# Additive Reducts of real closed fields and strongly bounded structures

Hind Abu Saleh  Department of Mathematics, University of Haifa, Haifa, Israel
<EMAIL_ADDRESS>and  Ya’acov Peterzil  Department of Mathematics,
University of Haifa, Haifa, Israel<EMAIL_ADDRESS>
###### Abstract.

Given a real closed field $R$, we identify exactly four proper reducts of $R$
which expand the underlying (unordered) $R$-vector space structure. Towards
this theorem we introduce a new notion, of strongly bounded reducts of
linearly ordered structures:

A reduct $\mathcal{M}$ of a linearly ordered structure $\langle
R;<,\cdots\rangle$ is called _strongly bounded_ if every
$\mathcal{M}$-definable subset of $R$ is either bounded or co-bounded in $R$.
We investigate strongly bounded additive reducts of o-minimal structures and
as a corollary prove the above theorem on additive reducts of real closed
fields.

The second author is partially supported by the Israel Science Foundation
grant 290/19.

##  1\. Introduction

The motivation behind the work here is a conjecture about reducts of real
closed fields, from [10]. Before stating the conjecture, let us clarify our
usage of the notion of “reduct” here.

######  Definition 1.1.

Given two structures $\mathcal{M}$ and $\mathcal{N}$, we say that
$\mathcal{M}$ is _reduct of_ $\mathcal{N}$ (or, $\mathcal{N}$ is _an expansion
of_ $\mathcal{M}$), denoted by $\mathcal{M}\dot{\subseteq}\mathcal{N}$, if
$\mathcal{M}$ and $\mathcal{N}$ have the same universe and every set that is
definable in $\mathcal{M}$ is also definable in $\mathcal{N}$ (where
definability allows parameters). We say that $\mathcal{M}$ and $\mathcal{N}$
are _interdefinable_ , denoted by $\mathcal{M}\dot{=}\mathcal{N}$, if
$\mathcal{M}$ is reduct of $\mathcal{N}$ and $\mathcal{N}$ is reduct of
$\mathcal{M}$.

We say $\mathcal{M}$ is _a proper reduct_ of $\mathcal{N}$ (or, $\mathcal{N}$
a proper expansion of $\mathcal{M}$) if
$\mathcal{M}\dot{\subseteq}\mathcal{N}$ and not
$\mathcal{M}\dot{=}\mathcal{N}$.

Below, we let $\Lambda_{R}$ be the family of all $R$-linear maps
$\lambda_{a}(x)=\alpha x$, for all $\alpha\in R$. Our ultimate goal here is to
prove:

######  Theorem 1.2.

Let $R$ be a real closed field. Then, the only reducts between the vector
space $\langle R;+,\Lambda_{R}\rangle$ and the field $\langle
R;<,+,\cdot\rangle$ are

$\mathcal{R}_{alg}:=\langle R;<,+,\cdot\rangle$

$\mathcal{R}_{sb}:=\langle R;<,+,\Lambda_{R},\mathfrak{B}_{sa}\rangle$

$\mathcal{R}_{semi}:=\langle R;<,+,\Lambda_{R}\rangle$
$\mathcal{R}_{bd}:=\langle R;<^{*},+,\Lambda_{R},\mathfrak{B}_{sa}\rangle$

$\mathcal{R}_{lin}^{*}:=\langle R;<^{*},+,\Lambda_{R}\rangle$

$\mathcal{R}_{lin}:=\langle R;+,\Lambda_{R}\rangle$,

where $<^{*}$ is the linear order on the interval $(0,1)$ and
$\mathfrak{B}_{sa}$ the collection of all bounded semialgebraic sets over $R$.

######  Remark 1.3.

  1. (1)

The definable sets in $\mathcal{R}_{alg}$ are called semialgebraic, while
those definable in $\mathcal{R}_{semi}$ are semilinear. The structure
$\mathcal{R}_{sb}$ above is called semibounded, as it expands the ordered
vector space by a collection of bounded sets. Semibounded structures were
studied in several articles, for example [3],[1],[11].

  2. (2)

Notice that because all the above structures expand the full underlying
$R$-vector space, then once $<^{*}$ is definable then the restriction of $<$
to every bounded interval is definable.

  3. (3)

A similar project, in the setting of Presburger Arithmetic, was carried out in
[2], where it was proven that there are no proper reducts between
$\langle\mathbb{Z};+\rangle$ and $\langle\mathbb{Z};<,+\rangle$. We expect
that in arbitrary models of Presburger arithmetic, an analoguous result to
Theorem 1.2 holds, with the intermediate reducts corresponding to possible
restrictions of $<$ to infinite subintervals.

Some of the work towards the proof of Theorem 1.2 can be read off earlier
results. In particular, the fact that the semibounded reduct
$\mathcal{R}_{sb}$ is the only proper reduct between $\mathcal{R}_{semi}$ and
$\mathcal{R}_{alg}$, was proven over $\mathbb{R}$ in [10] and can be deduced
for arbitrary real closed field from Edmundo’s [3] (see Fact 5.1 below).
However, the bulk of the work here is to show that if a reduct $\mathcal{M}$
of $\mathcal{R}_{alg}$ does not define the full order then it is necessarily a
reduct of $\mathcal{R}_{bd}$. Towards that, we introduce a new notion, of “a
strongly bounded structure” in a more general setting, and most of our results
here are about such structures:

######  Definition 1.4.

Let $\mathcal{R}=\langle R;<,\cdots\rangle$ be a linearly ordered structure. A
reduct $\mathcal{M}=\langle R;\cdots\rangle$ of $\mathcal{R}$ is called
_strongly bounded_ if every $\mathcal{M}$-definable $X\subseteq R$ is either
bounded or co-bounded (namely, $R\setminus X$ is bounded).

######  Remark 1.5.

  1. (1)

The term “strongly bounded” was chosen to reflect a combination of a
semibounded structure with a strongly minimal one. Almost all of our work here
concerns strongly bounded additive reducts of o-minimal structures, where the
underlying linear order is dense. Analogous definitions could be given for,
say, models of Presburger arithmetic if one wishes to study all reducts which
expand the underlying ordered group.

  2. (2)

The definition of a strongly bounded structure requires an ambient linear
order, thus it might not seem amenable to working in elementarily equivalent
structures. However, in practice we only work in sufficiently saturated
elementary extensions of a strongly bounded $\mathcal{M}$ as above, and thus
we may assume that this elementary extension is also a reduct of a linearly
ordered elementary extension of $\mathcal{R}$.

By definition, if $\mathcal{M}$ is a strongly bounded reduct of a linearly
ordered structure then the ordering $<$ is not definable in $\mathcal{M}$. We
prove several results about strongly bounded reducts of o-minimal structures
(see for example Theorem 4.5 and Theorem 4.27):

######  Theorem.

Let $\langle R;<,+,\cdots\rangle$ be an o-minimal expansion of an ordered
group and let $\mathcal{M}=\langle R,+,\cdots\rangle$ be a strongly bounded
reduct. Then

  1. (1)

Every $\mathcal{M}$-definable subset of $R^{n}$ is already definable in the
structure $\langle R;+,\Lambda_{\mathcal{M}},\mathfrak{B}^{*}\rangle$, where
$\Lambda_{M}$ is the collection of $\mathcal{M}$-definable endomorphisms of
$\langle R,+\rangle$ and $\mathfrak{B}^{*}$ is the collection of all
$\mathcal{M}$-definable bounded sets.

  2. (2)

For every $\mathcal{N}\equiv\mathcal{M}$, the model theoretic algebraic
closure equals the definable closure.

Acknowledgments The article are based on results from the M.Sc. thesis of the
first author, as part of her studies at the University of Haifa. We thank Itay
Kaplan and Assaf Hasson for reading an earlier version of this work and
commenting on it.

##  2\. Proper expansions of ${\mathcal{R}}_{lin}$

In this section we assume that $\mathcal{R}_{omin}$ is an o-minimal expansion
of a real closed field $R$ and $\mathcal{M}=\langle R;+,\cdots\rangle$ is an
additive reduct of $\mathcal{R}_{omin}$.

######  Theorem 2.1.

If $\mathcal{M}$ is not a redcut of $\mathcal{R}_{lin}=\langle
R;+,\Lambda_{R}\rangle$ then $<^{*}$ is definable in $\mathcal{M}$.

###### Proof.

It is sufficient to prove that some interval $[0,b]$ is
$\mathcal{M}$-definable, for $b>0$.

######  Claim 2.2.

$Th(\mathcal{M})$ is unstable

###### Proof.

This is based on the work of Hasson and Onshuus with the second author, [4].

Assume towards contradiction that $Th(\mathcal{M})$ is stable. By [4, Theorem
1], every 1-dimensional stable structure interpretable in an o-minimal
structure is necessarily 1-based. So $\mathcal{M}$ is 1-based. By [5, Theorem
4.1], it follows that every $\mathcal{M}$-definable set is a boolean
combination of cosets of definable subgroups of $R^{n}$. Every definable
subgroup of $\langle R^{n};+\rangle$ in an o-minimal structure is an $R$\-
vector subspace of $R^{n}$ and therefore every $\mathcal{M}$-definable set is
definable in $\mathcal{R}_{lin}$, a contradiction. Hence $\mathcal{M}$ is
unstable. ∎

Because $\mathcal{M}$ is unstable, it is in particular not strongly minimal.
This generally implies that in some elementary extension of $\mathcal{M}$, we
have an $\mathcal{M}$-definable subset in one variable which is infinite and
co-infinite. However, o-minimal structures eliminate $\exists^{\infty}$, and
therefore so does $\mathcal{M}$. It follows that there is some
$\mathcal{M}$-definable subset of $R$ itself which is infinite and co-
infinite. Call this set $Y$.

By o-minimality, $Y$ has the following form:

(1) $Y:=I_{1}\cup I_{2}\cup\cdots\cup I_{n}\cup L,$

such that for every $i\in\\{1,...,n\\},I_{i}:=(a_{i},b_{i})$, $L\>$ is a
finite set and in addition $-\infty\leq
a_{1}<b_{1}<a_{2}<...<a_{n}<b_{n}\leq+\infty$. Without loss of generality
$L=\emptyset$.

The following will be used in several places in this thesis.

######  Lemma 2.3.

Assume that $Y\subseteq R$ is definable in an o-minimal expansion of an
ordered group. If both $Y$ and $R\setminus Y$ are unbounded then full linear
order is definable in $\langle R;+,Y\rangle$.

###### Proof.

If both $Y$ and $R\setminus Y$ are unbounded then $Y$ has the form (1) above
and then without loss of generality, we may assume that
$I_{1}=(-\infty,b_{1})$, and $I_{i}=(a_{i},b_{i})$ for $i\in\\{2,...,n\\}$.

By replacing $Y$ by $Y-b_{1}$ we may assume that $b_{1}=0$ and then

$-Y\cap
Y=(-b_{n},-a_{n})\cup...\cup(-b_{2},-a_{2})\cup(a_{2},b_{2})\cup...\cup(a_{n},b_{n})$

So $(-Y\cap Y)\cap[(-Y\cap Y)+(a_{n}+b_{n})]$ equals the interval
$I_{n}=(a_{n},b_{n})$ in $Y$. Replace $Y$ by $Y_{1}:=Y\setminus I_{n}$, now
$Y_{1}$ contains a unbounded ray together with $n-2$ bounded intervals.
Continuing in this way we obtain a ray $(-\infty,0)$ that is definable, so we
can define $<$.∎

So we assume now that either $Y$ or $R\setminus Y$ are bounded. If $Y$ is
bounded then each of the intervals above in $Y$ is bounded. Let
$\alpha:=b_{n}-b_{1}$. The set $\>(Y+\alpha)\cap Y$ defines a single interval
whose right endpoint is $b_{n}$. So, we are done. If $Y$ is unbounded then
replace $Y$ by $R\setminus Y$ and finish as before. Hence, we showed that
$<^{*}$ is definable in $\mathcal{M}$.∎

##  3\. Reducts of $\mathcal{R}_{alg}$ which are not semilinear

Here $R$ is a real closed field and $\mathcal{R}_{alg}=\langle
R;<,+,\cdot\rangle$. Before the next theorem we recall previous work from [7]
(see a corrected and more general proof in [1]), which will be used in its
proof.

Given $a>0$ in $R$, let $I=(-a,a)$. Denote by $+^{*}$ the partial function
obtained by intersecting the graph of $+$ with $I^{3}$, and for each
$\alpha\in R$, let $\lambda_{\alpha}^{*}$ be the partial function obtained by
intersecting the graph of $\lambda_{\alpha}$ with $I^{2}$. Finally, let
$<^{*}$ be the restriction of $<$ to $I^{2}$. Notice that for each $X\subseteq
R^{n}$ such that $\langle R;<+,\cdot,X\rangle$ is o-minimal, the structure

$\mathcal{I}=\langle\mathrm{I};<^{*},+^{*},\\{\lambda^{*}_{\alpha}\\}_{\alpha\in
R},X\cap I^{n}\rangle$

is o-minimal as well.

In [7] the structure $\langle I;<^{*},+^{*}\rangle$ was called a group-
interval and its o-minimal expansions were studied there.

A partial endomorphism (p.e. in short) of this group-interval was a function
$f:I\rightarrow I$ which respects addition when defined: namely, if
$x,y,x+^{*}y\in I$ then $f(x+^{*}y)=f(x)+^{*}f(y)$.

Notice that in our setting every $\mathcal{I}$-definable p.e. is necessarily
the restriction of $\lambda_{\alpha}$ for some $\alpha\in R$. Indeed, if
$f:I\rightarrow I$ is an $\mathcal{I}$-definable p.e. then it is not hard to
verify that the following is a semialgebraic subgroup of $\langle R,+\rangle$
which contains all integers. Let

$H=\\{r\in R:\exists\varepsilon>0\forall
x\in(-\varepsilon,\varepsilon)f(rx)=rf(x)\\}.$

O-minimality of the real field implies that $H=R$ and therefore $f$ is the
restriction of an $R$-linear map, namely the restriction of $\lambda_{\alpha}$
for some $R$.

Now, without going through the precise definition of “a linear theory” from
[7], it was shown in [7, Proposition 4.2] that if $Th(\mathcal{I})$ is linear
then every $\mathcal{I}$-definable set is already defined in the structure
$\langle I;+^{*},<^{*},\\{\lambda_{\alpha}^{*}\\}_{\alpha\in R}\rangle,$
(together possibly with additional parameters). Thus if $Th(\mathcal{I})$ is
linear then $X\bigcap I^{n}$ is a semilinear set.

The following proposition seems to be obvious but for the sake of completion
we include a proof in the Appendix.

######  Fact 3.1.

Let $R$ be a real closed field and $X\subseteq R^{n}$ a definable set in an
o-minimal expansion of $\langle R;<,+,\cdot\rangle$. If $X$ is not semilinear
then, in the structure $\mathcal{M}=\langle R;<^{*},+,\Lambda_{R},X\rangle$,
there exists a definable bounded set which is not semilinear.

######  Theorem 3.2.

If $X\subseteq R^{n}$ is semialgebraic and not definable in
$\mathcal{R}_{semi}$, then every bounded $R$-semialgebraic set is definable in
$\langle R;+,\Lambda_{R},X\rangle$

###### Proof.

Let $\mathcal{M}:=\langle R;+,\Lambda_{R},X\rangle$. By Theorem 2.1, the
relation $<^{*}$ is definable in $\mathcal{M}$. Let us first see that
$\mathcal{M}$ defines a real closed field on some interval.

By Fact 3.1, we may assume that $X\cap I^{n}$ is not semilinear, for some
bounded interval $I=(-,a,a)$. Consider the o-minimal structure

$\mathcal{I}:=\langle I;<^{*},+^{*},\\{\lambda_{\alpha}^{*}\\}_{\alpha\in
R},X\cap I^{n}\rangle,$

as we described before stating the theorem. We noted that if $Th(\mathcal{I})$
is linear then the set $X\bigcap I^{n}$ must be semilinear set. Because
$I^{n}\cap X$ is not semilinear then $Th(\mathcal{I})$ is not linear in the
sense of [7] and therefore by [12, Theorem 1.2], a real closed field is
$\mathcal{I}$-definable, hence also $\mathcal{M}$-definable, on some interval
$J\subseteq I$.

Without loss of generality, assume that $J=(-a_{0},a_{0}),\>a_{0}>0$. Denote
the field by:

$\mathcal{J}=\langle\mathit{J},\varoplus,\varodot\rangle$

The structure $\mathcal{J}$ is $\mathcal{M}$-definable. By [10, Corollary
2.4], every $R$-semialgebraic subset of $J^{k}$, $k\in\mathbb{N}$, is
definable in $\mathcal{J}$, and therefore in $\mathcal{M}$.

Let $B\subseteq(-b,b)^{n}$ for some $b>0$ in $R$. Using scalar multiplication
from $\Lambda_{R}$, we can contract $(-b,b)$ into $(-a_{0},a_{0})$, so it is
definable in $\mathcal{J}$. It follows that $B$ is definable in $\mathcal{M}$
. ∎

##  4\. Strongly bounded structures

The ultimate goal of this section is to prove:

######  Theorem 4.1.

Let $R$ be a real closed field. If $X\subseteq R^{n}$ is semialgebraic and not
definable in $\mathcal{R}_{bd}=\langle
R;<^{*},+,\Lambda_{R},\mathfrak{B}_{sa}\rangle$ then $<$ is definable in the
structure $\langle R;+,\Lambda_{R},X\rangle$.

We are going to work in a more general setting than that of a real closed
field. Recall that a strongly bounded reduct of a linearly ordered $\langle
R;<,\cdots\rangle$ is one in which every definable subset of $R$ is bounded or
co-bounded. Below, we will mostly be interested in strongly bounded reducts of
o-minimal structures. By Lemma 2.3 we have:

######  Lemma 4.2.

Let $\mathcal{R}_{omin}=\langle R;<,+,\cdots\rangle$ be an o-minimal expansion
of an ordered group. If $\mathcal{M}=\langle R;+,\cdots\rangle$ is a reduct of
$\mathcal{R}_{omin}$ then $\mathcal{M}$ is strongly bounded if and only if $<$
is not definable in $\mathcal{M}$.

So in order to prove Theorem 4.1 it is sufficient to prove that if $X\subseteq
R^{n}$ is definable in a strongly bounded $\mathcal{M}=\langle
R;<,+,\cdots\rangle$ then $X$ is definable in $\langle
R;+,\Lambda_{\mathcal{M}},\mathfrak{B}_{\mathcal{M}}\rangle$, where
$\mathfrak{B}_{\mathcal{M}}$ is the collection of all $\mathcal{M}$-definable
bounded sets. A more precise and slightly stronger theorem will be proved
soon, Theorem 4.5. We first make a general observation which we shall exploit
repeatedly.

###  4.1. Definability of “boundedness”

For $X\subseteq T\times R^{n}$, $T\subseteq R^{m}$ and $t\in T$, we let

$X_{t}=\\{a\in R^{n}:\langle t,a\rangle\in X\\}$

The following general result will be very useful here.

######  Proposition 4.3.

Let $\mathcal{M}=\langle R;+,\cdots\rangle$ be any reduct of an o-minimal
expansion of an ordered group. If $\\{X_{t}:\>t\in T\\}$ is an
$\mathcal{M}$-definable family of subsets of $R^{n}$, then the set

$\\{t\in T:\>X_{t}\>\textrm{is\>bounded in $R^{n}$}\\}$

is definable in $\mathcal{M}$.

###### Proof.

Note that a set $Y\subseteq R^{n}$ is bounded if and only if for each $i$, the
image of $Y$ under the projection map $\pi_{i}:\left\langle
y_{1},...,y_{n}\right\rangle\longmapsto y_{i}$ is bounded in $R$. Thus, it is
sufficient to prove the result under the assumption that all $X_{t}$ are
subsets of $R$.

By o-minimality, each $X_{t}\subseteq R$ is unbounded if and only it contains
an unbounded ray. Thus, it is easy to see that

$\\{t\in T:\>X_{t}\textrm{ is bounded}\\}=\\{t\in T:\>\exists a\>\>a+X_{t}\cap
X_{t}=\emptyset\\}$

and hence the set is definable in $\mathcal{M}$. ∎

###  4.2. The strongly bounded setting

We first clarify and somewhat generalize our setting.

_Let $\mathcal{R}_{omin}=\langle R,<,+,\cdots\rangle$ denote an o-minimal
expansion of an ordered group in language $\mathcal{L}_{omin}$, and let
$\mathcal{M}=\langle R;+,\cdots\rangle$ denote a strongly bounded reduct of
$\mathcal{R}_{omin}$, in language $\mathcal{L}$, such that
$acl_{\mathcal{M}}(\emptyset)$ contains at least one nonzero element._

######  Definition 4.4.

An interval $(a,b)\subseteq R$ is called a _$\emptyset$ -interval in
$\mathcal{M}$_ if $a,b\in acl_{\mathcal{M}}(\emptyset)$. A subset $X\subseteq
R^{n}$ is called $\emptyset$-bounded in $\mathcal{M}$ if $X$ is contained in
some $I^{n}$, for $I$ a $\emptyset$-interval in $\mathcal{M}$.

Our standing assumption is that for every $\emptyset$-interval $I\subseteq R$,
the restricted order $<\upharpoonright\\!\\!I$ is $\emptyset$-definable in
$\mathcal{M}$. Notice that, using Theorem 2.1, this is true when $\mathcal{M}$
is elementarily equivalent to a reduct of a real closed field which properly
expands $\mathcal{R}_{lin}$.

We let $\Lambda_{\mathcal{M}}$ be the collection of all
$\mathcal{M}$-definable endomorphisms of $\langle R,+\rangle$, defined over
$\emptyset$. We let $\mathcal{L}_{bd}(\mathcal{M})$ be the language consisting
of $\\{+,\\{\lambda\\}_{\lambda\in\Lambda_{\mathcal{M}}}\\}$, augmented by a
predicate for every $\emptyset$-definable, $\emptyset$-bounded set in
$\mathcal{M}$.

By expanding $\mathcal{L}$ and $\mathcal{L}_{omin}$ by function symbols and
predicates for $\emptyset$-definable sets, we may assume that

$\mathcal{L}_{bd}\subseteq\mathcal{L}\subseteq\mathcal{L}_{omin}.$

We let $\mathcal{M}_{bd}$ be the reduct of $\mathcal{M}$ to
$\mathcal{L}_{bd}$.

Our ultimate goal in this section is to prove:

######  Theorem 4.5.

For $\mathcal{M}$ strongly bounded as above, every definable subset of $R^{n}$
is definable in $\mathcal{M}_{bd}$.

One of our main difficulties in working with strongly bounded structures is
the failure of global cell decomposition. E.g. the set $R\setminus\\{0\\}$
cannot be decomposed definably into definable cells in a strongly bounded
structure, because no ray is definable there.

Another difficulty is the fact that a-priori we do not know whether the model
theoretic algebraic closure equals the definable closure in strongly bounded
structures. However, we shall eventually show, see Theorem 4.27, that
$acl=dcl$ in this setting.

We assume from now on throughout this section that $\mathcal{M}$ is strongly
bounded as above.

###  4.3. Definable subsets of $R$ in strongly bounded structures

Notice that although the full order is not definable in $\mathcal{M}$, a basis
for the $<$-topology on $R$ and the product topology on $R^{n}$ is definable
in $\mathcal{M}$, using the restricted order. Thus we have:

######  Lemma 4.6.

If $\\{X_{t}:t\in T\\}$ is an $\mathcal{M}$-definable family of subsets of
$R^{n}$, then the families

$\\{Cl(X_{t}):t\in T\\}\,\,,\,\,\\{Int(X_{t}):t\in
T\\}\,\,,\,\,\\{Fr(X_{t}):t\in T\\}$

are definable in $\mathcal{M}$.

Every $\mathcal{M}$-definable $X\subseteq R$ is a union of finitely many
pairwise disjoint maximal open sub-intervals of $X$ (which are possibly not
$\mathcal{M}$-definable) and a finite set. Below, when we say that $I$ is an
interval in $X$ we mean that $I$ is one of these open components of $X$.

######  Definition 4.7.

Let $Y\subseteq R$ be an $\mathcal{M}$-definable set, we define:

$\partial^{-}(Y):=\\{y\in R:\>y\>\textrm{is a left endpoint of an interval in
}Y\\}.$

$\partial^{+}(Y):=\\{y\in R:\>y\>\textrm{is a right endpoint of an interval in
}Y\\}.$

######  Lemma 4.8.

If $\\{Y_{t}:\>t\in T\\}$ is an $\mathcal{M}$-definable family of bounded
subsets of $R$ then the families $\\{\partial^{-}(Y_{t}):\>t\in
T\\},\>\\{\partial^{+}(Y_{t}):\>t\in T\\}$ are
$\mathscr{\mathcal{M}}$-definable, over the same parameter set.

###### Proof.

We fix an $\mathcal{M}$-definable $<\upharpoonright\\!\\!(0,a_{0})$ for some
$a_{0}>0$. We define$\>\partial^{-}(Y_{t})$ by the formula :

$(x\notin Y_{t}\wedge\exists\epsilon<a_{0}\>(x,x+\epsilon)\subseteq Y_{t})$
$\vee$ $(x\in Y_{t}\wedge\exists\epsilon\leq a_{0}\>\wedge(x-\epsilon,x)\cap
Y_{t}=\emptyset\wedge(x,x+\epsilon)\subseteq Y_{t})).$

Because of the definability of $<^{*}$ in $\mathscr{\mathcal{M}}$,
$\\{\partial^{-}(Y_{t}):t\in T\\}$ is $\mathcal{M}$-definable. We similarly
handle $\partial^{+}(Y_{t})$. ∎

The next theorem is an important component of our analysis of strongly bounded
structures.

######  Theorem 4.9.

If $\\{X_{t}:\>t\in T\\}$ is an $\mathcal{M}$-definable family of bounded
subsets of $R$, then there is a uniform bound on the length of each interval
in $X_{t}$. Moreover, there exists such a bound in
$dcl_{\mathcal{M}}(\emptyset)$.

###### Proof.

By Proposition 4.3, every $\mathcal{M}$-definable family $\\{X_{t}:t\in T\\}$
of bounded subsets of $R$ is a sub-family of a $\emptyset$-definable family of
such sets. Namely, if $\phi(x,t,a)$ is the formula defining the $X_{t}$’s over
$a$, as $t$ varies, then we can consider the formula

$\psi(x,t,y):\phi(x,t,y)\wedge\mbox{ $\psi(R,t,y)$ is a bounded set}.$

Thus, it is sufficient to prove the result for $\emptyset$-definable families.

By Lemma 4.6, we may assume that each $X_{t}$ is an open set. We will use
induction on $n:=\textrm{the maximum number of intervals in }X_{t}$, for $t\in
T$.

For $n=1$, write $X_{t}=(a_{t},b_{t})$.

Consider the family $\\{X_{t}-a_{t}:t\in T\\}$. By Lemma 4.8, the family is
$\emptyset$-definable.

Thus, the set $Y=\underset{t\in T}{\bigcup}X_{t}-a_{t}$ is an
$\mathcal{M}$-definable interval, over $\emptyset$, whose left end-point is
$0$. Because $\mathcal{M}$ is strongly bounded, this interval must be bounded,
hence its right endpoint is some $K\in M$. By Lemma 4.8, the point $K$ is
definable over $\emptyset$.

Consider now the case $n=k+1$, namely each $Y_{t}$ consists of at most $k+1$
pairwise disjoint open intervals. For each $t\in T$, let
$D_{t}=\\{c_{{}_{1}}-c_{2}:\>c_{1},c_{2}\in\partial^{-}(X_{t})\\}$, an
$\mathcal{M}$-definable set by Lemma 4.8.

######  Claim 4.10.

For each $t\in T$, there exists $d\in D_{t}$ such that $(X_{t}+d)\cap X_{t}$
is one of the intervals in $X_{t}$.

###### Proof.

Let $X_{t}=I_{1,t}\cup I_{2,t}\cup...\cup I_{k+1,t},\textrm{ where each
}I_{m,t}:=(a_{m,t},b_{m,t})$, such that:

$a_{1,t}<b_{1,t}<a_{2,t}<b_{2,t}<...<a_{k+1,t}<b_{k+1,t}$

For an interval $I=(a,b)$, let $|I|=b-a$.

Let $d=a_{k+1,t}-a_{1,t}$. In the set $X_{t}+d$, for each $m$, the
interval$\textrm{ }I_{m,t}$ is shifted to$\textrm{ }I_{m,t}+d$. So
$(X_{t}+d)\cap X_{t}$ consists of either $I_{k+1,t}$ (when
$|I_{k+1,t}|<|I_{1,t}|$) or $I_{1,t}+d$ (when $|I_{k+1,t}|>|I_{1,t}|$) .

If it consists of $I_{k+1}$ we are done. Otherwise we take

$d^{\prime}=a_{1,t}-a_{k+1,t}\in D_{t}$

and then $(X_{t}+d^{\prime})\cap X_{t}=$ $I_{1,t}$.

So in both cases there exists $d\in D_{t}$ such that $X_{t}+d\cap X_{t}$ is
one of the intervals in $X_{t}$. ∎

We define the set:

$D^{\prime}_{t}:=\\{d\in D_{t}:\>(X_{t}+d)\cap X_{t}\>\textrm{is one of the
intervals in }X_{t}\\}$

######  Claim 4.11.

The family $\\{D^{\prime}_{t}\>:t\in T\\}$ is an
$\mathscr{\mathcal{M}}$-definable family of nonempty sets.

###### Proof.

For $t\in T$, $d\in D^{\prime}_{t}$ if and only if the following two hold:

  1. (1)

$\partial^{-}((X_{t}+d)\cap X_{t})\subseteq\partial^{-}(X_{t})\>\textrm{and
}|\partial^{-}((X_{t}+d)\cap X_{t})|=1$, and

  2. (2)

$\partial^{+}((X_{t}+d)\cap X_{t})\subseteq\partial^{+}(X_{t})\>\textrm{and
}|\partial^{+}((X_{t}+d)\cap X_{t})|=1$.

By Lemma 4.8, $(1),(2)$ are a definable properties in $\mathscr{\mathcal{M}}$.
By Claim 4.10, each $D_{t}^{\prime}$ is non-empty. ∎

We proceed with the proof of Theorem 4.9. Consider the $\mathcal{M}$-definable
family

$\\{\>Y_{t,d}:=X_{t}+d\cap X_{t}:\>d\in D^{\prime}_{t},\>t\in T\\},$

still defined in $\mathcal{M}$ over $\emptyset$. For every $t\textrm{ and
}d\in D^{\prime}_{t},$ the set $Y_{t,d}$ consists of a single interval which
is one of the intervals in $X_{t}.$ By case $n=1$ we know that there is a
uniform bound on length of each $Y_{t,d}$, call it $w_{1}$ which can be chosen
in $dcl_{\mathcal{M}}(\emptyset)$. We now define, still over $\emptyset$, the
following family

$\\{Z_{t,d}:=X_{t}\setminus\>Y_{t,d}:\>d\in D^{\prime}_{t},\>t\in T\\}$

Each subset $Z_{t,d}$ consists of at most $k$ intervals among the $k+1$
intervals of $X_{t}$. By the induction hypothesis, we know that there is a
uniform bound on the length of each interval, call it $w_{2}$ which we may
choose in $dcl_{\mathcal{M}}(\emptyset)$.

Thus the maximum of $w_{1},w_{2}$, which is in $dcl_{\mathcal{M}}(\emptyset)$,
is the bound on the length of each interval of $X_{t}$, as $t$ varies. This
ends the proof of Theorem 4.9.∎

As a corollary we can now match, definably in $\mathcal{M}$, each left
endpoint of an interval in $X_{t}$ with the corresponding right endpoint:

######  Proposition 4.12.

Let $\\{X_{t}:\>t\in T\\}$ be an $\mathcal{M}$-definable family of bounded
subsets of $R$, and let

$L_{t}=\\{\langle
a,b\rangle\in\partial^{-}(X_{t})\times\partial^{+}(X_{t})\textrm{: the
interval }(a,b)\textrm{ is one the intervals of }X_{t}\\}$

Then the family $\\{L_{t}:t\in T\\}$ is $\mathcal{M}$-definable.

###### Proof.

By Theorem 4.9, there is a bound, call it $K\in dcl_{\mathcal{M}}(\emptyset)$,
for the length of each interval in $X_{t}$, for all $t\in T$. For each $t\in
T\>$, we have

$(1)\langle a,b\rangle\in L_{t}\>\>\iff\>\>a\in\partial^{-}(X_{t})\textrm{ and
$b=min(\partial^{+}(X_{t})\cap[a,a+K])$ }$

By Lemma 4.8, $\partial^{-}(X_{t}),\>\partial^{+}(X_{t})$ are definable
families and since in $(1)\>$ we only use the order on $[0,K]$, the family
$\\{L_{t}:t\in T\\}$ is definable in $\mathscr{\mathcal{M}}$. ∎

######  Remark 4.13.

  1. (1)

Notice that Theorem 4.9 fails without the assumption that the $X_{t}$’s are
bounded sets. Namely, it is not true in general that the length of the bounded
components of $X_{t}$ is bounded in $t$. For example, the sets
$X_{t}=R\setminus\\{-t,t\\}$ has $(-t,t)$ as an open component, with unbounded
length as $t\to\infty$.

Also, even if each $X_{t}$’s is bounded it is not true that the diameter of
the $X_{t}$’s is uniformly bounded. For example, take the family
$\\{(-t,t-1)\cup(t,t+1):t\in R\\}$ that is definable using
$<\upharpoonright(0,1)$.

  2. (2)

We do not know whether Proposition 4.12 holds if we drop the assumption that
the $X_{t}$’s are bounded. Namely, can we still match definably the left and
right endpoints of the bounded components of $X_{t}$, when the $X_{t}$’s are
unbounded?

###  4.4. Affine sets and functions

Recall that $\mathcal{R}_{omin}$ is an o-minimal expansion of an ordered
divisible abelian group $R$, and we assume that $\mathcal{M}=\langle
R;+,\cdots\rangle$ is a strongly bounded reduct of $\mathcal{R}_{omin}$ in
which $<$ is $\emptyset$-definable on every $\emptyset$-interval. We let
$<^{*}$ denote the ordering on some fixed interval we call $(0,1)$.

######  Definition 4.14.

Let $\langle R;<,+\rangle$ be an abelian ordered divisible group.

  1. (1)

A map $f:R^{n}\to R^{k}$ is affine if it is of the form $\ell(x)+d$ for
$\ell:R^{n}\to R^{k}$ a homomorphism between $\langle R^{n},+\rangle$ and
$\langle R^{k},+\rangle$, and $d\in R^{k}$.

  2. (2)

A (partial) function $f:R\rightarrow R$ , is _eventually affine_ if there
exists $a>0$ such that $(a,\infty)\subseteq dom(f)$ and the restriction of $f$
to $(a,+\infty)$ is affine.

  3. (3)

$X\subseteq R^{n}$ is locally affine at $a\in X$ there is an open neighborhood
$U\ni a$ such that for all $x,y,z\in U\cap X$, $x-y+z\in X$. The affine part
of $X$ is the set;

$\mathscr{A}(X)=\\{x\in X:\>X\textrm{ is locally affine at }x\\}.$

Notice that if $X$ is the graph of an affine map then $\mathscr{A}(X)=X$.
Also, because a basis for the $R^{n}$-topology is definable in $\mathcal{M}$,
we immediately have:

######  Lemma 4.15.

Let $\\{X_{t}:t\in T\\}$ be an $\mathcal{M}$-definable family of subsets of
$R^{n}$, defined over $\emptyset$. Then the family
$\\{\mathscr{A}(X_{t}):\>t\in T\\}$ is $\mathcal{M}$-definable, over
$\emptyset$.

We now prove:

######  Proposition 4.16.

Every $\mathcal{M}$-definable endomorphism $f:R\to R$ is
$\emptyset$-definable.

###### Proof.

Assume that $f$ is defined by $\mathcal{M}$-formula $\phi(x,y,a)$, over the
parameter $a$. We will show that $f$ can be defined without parameters.

Since being an $R$-endomorphism is $\mathcal{M}$-definable, we may assume that
there is some $\mathcal{M}$-definable $T\subseteq R^{k}$, such that for all
$t\in T$. If $\phi(R^{2},t)$ is non-empty then it defines a non-zero
endomorphism $f_{t}$ of $\langle R;+\rangle$.

Assume first that the set of endomorphisms $f_{t}$’s defined by $\phi$ is
finite. Define $t_{1}Et_{2}$ iff $f_{t_{1}}=f_{t_{2}}$, an
$\mathcal{M}$-definable equivalence relation. Consider the functions near $0$,
and define $[t_{1}]_{E}<[t_{2}]_{E}$ if for all $x>0$ sufficiently small, we
have $f_{t_{1}}(x)<f_{t_{2}}(x)$. By o-minimality, we obtain a linear ordering
of the finitely many $E$-classes, and since $<$ is $\mathcal{M}$-definable in
a neighborhood of $0$, this ordering is $\mathcal{M}$-definable. Thus, each
$f_{t}$ in this finite family of endomorphisms is $\emptyset$-definable.

Assume now that the family $\\{f_{t}:t\in T\\}$ is infinite, and we shall
reach a contradiction. Consider the set $\\{f_{t}(1):t\in T\\}$. By
o-minimality it contains an open interval $(a,b)$, and by replacing each
$f_{t}$ with $f_{t}-f_{t_{0}}$, for some $t_{0}\in T$ for which
$f_{t_{0}}\in(a,b)$, we may assume that the interval $(a,b)$ contains $0$ and
the ordering on $(a,b)$ is $\mathcal{M}$-definable (we think of $f_{t}(a)$ as
“the slope” of $f_{t}$). Let $T_{0}=\\{t\in T:f_{t}(1)\in(0,b)\\}$.

We write $t_{1}\sim t_{2}$ if $f_{t_{1}}=f_{t_{2}}$, and let $[t]$ be the
equivalence class of $t$. In abuse of notation we let $f_{[t]}$ denote the
corresponding endomorphism of $R$.

By o-minimality, if $f_{t_{1}}(1)=f_{t_{1}}(1)$ then $f_{t_{1}}=f_{t_{2}}$,
thus we obtain an $\mathcal{M}$-definable function $t:(0,b)\to T_{0}/\sim$,
defined by $f_{[t(x)]}(1)=x$. Namely, $f_{[t(x)]}$ is the endomorphism whose
“slope” is $x$. Fix an element $d>0$, and define $\sigma:(0,b)\to R$ by:
$\sigma(x)=f_{[t(x)]}^{-1}(d)$. Namely, $\sigma(x)=y$ if there exists $t\in
T_{0}$ such that $f_{t}(1)=x$ and $f_{t}(y)=d$ (we may think of $\sigma(x)$ as
“$d/x$”). The function $\sigma$ is also $\mathcal{M}$-definable. For every
$t\in T_{0}$, we have $f_{t}(1)>0$, hence $f_{t}(x)>0$ if and only if $x>0$.
Threfore, $\sigma$ is positive on $(0,b)$.

Claim $Im(\sigma)$ is unbounded in $R$.

Indeed, assume towards contradiction that $K=sup(Im(\sigma))<\infty$. By our
observation, $K>0$. Choose $y_{0}\in Im(\sigma)$, $y_{0}<K$ and sufficient
close to $K$, such that $K<2y_{0}$. By assumption, there exists $t_{0}\in
T_{0}$ and $x_{0}>0$, such that $f_{t_{0}}(1)=x_{0}$ and $f_{t_{0}}(y_{0})=d$.

Let $t_{1}\in T_{0}$ be such that $[t_{1}]=t(x_{0}/2)$. Then
$f_{t_{1}}(1)=x_{0}/2=f_{t_{0}}(1)/2$. It follows that $f_{t_{1}}=f_{t_{0}}/2$
and hence

$f_{t_{1}}(2y_{0})=f_{t_{0}}(2y_{0})/2=f_{t_{0}}(y_{0})=d$

But then $f_{t_{1}}(1)=x_{0}/2$ and $f_{t_{1}}(2y_{0})=d$, so by definition,
$\sigma(x_{0}/2)=2y_{0}>K$, contradicting the assumption that $K$ bounds
$Im(\sigma)$.

Thus, $Im(\sigma)$ is an $\mathcal{M}$-definable set which is unbounded and
positive, contradicting the assumption that $\mathcal{M}$ is strongly
bounded.∎

######  Definition 4.17.

We denote by $\Lambda_{omin}$ the set of all $\mathcal{R}_{omin}$-definable
endomorphism $f:\langle R,+\rangle\to\langle R,+\rangle$. and we still let
$\Lambda_{\mathcal{M}}$ denote the set of all $\mathcal{M}$-definable
endomorphisms of $R$, which by Proposition 4.16, is necessarily
$\emptyset$-definable. Let $\Lambda_{omin}^{*}$ and
$\Lambda_{\mathcal{M}}^{*}$ denote those non-zero endomorphisms.

###  4.5. Definable functions of $1$-variable

Our goal is to describe definable functions in 1-variable, and prove that
$\mathcal{M}$ has no definable “poles”.

######  Proposition 4.18.

If $g:R\rightarrow R$ is an $\mathcal{M}$-definable partial function whose
domain is co-bounded and $Im(g)$ is bounded. Then $g$ is constant on a co-
bounded set.

###### Proof.

By o-minimality, there exists $L\in R$ such that
$\underset{x\rightarrow+\infty}{lim}g(x)=L$. We shall see that $g\equiv L$ on
a co-bounded set.

The function $g$ is definable in an o-minimal structure, thus there exists
$a_{1}\in R$ such that $g\upharpoonright(a_{1},+\infty)$ is either constant or
strictly monotone, and there exists $a_{2}$ such that $g$ is constant or
strictly monotone on $(-\infty,a_{2})$.

If $g$ is constant $L$ on $(a_{1},+\infty)$ then $\\{x\in R:g(x)=L\\}$ is
unbounded and since $\mathcal{M}$ strongly bounded the set must be co-bounded
and we are done. Assume towards contradiction that
$g\upharpoonright(a_{1},\infty)$ is strictly monotone.

Assume first that $g$ is strictly increasing on $(a_{1},\infty)$. Notice that
the property of being locally increasing in a neighborhood of $x\in R$ is
definable using $<^{*}$, thus the set

$\\{x\in R:g\mbox{ is locally increasing at $x$}\\}$

is $\mathcal{M}$-definable, contains $(a_{1},\infty)$ and hence must be co-
bounded. It follows that $g$ is strictly increasing on $(-\infty,a_{2})$.

Because $\lim_{x\to\infty}g(x)=L$ and $g$ is increasing, there exists $b\in R$
such that for all $x>b$, $L-1<g(x)<L$. Because $<^{*}$ is
$\mathcal{M}$-definable the set of all $x\in R$ such that $L-1<g(x)<L$ is
$\mathcal{M}$-definable so must be co-bounded. In particular, we may assume
that $L-1<g(x)<L$ for all $x<a_{2}$ and thus $g(x)$ has a limit $L_{1}\in R$
as $x\to-\infty$.

But since $g$ is increasing on $x<a_{2}$, it follows that $L_{1}<L$ and in
addition there exists $a_{2}^{\prime}\leq a_{2}$ and $\epsilon>0$, such that
for all $x<a_{2}^{\prime}$,

$L_{1}<g(x)<L_{1}+\epsilon<L.$

Using $<^{*}$ again, this is an $\mathcal{M}$-definable property of $x$ so
must hold also for all $x>a_{1}^{\prime}$, contradicting the fact that
$\lim_{x\to+\infty}g(x)=L$.

A similar argument works when $g$ is eventually decreasing.∎

######  Remark 4.19.

By [3], if $\mathcal{N}=\langle R;<,+,\cdots\rangle$ is an o-minimal expansion
of an ordered group in which every definable bounded function is eventually
constant then $\mathcal{N}$ is semibounded, namely every definable set is
definable using the underlying vector space, together with all the definable
bounded sets. This might suggest a fast deduction of Theorem 4.5 from
Proposition 4.18. The problem of this approach is that we do not know that the
definable functions in the strongly bounded $\mathcal{M}=\langle
R;+,<^{*},\cdots\rangle$ are the same as in its expansion by the full $<$.
Thus, we do not see how to apply Edmundo’s theorem here.

Next, using almost identical arguments to Edmundo’s [3] we shall show that
every $\mathcal{M}$-definable function $f:R\to R$ is affine on a co-bounded
set. For that, we recall some notation and facts, based on work of Miller and
Starchenko [8].

Notation For $\mathcal{R}_{omin}$-definable positive (partial) functions
$f,g:R\to R$, such that $(a,\infty)\subseteq\mbox{dom}(f),\mbox{dom}(g)$, we
write $f\leq g$ (or $f<g$) if $f(x)\leq g(x)$ (or, $f(x)<g(x)$) for all large
enough $x$.

We write $v(f)<v(g)$) if $|f|>|\lambda\circ g|$ for all
$\lambda\in\Lambda_{omin}^{*}$ such that $\lambda>0$. We also write
$v(f)=v(g)$ if there are $\lambda_{1},\lambda_{2}\in\Lambda_{omin}^{*}$, both
positive, such that

$|\lambda_{1}\circ g|\leq|f|\leq|\lambda_{2}\circ g|.$

This easily seen to be an equivalence relation.

Finally, we write $\Delta(f)=f(x+1)-f(x)$.

######  Fact 4.20.

[3] For every $\mathcal{R}_{omin}$-definable function on unbounded ray.

  1. (1)

If $v(f)>v(x)$ then $\lim_{x\to\infty}\Delta(f)=0$.

  2. (2)

If $v(f)<v(x)$ then $v(f^{-1})>v(x)$.

  3. (3)

If $v(f)=v(x)$ then $\Delta(f)(x)$ has a limit in $R$ as $x\to\infty$.

The following is just a warm-up towards Theorem 4.25. The proof follows
closely the proof of [3, Poposition 2.8], which uses results of Miller and
Starchenko [8]:

######  Lemma 4.21.

If $f:R\rightarrow R$ is $\mathcal{M}$-definable on a co-bounded set, then $f$
is eventually affine. Moreover, there exists a $\emptyset$-definable
endomorphism $\lambda\in\Lambda_{\mathcal{M}}$ and $A>0$ such that for all $x$
with $|x|>A$, we have $f(x)=\lambda(x)+d$, for some $d\in R$.

###### Proof.

Assume towards contradiction that $f:R\to R$ is not eventually affine. Without
loss of generality, $f$ is eventually increasing, and by Proposition 4.18, it
must approach $+\infty$. If $v(f)>v(x)$ then by Fact 4.20,
$\lim_{x\to\infty}\Delta(f)=0$. Since $\Delta(f):=f(x+1)-f(x)$ is definable in
$\mathcal{M}$, it follows from Proposition 4.18 that it must be eventually $0$
and therefore $f$ is eventually affine.

If $v(f)<v(x)$ then by 4.20, $v(f^{-1})>v(x)$, where $f^{-1}$ is taken to be
the eventual compositional inverse of $f$, which is also definable in
$\mathcal{M}$. Thus, as above, $f^{-1}$ is eventually affine so also $f$ is.

We are left with the case $v(f)=v(x)$. By Fact 4.20 (3), the
$\mathcal{M}$-definable function $\Delta(f)$ approaches a limit $c$ in $R$. By
Proposition 4.18, we have $\Delta(f)$ eventually constant, and thus, by
o-minimality, $f$ is eventually affine.

Thus, we showed so far that there exists a definable endomorphism
$\lambda\in\Lambda_{\mathcal{M}}$ such that $f(x)=\lambda(x)+d$ for all $x>0$
large enough. By Proposition 4.16, $\lambda$ is $\emptyset$-definable. The set

$\\{x\in R:f(x)=\lambda(x)+d\\}$

is $\mathcal{M}$-definable and contains an bounded ray so must be co-bounded.∎

Before the next proposition, we introduce a new notion.

######  Definition 4.22.

Given $X\subseteq R^{n}$, let

$Stab_{bd}(X):=\\{a\in R^{n}:\>(a+X)\triangle X\>is\>bounded\\},$

where $A\triangle B=A\cup B\setminus A\cap B$.

For a function $f$, we let $\Gamma(f)$ denote its graph.

By Proposition 4.3, if $X$ is definable in $\mathcal{M}$ over $A$ then so is
$Stab_{bd}(X)$. The following are easy to verify:

######  Fact 4.23.

  1. (1)

For every $X\subseteq R^{n}$, $Stab_{bd}(X)$ is a subgroup of $\langle
R^{n},+\rangle$.

  2. (2)

If $X\subseteq R^{2}$ is the graph of an affine function $f(x)=\lambda(x)+b$,
on a co-bounded subset of $R$, then

$Stab_{bd}(X)=\Gamma(\lambda).$

  3. (3)

If a definable set $X\subseteq R^{2}$ is a finite union of graphs of affine
functions, all of the form $\lambda+d$ for a fixed $\lambda$, and at least one
of the functions is defined on an unbounded set then
$Stab_{bd}(X)=\Gamma(\lambda)$.

The following statement would have been immediately true if definable sets in
$\mathcal{M}$ admitted definable cell decomposition (with respect to the
ambient ordering).

######  Proposition 4.24.

Assume that $X\subseteq R^{2}$ is $\mathcal{M}$-definable over $A$, and
$\dim(X)\leq 1$. Assume that there exists an $\mathcal{R}_{omin}$-definable
endomorphism $\lambda:R\to R$, and some $a,d\in R$ such that graph of
$\lambda(x)+d\upharpoonright(a,\infty)$ is contained in $X$. Then $\lambda$ is
$\mathcal{M}$-definable (necessarily over $\emptyset$).

###### Proof.

Recall that $\mathscr{A}(X)$, the affine part of $X$ is
$\mathcal{M}$-definable over $A$. For large enough $a$, it contains
$\Gamma(\lambda+d\upharpoonright\,(a,\infty))$. So, without loss of
generality, $X=\mathscr{A}(X)$.

We define for each $x,y\in X$, the relation $x\sim y$ iff there exist open
sets $U,V\ni 0$ in $R^{2}$, such that

$(y-x)+(x+U\cap X)=y+V\cap X.$

Said differently, up to translation, $X$ has the same germ at $x$ and at $y$.
Because a basis for the $R^{2}$ topology is definable in $\mathcal{M}$, the
relation $\sim$ is definable in $\mathcal{M}$.

Notice that for $x$ large enough, all elements on $\Gamma(\lambda+d)\cap X$
are in the same $\sim$-class, so we may replace $X$ by this $\sim$-class,
which is $\mathcal{M}$-definable.

Thus, we may assume that all elements of $X$ are $\sim$-equivalent, and $X$
contains $\Gamma(\lambda+d\upharpoonright(a,\infty))$. It follows that $X$ is
contained in finitely many translates of the graph of $\lambda$. Applying Fact
4.23(3), we conclude that $Stab_{bd}(X)$ is exactly the graph of $\lambda$,
thus the function $\lambda(x)$ is $\mathcal{M}$-definable. By Lemma 4.16,
$\lambda$ is $\emptyset$-definable.∎

###  4.6. Definable subsets of $R^{2}$

The next result is the main structure theorem of the paper.

######  Theorem 4.25.

Under our standing assumptions on $\mathcal{M}$.

Assume that $X\subseteq R^{2}$ is definable in $\mathcal{M}$ over a parameters
set $A\subseteq R$, with $\dim(X)\leq 1$.

Then, there are $\lambda_{1},\ldots,\lambda_{r}\in\Lambda_{\mathcal{M}}$, and
there are $\mathcal{M}$-definable finite set $D_{i}\subseteq R$,
$i=1,\ldots,r$, and $D\subseteq R$ all defined over $A$, such that

(i) for every $i=1,\ldots,r$, and $d\in D_{i}$,
$\Gamma(\lambda_{i}+d)\setminus X$ is bounded (i.e. $X$ contains the
restriction of $\lambda_{i}+d$ to a co-bounded set).

(ii) For every $d\in D$, $(\\{d\\}\times R)\setminus X$ is bounded.

(ii) The set

$X\setminus(\bigcup_{i=1}^{r}\bigcup_{d\in
D_{i}}\Gamma(\lambda_{i}+d)\,\cup\,\bigcup_{d\in D}\\{d\\}\times R)$

is bounded in $R^{2}$.

###### Proof.

If $X$ is bounded then there is nothing to prove so we assume $\dim(X)=1$ and
$X$ is unbounded. By the cell decomposition theorem in o-minimal structure,
$X$ can be decomposed into a finite union of cells of dimension 0 and 1.
However, these cells are not in general definable in $\mathcal{M}$.

Assume first that $X$ contains the graph of a function $f:(a,+\infty)\to R$,
and let $\Psi(x,y)$ be the $\mathcal{M}$-formula that defines $X$.

Case (i) $f$ is bounded at $\infty$.

In this case we prove a general statement:

######  Claim 4.26.

If $\dim X\leq 1$ and $X$ contains the graph of a bounded function
$f:(a,\infty)\to R$ then $f$ is eventually constant.

###### Proof.

By o-minimality, $\lim_{x\to+\infty}f(x)=L$ for some $L\in R$.

By our standing assumption, $<\upharpoonright(0,a_{0})$ is
$\mathcal{M}$-definable, for some $a_{0}>0$, and thus $<$ is definable on
every interval of length $\leq a_{0}$. Let
$X_{L}:=R\times[L-a_{0},L+a_{0}])\cap X$. By o-minimality, there exists
$m\in\mathbb{N}$, such that for all large enough $a\in R$, we have
$|X_{a}|\leq m$. The set $Z=\\{a\in R:|X_{a}|\leq m\\}$ is definable in
$\mathcal{M}$ and unbounded, thus we may replace $X_{L}$ by $X_{L}\cap Z\times
R$, containing the graph of $f$. We call it $X_{L}$ again.

Using the restricted order, we can partition $X_{L}$, definably in
$\mathcal{M}$, into finitely many graphs of functions
$g_{1},g_{2},\dots,g_{k}$, $k\leq m$. E.g, we let

$g_{1}(x)=min\\{y\in[L-a_{0},L+a_{0}]:\>\langle x,y\rangle\in X_{L}\\}$

and continue similarly to obtain the other $g_{i}$’s. For $x$ large enough,
the function $f$ is one of those $g_{i}$’s, therefore it is
$\mathcal{M}$-definable. Using Proposition 4.18 we get that $f$ is eventually
constant.∎

Case (ii): $\underset{x\rightarrow+\infty}{lim}f(x)=+\infty$:

We recall the proof of Lemma 4.21, and consider three cases: $v(f)>v(x)$,
$v(f)<v(x)$ and $v(f)=v(x)$ (remembering though that we do not know yet that
$f$ is an $\mathcal{M}$-definable function).

Assume first that $v(f)>v(x)$. By Fact 4.20, $f(x+1)-f(x)\to 0$, as
$x\to\infty$. We want to capture $\Delta(f)=f(x+1)-f(x)$ within an
$\mathcal{M}$-definable set.

The formula

$\varphi(x,y):=\exists z_{1}\exists
z_{2}(\Psi(x+1,z_{1})\wedge\Psi(x,z_{2})\wedge(y=z_{1}-z_{2})),$

defines in $\mathcal{M}$ a new subset of $R^{2}$ call it $\Delta(X)$ which
contains the graph of $\Delta(f)$ (but possibly more functions).

We first note that $\dim(\Delta(X))=1$: Indeed, for $a\in R$, $\Delta(X)_{a}$
is infinite if either $X_{a}$ or $X_{a+1}$ is infinite. Since only finitely
many $X_{a}$’s are infinite the same is true for $\Delta(X)$. Thus, the graph
of $\Delta(f)$ is contained in the one-dimensional $\mathcal{M}$-definable set
$\Delta(X)$, so by Claim 4.26, $\Delta(f)$ must be eventually constant,
implying that $f$ is eventually affine.

Assume now that $v(f)<v(x)$. The formula $\Upsilon(x,y):=\Psi(y,x)$ defines in
$\mathcal{M}$ a new set $X^{-1}$ containing in it the graph of $f^{-1}$ (a
partial function). The graph of $f^{-1}$ is still contained in $X^{-1}$ and we
have $v(f^{-1})>v(x)$. Thus, applying the case we already handled, we see that
$f^{-1}$, and hence also $f$, is eventually affine.

We are left with the case $v(f)=v(x)$. Using Fact 4.20 (3), the function
$\Delta(f)$ tends to a constant. Thus, as above, we may use the
$\mathcal{M}$-definable set $\Delta(X)$ to deduce that $\Delta(f)$ is
eventually constant and thus $f$ is eventually affine.

So far we handled all cases where the bounded cell in $X$ has is the graph of
some function on a ray $(a,\infty)$. The same reasoning applies to rays
$(-\infty,a)$. Applying this reasoning to $X^{-1}$, we obtain in addition
those functions which are eventually constant in $X^{-1}$, namely sets of the
form $\\{d\\}\times R$ whose intersection with $X$ is co-unbounded in
$\\{d\\}\times R$. The set of such $d$’s is clearly definable over $A$.

To summarize, we showed that every unbounded cell in $X$ is either contained
in the graph of an eventually affine function $f$ definable in $\mathcal{M}$,
or in $\\{d\\}\times R$ for some $d$. By Proposition 4.24, the function $f$
has the form $\lambda(x)+d$, for $\lambda\in\Lambda_{\mathcal{M}}$. Thus, we
have $\lambda_{1},\ldots,\lambda_{k}\in\Lambda_{\mathcal{M}}$, and for each
such $i=1,\ldots,k$, the set $D_{i}$ of $d\in R$ such that
$\Gamma(\lambda_{i}+d)\cap X$ is unbounded, is $\mathcal{M}$-definable over
$A$, and must be finite. For every such $d$, $\Gamma(\lambda_{i}+d)\setminus
X$ is bounded.

The above proof handles all unbounded cells, so the set

$X\setminus(\bigcup_{i=1}^{r}\bigcup_{d\in
D_{i}}\Gamma(\lambda_{i}+d)\,\cup\,\bigcup_{d\in D}\\{d\\}\times R)$

is bounded. ∎

###  4.7. The algebraic closure and definable closure in strongly bounded
structures

Even though the full ordering on $R$ is not definable, we can still prove:

######  Theorem 4.27.

The algebraic closure in $\mathcal{M}$ equals the definable closure. Moreover,
if $a\in acl_{\mathcal{M}}(\bar{b})$ then $b$ is in the
$\mathcal{L}_{bd}$-definable closure of $\bar{b}$.

###### Proof.

We use $acl$, $dcl$ and $acl_{bd}$, $dcl_{bd}$, to denote the corresponding
operations in $\mathcal{M}$ and $\mathcal{M}_{bd}$, respectively. We shall
prove by induction on $n$: If $a\in acl(b_{1},\ldots,b_{n})$, for some
$a,b_{i}\in R$, then $a\in dcl_{bd}(b_{1},\ldots b_{n})$.

We first handle the case $n=0$, namely $a\in acl(\emptyset)$. In this case,
there is a finite $\emptyset$-definable set $A\subseteq M$ such that $a\in A$.
Viewing the set $A$ in $\mathcal{R}_{omin}$, we can order the elements,
$a_{1}<\cdots<a_{n}$. The interval $(a_{1},a_{n})$ is a $\emptyset$-interval,
and $<\upharpoonright(a_{1},a_{n})$ is $\mathcal{M}_{bd}$-definable over
$\emptyset$, hence each $a_{i}\in dcl_{bd}(\emptyset)$.

We proceed by induction, and assume that we proved the result for $n-1$.
Assume now that $a\in acl(b_{1},\ldots,b_{n-1},b_{n})$. Let $X\subseteq
R^{n+1}$ be a $\emptyset$-definable set such that $\langle
b_{1},\ldots,b_{n},a\rangle$ and $X_{b_{1},\ldots,b_{n}}$ has size $m$.
Without loss of generality, for every $b_{n}^{\prime}$, the set
$X_{b_{1},\ldots,b_{n-1},b_{n}^{\prime}}$ has size $m$.

Let $b^{\prime}=(b_{1},\ldots,b_{n_{-}1})$ and consider the set
$X_{b^{\prime}}=\\{\langle x,y\rangle\in R^{2}:\langle
b^{\prime},x,y\rangle\in X\\}$. By our assumption, $\dim(X_{b^{\prime}})\leq
1$, and $\langle b_{n},a\rangle\in X_{b^{\prime}}$.

We now apply Theorem 4.25. We obtain finitely many $\emptyset$-definable
endomorphisms $\lambda_{1},\ldots,\lambda_{k}\in\Lambda_{\mathcal{M}}$ and for
each $i=1,\ldots,k$, we have a $b^{\prime}$-definable finite set $A_{i}$, such
that

$X^{bd}_{b^{\prime}}=X_{b^{\prime}}\setminus(\bigcup_{i=1}^{k}\bigcup_{d\in
D_{i}}\Gamma(\lambda_{i}+d))$

is bounded in $R^{2}$.

Since $|b^{\prime}|=n-1$, it follows by induction that every $d\in A_{i}$ is
in $dcl_{bd}(b^{\prime})$. Assume first that $\langle b_{n},a\rangle$ is in
the graph of one of the $\lambda_{i}+d$, $d\in D_{i}$, namely
$a=\lambda_{i}(b_{n})+d$. Because $\lambda_{i}$ is $\emptyset$-definable and
$d\in dcl_{bd}(b^{\prime})$ it follows that $a\in
dcl_{bd}(b_{1},\ldots,b_{n})$.

We are left with the case $\langle b_{n},a\rangle\in X^{bd}_{b^{\prime}}$. The
set $X^{bd}_{b^{\prime}}$ is $b^{\prime}$-definable so we may assume that
$X_{b^{\prime}}=X^{bd}_{b^{\prime}}$ is bounded (but possibly not
$\emptyset$-bounded). Let $\pi_{1}$, $\pi_{2}$, be the projection of
$X_{b^{\prime}}$ onto the first and second coordiantes. Each of these is a
finite union of points and pairwise disjoint bounded open intervals. Let

$\pi_{1}(X_{b^{\prime}})=F_{1}\cup\bigcup_{i=1}^{k}(a_{i},b_{i}),\mbox{ for
$F_{1}$ finite and $a_{1}<b_{1}\cdots<a_{k}<b_{k}.$}.$

and

$\pi_{2}(X_{b^{\prime}})=F_{2}\cup\bigcup_{j=1}^{r}(c_{j},d_{j}),\mbox{ for
$F_{2}$ finite and $c_{1}<d_{1}\cdots<c_{r}<d_{r}.$}$

By Theorem 4.9, there is a fixed $K\in dcl(\emptyset)$ such that for all
$i,=1,\ldots,k$ and $j=1,\ldots,r$, we have $b_{i}-a_{i},d_{j}-c_{j}\leq K$.

By Proposition 4.8, the sets $\\{a_{i}\\},\\{b_{i}\\},\\{c_{j}\\},\\{d_{j}\\}$
are all finite and $\mathcal{M}$-definable over $b^{\prime}$, and thus, by
induction each of these endpoints is in $dcl_{bd}(b^{\prime})$. Assume that
$\langle b_{n},a\rangle\in X\cap(a_{i},b_{i})\times(c_{j},d_{j})$, for some
$i=1,\ldots,k$ and $j=1,\ldots,r$. We replace $X$ by the
$b^{\prime}$-definable set $X_{1}=X-\langle
a_{i},c_{j}\rangle\cap(0,b_{i}-a_{i})\times(0,d_{j}-c_{j})\subseteq(0,K)^{2}$.
Notice that $\langle b_{n}-a_{i},a-c_{j}\rangle\in X^{\prime}$, and the fiber
in $X^{\prime}$ over $b_{n}-a_{i}$ is finite. Because the ordering on $(0,K)$
is $\mathcal{M}_{bd}$-definable over $\emptyset$, we have $a-c_{j}\in
dcl_{bd}(b^{\prime},b_{n}-a_{i})$, but since $a_{i},c_{j}\in
dcl_{bd}(b^{\prime})$ we have $a\in dcl_{bd}(b^{\prime},b_{n})$. This ends the
proof that $acl=dcl_{bd}$ in $\mathcal{M}$.∎

###  4.8. Definable subsets of $R^{n}$

We are now ready to prove the main theorem, under the assumptions of Section
4.2.

######  Theorem 4.28.

If $X\subseteq R^{n}$ is $\mathcal{M}$-definable over $A\subseteq R$ then $X$
is definable in $\mathcal{M}_{bd}$ over $A$.

###### Proof.

It is sufficient to prove the result in $\mathcal{N}\succ\mathcal{M}$, so by
replacing $\mathcal{R}_{omin}$ (thus also its reducts) by a sufficiently
saturated extension, we may assume that $\mathcal{M}$ is $\omega$-saturated.

We prove the result by induction on $n$. For $X\subseteq R$, the set $X$ is
either bounded or co-bounded, so we may assume that it is bounded. Thus, it
can be written as a disjoint union

$(a_{1},b_{1})\cup\cdots\cup(a_{n},b_{n})\cup F,$

with $a_{1}<b_{1}<\cdots<a_{n}<b_{n}$ and $F$ finite. By Lemma 4.8, each
$a_{i}$ and $b_{i}$ is in $acl_{\mathcal{M}}(A)$, so by Theorem 4.27, it
belong to $dcl_{bd}(A)$. Similarly, $F\subseteq dcl_{bd}(A)$. By Theorem 4.9,
there is $K\in dcl_{bd}(\emptyset)$ such that all intervals $(a_{i},b_{i})$
are of length at most $K$. But then each interval $(0,b_{i}-a_{i})$ is
contained in a $\emptyset$-interval, hence definable in $\mathcal{M}_{bd}$
over $A$, so also $(a_{i},b_{i})$ is $\mathcal{M}_{bd}$-definable over $A$. It
follows that $X$ is definable in $\mathcal{M}_{bd}$.

We now use induction on $n$: Given $X\subseteq R^{n+1}$ that is
$\mathcal{M}$-definable over $A$, we consider, for each $t\in R^{n}$, the set
$X_{t}=\\{b\in R:\langle t,b\rangle\in X\\}\subseteq R$. By the case $n=1$,
each $X_{t}$ is $\mathcal{M}_{bd}$-definable over $At$. Thus, by compactness
and saturation, we can find $\mathcal{L}_{bd}$-formulas over $A$,
$\phi_{1}(t,x),\ldots,\phi_{k}(t,x)$ such that for every $t\in R^{n}$, one of
the $\phi_{i}(t,x)$ defines $X_{t}$. Let

$T_{i}=\\{t\in R^{n}:\exists x(\langle t,x\rangle\in X\wedge\,\,\forall x(x\in
X_{t}\leftrightarrow\phi_{i}(t,x)))\\}.$

The set $T_{i}$ is $\mathcal{M}$-definable, over $A$, thus, by induction, it
is $\mathcal{M}_{bd}$-definable over $A$, by some $\psi_{i}(t)$. The formula
$\phi_{i}(t,x)\wedge\psi_{i}(t)$ defines $X\cap T_{i}\times R$, thus $X$ is
definable in $\mathcal{M}_{bd}$ over $A$.∎

###  4.9. A comment on failure of Definable Choice in strongly bounded
$\mathcal{M}$

Recall that a structure $\mathcal{M}$ has Definable Choice if for every
definable family $\\{X_{t}:t\in T\\}$ of sets, there is a definable function
$f:T\to\bigcup X_{t}$ such that $f(t)\in X_{t}$ and if $t_{1}=t_{2}$ then
$f(t_{1})=f(t_{2})$. Equivalently, every definable equivalence relation has a
definable set of representatives. This fails in strongly bounded
$\mathcal{M}$, because the relation $xEy\Leftrightarrow y=-x$ on $R$ cannot
have a definable set of representatives. If it did then it will contain either
a positive or a negative ray (without its inverse).

We believe that Elimination of Imaginaries similarly fails.

##  5\. Conclusion: The proof of Theorem 1.2

We are now ready to collect the results proved thus far in order to prove
Theorem 1.2.

Recall that now want to prove that the only reducts between
$\mathcal{R}_{lin}$ and $\mathcal{R}_{alg}$ are:

$\mathcal{R}_{alg}=\langle R;+,\cdot,<\rangle$

$\mathcal{R}_{sb}=\langle R;+,<,\Lambda_{R},\mathfrak{B}\rangle$

$\mathcal{R}_{semi}=\langle R;+,<,\Lambda_{R}\rangle$ ,
$\mathcal{R}_{bd}=\langle R;+,<^{*},\Lambda_{R},\mathfrak{B}\rangle$

$\mathcal{R}_{lin}^{*}=\langle R;+,<^{*},\Lambda_{R}\rangle$

$\mathcal{R}_{lin}=\langle R;+,\Lambda_{R}\rangle$.

First, we note that using [3] we can generalize [10, Theorem 1.1] from
$\mathbb{R}$ to arbitrary real closed fields, and show:

######  Fact 5.1.

Let $R$ be a real closed field. The only reduct between $\mathcal{R}_{semi}$
and $\mathcal{R}_{alg}$ is $\mathcal{R}_{sb}$.

###### Proof.

Assume that $\mathcal{M}$ is a reduct of $R_{alg}$ which properly expands
$\mathcal{R}_{semi}$. By [3, Fact 1.6], either $\mathcal{M}$ is a reduct of
$\mathcal{R}_{sb}$ or a real closed field $F=\langle R;\oplus,\odot\rangle$
whose universe is $R$ is definable in $\mathcal{M}$. Assume the latter, and
then since the field is semialgebraic then, again by [10, Corollary 2.4],
every semialgebraic subset of $R$ is definable in $F$ and hence in
$\mathcal{M}$. Thus, $\mathcal{M}\dot{=}\mathcal{R}_{alg}$.

If $\mathcal{M}$ is a reduct of $\mathcal{R}_{sb}$ which is not semilinear
then by Theorem 3.2, every bounded $R$-semialgebraic set is definable in
$\mathcal{M}$, thus $\mathcal{M}\dot{=}\mathcal{R}_{sb}$.∎

We now consider an arbitrary reduct $\mathcal{M}$ of $\mathcal{R}_{alg}$. Our
goal is to show that $\mathcal{M}$ is one of the reducts in the above list.

First, if $\mathcal{M}$ is stable then by Claim 2.2,
$\mathcal{R}_{lin}\dot{=}\mathcal{M}.$ If $\mathcal{M}$ is unstable then by
Theorem 2.1, $<^{*}$ is definable in $\mathcal{M}$. So
$\mathcal{R}_{lin}^{*}\dot{\subseteq}\mathcal{M}.$ So, we may assume that
$<^{*}$ is definable in $\mathcal{M}$, thus
$\mathcal{R}_{lin}^{*}\dot{\subseteq}\mathcal{M}$.

_Case 1:_ $\mathcal{M}$ is strongly bounded and
$\mathcal{M}\dot{\subseteq}\mathcal{R}_{semi}$.

We claim that $\mathcal{M}\dot{=}\mathcal{R}_{lin}^{*}$: Indeed, because
$\mathcal{M}$ is strongly bounded then, by Theorem 4.5,
$\mathcal{M}\dot{=}\mathcal{M}_{bd}$. Because
$\mathcal{M}\dot{\subseteq}\mathcal{R}_{semi}$, every $\mathcal{M}$-definable
set is semilinear, and in particular this is true for each of the
$\emptyset$-bounded sets in $\mathcal{M}_{bd}$. However, it is easy to verify
that every bounded semilinear set is definable in $\mathcal{R}_{lin}^{*}$, so
the whole structure $\mathcal{M}_{bd}$ is a reduct of $\mathcal{R}_{lin}^{*}$,
thus so is $\mathcal{M}$ as well. The converse
$\mathcal{R}_{lin}^{*}\dot{\subseteq}\mathcal{M}$ is already assumed.

_Case 2:_ $\mathcal{M}$ is strongly bounded and
$\mathcal{M}\dot{\nsubseteq}\mathcal{R}_{semi}$. We claim that
$\mathcal{M}=\mathcal{R}_{bd}$.

As in Case 1, every $\mathcal{M}$-definable set is definable in
$\mathcal{M}_{bd}$. Because $\mathcal{M}$ is a reduct of $\mathcal{R}_{alg}$
then $\mathcal{M}_{bd}$ is a reduct of $\mathcal{R}_{bd}$ and so
$\mathcal{M}\dot{\subseteq}\mathcal{R}_{bd}$. By the assumption that
$\mathcal{M}\dot{\nsubseteq}\mathcal{R}_{semi}$, we know that there is an
$\mathcal{M}$-definable semialgebraic set which is not semilinear so by
Theorem 3.2, we get that every bounded semialgebraic set is definable in
$\mathcal{M}$, hence $\mathcal{R}_{bd}\dot{\subseteq}\mathcal{M}$.

Next we assume that $\mathcal{M}$ is not strongly bounded.

_Case 3:_ $\mathcal{M}$ is not strongly bounded and
$\mathcal{M}\dot{\subseteq}\mathcal{R}_{semi}$. By Lemma 2.3, the linear order
$<$ is definable in $\mathcal{M}$, so, since
$\mathcal{R}_{semi}^{*}\dot{\subseteq}\mathcal{M}$, we have
$\mathcal{R}_{semi}\dot{=}\mathcal{M}$.

_Case 4:_ $\mathcal{M}$ is not strongly bounded and
$\mathcal{M}\dot{\nsubseteq}\mathcal{R}_{semi}$. As in Case 3, the linear
order $<$ is definable in $\mathcal{M}$, so
$\mathcal{R}_{semi}\dot{\subseteq}\mathcal{M}$ . So we know that $\mathcal{M}$
is a reduct of $\mathcal{R}_{alg}$ which properly expands
$\mathcal{R}_{semi}$. By Fact 5.1, either
$\mathcal{M}\dot{=}\mathcal{R}_{alg}$ or $\mathcal{M}\dot{=}\mathcal{R}_{bd}$.

This completes the proof that if $\mathcal{M}$ is a reduct of
$\mathcal{R}_{alg}$ expanding $\mathcal{R}_{lin}$, then it is one of the
reducts in the above diagram.

It is left to see that all reducts in the above diagram are distinct. Because
$\mathcal{R}_{lin}$ is stable and $\mathcal{R}_{lin}^{*}$ is unstable, these
two are distinct. Also, the fact that $\mathcal{R}_{lin}^{*}$ and
$\mathcal{R}_{bd}$ are distinct is easy to verify (e.g., the unit circle is
definable in $\mathcal{R}_{bd}$ but not in $\mathcal{R}_{lin}^{*}$). The fact
that $\mathcal{R}_{bd}$ is different than $\mathcal{R}_{sb}$ and
$\mathcal{R}_{semi}$ follows from the next lemma.

######  Lemma 5.2.

Let $R$ be a real closed field. If $\mathcal{B}^{*}$ is any collection of
bounded subsets of $R^{n}$, $n\in\mathbb{N}$, then $<$ is not definable in
$\mathcal{M}=\langle R;+,\Lambda_{R},\mathcal{B}^{*}\rangle$.

###### Proof.

We use a similar idea to [9] Assume towards contradiction that $<$ is
definable in $\mathcal{M}$, and let $\mathcal{N}=\langle
R:+,<,\Lambda_{R},\mathfrak{B}^{*}\rangle$.

Let $\psi(x,y,\bar{a})$, $\bar{a}\in R$, be the $\mathcal{M}$-formula that
defines $<$. Namely

$\mathcal{N}\models\forall x\forall y\>(\psi(x,y,\bar{a})\leftrightarrow
x<y).$

Let
$\tilde{\mathcal{N}}=\langle\tilde{R};+,<,\Lambda_{R},\mathfrak{B}^{*}\rangle\succ\mathcal{N}$
be an $|N|^{+}$-saturated elementary extension, whose reduct to the
$\mathcal{M}$-language is $\tilde{\mathcal{M}}$. It follows that
$\psi(x,y,\bar{a})$ defines $<$ in $\tilde{\mathcal{N}}$ as well.

We will show that there is an automorphism of $\tilde{\mathcal{M}}$ which
fixes $\bar{a}$, thus leaving $\psi(\tilde{R}\times\tilde{R},\bar{a})$
invariant, and yet not respecting $<$, leading to a contradiction.

The group $\langle\tilde{R},+\rangle$ is a vector space over $R$. We define an
$R$-vector subspace of $\tilde{R}$ by

$\mathcal{A}=\\{x\in\tilde{R}:\exists\alpha\in
R,\>|x|<\lambda_{\alpha}(1)\\}.$

So, by Zorn’s Lemma, there exists an $R$-vector space $V\subseteq\tilde{R}$
such that $\tilde{R}=\mathcal{A}\varoplus V$, and by the saturation
assumption, $V$ is non-trivial. Now we define the following automorphism of
the $R$-vector space $\tilde{R}$: On $\mathcal{A}$ we define $\tau_{1}(v)=v$,
on $\left\langle V,+\right\rangle$ we define $\tau_{2}(v)=-v$, and we let
$\tau$: $\tilde{R}\longrightarrow\tilde{R}$ be:

$\tau(v_{1}+v_{2})=\tau_{1}(v_{1})+\tau_{2}(v_{2})=v_{1}-v_{2}.$

This automorphism fixes all elements in $\mathcal{A}$ and in particular fixes
all sets in $\mathfrak{B^{*}}$ pointewise, but does not respect $<$ (as
positive elements in $V$ are sent to negative ones). In model theoretic
language $\tau$ is an automorphism of the structure $\tilde{\mathcal{M}}$,
which fixes $\overline{a}$ (since $\overline{a}\in\mathcal{A}$). However,
$\tau$ does not preserve $<$, contradiction. ∎

This ends the proof of Theorem 1.2.∎

##  6\. Appendix: The proof of Fact 3.1

We now prove Fact 3.1:

######  Fact 6.1.

Let $R$ be a real closed field and $X\subseteq R^{n}$ a definable set in an
o-minimal expansion of $\langle R;<,+,\cdot\rangle$. If $X$ is not definable
in $\mathcal{R}_{semi}$ then, in the structure $\mathcal{M}=\langle
R;<^{*},+,\Lambda_{R},X\rangle$ there exists a definable bounded set which is
not definable in $\mathcal{R}_{semi}$.

###### Proof.

We believe that this is known so we shall be brief. We prove the result by
induction on $\dim(X)$, where the case $\dim X=0$ is trivially true. Consider
the affine part of $X$, $\mathscr{A}(X)$, which is definable in $\mathcal{M}$.

Assume first ${\mathscr{A}}{(X)}$ is not dense in $X$. Then there is an open
box $U\subseteq R^{n}$ such that $U\cap X\neq\emptyset$ and
$U\cap\mathscr{A}(X)=\emptyset$. We claim that $U\cap X$ is not semilinear.
Indeed, if it were then ${\mathscr{A}}{(U\cap X)}$ must be nonempty, but
because $U\cap X$ is relatively open in $X$ then ${\mathscr{A}}{(U\cap
X)}=U\cap{\mathscr{A}}{(X)}=\emptyset$, contradiction.

Thus, $U\cap X$ above is not semilinear. and this gives the desired box when
${\mathscr{A}}{(X)}$ is not dense in $X$.

We assume then that ${\mathscr{A}}{(X)}$ is dense in $X$, and consider two
cases: ${\mathscr{A}}{(X)}$ is either semilinear or not. If it were semilinear
then necessarily $X\setminus{\mathscr{A}}{(X)}$ is not semilinear, and because
of the density assumption, $\dim(X\setminus{\mathscr{A}}{(X)})<\dim(X)$ and we
can finish by induction.

Thus, we are left with the case that ${\mathscr{A}}{(X)}$ is not semilinear.
For simplicity, we may assume now that $X={\mathscr{A}}{(X)}$. We recall the
$\mathcal{M}$-definable relation $a\sim b$ from the proof of Proposition 4.24,
defined by: $X$ has the same germ at $a$ and $b$, up to translation.

Because $X={\mathscr{A}}{(X)}$, each $\sim$-class is open in $X$, thus there
are finitely many classes, at least one of which is not semilinear. Thus, we
may assume that $X={\mathscr{A}}{(X)}$ consists of a single $\sim$-class. It
follows that there is some $R$-subspace $L\subseteq R^{n}$, $\dim L=\dim X$,
such that $X$ is contained in a finite union of cosets of $L$. Thus each
definably connected component of $X$ is contained in a single such coset of
$L$.

Each $L$ is definable in $\mathcal{M}$ using $\Lambda_{R}$, thus the
intersection of $X$ with each of these cosets is definable in $\mathcal{M}$.
One of these intersections is not semilinear so we may assume that $X\subseteq
c+L$, for some $c$. Because $\dim X=\dim L$, and ${\mathscr{A}}{(X)}=X$, then
$X$ is open in $c+L$. We claim that $Fr(X)\subseteq c+L$ is not semilinear:
Indeed, $Fr(X)$ is a closed subset of $c+L$, and $X$ consists of finitely many
components of $c+L\setminus Fr(X)$. If $Fr(X)$ were semilinear then each of
its components will also be, so $X$ would be semilinear.

Thus, $Fr(X)$ is not semilinear, and definable in $\mathcal{M}$. By
o-minimality, $\dim(Fr(X))<\dim(X)$, thus by induction we may find an
$\mathcal{M}$-definable bounded set which is not semilinear.∎

In fact, a stronger result is true: If $X\subseteq R^{n}$ is definable in an
o-minimal expansion of the field $R$, and not semilinear then there is some
bounded open box $U\subseteq R^{n}$ such that $U\cap X$ is not semilinear (we
omit the proof here as we do not need it). Notice that this last statement
fails if we replace “not semilinear” by “not semialgebraic”, as Rolin’s
example from [6] shows: There exists a definable function
$f:\mathbb{R}\to\mathbb{R}$ in an o-minimal expansion of the real field, such
that the restriction of $f$ to every bounded interval is semialgebraic but $f$
itself is not semialgebraic.

## References

  * [1] Oleg Belegradek.  Semi-bounded relations in ordered modules.  J. Symbolic Logic, 69(2):499–514, 2004.
  * [2] Gabriel Conant.  There are no intermediate structures between the group of integers and Presburger arithmetic.  J. Symb. Log., 83(1):187–207, 2018.
  * [3] Mario J. Edmundo.  Structure theorems for o-minimal expansions of groups.  Ann. Pure Appl. Logic, 102(1-2):159–181, 2000.
  * [4] Assaf Hasson, Alf Onshuus, and Ya’acov Peterzil.  Definable one dimensional structures in o-minimal theories.  Israel J. Math., 179:297–361, 2010.
  * [5] U. Hrushovski and A. Pillay.  Weakly normal groups.  In Logic colloquium ’85 (Orsay, 1985), volume 122 of Stud. Logic Found. Math., pages 233–244. North-Holland, Amsterdam, 1987.
  * [6] Olivier Le Gal and Jean-Philippe Rolin.  An o-minimal structure which does not admit $C^{\infty}$ cellular decomposition.  Ann. Inst. Fourier (Grenoble), 59(2):543–562, 2009.
  * [7] James Loveys and Ya’acov Peterzil.  Linear o-minimal structures.  Israel J. Math., 81(1-2):1–30, 1993.
  * [8] Chris Miller and Sergei Starchenko.  A growth dichotomy for o-minimal expansions of ordered groups.  Trans. Amer. Math. Soc., 350(9):3505–3521, 1998.
  * [9] Ya’acov Peterzil.  A structure theorem for semibounded sets in the reals.  J. Symbolic Logic, 57(3):779–794, 1992.
  * [10] Ya’acov Peterzil.  Reducts of some structures over the reals.  J. Symbolic Logic, 58(3):955–966, 1993.
  * [11] Ya’acov Peterzil.  Returning to semi-bounded sets.  J. Symbolic Logic, 74(2):597–617, 2009.
  * [12] Ya’acov Peterzil and Sergei Starchenko.  A trichotomy theorem for o-minimal structures.  Proc. London Math. Soc. (3), 77(3):481–523, 1998.