G$ be a definably compact group. Then the map $p\mapsto[p]_{\mu}$ from
$\beta_{G}(\mathcal{R})$ to $S^{\mu}_{G}(\mathcal{R})$ is a homeomorphism, so
the action of $G$ on $\beta_{G}(\mathcal{R})$ is continuous.

###### Proof.

By Lemma 3.10, the natural projection $\beta_{G}(\mathcal{R})\rightarrow
S^{\mu}_{G}(\mathcal{R})$ is continuous. In addition, it is bijective by
Theorem 3.13 and therefore it is a homeomorphism. Hence, the action of $G$ on
$\beta_{G}(\mathcal{R})$ is continuous, which follows from the proof of [19,
Claim A.5]. ∎

In general, if $p$ is not bounded then $[p]_{\mu}\cap\beta_{X}(\mathcal{R})$
is not a singleton. We refer to the following example from [19].

######  Example 3.15.

Let $G=\mathbb{R}^{2}$ and let $\mathcal{R}$ be an elementary extension of the
field of real numbers $\mathbb{R}$. Fix some infinite element $a\in R$ and
consider the types $p=\operatorname{tp}(a,0/\mathbb{R})$ and
$q=\operatorname{tp}(a,a^{-1}/\mathbb{R})$. Both types are closed in
$S^{\mathrm{t}}_{G}(\mathbb{R})$ but $p\sim_{\mu}q$. In particular, the action
$\mathbb{R}^{2}\times\beta_{G}(\mathbb{R})\rightarrow\beta_{G}(\mathbb{R})$ is
not continuous by Lemma 3.11.

######  Remark 3.16.

An inspection of the proof of Lemma 3.10 yields that: If $G$ is a definable
o-minimal group, the set $X\subset R^{n}$ is a definable $G$-space and the
action $(g,x)\mapsto g\cdot x$ is definable, continuous and for every $x\in X$
the map $g\mapsto g\cdot x$ is open, then the natural projection map
$\beta_{X}(\mathcal{R})\rightarrow S^{\mu}_{X}(\mathcal{R})$ is continuous, so
it is a homeomorphism. It is enough to adapt the proofs of Theorem 3.13 and
Corollary 3.14.

##  4\. Closed, finitely satisfiable and invariant types

Fix an o-minimal expansion $\mathcal{R}$ of a real closed field and let
$\bar{\mathcal{R}}$ be a sufficiently saturated elementary extension. Given a
definable subset $X\subset\bar{R}^{n}$ definable over $\mathcal{R}$, we can
consider the space of closed points $\beta_{X}(\bar{\mathcal{R}})$ within
$S^{\mathrm{t}}_{X}(\bar{\mathcal{R}})$, as well as the usual space of types
$S_{X}(\bar{\mathcal{R}})$. In $S_{X}(\bar{\mathcal{R}})$ we have two natural
subsets with respect to the structure $\mathcal{R}$. Namely, the set of types
$S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}$ which are finitely satisfiable in $R$
and the set of $R$-invariant types $S_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}$.
Recall that a type $p\in S_{X}(\bar{\mathcal{R}})$ is:

  1. i)

$R$-_invariant_ if for every automorphism $\sigma\in{\rm
Aut}_{R}(\bar{\mathcal{R}})$ of $\bar{\mathcal{R}}$ that fixes $R$ pointwise
we have $\sigma(p)=p$, where $\sigma(p):=\\{\psi(x,\sigma(b)):\psi(x,b)\in
p\\}$, and

  2. ii)

_finitely satisfiable_ in $R$ if for every formula $\psi(x,b)\in p$ the set of
realizations $\psi(R,b):=\psi(\bar{R},b)\cap R$ of $\psi(x,b)$ is non-empty.

We briefly recall some well-known facts on invariant and finitely satisfiable
types.

######  Remark 4.1.

It is very easy to verify that finitely satisfiable types in $R$ are
$R$-invariant. Indeed, given a formula $\psi(x,b)$ and some $\sigma\in{\rm
Aut}_{R}(\bar{\mathcal{R}})$, if $p$ is finitely satisfiable in $R$, then the
relation $\psi(a,b)\leftrightarrow\psi(a,\sigma(b))$ holds for every $a\in R$.
Thus $\psi(x,\sigma(b))\in p$ whenever $\psi(x,b)\in p$.

The sets $S_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}$ and
$S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}$ are both closed in
$S_{X}(\bar{\mathcal{R}})$. For the latter, notice that

$S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}=\bigcap\left\\{[\neg\psi]:\psi\text{ is
an $\mathcal{L}_{\bar{R}}$-formulas with $\psi(R)=\emptyset$}\right\\}.$

For the former, given any $\sigma\in{\rm Aut}_{R}(\bar{\mathcal{R}})$, the map
$S_{X}(\bar{\mathcal{R}})\rightarrow S_{X}(\bar{\mathcal{R}})\times
S_{X}(\bar{\mathcal{R}})$ defined by $p\mapsto(p,\sigma(p))$ is continuous.
Thus, the preimage $C_{\sigma}$ of the diagonal of
$S_{X}(\bar{\mathcal{R}})\times S_{X}(\bar{\mathcal{R}})$ under this map is a
closed subset of $S_{X}(\bar{\mathcal{R}})$, because
$S_{X}(\bar{\mathcal{R}})$ is Hausdorff. Hence $S_{X}(\bar{\mathcal{R}})^{\rm
inv}_{R}=\bigcap_{\sigma\in{\rm Aut}_{R}(\bar{\mathcal{R}})}C_{\sigma}$ is a
closed subset of $S_{X}(\bar{\mathcal{R}})$.

It will also be useful to regard $S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}$ as
the space of types of the externally definable sets. Recall that a set
$Z\subset R^{n}$ is called externally definable if there exists some
$\mathcal{L}_{\bar{R}}$-formula $\phi(x)$ such that $Z=\phi(R)$. If
$S_{X}(\mathcal{R})^{\text{ext}}$ denotes the set of ultrafilters of
externally definable subsets of $X$ with the Stone topology, then the map

$S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}\rightarrow
S_{X}(\mathcal{R})^{\text{ext}},\ p\mapsto\\{\psi(R):\psi(x)\in p\\}$

is certainly a homeomorphism.

###  4.1. Finitely satisfiable types are closed

We aim to prove that $S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}$ is contained and
closed in $\beta_{X}(\mathcal{R})$. In order to see this, we obtain some
results on externally definable sets which are interesting by themselves. We
show that every externally definable subset of $R^{n}$ is given by the trace
of an open definable subset of $\bar{R}^{n}$.

The cell decomposition [12, Ch. 3] is a fundamental result for the development
of o-minimal structures. A cell in the o-minimal structure $\bar{\mathcal{R}}$
is defined inductively. A subset of $\bar{R}$ is a _cell_ if it is a point or
an open interval of $\bar{R}$ with endpoints in $\bar{R}\cup\\{\pm\infty\\}$.
More generally, a subset $C\subset\bar{R}^{n+1}$ is a _cell_ if there is a
cell $D\subset\bar{R}^{n}$ an either there is a continuous $\bar{R}$-definable
function $f:D\rightarrow\bar{R}$ such that $C$ is the graph of $f$, or there
are two continuous $\bar{R}$-definable functions
$f_{1},f_{2}:D\rightarrow\bar{R}$ with $f_{1}(x)<f_{2}(x)$ for all $x\in D$
such that $C=\\{(x,y):x\in D\text{ and }f_{1}(x)<y<f_{2}(x)\\}$.

######  Lemma 4.2.

Let $C\subset\bar{R}^{n}$ be a cell. Then there exists some open cell
$U\subset\bar{R}^{n}$ such that $C\subset U$ and $C\cap R^{n}=U\cap R^{n}$.
Moreover, there is also a definable continuous retraction $r:U\to C$.

###### Proof.

We proceed by induction on $n$. For $n=1$, suppose first that $C=\\{a\\}$. In
this case set $U=(a-\delta,a+\delta)$ for some $\delta>0$ infinitesimal with
respect to a model containing $R$ and $a$. It is clear that $U$ and the
definable constant map $r:U\to C$ defined by $u\mapsto a$ satisfy the
requirements. On the other hand, if $C$ is an open interval, then it suffices
to set $U=C$ and $r$ as the identity map.

Assume that the statement holds for $n$ and let $C$ be a cell of
$\bar{R}^{n+1}$. Suppose first that $C$ is the graph $\Gamma(f)$ of a
definable continuous function $f:D\to\bar{R}$ with $D\subset\bar{R}^{n}$ a
cell. By induction, there is a definable open cell $U$ of $\bar{R}^{n}$
containing $D$ such that $D\cap R^{n}=U\cap R^{n}$ and a definable continuous
retraction $r:U\to D$. Set $g=f\circ r$, which is a definable continuous
function such that $g|_{D}=f$. Set $V=(g-\delta,g+\delta)_{U}$ for some
$\delta>0$ infinitesimal with respect to $R$, that is,

$V=\\{(\bar{x},y)\in
U\times\bar{R}\,|\,y\in(g(\bar{x})-\delta,g(\bar{x})+\delta)\\}.$

It is clear that $V\subset\bar{R}^{n+1}$ is an open cell such that $C\subset
V$. Moreover, since $U\cap R^{n}=D\cap R^{n}$ and $g|_{D}=f$, it holds $V\cap
R^{n+1}\subset C\cap R^{n+1}$. To obtain a definable continuous retraction
$\tilde{r}:V\to C$ it is enough to consider the map
$(\bar{x},y)\mapsto(r(\bar{x}),g(\bar{x}))$.

Finally, suppose that $C=(f_{1},f_{2})_{D}$ for some cell
$D\subset\bar{R}^{n}$ and some definable functions $f_{1},f_{2}:D\to\bar{R}$
with $f_{1}<f_{2}$. By induction hypothesis, there exists a definable open
cell $U$ of $\bar{R}^{n}$ containing $D$ such that $D\cap R^{n}=U\cap R^{n}$
and a definable continuous retraction $r:U\to D$. For $i=1,2$ set
$g_{i}=f_{i}\circ r:U\to\bar{R}$ and note that $g_{i}|_{D}=f_{i}$ for $i=1,2$.
Now, take $V=(g_{1},g_{2})_{U}$, that is,

$V=\\{(\bar{x},y)\in\bar{U}\times\bar{R}\,|\,y\in(g_{1}(\bar{x}),g_{2}(\bar{x}))\\}.$

It holds $C\subset V$ and that $V\cap R^{n+1}\subset C\cap R^{n+1}$. Now, to
obtain a definable retraction $\tilde{r}:V\to C$ it is enough to consider the
map $(\bar{x},y)\mapsto(r(\bar{x}),y)$, as required. ∎

######  Proposition 4.3.

Let $Z\subset R^{n}$ be an externally definable set given by a set $W$ of
$\bar{R}^{n}$, i.e. $W\cap R^{n}=Z$. Then there is a definable open set $U$ of
$\bar{R}^{n}$ such that $W\subset U$ and $Z=U\cap R^{n}$.

###### Proof.

Let $\mathcal{C}=\\{C_{1},\ldots,C_{r}\\}$ be a cell decomposition of
$\bar{R}^{n}$ compatible with $W$. For each $i$ let $U_{i}$ be the open cell
given by the previous lemma containing $C_{i}$ and note that

$Z=W\cap R^{n}=\bigsqcup_{i=1}^{r}C_{i}\cap R^{n}=\bigsqcup_{i=1}^{r}U_{i}\cap
R^{n}.$

Thus, the definable open set $U:=U_{1}\cup\ldots\cup U_{r}$ yields the result.
∎

As a consequence, we obtain:

######  Corollary 4.4.

The collection

$\left\\{[\phi]\cap S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}\ |\
\phi\in\mathcal{L}_{\bar{R}},\ \phi(\bar{R})\subset X\text{ is open and
definable}\right\\}$

is a basis of the space $S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}$ with respect
to the Stone topology. In particular, the subspace topologies on
$S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}$ induced from both the Stone and the
spectral topologies coincide.

We also deduce the following:

######  Corollary 4.5.

The set $S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}$ is closed in
$S^{\mathrm{t}}_{X}(\bar{\mathcal{R}})$ and contained in
$\beta_{X}(\bar{\mathcal{R}})$. Namely,

  1. (i)

$\mathrm{cl}^{\rm t}(S_{X}(\bar{\mathcal{R}})^{\rm
fs}_{R})=S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}$ and

  2. (ii)

for every $p\in S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}$ we have that $r(p)=p$.

###### Proof.

(i) We prove that

$S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}=\bigcap\left\\{[\neg\psi]\ |\
\psi\in\mathcal{L}_{\bar{R}},\psi(\bar{R})\subset X\text{ open and
}\psi(R)=\emptyset\right\\}.$

It is clear that the inclusion $\subset$ holds. For the other inclusion,
suppose $p\notin S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}$ and let $\varphi(x)$
be an $\mathcal{L}_{\bar{R}}$-formula witnessing this. By Proposition 4.3
there is some $\mathcal{L}_{\bar{R}}$-formula $\psi(x)$ such that
$\psi(\bar{R})$ is open with $\varphi(\bar{R})\subset\psi(\bar{R})$ and
$\varphi(R)=\psi(R)$. Hence, $p\in[\psi]$, which yields the desired equality.

(ii) By (i) the set $S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}$ is closed in
$S^{\mathrm{t}}_{X}(\bar{\mathcal{R}})$. Thus, a point of
$S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}$ is closed in
$S_{X}(\bar{\mathcal{R}})_{\mathcal{R}}^{\rm fs}$ if and only if it is closed
in $S^{\mathrm{t}}_{X}(\bar{\mathcal{R}})$. On the other hand, the previous
corollary asserts that the subspace topologies on
$S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}$ induced from both the Stone and the
spectral topology coincide. Consequently, a point of
$S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}$ is always closed in
$S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}$, as required. ∎

###  4.2. Invariant types

Once we have seen that finitely satisfiable types are closed within the
spectral topology, it is natural to ask which is the relation between them and
closed invariant types, as well as the relation between invariant and closed
types. We first prove:

######  Lemma 4.6.

The set $S_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}$ is a closed subspace of
$S_{X}^{\rm t}(\bar{\mathcal{R}})$. In particular, the set $S^{\rm
t}_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}$ of $R$-invariant types with the
induced spectral topology from $S^{\rm t}_{X}(\bar{\mathcal{R}})$ is a normal
spectral space.

###### Proof.

It is enough to prove that for $p\in S_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}$
its closure ${\rm cl}^{\rm t}(p)$ in $S^{\rm t}_{X}(\bar{\mathcal{R}})$ is
contained in $S_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}$. Indeed, if $q\in{\rm
cl}^{\rm t}(S_{X}(\bar{\mathcal{R}})^{\rm inv}_{R})$, then for each open
$\bar{R}$-definable subset $U\subset X$ with $q\in[U]$ we have $[U]\cap
S_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}\neq\emptyset$. Since
$S_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}$ is a closed subset of
$S_{X}(\bar{\mathcal{R}})$ with the Stone topology, by quasi-compactness there
is a type $p\in S_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}$ that is in all the
open neighbourhoods in $S^{\rm t}_{X}(\bar{\mathcal{R}})$ of $q$ and so
$q\in{\rm cl}^{\rm t}(p)\subset S_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}$.

Now, let $p\in S_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}$. Consider
$\sigma\in{\rm Aut}_{R}(\bar{\mathcal{R}})$ and note that $S^{\rm
t}_{X}(\bar{\mathcal{R}})\rightarrow S^{\rm t}_{X}(\bar{\mathcal{R}})$ given
by $p\mapsto\sigma(p)$ is a homeomorphism. Thus $\sigma({\rm cl}^{\rm
t}(p))={\rm cl}^{\rm t}(\sigma(p))={\rm cl}^{\rm t}(p)$. It is also clear that
$\sigma(q_{1})\in\mathrm{cl}^{\rm t}(\sigma(q_{2}))$ whenever
$q_{1}\in\mathrm{cl}^{\rm t}(q_{2})$. Therefore $\sigma$ respects the
specialization order, so by Lemmas 2.5 and 2.16 the set ${\rm cl}^{\rm t}(p)$
is a finite set totally ordered under specialization. We deduce that $\sigma$
fixes ${\rm cl}^{\rm t}(p)$ pointwise, as required.

The fact that $S^{\rm t}_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}$ is a spectral
space follows from [8, Thm. 2.1.3], as $S_{X}(\bar{\mathcal{R}})^{\rm
inv}_{R}$ is closed in $S_{X}(\bar{\mathcal{R}})$ with respect to the Stone
topology. Since $S^{\rm t}_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}$ is closed in
$S^{\rm t}_{X}(\bar{\mathcal{R}})$ and the latter is normal, we deduce that
$S^{\rm t}_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}$ is normal. ∎

######  Remark 4.7.

The statement above yields that $S_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}$ is
closed under images of the map
$r:S^{\mathrm{t}}_{X}(\bar{\mathcal{R}})\to\beta_{X}(\bar{R})$ and that the
set of closed points of $S^{\rm t}_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}$
agrees with $S^{\rm t}_{X}(\bar{\mathcal{R}})^{\rm
inv}_{R}\cap\beta_{X}(\bar{\mathcal{R}})$. As a consequence, the natural
retraction from $S^{\rm t}_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}$ onto its
closed points coincides with $r|_{S^{\rm t}_{X}(\bar{\mathcal{R}})^{\rm
inv}_{R}}$.

In general, to be a closed point of $S^{\rm t}_{X}(\bar{\mathcal{R}})^{\rm
inv}_{R}$ depends on the set $X$, see Example 4.9 below. However, when $X$ is
closed, we prove the following:

######  Theorem 4.8.

Assume that $X\subset\bar{R}^{n}$ is closed. A type is $R$-invariant, bounded
and closed if and only if it is finitely satisfiable in $R$, that is:

$S_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}\cap\beta_{X}(\bar{\mathcal{R}})\cap
S_{X}(\bar{\mathcal{R}})^{\rm bdd}=S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R},$

where $S_{X}(\bar{\mathcal{R}})^{\rm bdd}$ is the set of bounded types.

###### Proof.

By Corollary 4.5 we have $S_{X}(\bar{\mathcal{R}})^{\rm
fs}_{R}\subset\beta_{X}(\bar{\mathcal{R}})$. Hence to prove that the inclusion
$\supset$ of the statement holds, it suffices to see that any $p\in
S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}$ is bounded. Choose some $r\in\bar{R}$
with $r>\|m\|$ for every $m\in X\cap R^{n}$. Thus the definable subset

$V=\\{x\in X\ |\ \|x\|>r\\}$

is not realized in $\mathcal{R}$, hence $p\in[V^{c}]$, which implies that $p$
is bounded.

To prove the other inclusion, fix some $p\in S_{X}(\bar{\mathcal{R}})^{\rm
inv}_{R}\cap\beta_{X}(\bar{\mathcal{R}})\cap S_{X}(\bar{\mathcal{R}})^{\rm
bdd}$. Let $\varphi(x)$ be an arbitrary $\mathcal{L}_{\bar{R}}$-formula such
that $p\in[\varphi]$, and let us show that $\varphi(R)$ is non-empty. Clearly,
we may take $\varphi(\bar{R})$ to be bounded. As $\mathrm{cl}^{\rm
t}(p)=\\{p\\}$, there exist some $\mathcal{L}_{\bar{R}}$-formulas
$\psi_{i}(x)$ such that each $\psi_{i}(\bar{R})\subset X$ is closed in $X$ and

$\\{p\\}=\bigcap_{i\in I}[\psi_{i}]$

in $S^{\mathrm{t}}_{X}(\bar{\mathcal{R}})$. Thus, we have that
$\\{\psi_{i}(x)\ |\ i\in I\\}\cup\\{\neg\varphi(x)\\}$ is inconsistent with
$\mathrm{Th}(\bar{\mathcal{R}})$, so there is some finite $I_{0}\subset I$
such that, for

$\psi(x):=\bigwedge_{i\in I_{0}}\psi_{i}(x),$

we have $\mathrm{Th}(\bar{\mathcal{R}})\cup\\{\psi(x)\\}\vdash\varphi(x)$ .
Note that $p\in[\psi]$ and $\psi(\bar{R})\subset\varphi(\bar{R})$ is closed in
$\bar{R}^{n}$ as so is $X$. Thus, as $p$ is $R$-invariant, by [18, Thm. 6.5]
(see also [9, Thm. 3.5] or [25]) we deduce that $\psi(R)\neq\emptyset$, so
$\varphi(R)\neq\emptyset$ and consequently $p\in S_{X}(\bar{\mathcal{R}})^{\rm
fs}_{R}$. ∎

The following easy example shows that in general the statement above fails for
unbounded types, as well as for non-closed sets:

######  Example 4.9.

Let $\mathcal{R}$ be the real closed field and let $\bar{\mathcal{R}}$ denote
a sufficiently saturated and homogeneous elementary extension.

  1. i)

Set $X=\bar{R}$ and let $p_{1}\in S_{X}(\bar{\mathcal{R}})$ be the type at
infinity, that is, the type $p_{1}(x)$ is determined by all formulas $a<x$ for
$a\in\bar{R}$. Then $p_{1}$ is unbounded, $R$-invariant, closed in
$S^{\mathrm{t}}_{X}(\bar{\mathcal{R}})$, but it is not finitely satisfiable in
$R$.

  2. ii)

Set $Y=(0,\infty)$ and let $p_{2}\in S_{Y}(\bar{\mathcal{R}})$ be the
infinitesimal type, that is, the type $p_{2}(x)$ is determined by formulas of
the form $0<x\wedge x<a$ for $a\in\bar{R}$. Then $p_{2}$ is bounded,
$R$-invariant, closed in $S^{\mathrm{t}}_{Y}(\bar{\mathcal{R}})$, but it is
not finitely satisfiable in $R$.

When the set $X\subset R^{n}$ is bounded and closed, Remark 4.7 and Theorem
4.8 yield that the retraction $r|_{S_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}}$
maps invariant types to finitely satisfiable ones. Summarizing, we get the
following statement:

######  Corollary 4.10.

Assume $X\subset R^{n}$ is definably compact (i.e. definable, closed and
bounded). Then

$S_{X}(\bar{\mathcal{R}})^{\rm
inv}_{R}\cap\beta_{X}(\bar{\mathcal{R}})=S_{X}(\bar{\mathcal{R}})^{\rm
fs}_{R},$

and therefore there is a continuous retraction $r|_{S^{\rm
t}_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}}:S^{\rm t}_{X}(\bar{\mathcal{R}})^{\rm
inv}_{R}\rightarrow S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}$.

We stress out that by Corollary 4.5 the spectral and the Stone topologies
coincide on $S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}$ and hence there is no need
to distinguish them.

To conclude the section we prove that the retraction $r|_{S^{\rm
t}_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}}$ is essentially unique when $X\subset
R^{n}$ is definably compact.

######  Proposition 4.11.

Assume $X\subset R^{n}$ is definably compact. There is a unique continuous
retraction from $S^{\rm t}_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}$ onto
$S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}$. In particular, it is
$r|_{S_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}}$.

###### Proof.

Let $f$ denote a continuous retraction from $S^{\rm
t}_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}$ onto $S_{X}(\bar{\mathcal{R}})^{\rm
fs}_{R}$. By continuity $f$ preserves specializations, that is, if
$q\in\mathrm{cl}^{\rm t}(p)$ then $f(q)\in\mathrm{cl}^{\rm t}(f(p))$ for any
$p,q\in S^{\rm t}_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}$. Note that
$r(p)\in\mathrm{cl}^{\rm t}(p)$, so

$f(r(p))\in\mathrm{cl}^{\rm t}(f(p))=\\{f(p)\\}$

since $f(p)\in\beta_{X}(\bar{\mathcal{R}})$ by Corollary 4.5. As $f$ is a
retraction and we have by Theorem 4.8 that $r(p)\in
S_{X}(\bar{\mathcal{R}})^{\rm fs}_{R}$, we deduce $f(r(p))=r(p)$. Altogether
we conclude $r(p)=f(p)$, as desired. ∎

##  5\. An honest topology

The results of the previous section yield the existence of a continuous
retraction

$r|_{S^{\rm t}_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}}:S^{\rm
t}_{R}(\bar{\mathcal{R}})^{\rm inv}_{R}\to S_{X}(\bar{\mathcal{R}})^{\rm
fs}_{R},$

when $X\subset\bar{R}^{n}$ is a definably compact set definable in an
o-minimal saturated extension $\bar{\mathcal{R}}$ of a real closed field
$\mathcal{R}$. This retraction comes from considering the spectral topology on
the space of types and it captures in a way the fact of being infinitesimally
close. It does not seem feasible to transfer this construction for arbitrary
NIP structures $\mathcal{M}\preceq\bar{\mathcal{M}}$ due to the absence of a
natural topology on $\bar{\mathcal{M}}$. Nonetheless, Simon [23] is able to
construct for NIP structures a canonical continuous retraction
$F_{M}:S_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}\to S_{X}(\bar{\mathcal{M}})^{\rm
fs}_{M}$. His construction lacks a priori of a topological interpretation. The
purpose of this final section is to find a topological interpretation of
$F_{M}$ within the context of spectral spaces.

Let $T$ be a complete $\mathcal{L}$-theory with NIP. Henceforth we fix a model
$\mathcal{M}$ of $T$ and a saturated elementary extension $\bar{\mathcal{M}}$
of $\mathcal{M}$. For a subset $X\subset M^{n}$ definable in $\mathcal{M}$,
the retraction

$F_{M}:S_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}\to S_{X}(\bar{\mathcal{M}})^{\rm
fs}_{M}$

is continuous with respect to the Stone topology and satisfies
$F_{M}(p)|_{M}=p|_{M}$. We refer to [23, Sec. 3] for further details (cf.
[7]), but we briefly recall some details of Simon’s construction.

######  Remark 5.1.

Let $\chi(x)$ be an $\mathcal{L}_{\bar{M}}$-formula and fix a saturated model
$M^{\prime}$ of $M$ inside $\bar{\mathcal{M}}$ that contains the parameters of
$\chi$. Consider a new unary predicate $P$ and set
$\mathcal{L}_{P}=\mathcal{L}\cup\\{P\\}$ and fix the
$\mathcal{L}_{P}$-structure $(\mathcal{M}^{\prime},\mathcal{M})$ with
$P(M^{\prime})=M$. Choose inside $\bar{\mathcal{M}}$ a sufficiently saturated
elementary extension $(\mathcal{N}^{\prime},\mathcal{N})$ of
$(\mathcal{M}^{\prime},\mathcal{M})$. In [23, Sec. 3.1] it is proven that for
every $p\in S_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$ both
$p|_{N}(x)\cup\\{P(x),\chi(x)\\}$ and $p|_{N}(x)\cup\\{P(x),\neg\chi(x)\\}$
cannot be consistent. In particular, either

($\ast$) $p|_{N}(x)\cup\\{P(x)\\}\vdash\chi(x)\qquad\text{or}\qquad
p|_{N}(x)\cup\\{P(x)\\}\vdash\neg\chi(x).$

Therefore there is a unique type $F_{M}(p)\in S_{X}(\mathcal{M}^{\prime})$
such that $p|_{N}\cup\\{P(x)\\}\cup F_{M}(p)$ is consistent. Note that
$F_{M}(p)$ is finitely satisfiable in $M$ as it is consistent with $P$.

The following example shows that the retraction $F_{M}$ is certainly different
from $r|_{S^{\rm t}_{X}(\bar{\mathcal{R}})^{\rm inv}_{R}}$.

######  Example 5.2.

Let $\mathcal{R}$ be the real field and consider some sufficiently saturated
model $\bar{\mathcal{R}}$ of its theory. Let $X=[0,1]$ and let $p(x)$ be the
complete type determined by the formulas $0<x$ and $x<a$ for every
$a\in\bar{R}_{>0}$. This is an $R$-invariant type and satisfies that
$r(p)=\operatorname{tp}(0/\bar{R})$. On the other hand, the formula $x=0$ does
not belong to $p|_{M}$, so $r(p)|_{M}\neq p|_{M}=F_{M}(p)|_{M}$. Hence, both
retractions $r$ and $F_{M}$ are distinct.

Our goal is to equip $S_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$ with a topology
of normal spectral space and to show that the natural retraction to the closed
points is Simon’s retraction $F_{M}$.

######  Definition 5.3.

An $\mathcal{L}_{\bar{M}}$-formula $\theta(x)$ is honest over $M$ if for any
$\mathcal{L}_{M}$-formula $\varphi(x)$ we have
$\theta(\bar{M})\subset\varphi(\bar{M})$ whenever
$\theta(M)\subset\varphi(M)$.

By definition if $\theta(M)$ is empty, then $\theta(\bar{M})$ must be also
empty. Each $\mathcal{L}_{M}$-formula, and each
$\mathcal{L}_{\bar{M}}$-formula $\theta(x)$ with $\theta(M)=M^{|x|}$, is
honest. Furthermore, the following result of Chernikov and Simon yields the
existence of honest formulas (the third clause appears in [7, Prop. 3.11]):

######  Fact 5.4.

[6, Prop. 1.7] For any $\mathcal{L}_{\bar{M}}$-formula $\psi(x)$ there exists
an $\mathcal{L}_{\bar{M}}$-formula $\theta(x)$ honest over $M$ such that
$\theta(M)=\psi(M)$. Moreover, there is an honest formula
$\theta^{\prime}(x)\in\mathcal{L}_{\bar{M}}$ such that

  1. i)

$\theta(M)=\psi(M)=\neg\theta^{\prime}(M)$,

  2. ii)

$\theta(\bar{M})\subset\neg\theta^{\prime}(\bar{M})$ and

  3. iii)

no $M$-invariant type contains the formula
$\neg\theta^{\prime}\wedge\neg\theta$.

Since it is clear from the context, we shall omit the reference “over $M$”
when talking about honestity. As a finite disjunction of honest formulas is
again honest, we can equip the set $S_{n}(\bar{\mathcal{M}})$ with a new
topology.

######  Definition 5.5.

The honest topology over $M$ on $S_{X}(\bar{\mathcal{M}})$ has as a basis of
closed sets the sets of the form $[\theta]$ where $\theta(x)$ is honest over
$M$. We denote the space of types with the honest topology by
$S^{\mathrm{h}_{M}}_{X}(\bar{\mathcal{M}})$.

######  Proposition 5.6.

The topological space $S^{\mathrm{h}_{M}}_{X}(\bar{\mathcal{M}})$ is spectral.

###### Proof.

To see that it is $\mathrm{T}_{0}$, let $p$ and $q$ be different types in
$S_{X}(\bar{\mathcal{M}})$. We show that there is an honest
$\mathcal{L}_{\bar{M}}$-formula $\theta$ so that $p\in[\theta]$ and
$q\notin[\theta]$, or $p\notin[\theta]$ and $q\in[\theta]$. Since $p\neq q$,
there is an $\mathcal{L}_{\bar{M}}$-formula $\psi$ such that $\psi\in p$ and
$\neg\psi\in q$. We first consider the cases when $p$ is not finitely
satisfiable in $M$. Then we can find an $\mathcal{L}_{\bar{M}}$-formula
$\psi_{0}\in p$ such that $\psi_{0}(M)=\emptyset$ and
$\psi_{0}(\bar{M})\subset\psi(\bar{M})$. Thus $\neg\psi_{0}$ is honest and
clearly $q\in[\neg\psi_{0}]$, which yields the result. Likewise, we obtain the
result whenever $q$ is not finitely satisfiable. Therefore we may assume that
both $p$ and $q$ are finitely satisfiable in $M$, in which case, it is enough
to take honest definitions $\theta_{1}$ and $\theta_{2}$ of $\psi$ and
$\neg\psi$ respectively. It follows that
$p\in[\theta_{1}]\setminus[\theta_{2}]$ and
$q\in[\theta_{2}]\setminus[\theta_{1}]$, as desired.

The map $S_{X}(\bar{\mathcal{M}})\to
S^{\mathrm{h}_{M}}_{X}(\bar{\mathcal{M}})$ given by $p\mapsto p$ is continuous
and thus $S^{\mathrm{h}_{M}}_{X}(\bar{\mathcal{M}})$ is quasi-compact. Hence
we get (S1). Moreover, the continuity of the map above yields that every basic
open set is quasi-compact. Hence, the set of all quasi-compact open sets is
the set of finite unions of basic open sets. Thus (S2) and (S3) follow.
Finally, to show that $S^{\mathrm{h}_{M}}_{X}(\bar{\mathcal{M}})$ is sober,
let $C$ be a nonempty closed and irreducible subset of
$S^{\mathrm{h}_{M}}_{X}(\bar{\mathcal{M}})$. It is straightforward to check
that the set

$\Sigma(x)=\\{\theta(x)\in\mathcal{L}_{\bar{M}}\ |\ C\subset[\theta],\
[\theta]\text{ honest}\\}\cup\\{\neg\theta(x)\in\mathcal{L}_{\bar{M}}\ |\
C\not\subset[\theta],\ [\theta]\text{ honest}\\}$

is consistent. Therefore we can complete it into a type $p\in
S_{X}(\bar{\mathcal{M}})$. Clearly $p\in C$ because $p$ belongs to all the
basic closed sets containing $C$, so $\text{cl}^{\text{h}}(p)\subset C$.
Suppose there is $q\in C$ such that $q\notin\text{cl}^{\text{h}}(p)$. Then
there is an honest formula $\theta_{0}$ such that $q\in[\neg\theta_{0}]$ and
$p\in[\theta_{0}]$. By definition of $\Sigma(x)$ and since $p\in[\theta_{0}]$,
we deduce $C\subset[\theta_{0}]$, so $q\in[\theta_{0}]$, which is a
contradiction. It follows that $\text{cl}^{\text{h}}(p)=C$, as required. ∎

######  Remark 5.7.

The induced honest topology and the constructible topology on
$S_{X}(\bar{\mathcal{M}})^{\rm fs}_{M}$ coincide and hence there is no need to
distinguish them. In fact, one can prove that every basic closed set in
$S_{X}(\bar{\mathcal{M}})^{\rm fs}_{M}$ is open. Indeed, a basic closed subset
is of the form $S_{X}(\bar{\mathcal{M}})^{\rm fs}_{M}\cap[\theta]$, for some
honest formula $\theta$. We may assume $S_{X}(\bar{\mathcal{M}})^{\rm
fs}_{M}\not\subset[\theta]$. Thus, $\neg\theta(M)\neq\emptyset$, so we can
find an honest definition $\psi$ for $\neg\theta$. Since
$\psi(M)=\neg\theta(M)$, we deduce $S_{X}(\bar{\mathcal{M}})^{\rm
fs}_{M}\cap[\neg\theta]=S_{X}(\bar{\mathcal{M}})^{\rm fs}_{M}\cap[\psi]$ is
also closed.

######  Lemma 5.8.

Every finitely satisfiable type in $M$ is a closed point of
$S^{\mathrm{h}_{M}}_{X}(\bar{\mathcal{M}})$.

###### Proof.

Let $p$ be finitely satisfiable in $M$. We show that it is closed. Pick
$q\in\text{cl}^{\text{h}}(p)$ and suppose that $q$ is not finitely satisfiable
in $M$. Thus, there is some formula $\psi\in q$ such that $\psi(M)=\emptyset$.
Since $\neg\psi$ is honest, the type $q$ belongs to the open set $[\psi]$ and
$p\notin[\psi]$, we get a contradiction. Hence, the type $q$ is finitely
satisfiable in $M$ as well. This yields that $p=q$, since
$S_{X}(\bar{\mathcal{M}})^{\rm fs}_{M}$ with the inherited honest topology is
a Hausdorff space. ∎

######  Example 5.9.

In general, it is not true that the closed points coincide with the finitely
satisfiable types in $M$. For example, if $M=\mathbb{R}_{\text{alg}}$ and
$\bar{\mathcal{M}}$ is a saturated elementary extension containing
$\mathbb{R}$, the type $\text{tp}(\pi/{\bar{M}})$ is not finitely satisfiable
in $M$. However, both intervals $[0,\pi]$ and $[\pi,4]$ of $\bar{M}$ are
defined by honest (over $M$) $\mathcal{L}_{\bar{\mathcal{M}}}$-formulas, and
$[0,\pi]\cap[\pi,4]=\\{\pi\\}$, so that $\text{tp}(\pi/{\bar{M}})$ is closed
in $S^{\mathrm{h}_{M}}_{X}(\bar{\mathcal{M}})$. One can check that the honest
topology is normal in this example.

In the light of the example above, we ask:

######  Question.

Is in general $S^{\mathrm{h}_{M}}_{X}(\bar{\mathcal{M}})$ a normal topological
space?

Henceforth, the space $S_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$ with the
subspace honest topology is denoted by $S^{{\rm
h}_{M}}_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$. Next, we prove that it is a
spectral topological space. In addition, we will also see in Theorem 5.13 that
it is normal.

######  Proposition 5.10.

The topological space $S^{{\rm h}_{M}}_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$
is spectral.

###### Proof.

This follows from [8, Thm. 2.1.3]. Nevertheless, we give a direct proof for
completeness. Properties (S1)-(S3) follow from the fact that
$S^{\mathrm{h}_{M}}_{X}(\bar{\mathcal{M}})$ is a spectral space and
$S_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$ is closed in
$S_{X}(\bar{\mathcal{M}})$ with the Stone topology. Let us show (S4). Pick a
closed and irreducible set $C_{0}$ of $S^{{\rm
h}_{M}}_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$. Consider its closure
$C:=\text{cl}^{\text{h}}(C_{0})$ in
$S^{\mathrm{h}_{M}}_{X}(\bar{\mathcal{M}})$. We clearly have that the closed
and irreducible subset $C$ of $S^{\mathrm{h}_{M}}_{X}(\bar{\mathcal{M}})$
satisfies $C\cap S_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}=C_{0}$. Hence, since
$S^{\mathrm{h}_{M}}_{X}(\bar{\mathcal{M}})$ is a spectral space, there is a
unique $p\in S^{\mathrm{h}_{M}}_{X}(\bar{\mathcal{M}})$ such that
$\text{cl}^{\text{h}}(p)=C$. Let us see that $p\in
S_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$. For each $\sigma\in{\rm
Aut}_{M}(\bar{\mathcal{M}})$ the induced map
$S^{\mathrm{h}_{M}}_{X}(\bar{\mathcal{M}})\rightarrow
S^{\mathrm{h}_{M}}_{X}(\bar{\mathcal{M}})$ given by $p\mapsto\sigma(p)$ is a
homeomorphism. Indeed, it is enough to note that if $\theta(x,d)$ is honest,
then so is $\theta(x,\sigma(d))$. Hence, it follows that

$\textrm{cl}^{\rm h}(\sigma(p))=\sigma(\textrm{cl}^{\rm
h}(p))=\sigma(C)=\sigma(\textrm{cl}^{\rm h}(C_{0}))=\textrm{cl}^{\rm
h}(\sigma(C_{0}))=\textrm{cl}^{\rm h}(C_{0})=C$

and by uniqueness of $p$ we deduce that $\sigma(p)=p$. Thus $p\in C\cap
S_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}=C_{0}$, so $C_{0}$ is the closure of
$p$ in $S^{{\rm h}_{M}}_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$, as required. ∎

We next prove normality of $S^{{\rm h}_{M}}_{X}(\bar{\mathcal{M}})^{\rm
inv}_{M}$. To that end we need to characterize Simon’s $F_{M}$ retraction.

######  Proposition 5.11.

Let $p\in S_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$. An
$\mathcal{L}_{\bar{M}}$-formula $\psi(x)$ belongs to $F_{M}(p)$ if and only if
there is an honest $\mathcal{L}_{\bar{M}}$-formula $\theta(x)\in p$ such that
$\theta(M)=\psi(M)$.

###### Proof.

($\Rightarrow$) Let $\psi(x)$ be an $\mathcal{L}_{\bar{M}}$-formula and
suppose $\psi(x)\in F_{M}(p)$. Let
$\mathcal{M}^{\prime}\preceq\bar{\mathcal{M}}$ be an $|M|^{+}$-saturated
elementary extension of $\mathcal{M}$ containing the parameters of $\psi(x)$.
Now, working in the language $\mathcal{L}_{P}=\mathcal{L}\cup\\{P\\}$,
consider a sufficiently saturated elementary extension
$(\mathcal{N}^{\prime},\mathcal{N})$ of $(\mathcal{M}^{\prime},\mathcal{M})$.
Set

$S:=\left\\{q\in S_{X}(\bar{\mathcal{M}})^{\rm
inv}_{M}:q|_{N}(x)\cup\\{P(x)\\}\cup\\{\psi(x)\\}\text{ is
consistent}\right\\}.$

By ($\ast$ ‣ 5.1) in Remark 5.1 for every $q\in S$ there is some
$\mathcal{L}_{N}$-formula $\theta_{q}$ such that $\theta_{q}(x)\wedge
P(x)\vdash\psi(x)$. Since $S$ is a closed subset of
$S_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$ and $S\subset\bigcup_{q\in
S}[\theta_{q}]$, by quasi-compactness there is an $\mathcal{L}_{N}$-formula
$\theta(x)$ such that $\theta(x)\wedge P(x)\vdash\psi(x)$ and
$S\subset[\theta]$. In particular, there is no type $q\in
S_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$ for which the set
$q|_{N}(x)\cup\\{P(x)\\}\cup\\{\psi(x)\wedge\neg\theta(x)\\}$ is consistent.

By definition of $F_{M}(p)$ we have $p\in S$, so $\theta\in p$. Therefore, it
is enough to show that $\theta$ is an honest and $\theta(M)=\psi(M)$. For the
latter, if $a\in\psi(M)$ then $\operatorname{tp}(a/\bar{M})\in S$ and so
$a\in\theta(M)$. On the other hand, if $a\in\theta(M)$, then, since
$\theta(x)\wedge P(x)\vdash\psi(x)$, we deduce $a\in\psi(M)$. Hence
$\theta(M)=\psi(M)$. Finally, to prove that $\theta(x)$ is honest, let
$\varphi(x)$ be an $\mathcal{L}_{M}$-formula with
$\psi(M)=\theta(M)\subset\varphi(M)$. Then
$(\mathcal{M}^{\prime},\mathcal{M})$ and so
$(\mathcal{N}^{\prime},\mathcal{N})$ satisfy

$\forall x(\psi(x)\wedge P(x)\rightarrow\varphi(x)).$

As $\theta(x)\wedge P(x)\vdash\psi(x)$, we deduce that $(N^{\prime},N)$
satisfies $\forall x(\theta(x)\wedge P(x)\rightarrow\varphi(x))$ and, since
$\theta$ has parameters in $N$, we deduce
$\theta(\bar{M})\subset\varphi(\bar{M})$.

($\Leftarrow$) Fix $\psi(x)$ and let $\theta(x,b)\in p$ be an honest
$\mathcal{L}_{\bar{M}}$-formula such that $\theta(M,b)=\psi(M)$. As before,
consider an $|M|^{+}$-saturated elementary extension
$\mathcal{M}^{\prime}\preceq\bar{\mathcal{M}}$ of $\mathcal{M}$ containing the
parameters of $\psi(x)$ and $\theta(x,b)$. Also, consider a sufficiently
saturated elementary extension $(\mathcal{N}^{\prime},\mathcal{N})$ of
$(\mathcal{M}^{\prime},\mathcal{M})$.

Suppose that $\psi(x)\not\in F_{M}(p)$. As $\psi(M)=\theta(M,b)$, we deduce
$\neg\theta(x,b)\in F_{M}(p)$. By the implication ($\Rightarrow$), there is
some honest $\theta_{0}(x,c)\in p|_{N}$ such that
$\theta_{0}(M,c)=\neg\theta(M,b)$ and $\theta_{0}(x,c)\wedge
P(x)\vdash\neg\theta(x,b)$. Since $p$ is $M$-invariant and
$p\in[\theta(x,b)\wedge\theta_{0}(x,c)]$, the formula
$\theta(x,b)\wedge\theta_{0}(x,c)$ is consistent. Set

$\varphi(y,z):=\exists x(\theta(x,y)\wedge\theta_{0}(x,z)).$

The pair $(N^{\prime},N)$ satisfies the $\mathcal{L}_{M^{\prime}}$-sentence

$\exists z(P(z)\wedge\varphi(b,z)\wedge\forall x\big{(}\theta_{0}(x,z)\wedge
P(x)\rightarrow\neg\theta(x,b)\big{)},$

and so does $(M^{\prime},M)$. Therefore there exists $m\in M$ such that
$\varphi(b,m)$ holds and $\theta_{0}(M,m)\subset\neg\theta(M,b)$. In
particular, we obtain that
$\theta(\bar{M},b)\cap\theta_{0}(\bar{M},m)\neq\emptyset$ by the definition of
$\varphi(y,z)$ and also that $\theta(M,b)\subset\neg\theta_{0}(M,m)$. However,
as $\theta(x,b)$ is honest and $\theta_{0}(x,m)$ is an
$\mathcal{L}_{M}$-formula, the latter yields
$\theta(\bar{M},b)\subset\neg\theta_{0}(\bar{M},m)$, which is a contradiction.
∎

As an immediate consequence, we obtain the following:

$F_{M}(p)\in\bigcap\left\\{[\theta]:{\theta\in p\text{ is
honest}}\right\\}=\mathrm{cl}^{\rm h}(p).$

We will use the following notation: given an $\mathcal{L}_{\bar{M}}$-formula
$\psi(x)$, we write

$[\psi]^{\rm inv}:=\\{p\in S_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}:\psi\in
p\\}.$

######  Corollary 5.12.

If $\theta_{1},\ldots,\theta_{\ell}$ are honest over $M$ formulas and
$[\theta_{1}]^{{\rm inv}}\cap\cdots\cap[\theta_{\ell}]^{{\rm inv}}$ is non-
empty, then

$(\theta_{1}\wedge\cdots\wedge\theta_{\ell})(M)\neq\emptyset.$

###### Proof.

By assumption, there is an $M$-invariant type $p\in[\theta_{i}]^{{\rm inv}}$
for $i=1,\ldots,\ell$, so by Proposition 5.11 we have
$\theta_{1},\ldots,\theta_{\ell}\in F_{M}(p)$. Since $F_{M}(p)$ is finitely
satisfiable in $M$, we deduce
$(\theta_{1}\wedge\cdots\wedge\theta_{\ell})(M)\neq\emptyset$, as required. ∎

Recall that if $X$ is a normal spectral space, then the set of the closed
points $\text{Max}(X)$ is a quasi-compact Hausdorff topological space.
Moreover, for every $x\in X$ there is a unique closed point
$r(x)\in\text{cl}(x)$ and the map $r:X\rightarrow\text{Max}(X)$ is a
continuous retraction (see Fact 2.6). In the following result we show that
$S^{{\rm h}_{M}}_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$ is a normal spectral
space and that $F_{M}$ coincides with the natural retraction to the closed
points.

######  Theorem 5.13.

The closed points of the spectral topological space $S^{{\rm
h}_{M}}_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$ are exactly the finitely
satisfiable types. Moreover, the space $S^{{\rm
h}_{M}}_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$ is normal and the natural
retraction to the closed points

$r_{M}^{\rm h}:S^{{\rm h}_{M}}_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}\rightarrow
S_{X}(\bar{\mathcal{M}})^{\rm fs}_{M}$

coincides with $F_{M}$.

###### Proof.

Let $p\in S_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$ be a closed point and let us
show that $p$ is finitely satisfiable in $M$. The converse follows from Lemma
5.8. Since

$\bigcap_{\theta\in p\text{ honest}}[\theta]^{\rm inv}=\mathrm{cl}^{\rm
h}(p)\cap S_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}=\\{p\\},$

for any $\psi\in p$ there are finitely many honest formulas
$\theta_{1},\ldots,\theta_{\ell}\in p$ such that

$[\theta_{1}]^{\rm inv}\cap\cdots\cap[\theta_{\ell}]^{\rm
inv}\subset[\psi]^{\rm inv}.$

Thus, by Corollary 5.12 we have
$\emptyset\neq(\theta_{1}\wedge\cdots\wedge\theta_{\ell})(M)\subset\psi(M)$.

To verify that the space is normal, consider two closed non-empty disjoint
subsets $C_{1}$ and $C_{2}$ of $S^{{\rm h}_{M}}_{X}(\bar{\mathcal{M}})^{\rm
inv}_{M}$. We can assume that $C_{1}=[\psi_{1}]^{\rm inv}$ and
$C_{2}=[\psi_{2}]^{\rm inv}$ where both $\psi_{1}$ and $\psi_{2}$ are a finite
conjunction of honest formulas. By Corollary 5.12, we have
$\psi_{1}(M)\neq\emptyset$ and $\psi_{2}(M)\neq\emptyset$. Thus, by Fact 5.4
there are two honest $\mathcal{L}_{\bar{M}}$-formulas $\theta_{1}(x)$ and
$\theta_{2}(x)$ such that:

  * •

$\theta_{1}(M)=\psi_{1}(M)=\neg\theta_{2}(M)$,
$\theta_{1}(\bar{M})\subset\neg\theta_{2}(\bar{M})$ and

  * •

no $M$-invariant type contains the formula
$\neg\theta_{2}\wedge\neg\theta_{1}$.

In particular, the set $[\theta_{1}]^{\rm inv}=[\neg\theta_{2}]^{\rm inv}$ is
open and closed in $S_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$. In addition, we
check that $[\psi_{i}]^{\rm inv}\subset[\theta_{i}]^{\rm inv}$ for $i=1,2$.
Indeed, if $[\psi_{1}]^{\rm inv}\cap[\theta_{2}]^{\rm inv}$ were non-empty,
then it would follow by Corollary 5.12 that
$\psi_{1}(M)\cap\theta_{2}(M)\neq\emptyset$, which is a contradiction.
Therefore $[\psi_{1}]^{\rm inv}\subset[\neg\theta_{2}]^{\rm
inv}=[\theta_{1}]^{\rm inv}$. Likewise, we get that $[\psi_{2}]^{\rm
inv}\cap[\theta_{1}]^{\rm inv}=\emptyset$, as otherwise
$\psi_{2}(M)\cap\theta_{1}(M)$ and hence $\psi_{1}(M)\cap\psi_{2}(M)$ would be
non-empty. Consequently $[\psi_{2}]^{\rm inv}\subset[\neg\theta_{1}]^{\rm
inv}=[\theta_{2}]^{\rm inv}$, as required.

Once we have shown that $S^{{\rm h}_{M}}_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$
is a normal spectral space, the retraction $r_{M}^{\rm h}(p)$ is by definition
the unique closed point in $\text{cl}^{\rm h}(p)$, so $r_{M}^{\rm
h}(p)=F_{M}(p)$ for every $p\in S^{{\rm h}_{M}}_{X}(\bar{\mathcal{M}})^{\rm
inv}_{M}$. In particular, the map $F_{M}$ is continuous for the honest
topology. ∎

######  Remark 5.14.

We write down an alternative proof of the normality of $S^{{\rm
h}_{M}}_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$ by spectral topological means.
By [4, Prop. 2], to prove that $S^{{\rm h}_{M}}_{X}(\bar{\mathcal{M}})^{\rm
inv}_{M}$ is normal it is enough to show for any $p\in S^{{\rm
h}_{M}}_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$ that $\text{cl}^{\rm h}(p)\cap
S_{X}(\bar{\mathcal{M}})^{\rm fs}_{M}=\\{F_{M}(p)\\}$. Fix $q\in
S_{X}(\bar{\mathcal{M}})^{\rm fs}_{M}$ such that $q\in\text{cl}^{\rm h}(p)$
and $q\neq F_{M}(p)$. Then there is an honest formula $\theta(x)\in q$ such
that $\neg\theta(x)\in F_{M}(p)$. By Proposition 5.11 there is an honest
formula $\theta_{0}\in p$ with $\theta_{0}(M)=\neg\theta(M)$. Since
$p\in[\theta_{0}]^{\rm inv}$, we get $q\in[\theta_{0}]^{\rm inv}$, so
$\theta\wedge\theta_{0}\in q$, which is a contradiction since
$(\theta\wedge\theta_{0})(M)=\emptyset$ and $q$ is finitely satisfiable in
$M$.

As in the case of the spectral topology, the closure of a point in $S^{{\rm
h}_{M}}_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$ is totally ordered under
specialization. However, the reason why this is true in the honest topology is
radically different and concerns the small size of such closure:

######  Proposition 5.15.

For every $p\in S^{{\rm h}_{M}}_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$ we have
that $\mathrm{cl}^{\rm h}(p)=\\{p,F_{M}(p)\\}$.

###### Proof.

Assume $p\notin S_{X}(\bar{\mathcal{M}})^{\rm fs}_{M}$, since the points of
the latter set are closed. Suppose that there is some $p_{1}\in\text{cl}^{\rm
h}(p)$ with $p_{1}\neq p$ and $p_{1}\neq F_{M}(p)$. Note that

$F_{M}(p)=r_{M}^{\rm h}(p)=r_{M}^{\rm h}(p_{1})\in\text{cl}^{\rm h}(p_{1}).$

Since the honest topology is $\mathrm{T}_{0}$ and $p_{1}\neq F_{M}(p)$, there
is an honest formula $\theta_{1}(x)$ such that $\theta_{1}\in F_{M}(p)$ but
$\theta_{1}\notin p_{1}$. We claim: $\text{cl}^{h}(p)=\text{cl}^{h}(p_{1})$.

It is enough to show that $\text{cl}^{h}(p)\subset\text{cl}^{h}(p_{1})$, so
pick an honest formula $\theta(x)$ with $p_{1}\in[\theta]$. Thus,
$\theta\wedge\theta_{1}\in F_{M}(p)$ and so
$(\theta\wedge\theta_{1})(M)\neq\emptyset$. By Proposition 5.11 there is an
honest formula $\theta^{\prime}\in p$ with
$(\theta\wedge\theta_{1})(M)=\theta^{\prime}(M)$. Thus
$\theta\wedge\theta^{\prime}\wedge\neg\theta_{1}\in p_{1}$, as
$p_{1}\in\text{cl}^{\rm h}(p)$. Note that
$(\theta\wedge\theta^{\prime}\wedge\neg\theta_{1})(M)=\emptyset$, so
$\theta\wedge\theta^{\prime}\wedge\neg\theta_{1}$ is an open set in the honest
topology. Hence, as
$p_{1}\in[\theta\wedge\theta^{\prime}\wedge\neg\theta_{1}]$, it follows
$p\in[\theta\wedge\theta^{\prime}\wedge\neg\theta_{1}]$ and so $p\in[\theta]$,
as required.

However, once we have shown that $\text{cl}^{h}(p)=\text{cl}^{h}(p_{1})$,
since the honest topology is $\mathrm{T}_{0}$, it follows that $p=p_{1}$,
which is a contradiction. ∎

We finish the paper pointing out two more differences between the spectral and
the honest topology (see Lemma 3.7 and Lemma 2.16).

######  Lemma 5.16.

The spectral spaces $S^{\mathrm{h}_{M}}_{X}(\bar{\mathcal{M}})$ and $S^{{\rm
h}_{M}}_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$ are not locally connected.

###### Proof.

Let $S^{\mathrm{h}_{M}}_{X}(\bar{\mathcal{M}})^{\diamond}$ denote one of the
two topological space. Suppose that
$S^{\mathrm{h}_{M}}_{X}(\bar{\mathcal{M}})^{\diamond}$ is locally connected
and fix a connected component $C$ of
$S^{\mathrm{h}_{M}}_{X}(\bar{\mathcal{M}})^{\diamond}$ containing a non-
realized $M$-invariant type. As $C$ is open and closed, there are honest
$\mathcal{L}_{\bar{M}}$-formulas $\theta_{1},\ldots,\theta_{\ell}$ such that
$C=[\theta_{1}]^{\diamond}\cap\cdots\cap[\theta_{\ell}]^{\diamond}$, where
$[\theta_{i}]^{\diamond}$ denotes $[\theta_{i}]\cap
S^{\mathrm{h}_{M}}_{X}(\bar{\mathcal{M}})^{\diamond}$. By Corollary 5.12 there
is some $m\in(\theta_{1}\wedge\cdots\wedge\theta_{\ell})(M)$. Both the
formulas $x=m$ and $\neg(x=m)$ are $M$-definable and therefore they are
honest. Since

$C=[\theta_{1}]^{\diamond}\cap\cdots\cap[\theta_{\ell}]^{\diamond}\subset[x=m]\cup[\neg(x=m)]$

and $C$ is connected, we deduce that $C$ equals $[x=m]$, which is a
contradiction. ∎

######  Lemma 5.17.

Assume that for every model of the theory the algebraic closure provides a
geometry (so that we have a well-defined dimension). Then
$\dim(p)=\dim(F_{M}(p))$ for all $p\in S^{{\rm
h}_{M}}_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$.

###### Proof.

For each $p\in S^{{\rm h}_{M}}_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$ we have
$p|_{M}=F_{M}(p)|_{M}$. So, clearly it is enough to prove: if $q\in S^{{\rm
h}_{M}}_{X}(\bar{\mathcal{M}})^{\rm inv}_{M}$ with $\dim(q)=d$, then
$\dim(q|_{M})=d$.

Let $a$ be a realization of $q$ in some elementary extension of
$\bar{\mathcal{M}}$ and let $b=(b_{1},\ldots,b_{\ell})\in\bar{M}^{\ell}$ be
such that $\dim(b/M)=\ell$ and $\dim(a/M,b)=d$. Suppose $\dim(a/M)>d$. Since

$\dim(a,b/M)=\dim(a/M,b)+\dim(b/M)=\dim(b/M,a)+\dim(a/M),$

it follows that $\dim(b/M,a)<\dim(b/M)=\ell$. Then there exists an
$\mathcal{L}_{M}$-formula $\psi(x,y_{1},\ldots,y_{\ell})$ such that
$\psi(a,b_{1},\ldots,b_{\ell-1},y_{\ell})$ defines a finite set and
$\psi(a,b_{1},\ldots,b_{\ell})$ holds, so $\psi(x,b_{1},\ldots,b_{\ell})\in
q$. By $M$-invariance of $q$, for each automorphism $\sigma$ of
$\bar{\mathcal{M}}$ that fixes $M,b_{1},\ldots,b_{\ell-1}$ pointwise we have
that $\psi(x,b_{1},\ldots,b_{\ell-1},\sigma(b_{\ell}))$ belongs to $q$, so $a$
realizes each $\psi(x,b_{1},\ldots,b_{\ell-1},\sigma(b_{\ell}))$. Thus, since
$\psi(a,b_{1},\ldots,b_{\ell-1},y_{\ell})$ only has a finite number of
realizations, we deduce that $b_{\ell}$ has a finite orbit, so it is in the
algebraic closure of $M,b_{1},\ldots,b_{\ell-1}$ and hence $\dim(b/M)<\ell$,
which is a contradiction. ∎

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