[AF11], one may show that every definable family of convex subsets of $M^{n}$
that is $(n+1)$-consistent extends to a definable type.

######  Remark 4.8.

In [AGTW21, Section 6], Thomas, Walsberg and the author introduce the notion
of _definable net_ $\gamma:(\Omega,\preccurlyeq)\rightarrow(X,\tau)$ to mean a
definable map from a definable (downwards) directed set
$(\Omega,\preccurlyeq)$ into a definable topological space $(X,\tau)$. A
_subnet_ (a _Kelley subnet_) of $\gamma$ is a net of the form
$\gamma^{\prime}=\gamma\circ\mu$ where
$\mu:(\Omega^{\prime},\preccurlyeq^{\prime})\rightarrow(\Omega,\preccurlyeq)$
is a downward cofinal map on some directed set
$(\Omega^{\prime},\preccurlyeq^{\prime})$. We say that such a net
$\gamma^{\prime}$ is definable if all of
$(\Omega^{\prime},\preccurlyeq^{\prime})$, $\mu$ and $\gamma$ are definable.

Classically, a topological space is compact if and only if every net in it has
a converging subnet. Following the classical proof of this result one may show
that, in any model theoretic structure (regardless of the axiom of
o-minimality), definable compactness implies that every definable net has a
definable converging subnet. The reverse implication follows whenever the
structure has definable choice. In our o-minimal $\mathcal{M}$ we have that,
by the relationship between (1) and (7) in Theorem 4.6 (and more specifically
by Lemma 4.12 below), this reverse implication also holds whenever the
topology is Hausdorff. See [AGTW21, Corollary 44] for a proof of the
equivalence between definable compactness, curve-compactness, and the property
that every definable net has a convergent definable subnet, in o-minimal
expansions of ordered groups.

We divide the proof of Theorem 4.6 into a number of propositions. Note that
the equivalence (2)$\Leftrightarrow$(3) is given by Proposition 4.4. In
Proposition 4.9 we prove, using results from previous sections, the
equivalence between (1), (3), (4), (5) and (6). In Proposition 4.13 we prove
the implication (1)$\Rightarrow$(7), and reverse implication when $\tau$ is
Hausdorff or $\mathcal{M}$ has definable choice.

######  Proposition 4.9.

Let $(X,\tau)$ be a definable topological space. The following are equivalent.

  1. (1)

The topology $\tau$ is definably compact.

  2. (2)

Any definable family of $\tau$-closed sets that extends to a definable type
has nonempty intersection.

  3. (3)

Any consistent definable family of $\tau$-closed sets has a finite transversal
in $X$.

  4. (4)

Any definable family $\mathcal{C}$ of nonempty $\tau$-closed sets with the
$(m,n)$-property, where $m\geq n>\dim\cup\mathcal{C}$, has a finite
transversal in $X$.

  5. (5)

Any definable family $\mathcal{C}$ of nonempty $\tau$-closed sets with the
$(m,n)$-property, where $m\geq n$ and $n$ is greater than the VC-codensity of
$\mathcal{C}$, has a finite transversal in $X$.

###### Proof.

We show that
$\eqref{itm:compactness_transversals_1}\Rightarrow\eqref{itm:compactness_transversals_1.5}$,
$\eqref{itm:compactness_transversals_1.5}\Rightarrow\eqref{itm:compactness_transversals_2}\wedge\eqref{itm:compactness_transversals_3}\wedge\eqref{itm:compactness_transversals_4}$,
and
$\eqref{itm:compactness_transversals_2}\vee\eqref{itm:compactness_transversals_3}\vee\eqref{itm:compactness_transversals_4}\Rightarrow\eqref{itm:compactness_transversals_1}$.

Note that, if a downward directed family of sets has a finite transversal,
then, by Fact 2.5, it has nonempty intersection. Hence (3), (4) and (5) each
imply (1). The implications (2)$\Rightarrow$(4) and (2)$\Rightarrow$(5) follow
from Theorem 3.5 and Corollary 3.14 respectively. The implication
(2)$\Rightarrow$(3) follows because (4)$\Rightarrow$(3) is obvious. It remains
thus to show that (1)$\Rightarrow$(2).

Suppose that $(X,\tau)$ is definably compact and let $\mathcal{C}$ be a
definable family of $\tau$-closed sets that extends to a definable type. By
Lemma 2.10, let $\mathcal{F}\subseteq p$ be a definable downward directed
family of sets that is finer than $\mathcal{C}$. Note that the family
$\\{cl_{\tau}(F):F\in\mathcal{F}\\}$ is definable, downward directed and finer
that $\mathcal{C}$. By definable compactness there exists
$x\in\bigcap\\{cl_{\tau}(F):F\in\mathcal{F}\\}$. In particular
$x\in\cap\mathcal{C}$. ∎

######  Remark 4.10.

It is worth noting which results from Sections 2 and 3 are used in proving the
non-trivial implications in Proposition 4.9. By virtue of Proposition 4.4, let
us refer to condition (2) in Proposition 4.9 as “type-compactness”.

We prove that definable compactness implies type-compatness
((1)$\Rightarrow$(2)) using Lemma 2.10. It is easy to see that the converse
implication can be proved using Lemma 2.7, and in particular the fact that
every definable downward directed family extends to a definable type, which,
as noted in Remark 3.15, is a condition weaker than Theorem 3.5.

Furthermore, we prove that type-compactness implies (3), (4) and (5) using
Theorem 3.5 and Corollary 3.14. Alternatively this can be done, and
generalized to a broader dp-minimal setting, using Theorem 3.20 due to Simon
and Starchenko, as well as VC-theory, in particular the Alon-Kleitman-Matoušek
$(p,q)$-theorem (Theorem 1.2) and known bounds on VC-density in NIP theories
(see [ADH+16]) such as Theorem 1.3.

The following Corollary is a direct consequence of Propositions 4.4 and 4.9.

######  Corollary 4.11.

Suppose that $\mathcal{M}$ extends $(\mathbb{R},<)$, the linear order of
reals, and let $(X,\tau)$ be a definable topological space. Then $(X,\tau)$ is
definably compact if and only if it is compact.

One may use Corollary 4.11 and to show that, whenever $\mathcal{M}$ expands
the real line, the chatacterization of definable compactness given by Theorem
4.6 (6) follows directly from the Alon-Kleitman-Matoušek $(p,q)$-theorem
(Theorem 1.2).

We now prove the connection between definable compactness and curve-
compactness stated in Theorem 4.6. That is, that definable compactness implies
curve-compactness in general, and that both notions are equivalent when the
topology is Hausdorff or when the underlying structure $\mathcal{M}$ has
definable choice. This is Proposition 4.13. We follow the proposition with an
example of a non-Hausdorff topological space definable in the trivial
structure $(M,<)$ that is curve-compact but not definably compact.

In the case where $\mathcal{M}$ extends an ordered field, the equivalence
between curve- and type-compactness, as well as their equivalence with
classical compactness whenever $\mathcal{M}$ expands the real line, can be
derived using the theory of tame pairs. This was done for definable metric
spaces in [Wal15, Proposition 6.6] and in general in [AGTW21, Corollaries 47
and 48].

The next lemma allows us to apply definable choice in certain instances even
when the underlying structure $\mathcal{M}$ does not have the property.

######  Lemma 4.12 (Definable choice in compact Hausdorff spaces).

Let $C\subseteq M^{m}$ be a definable nonempty $\tau$-closed set in a curve-
compact Hausdorff definable topological space $(X,\tau)$. Suppose that $\tau$
and $C$ are $A$-definable. Then there exists a point $x\in C\cap dcl(A)$,
where $dcl(A)$ denotes de definable closure of $A$.

Consequently, for any $A$-definable family $\\{\varphi(u,M^{m}):u\in\Omega\\}$
of nonempty subsets of $X$, there exists an $A$-definable selection function
$s:\Omega\rightarrow X$ such that $s(u)\in cl_{\tau}(\varphi(u,M^{m}))$ for
every $u\in\Omega$.

###### Proof.

We prove the first paragraph of the lemma. The uniform result is derived in
the usual way by the use of first-order logic compactness.

For this proof we adopt the convention of the one point euclidean space
$M^{0}=\\{0\\}$. In particular any projection $M^{k}\rightarrow M^{0}$ is
simply the constant function $0$ and any relation $E\subseteq M^{0}\times
M^{k}$ is definable if and only if its restriction to $M^{k}$ is.

Let $C$ and $\tau$ be as in the Lemma. Let $0\leq n\leq m$ be such that there
exists an $A$-definable function $f:D\subseteq M^{n}\rightarrow C$, for $D$ a
nonempty set. If $n$ can be chosen to be zero then the lemma follows. We prove
that this is the case by backwards induction on $n$.

Note that $n$ can always be chosen equal to $m$ by letting $f$ be the identity
on $C$. Suppose that $0<n\leq m$. For any $x\in M^{n-1}$ let $D_{x}$ denote
the fiber $\\{t\in M:\langle x,t\rangle\in D\\}$. For each $x\in\pi(D)$ let
$s_{x}=\sup D_{x}$, and consider the $A$-definable set
$H=\\{x\in\pi(D):s_{x}\in D_{x}\\}$.

If $H\neq\emptyset$ then let $g$ be the map $x\mapsto f(s_{x}):H\rightarrow
C$. If $H=\emptyset$ then let $g$ be the map
$x\mapsto\tau\text{-}\lim_{t\rightarrow s_{x}^{-}}f(x,t):\pi(D)\rightarrow C$
which, by curve-compactness and Hausdorffness, is well defined. In both cases
$g$ is an $A$-definable nonempty partial function $M^{n-1}\rightarrow C$. ∎

Lemma 4.12 implies that, even when $\mathcal{M}$ does not have definable
choice, if an $A$-definable family of nonempty closed sets $\mathcal{C}$ in a
curve-compact Hausdorff $A$-definable topology has a finite transversal, then
it also has one in $dcl(A)$. To prove this it suffices to note that, for any
$k\geq 1$, the set of $k$-tuples of points corresponding to a transversal of
$\mathcal{C}$ is $A$-definable and closed in the product topology.

######  Proposition 4.13.

Let $(X,\tau)$ be a definable topological space. If $(X,\tau)$ is definably
compact then it is curve-compact.

Suppose that either $\tau$ is Hausdorff or $\mathcal{M}$ has definable choice.
Then $(X,\tau)$ is definably compact if and only if it is curve-compact.

We prove the left to right direction through a short lemma.

######  Lemma 4.14.

Let $(X,\tau)$ be a definably compact definable topological space. Then
$(X,\tau)$ is curve-compact.

###### Proof.

Let $\gamma:(a,b)\rightarrow X$ be a definable curve in $X$. Consider the
definable family of $\tau$-closed nested sets
$\mathcal{C}_{\gamma}=\\{cl_{\tau}(\gamma[(a,t)]):a<t<b\\}$. By definable
compactness there exists $x\in\cap\mathcal{C}_{\gamma}$. Clearly
$x\in\tau\text{-}\lim_{t\rightarrow a^{-}}\gamma(t)$. Similarly one shows that
$\gamma$ $\tau$-converges as $t\rightarrow b$. ∎

We now prove a simpler case of the left to right implication.

######  Lemma 4.15.

Let $(X,\tau)$ be a definable topological space. Suppose that either $\tau$ is
Hausdorff or $\mathcal{M}$ has definable choice. Let $\mathcal{C}$ be a nested
definable family of $\tau$-closed nonempty subsets of $X$. If $(X,\tau)$ is
curve-compact then $\cap\mathcal{C}\neq\emptyset$.

###### Proof.

Let $\mathcal{C}=\\{\varphi(u,M^{m}):u\in\Omega\\}$, $\Omega\subseteq M^{n}$,
and $(X,\tau)$ be as in the lemma. We proceed by induction on $n$.

Base case: $n=1$.

Consider the definable total preorder $\preccurlyeq$ in $\Omega$ given by
$u\preccurlyeq v$ if and only if $\varphi(u,M^{m})\subseteq\varphi(v,M^{m})$.
If $\Omega$ has a minimum $u$ with respect to $\preccurlyeq$ then
$\varphi(u,M^{n})\subseteq C$ for every $C\in\mathcal{C}$ and the result is
obvious. We suppose that $(\Omega,\preccurlyeq)$ does not have a minimum and
consider the definable nested family of infinite sets
$\\{(-\infty,u)_{\preccurlyeq}:u\in\Omega\\}$. By the base case in the proof
of Lemma 2.7, this family extends to a definable type with a basis of sets
$\\{(a,t):t>a\\}$ for some $a\in M\cup\\{-\infty\\}$ or $\\{(t,a):t<a\\}$ for
some $a\in M\cup\\{+\infty\\}$. We consider the former case, being the proof
for the latter analogous. This means that, for every $u\in\Omega$, there
exists $t(u)>a$ such that $v\prec u$ for every $a<v<t(u)$.

If $\mathcal{M}$ has definable choice or if $\tau$ is Hausdorff (Lemma 4.12)
there exists a definable function $f:\Omega\rightarrow\cup\mathcal{C}$
satisfying that $f(u)\in\varphi(u,M^{m})$ for every $u\in\Omega$. Recall that,
for every $u,v\in\Omega$, if $v\preccurlyeq u$ then
$\varphi(v,M^{m})\subseteq\varphi(u,M^{m})$, and in particular
$f(v)\in\varphi(u,M^{n})$. It follows that, for every $u\in\Omega$, there
exists $t(u)>a$ such that $f(v)\in\varphi(u,M^{n})$ for every $a<v<t(u)$. Let
$b>a$ be such that $(a,b)\subseteq\Omega$ and $\gamma$ be the restriction of
$f$ to $(a,b)$. We conclude that, for every $C\in\mathcal{C}$,
$\tau\text{-}\lim_{t\rightarrow a}\gamma(t)\in C$.

Inductive step: $n>1$.

For any $x\in\pi(\Omega)$ let $\mathcal{C}(x)$ denote the family
$\\{\varphi(x,t,M^{m}):t\in\Omega_{x}\\}$, and set $C(x):=\cap\mathcal{C}(x)$.
By the base case the definable family of $\tau$-closed sets
$\mathcal{D}=\\{C(x):x\in\pi(\Omega)\\}$ does not contain the empty set.
Clearly $\cap\mathcal{D}=\cap\mathcal{C}$. We observe that the family
$\mathcal{D}$ is nested and the result follows from the induction hypothesis.

Given $x,y\in\pi(\Omega)$, if for every $C\in\mathcal{C}(x)$ there is
$C^{\prime}\in\mathcal{C}(y)$ with $C^{\prime}\subseteq C$ then
$\cap\mathcal{C}(y)\subseteq\cap\mathcal{C}(x)$. Otherwise there is
$C\in\mathcal{C}(x)$ such that, for every $C^{\prime}\in\mathcal{C}(y)$, it
holds that $C\subseteq C^{\prime}$, in which case $\cap\mathcal{C}(x)\subseteq
C\subseteq\cap\mathcal{C}(y)$. ∎

We may now prove Proposition 4.13.

###### Proof of Proposition 4.13.

By Lemma 4.14 we must only prove the right to left implication. Let
$(X,\tau)$, $X\subseteq M^{m}$, be a curve-compact Hausdorff definable
topological space. Let $\mathcal{C}\subseteq\mathcal{P}(X)$ be a definable
downward directed family of subsets of $(X,\tau)$, not necessarily
$\tau$-closed. We show that
$\bigcap\\{cl_{\tau}(C):C\in\mathcal{C}\\}\neq\emptyset$.

We proceed by induction on $n=\min\\{\dim C:C\in\mathcal{C}\\}$. By Lemma 2.7,
after passing to a finer family if necessary we may assume that $\mathcal{C}$
is a complete family of cells. For the length of this proof, given two partial
$M_{\pm\infty}$-valued functions $f$ and $g$ let $(f,g)=\\{\langle
x,t\rangle:x\in dom(f)\cap dom(g)\text{ and }f(x)<t<g(x)\\}$, relaxing thus
the classical notation by allowing that $f$ and $g$ have different domains.

If $n=0$ there exists a finite set in $\mathcal{C}$ and so (Fact 2.5)
$\bigcap\mathcal{C}\neq\emptyset$. We now prove the case $n=m>0$. Hence
suppose that every $C\in\mathcal{C}$ is an open cell $C=(f_{C},g_{C})$, for
definable continuous functions $f_{C},g_{C}:\pi(C)\rightarrow M_{\pm\infty}$
with $f_{C}<g_{C}$. For any $C\in\mathcal{C}$ let
$D(C)=\bigcap\\{cl_{\tau}(f_{C},g_{C^{\prime}}):C^{\prime}\in\mathcal{C}\\}$.
We first show that these sets are nonempty.

Let us fix $C=(f,g)$ and, for any $x\in\pi(C)$, let
$C^{0}(x)=\tau\text{-}\lim_{t\rightarrow f(x)^{+}}\langle x,t\rangle$. If
$\tau$ is Hausdorff then this point is unique, otherwise we use definable
choice to pick one such point definably on $x$. The definable set
$C^{0}=\\{C^{0}(x):x\in\pi(C)\\}$ has dimension $\dim(C)-1=n-1$. For any
$C^{\prime}=(f^{\prime},g^{\prime})\in\mathcal{C}$, the definable set
$\\{x\in\pi(C)\cap\pi(C^{\prime}):f(x)<g^{\prime}(x)\\}$ is nonempty, since
otherwise we would have $C\cap C^{\prime}=\emptyset$. It follows that
$C^{0}\cap cl_{\tau}(f,g^{\prime})\neq\emptyset$. Note that the definable
family $\\{C^{0}\cap cl_{\tau}(f,g_{C^{\prime}}):C^{\prime}\in\mathcal{C}\\}$
is downward directed. By inductive hypothesis there is a point that belongs in
the $\tau$-closure of $C^{0}\cap cl_{\tau}(f,g_{C^{\prime}})$, and in
particular in $cl_{\tau}(f,g_{C^{\prime}})$, for all
$C^{\prime}\in\mathcal{C}$. So $D(C)\neq\emptyset$.

Note that, for every $C\in\mathcal{C}$, $D(C)\subseteq cl_{\tau}(C)$. We now
note that the definable family of nonempty $\tau$-closed sets
$\\{D(C):C\in\mathcal{C}\\}$ is nested. Then, by Lemma 4.15,
$\bigcap\\{D(C):C\in\mathcal{C}\\}\neq\emptyset$, and thus
$\bigcap\\{cl_{\tau}(C):C\in\mathcal{C}\\}\neq\emptyset$.

Let us fix $C_{1}=(f_{1},g_{1}),C_{2}=(f_{2},g_{2})\in\mathcal{C}$. We may
partition $B=\pi(C_{1})\cap\pi(C_{2})$ by the definable sets

$B_{1}=\\{x\in B:f_{1}(x)\leq f_{2}(x)\\}\text{ and }B_{2}=\\{x\in
B:f_{1}(x)>f_{2}(x)\\}.$

Since $\mathcal{C}$ is complete there exists some $i\in\\{1,2\\}$ and
$C\in\mathcal{C}$ such that $\pi(C)\subseteq B_{i}$. Without loss of
generality suppose that $i=1$, and let us fix $C_{3}\in\mathcal{C}$ with
$\pi(C_{3})\subseteq B_{1}$. For any $C=(f,g)\in\mathcal{C}$ let
$C^{\prime}=(f^{\prime},g^{\prime})\in\mathcal{C}$ be such that
$C^{\prime}\subseteq C\cap C_{3}$. Clearly
$(f_{2},g^{\prime})\subseteq(f_{1},g^{\prime})\subseteq(f_{1},g)$. It follows
that $D(C_{2})\subseteq D(C_{1})$. This completes the proof of the case
$n=m>0$.

Finally we prove the case $0<n<m$ by adapting the arguments above. Fix
$\hat{C}\in\mathcal{C}$ with $\dim\hat{C}=n<m$ and a projection
$\hat{\pi}:M^{m}\rightarrow M^{n}$ such that
$\hat{\pi}|_{\hat{C}}:\hat{C}\rightarrow\hat{\pi}(\hat{C})$ is a bijection. By
passing to a finner subfamily if necessary we may assume that every set in
$\mathcal{C}$ is contained in $\hat{C}$. It follows that the definable
downward directed family
$\hat{\pi}(\mathcal{C})=\\{\hat{\pi}(C):C\in\mathcal{C}\\}$ contains only open
cells in $M^{n}$. Note moreover that this family is complete.

Set $h:=(\hat{\pi}|_{\hat{C}})^{-1}$. Then the proof in the case $n=m$ can be
applied to $\hat{\pi}(\mathcal{C})$ instead of $\mathcal{C}$, making sure to
write throughout $cl_{\tau}h(\cdot)$ in place of $cl_{\tau}(\cdot)$, and
letting $\hat{\pi}(C)^{0}=\\{\tau\text{-}\lim_{t\rightarrow
f(x)^{+}}h(x,t):x\in\pi(\hat{\pi}(C))\\}$ for any
$\hat{\pi}(C)=(f,g)\in\hat{\pi}(\mathcal{C})$. It follows that
$\bigcap\\{cl_{\tau}(C):C\in\mathcal{C}\\}\neq\emptyset$. This completes the
proof of the proposition. ∎

The following is an example of a non-Hausdorff definable topological space in
$(M,<)$ that is curve-compact but not definably compact.

######  Example 4.16.

Let $X=\\{\langle x,y\rangle\in M^{2}:y<x\\}$. Consider the family
$\mathcal{B}$ of sets

$\displaystyle
A(x^{\prime},x^{\prime\prime},x^{\prime\prime\prime},y^{\prime},y^{\prime\prime},y^{\prime\prime\prime})=$
$\displaystyle\\{\langle x,y\rangle:y<y^{\prime},y<x\\}\cup$
$\displaystyle\\{\langle
x,y\rangle:y^{\prime\prime}<y<y^{\prime\prime\prime}\wedge(y<x<y^{\prime\prime\prime}\vee
x^{\prime}<x<x^{\prime\prime}\vee x^{\prime\prime\prime}<x)\\}$

definable uniformly over
$y^{\prime}<y^{\prime\prime}<y^{\prime\prime\prime}<x^{\prime}<x^{\prime\prime}<x^{\prime\prime\prime}$.

Figure 1: Depicting (in blue) the set
$A(x^{\prime},x^{\prime\prime},x^{\prime\prime\prime},y^{\prime},y^{\prime\prime},y^{\prime\prime\prime})$.

Given any
$A_{0}=A(x_{0}^{\prime},x_{0}^{\prime\prime},x_{0}^{\prime\prime\prime},y_{0}^{\prime},y_{0}^{\prime\prime},y_{0}^{\prime\prime\prime})$
and
$A_{1}=A(x_{1}^{\prime},x_{1}^{\prime\prime},x_{1}^{\prime\prime\prime},y_{1}^{\prime},y_{1}^{\prime\prime},y_{1}^{\prime\prime\prime})$
in $\mathcal{B}$ and any $\langle x,y\rangle\in A_{1}\cap A_{2}$, since every
set in $\mathcal{B}$ is open in the euclidean topology, we may find
$y^{\prime\prime}<y<y^{\prime\prime\prime}<x^{\prime}<x<x^{\prime\prime}$ such
that the box
$(x^{\prime},x^{\prime\prime})\times(y^{\prime\prime},y^{\prime\prime\prime})$
is a subset of $A_{1}\cap A_{2}$. Let
$y^{\prime}<\min\\{y^{\prime\prime},y^{\prime}_{0},y^{\prime}_{1}\\}$ and
$x^{\prime\prime\prime}>\max\\{x^{\prime\prime},x^{\prime\prime\prime}_{0},x^{\prime\prime\prime}_{1}\\}$.
Then $\langle x,y\rangle\in
A(x^{\prime},x^{\prime\prime},x^{\prime\prime\prime},y^{\prime},y^{\prime\prime},y^{\prime\prime\prime})\subseteq
A_{1}\cap A_{2}$. Hence the family $\mathcal{B}$ is a definable basis for a
topology $\tilde{\tau}$.

This topology is $T_{1}$, i.e. every singleton is closed. For every $y\in M$,
$\tilde{\tau}\text{-}\lim_{t\rightarrow y^{+}}\langle
t,y\rangle=\tilde{\tau}\text{-}\lim_{t\rightarrow+\infty}\langle
t,y\rangle=(M\times\\{y\\})\cap X$, and, for every $x\in M$,
$\tilde{\tau}\text{-}\lim_{t\rightarrow x^{-}}\langle
x,t\rangle=(M\times\\{x\\})\cap X$ and
$\tilde{\tau}\text{-}\lim_{t\rightarrow-\infty}\langle x,t\rangle=X$. In
particular $\tilde{\tau}$ is not Hausdorff.

Now suppose that $\mathcal{M}=(M,<)$. By quantifier elimination we know that
in this structure any definable partial map $M\rightarrow M$ is piecewise
either constant or the identity. Let
$\gamma=(\gamma_{0},\gamma_{1}):(a,b)\rightarrow X$ be an injective definable
curve in $X$. Let $I=(a,c)\subseteq(a,b)$ be an interval where $\gamma_{0}$
and $\gamma_{1}$ are either constant or the identity. Since the graph of the
identity is disjoint from $X$ and $\gamma$ is injective it must be that
$\gamma_{i}$ is constant and $\gamma_{1-i}$ is the identity on $I$ for some
$i\in\\{0,1\\}$.

Suppose that $\gamma_{1}|_{I}$ has a constant value $y$. Then, by the
observations made above about the topology, $\gamma$ $\tilde{\tau}$-converges
as $t\rightarrow a$ to either $\langle a,y\rangle$ (if $y<a$) or
$(M\times\\{y\\})\cap X$ (if $a=y$). On the other hand, if $\gamma_{0}|_{I}$
has a constant value $x$, then $\gamma$ $\tilde{\tau}$-converges as
$t\rightarrow a$ to either $\langle x,a\rangle$ (if $a>-\infty$) or the whole
space $X$ (if $a=-\infty$). Treating the limit as $t\rightarrow b$ similarly
allows us to conclude that $\gamma$ is $\tilde{\tau}$-completable. Hence the
space $(X,\tilde{\tau})$ is curve-compact.

Meanwhile, the definable downward directed family of $\tilde{\tau}$-closed
sets $\\{X\cap(R\times[u,+\infty)):u\in M\\}$ has empty intersection. So
$(X,\tilde{\tau})$ is not definably compact.

We end the paper with a question. Notice that all the characterizations of
definable compactness in Theorem 4.6 are upfront expressible with infinitely
many sentences (possibly with parameters) in the language of $\mathcal{M}$
(you have to check all relevant definable families of closed sets or all
definable curves). In [PS99] it is shown that, in the euclidean topology, a
set is curve-compact if and only if it is closed and bounded. Being closed and
bounded is expressible by a single formula. Moreover, given a definable family
of sets with the euclidean topology, the subfamily of those that are closed
and bounded is definable. We ask the following.

######  Question 4.17.

  1. (i)

Can definable compactness of an arbitrary definable topology be expressed with
a single formula in the language of $\mathcal{M}$?

  2. (ii)

More generally, given a definable family of definable topological spaces, is
the subfamily of all which are definably compact definable?

##  Appendix A Appendix: Proof of The Marker-Steinhorn Theorem

In this appendix we use the approach of various proofs in this paper, based on
the the preorder described in Subsection 1.3, to extract from [MS94] a short
proof of the Marker-Steinhorn Theorem (that avoids a treatment by cases). It
is worth noting that a similar approach via definable linear orders was used
by Walsberg [Wal19] to prove the theorem in o-minimal expansions of ordered
groups.

Recall that $\mathcal{M}$ denotes an o-minimal structure and that, unless
stated otherwise, definable means in $\mathcal{M}$ over $M$. Recall that an
elementary extension $\mathcal{N}$ of $\mathcal{M}$ is _tame_ if, for every
$s\in N$, the set $\\{t\in M:t<s\\}$ has a supremum in $M\cup\\{+\infty\\}$.
This is also referred to as $\mathcal{M}$ being _Dedekind complete_ in
$\mathcal{N}$ [MS94].

In full generality the Marker-Steinhorn theorem shows that a type $p\in
S_{n}(M)$ is definable if and only if it is realized in some tame extension.
We state and prove the “if” direction, and direct the reader to [MS94, Theorem
2.1] for the full result.

The approach of the proof below involves using cell decomposition to reduce
the problem to showing that any cut in a definable preordered set that is
realized in a tame extension is definable. This of course is obvious for the
preordered set $(M,\leq)$, by definition of tameness.

######  Theorem A.1 (Marker-Steinhorn Theorem).

Let $\mathcal{N}=(N,\ldots)$ be a tame extension of $\mathcal{M}$. For every
$a\in N^{n}$, the type $\operatorname{tp}(a/M)$ is definable.

###### Proof.

We must prove that, for every $a\in N^{n}$ and formula $\varphi(u,a)$,
$l(u)=m$, the set $\varphi(M^{m},a)=\\{u\in
M^{m}:\mathcal{N}\models\varphi(u,a)\\}$ is definable. We do this by induction
on $n$ and $m$, where in the inductive step we assume that it holds for any
$\langle n^{\prime},m^{\prime}\rangle$ smaller than $\langle n,m\rangle$ in
the lexicographic order. We may assume that $a\notin M^{n}$.

The case $n=1$ (for any $m$) follows easily from o-minimality and tameness. In
particular, let $s_{a}$ be the supremum in $M\cup\\{\pm\infty\\}$ of
$(-\infty,a)\cap M$. If $s_{a}<a$ then $\operatorname{tp}(a/M)$ has a
definable basis of the form $\\{(s_{a},t):s_{a}<t\in M\\}$, otherwise it has a
definable basis of the form $\\{(t,s_{a}):s_{a}>t\in M\\}$.

Suppose that $n>1$ and let $a=\langle c,d\rangle\in N^{n-1}\times N$. Let
$\psi_{i}(N^{m},N^{n})$, $0\leq i\leq l$, be a ($0$-definable) cell
decomposition of $\varphi(N^{m},N^{n})$. For every $u\in M^{m}$, the set
$\varphi(u,N^{n})$ is partitioned by the cells $\psi_{i}(u,N^{n})$, $0\leq
i\leq l$. In particular $\mathcal{N}\models\varphi(u,a)$ if and only if
$\mathcal{N}\models\psi_{i}(u,a)$ for some $i$. So to prove the theorem it
suffices to pass to an arbitrary $0\leq i\leq l$ and show that
$\psi_{i}(M^{m},a)$ is definable. Hence, without loss of generality we assume
that all the sets $\varphi(u,N^{n})$ are cells.

By induction hypothesis $\operatorname{tp}(c/M)$ is definable. Suppose that
there exists an $M$-definable partial function $f:N^{n-1}\rightarrow N$ with
$f(c)=d$. Then, for every $u\in N^{m}$, we have that
$\mathcal{N}\models\varphi(u,a)$ if and only if $\mathcal{N}\models\exists
t\,(\varphi(u,c,t)\wedge(f(c)=t))$. And so the result follows.

Hence onwards we assume that there exists no $M$-definable partial function
$f$ with $f(c)=d$. In particular we have that every $u\in M^{m}$ with
$\mathcal{N}\models\varphi(u,a)$ satisfies that $\varphi(u,N^{n})$ is a cell
of the form $(f_{u},g_{u})$ for two definable continuous functions $f_{u}$ and
$g_{u}$. Let

$\Omega=\\{u\in M^{m}:\varphi(u,N^{n})\text{ is of the form
}(f_{u},g_{u})\text{ and }\mathcal{N}\models\exists t\,\varphi(u,c,t)\\}.$

Clearly $\varphi(M^{m},a)\subseteq\Omega$. Since $\operatorname{tp}(c/M)$ is
definable the set $\Omega$ is definable. We prove that the sets
$\\{u\in\Omega:f_{u}(c)<d\\}$ and $\\{u\in\Omega:g_{u}(c)>d\\}$ are definable.
Then their intersection equals $\varphi(M^{m},a)$. The proof of definability
is the same for both sets, so we show it only for the former.

Let $\preccurlyeq$ be the total preorder on $\Omega$ induced by
$\\{f_{u}(c):u\in\Omega\\}$ described in Subsection 1.3, i.e. for
$u,w\in\Omega$, $u\preccurlyeq w$ if and only if $f_{u}(c)\leq f_{w}(c)$. By
definability of $\operatorname{tp}(c/M)$ the preordered set
$(\Omega,\preccurlyeq)$ is definable. Let $P=\\{u\in\Omega:f_{u}(c)<d\\}$ and
$Q=\\{u\in\Omega:f_{u}(c)>d\\}=\Omega\setminus P$. We must prove that $P$
(equivalently $Q$) is definable. If $P$ has a supremum in
$\Omega\cup\\{\pm\infty\\}$ with respect to $\preccurlyeq$ then the result is
immediate, so we assume otherwise. In particular we have that, for every $u\in
P$ and $v\in Q$, $\dim(u,v)_{\preccurlyeq}>0$.

Note that, to prove the definability of $P$, it suffices to show that there
exists a definable set $P^{\prime}\subseteq P$ that is cofinal in $P$, since
then $P=\\{u\in\Omega:u\preccurlyeq v\text{ for some }v\in P^{\prime}\\}$.
Similarly it is enough to show the existence of a defiable
$Q^{\prime}\subseteq Q$ coinitial in $Q$. So we may always pass to a subset
$\Omega^{\prime}\subseteq\Omega$ such that either $\Omega^{\prime}\cap P$ is
cofinal in $P$ or $\Omega^{\prime}\cap Q$ is coinitial in $Q$, and then prove
definability of $\Omega^{\prime}\cap P$ or $\Omega^{\prime}\cap Q$. Hence, by
passing to a subset if necessary, we may assume that, for any $u\in P$ and
$v\in Q$,

$\dim(u,v)_{\preccurlyeq}=\dim\Omega.$ ($\star$)

Moreover, note that we may also pass to a set in any given definable finite
partition of $\Omega$. In particular, by cell decomposition, we assume that
$\Omega$ is a cell.

Suppose that $m=1$. For each $u\in\Omega(\mathcal{N})$ with $c\in dom(f_{u})$
(note that this includes every $u$ in $\Omega$) let $\hat{f}(u)=f_{u}(c)$.
Then $\hat{f}$ is definable in $\mathcal{N}$ over $M\cup\\{c\\}$. By
o-minimality there exists a partition (definable over $M\cup\\{c\\}$) of the
domain of $\hat{f}$ into points and intervals such that, on each interval,
$\hat{f}$ is continuous and either constant or strictly monotonic. Since
$\operatorname{tp}(c/M)$ is definable the intersections of these cells with
$\Omega$ are definable. Note that, on any such intersection, the restriction
of the preorder $\preccurlyeq$ is either $\leq$, $\geq$, or $\leq\cup\geq$
(the trivial relation where any two points are indistinguishable), depending
respectively on whether $\hat{f}$ is strictly increasing, decreasing or
constant. We fix one such interval $I$ and show that $I\cap P$ is definable.

If $I\cap P=I\cap\Omega$ or $I\cap Q=I\cap\Omega$ then the result is
immediate. Otherwise there exist $u,v\in I\cap\Omega$ such that $\hat{f}(u)<d$
and $\hat{f}(v)>d$. By continuity there must exist $r$ in the subinterval of
$I$ with endpoints $u$ and $v$ with $\hat{f}(r)=d$. By tameness
$J=(-\infty,d)\cap M$ is definable. Finally note that $\hat{f}|_{I}$ is not
constant, and thus it is strictly monotonic. If it is increasing then it must
be that $J\cap I=P\cap I$ and otherwise $J\cap I=Q\cap I$.

Now suppose that $m>1$. For every $x$ in the projection $\pi(\Omega)$ to the
first $m-1$ coordinates, let $\preccurlyeq_{x}$ be the definable preorder on
the fiber $\Omega_{x}$ given by $s\preccurlyeq_{x}t$ if and only if $\langle
x,s\rangle\preccurlyeq\langle x,t\rangle$. Note that, following the arguments
in the case $m=1$, the sets $P_{x}$ and $Q_{x}$ are definable, and moreover
$\Omega_{x}$ can be partitioned into finitely many points and intervals where
the restriction of $\preccurlyeq_{x}$ is either $\leq$, $\geq$, or
$\leq\cup\geq$.

If there exists $x\in\pi(\Omega)$ such that $\\{x\\}\times P_{x}$ is cofinal
in $P$ or $\\{x\\}\times Q_{x}$ is coinitial in $Q$, then we are done. Suppose
otherwise. We complete the proof by partitioning $\Omega$ into finitely many
definable sets with the following property. For each set $\Sigma$ in the
partition and $x\in\pi(\Sigma)$, either $\Sigma_{x}\subseteq P_{x}$ or
$\Sigma_{x}\subseteq Q_{x}$. Observe that the set $\Theta$ of all
$x\in\pi(\Sigma)$ such that $\Sigma_{x}\subseteq P_{x}$ is described by

$x\in\pi(\Sigma),\,\forall t\in\Sigma_{x},\,f_{(x,t)}(c)<d,$

so, by induction hypothesis (applied in the case $\langle n,m-1\rangle$), this
set is definable. It follows that $\Sigma\cap
P=\bigcup_{x\in\Theta}(\\{x\\}\times\Sigma_{x})$ is definable, and we may
conclude that $P$ is definable.

Recall that $\Omega$ is a cell. If it is defined as the graph of a function
then, by taking $\Sigma$ in the above paragraph to be $\Omega$, we are done,
so we assume otherwise. Let $x\in\pi(\Omega)$. By ($\star$ ‣ A), since
$\\{x\\}\times P_{x}$ is not cofinal in $P$ and $\\{x\\}\times Q_{x}$ is not
coinitial in $Q$, the dimension of

$\\{u\in\Omega:\\{x\\}\times P_{x}\prec u\prec\\{x\\}\times Q_{x}\\}$

equals $\dim\Omega$. Consider $r\in\Omega_{x}$ to be the right endpoint of
some maximal subinterval $I^{\prime}$ of $P_{x}$. Then in particular $r$ is
the left endpoint of a subinterval $I^{\prime\prime}$ of $Q_{x}$. If $r\notin
P_{x}$ then the set $\\{u\in\Omega:\\{x\\}\times I^{\prime}\prec u\prec\langle
x,r\rangle\\}$ has dimension $\dim(\Omega)$. If however $r\in P_{x}$, then the
set $\\{u\in\Omega:\langle x,r\rangle\prec u\prec\\{x\\}\times
I^{\prime\prime}\\}$ has dimension $\dim(\Omega)$. If $r$ is the right
endpoint in $\Omega_{x}$ of a a maximal subinterval of $Q_{x}$ then the
analogous holds. Note that, if there exists $s,t\in\Omega_{x}$ with $s\in
P_{x}$ and $t\in Q_{x}$, there will always be some $r$ in the closed interval
between $s$ and $t$ with the described properties.

For any $u=\langle x,t\rangle\in\Omega$, let $L(x,t)$ be the set of
$v\in\Omega$ such that either $\\{x\\}\times I^{\prime}\prec v\prec u$ or
$u\prec v\prec\\{x\\}\times I^{\prime}$ for some interval
$I^{\prime}\subseteq\Omega_{x}$ with right endpoint $t$. Similarly let
$U(x,t)$ be the set of $v\in\Omega$ such that either $\\{x\\}\times
I^{\prime\prime}\prec v\prec u$ or $u\prec v\prec\\{x\\}\times
I^{\prime\prime}$ for some interval $I^{\prime\prime}\subseteq\Omega_{x}$ with
left endpoint $t$. These sets are definable uniformly on $u\in\Omega$. Let
$\Lambda$ be the definable set of all $u\in\Omega$ such that $U(u)\cup L(u)$
has dimension $\dim(\Omega)$. By the above paragraph, for every
$x\in\pi(\Omega)$ and $s,t\in\Omega_{x}$, if $s\in P_{x}$ and $t\in Q_{x}$,
then there is some $r$ in the closed interval between $s$ and $t$ such that
$\langle x,r\rangle\in\Lambda$.

We now show that, for every $x\in\pi(\Omega)$, the fiber $\Lambda_{x}$ is
finite. Then the proof is completed by taking any finite cell partition of
$\Omega$ compatible with $\Lambda$, since, for any $\Sigma$ in said partition
and $x\in\pi(\Sigma)$, the fiber $\Sigma_{x}$ is going to be either a point or
an interval contained in $P_{x}$ or $Q_{x}$.

Towards a contradiction suppose that $\Lambda_{x}$ is infinite for some
$x\in\Omega$. Let $J^{\prime}$ be a subinterval of $\Lambda_{x}$ where
$\preccurlyeq_{x}$ is either $\leq$, $\geq$ or $\leq\cup\geq$. If
$\preccurlyeq_{x}$ equals $\leq\cup\geq$ then, for every $t\in J^{\prime}$,
the sets $U(x,t)$ and $L(x,t)$ are empty, contradicting that $\dim(L(x,t)\cup
U(x,t))=\dim\Omega$. Suppose that $\preccurlyeq_{x}$ is either $\leq$ or
$\geq$. Note that, for any distinct $s,t\in J^{\prime}$, the sets $L(x,s)$,
$U(x,s)$, $L(x,t)$ and $U(x,t)$ are pairwise disjoint. Using the fact that
$\dim(L(x,t)\cup U(x,t))=\dim\Omega$ for every $t\in J^{\prime}$, and applying
the Fiber Lemma for o-minimal dimension [vdD98, Chapter 4, Proposition 1.5 and
Corollary 1.6], we derive that

$\dim\left(\bigcup_{t\in J^{\prime}}(L(x,t)\cup U(x,t))\right)>\dim(\Omega),$

contradiction. ∎

##  Appendix B Appendix (by Will Johnson): $A$-definability in Lemma 2.7

Let $M$ be an o-minimal structure666In this appendix we resort to the
convention of using the same terminology for a structure and its universe..
Let $\mathcal{S}\subseteq\mathcal{P}(M^{n})$ be an $A$-definable downward
directed family.

  * •

By Theorem 2.12, $\mathcal{S}$ can be extended to a definable type $p$ with a
definable basis.

  * •

By Proposition 2.17, $\mathcal{S}$ can be extended to an $A$-definable type.

It is natural to ask whether we can do both at the same time. Can
$\mathcal{S}$ be extended to an $A$-definable type with a definable basis? As
we will see below, the answer is no in general.

This shows that a parameter version of Lemma 2.7 is not possible: in our
counterexample, there is no _complete_ $A$-definable downward directed family
$\mathcal{S}^{\prime}$ finer than $S$. Otherwise, $\mathcal{S}^{\prime}$ would
be a definable basis for an $A$-definable type $p$ extending $S$.

###  B.1 Types with definable bases

Let $M$ be any structure, not necessarily o-minimal.

Recall that, for any $M^{\prime}\succeq M$, a type $p\in S_{x}(M^{\prime})$ is
an _heir_ of $p|_{M}$ if, for every formula $\varphi(x,a,c)\in p$, where $a$
and $c$ are tuples of parameters from $M^{\prime}$ and $M$ respectively, there
exists $b\in M^{l(a)}$ such that $\varphi(x,b,c)\in p$. It is easy to see that
the unique heir of a definable type is its canonical extension using the same
defining scheme.

######  Lemma B.1.

Let $p(x)$ be an $n$-type over $M$ with a definable basis.

  1. 1.

$p(x)$ is a definable type.

  2. 2.

If $M^{\prime}\succeq M$ and $p^{M^{\prime}}$ is the heir of $p$, then
$p^{M^{\prime}}$ has a definable basis.

  3. 3.

Suppose $M$ is $\aleph_{1}$-saturated. Let
$\mathcal{S}_{1},\mathcal{S}_{2},\ldots$ be a sequence of definable downward
directed families of subsets of $M^{n}$. Suppose that $\mathcal{S}_{i+1}$ is
finer than $\mathcal{S}_{i}$ for all $i$, and suppose
$\bigcup_{i=1}^{\infty}\mathcal{S}_{i}$ generates $p(x)$. Then there is some
$i_{0}\in\mathbb{N}$ such that $\mathcal{S}_{i_{0}}$ is finer than
$\mathcal{S}_{i}$ for all $i$. In particular $\mathcal{S}_{i_{0}}$ generates
$p(x)$.

###### Proof.

  1. 1.

Easy.

  2. 2.

For any definable family $\mathcal{F}$ in $M$, let $\mathcal{F}(M^{\prime})$
denote its interpretation in $M^{\prime}$. Let $\mathcal{D}$ be the definable
downward directed family generating $p(x)$. For any definable family
$\mathcal{F}$ in $M$, we have

$\forall X\in\mathcal{F}\leavevmode\nobreak\ \exists
Y\in\mathcal{D}:(Y\subseteq X\leavevmode\nobreak\ \vee\leavevmode\nobreak\
Y\subseteq M^{n}\setminus X).$

This is expressed by a first-order formula, so it remains true in
$M^{\prime}$. Therefore $\mathcal{D}(M^{\prime})$ generates an $n$-type over
$M^{\prime}$, which is clearly $p^{M^{\prime}}$.

  3. 3.

The families $\mathcal{D}$ and $\bigcup_{i=1}^{\infty}\mathcal{S}_{i}$
generate the same type, so $\mathcal{D}$ is finer than
$\bigcup_{i=1}^{\infty}\mathcal{S}_{i}$. Therefore

$\forall i\in\mathbb{N}:(\mathcal{D}\text{ is finer than }\mathcal{S}_{i}).$
(1)

Additionally, the union $\bigcup_{i=1}^{\infty}\mathcal{S}_{i}$ is finer than
$\mathcal{D}$. Therefore

$\forall X\in\mathcal{D}\leavevmode\nobreak\ \exists
i\in\mathbb{N}\leavevmode\nobreak\ \exists Y\in\mathcal{S}_{i}:Y\subseteq X.$

By $\aleph_{1}$-saturation, there is some finite $N$ such that

$\forall X\in\mathcal{D}\leavevmode\nobreak\ \exists i\leq
N\leavevmode\nobreak\ \exists Y\in\mathcal{S}_{i}:Y\subseteq X.$

Thus $\bigcup_{i=1}^{N}\mathcal{S}_{i}$ is finer than $\mathcal{D}$. Now
$\mathcal{S}_{N}$ is finer than $\mathcal{S}_{i}$ for all $i\leq N$, so
$\mathcal{S}_{N}$ is finer than $\bigcup_{i=1}^{N}\mathcal{S}_{i}$. Then

$\mathcal{S}_{N}\text{ is finer than
}\left(\bigcup_{i=1}^{N}\mathcal{S}_{i}\right)\text{ is finer than
}\mathcal{D}\text{ is finer than
}\left(\bigcup_{i=1}^{\infty}\mathcal{S}_{i}\right),$

by (1). So $\mathcal{S}_{N}$ is finer than
$\bigcup_{i=1}^{\infty}\mathcal{S}_{i}$, which implies that $\mathcal{S}_{N}$
is finer than $\mathcal{S}_{i}$ for all $i$. ∎

###  B.2 Divisible ordered abelian groups

For any linearly ordered set $(S,\leq)$, let $\operatorname{Cut}(S)$ be the
set of cuts in $S$, i.e., downward-closed subsets $\xi\subseteq S$. Let
$\overline{S}$ denote the disjoint union $S\sqcup\operatorname{Cut}(S)$. The
linear order on $S$ extends to a linear order on $\overline{S}$ in which

$\displaystyle x<\xi$ $\displaystyle\iff x\in\xi$ $\displaystyle\text{ for
}x\in S,\leavevmode\nobreak\ \xi\in\operatorname{Cut}(S)$
$\displaystyle\xi_{1}\leq\xi_{2}$ $\displaystyle\iff\xi_{1}\subseteq\xi_{2}$
$\displaystyle\text{ for }\xi_{1},\xi_{2}\in\operatorname{Cut}(S).$

We let $-\infty$ and $+\infty$ denote the least and greatest elements of
$\operatorname{Cut}(S)$, i.e., $\emptyset$ and $S$. If $x\in S$, we let
$x^{+}$ and $x^{-}$ denote the cuts immediately above and below $x$:

$\displaystyle x^{+}$ $\displaystyle:=\\{y\in S:y\leq x\\}$ $\displaystyle
x^{-}$ $\displaystyle:=\\{y\in S:y<x\\}.$

When $S=\mathbb{Q}$, we identify an irrational number
$\alpha\in\mathbb{R}\setminus\mathbb{Q}$ with the corresponding cut. Thus

$\overline{\mathbb{Q}}=\mathbb{R}\cup\\{\pm\infty\\}\cup\\{q^{+},\leavevmode\nobreak\
q^{-}:q\in\mathbb{Q}\\}.$

Fix a divisible ordered abelian group $(M,+,\leq)$. Embed $M$ into a monster
model $\mathbb{U}$.

######  Lemma B.2.

For any $\xi\in\operatorname{Cut}(\mathbb{Q})$, there is a 2-type
$p_{\xi}(x,y)$ over $M$ generated by the following formulas:

  * •

$x>t$, for each $t\in M$.

  * •

$y>qx+s$, for each $q\in\mathbb{Q}$ with $q<\xi$ and $s\in M$.

  * •

$y<qx+s$, for each $q\in\mathbb{Q}$ with $q>\xi$ and $s\in M$.

###### Proof.

Take $a\in\mathbb{U}$ with $a>M$. The prime model $M[a]$ is then
$\mathbb{Q}\cdot a\oplus M$, which is isomorphic to $\mathbb{Q}\times M$ with
the lexicographic order. Take $b\in\mathbb{U}$ with

$\displaystyle b$ $\displaystyle>qa+M\text{ for }q<\xi$ $\displaystyle b$
$\displaystyle<qa+M\text{ for }q>\xi.$

Then $(a,b)$ realizes $p_{\xi}(x,y)$, so $p_{\xi}(x,y)$ is a consistent
partial type. For completeness, let $(a^{\prime},b^{\prime})$ be another pair
realizing $p_{\xi}(x,y)$. Then $a^{\prime}>M$, because of the formulas $x>t$.
By o-minimality, $\operatorname{tp}(a^{\prime}/M)=\operatorname{tp}(a/M)$.
Moving $(a^{\prime},b^{\prime})$ by an element of
$\operatorname{Aut}(\mathbb{U}/M)$, we may assume $a^{\prime}=a$. Then for any
$q\in\mathbb{Q}$ and $s\in M$, we have

$\displaystyle b>qa+s$ $\displaystyle\iff\xi>q\iff b^{\prime}>qa+s$
$\displaystyle b<qa+s$ $\displaystyle\iff\xi<q\iff b^{\prime}<qa+s.$

Therefore $b$ and $b^{\prime}$ determine the same cut in $M[a]$. By
o-minimality, $\operatorname{tp}(b/Ma)=\operatorname{tp}(b^{\prime}/Ma)$. Thus
$\operatorname{tp}(a,b/M)=\operatorname{tp}(a^{\prime},b^{\prime}/M)$, as
desired. So $p_{\xi}$ is a (complete) type. ∎

######  Lemma B.3.

For any $q\in\mathbb{Q}$ and $\xi\in\overline{M}$, there is a 2-type
$r_{q,\xi}(x,y)$ over $M$ generated by the following formulas:

  * •

$x>t$, for each $t\in M$.

  * •

$y\leq qx+s$ for $s\in M$ with $s\geq\xi$.

  * •

$y\geq qx+s$ for $s\in M$ with $s\leq\xi$.

###### Proof.

Similar to Lemma B.2. ∎

######  Lemma B.4.

Let $p(x,y)$ be a 2-type over $M$ containing the formulas $x>t$ for all $t\in
M$. Then $p(x,y)$ is one of the types $p_{\xi}$ or $r_{q,\xi}$ from Lemma
B.2,B.3.

###### Proof.

Take $(a,b)\in\mathbb{U}^{2}$ realizing $p(x,y)$. Then $a>M$, so the prime
model $M[a]=\mathbb{Q}\cdot a\oplus M$ is isomorphic to $\mathbb{Q}\times M$
with the lexicographic order. Then $b$ determines a cut in $\mathbb{Q}\times
M$. If $b$ lies in the middle of some coset $q\cdot a+M$, then $b$ realizes
$r_{q,\xi}$ for some cut $\xi\in\overline{M}$. Otherwise, $b$ realizes
$p_{\xi}$ for some $\xi\in\operatorname{Cut}(\mathbb{Q})$. ∎

######  Remark B.5.

There is some overlap between the $p_{\xi}$ and $r_{q,\xi}$. Specifically,
$p_{q^{+}}=r_{q,+\infty}$ and $p_{q^{-}}=r_{q,-\infty}$.

######  Remark B.6.

In the notation of Definition 2.15, the types $r_{q,m}$, $r_{q,m^{+}}$, and
$r_{q,m^{-}}$ are $f|_{p}$, $f^{+}|_{p}$, and $f^{-}|_{p}$ respectively, where
$f(x)=qx+m$ and $p$ is the 1-type at $+\infty$.

When we need to make the dependence on $M$ explicit, we write $p_{\xi}$ and
$r_{q,\xi}$ as $p^{M}_{\xi}$ and $r^{M}_{q,\xi}$.

######  Proposition B.7.

Let $M^{\prime}$ be an elementary extension of $M$.

  1. 1.

For $\xi\in\operatorname{Cut}(\mathbb{Q})$, $p^{M^{\prime}}_{\xi}$ is the
unique heir of $p^{M}_{\xi}$.

  2. 2.

For $q\in\mathbb{Q}$ and $m\in M$,

     * •

$r^{M^{\prime}}_{q,m}$ is the unique heir of $r^{M}_{q,m}$.

     * •

$r^{M^{\prime}}_{q,m^{+}}$ is the unique heir of $r^{M}_{q,m^{+}}$.

     * •

$r^{M^{\prime}}_{q,m^{-}}$ is the unique heir of $r^{M}_{q,m^{-}}$.

###### Proof.

Let $p^{\prime}(x,y)$ be an heir of $p^{M}_{\xi}$ over $M^{\prime}$. Then
$p^{\prime}(x,y)$ contains the formulas

$\displaystyle\\{x>t:t\in M^{\prime}\\}$
$\displaystyle\\{y>qx+s:q\in\mathbb{Q},\leavevmode\nobreak\
q<\xi,\leavevmode\nobreak\ s\in M^{\prime}\\}$
$\displaystyle\\{y<qx+s:q\in\mathbb{Q},\leavevmode\nobreak\
q>\xi,\leavevmode\nobreak\ s\in M^{\prime}\\}$

because it is an heir. By Lemma B.2, $p^{\prime}$ must be
$p^{M^{\prime}}_{\xi}$. This proves part (1), and part (2) is similar. ∎

A type $p$ is definable if and only if it has a unique heir [Poi00, Theorem
11.7], in which case the unique heir is the canonical extension using the same
defining scheme as $p$. Therefore Proposition B.7 implies that $p_{\xi}$,
$r_{q,m}$, $r_{q,m^{+}}$, and $r_{q,m^{-}}$ are definable types, and specifies
their canonical extensions.

######  Proposition B.8.

Let $p(x,y)$ be a 2-type over $M$ containing the formulas $x>t$ for all $t\in
M$. Suppose $p(x,y)$ has a definable basis. Then $p(x,y)$ is $r_{q,m}$,
$r_{q,m^{+}}$, or $r_{q,m^{-}}$ for some $q\in\mathbb{Q}$ and $m\in M$.

###### Proof.

By Lemma B.1.1, the type $p$ is definable. By Lemma B.4 and Remark B.5, $p$
must be one of the following:

  * •

$p_{\xi}$ for some $\xi\in\operatorname{Cut}(\mathbb{Q})$.

  * •

$r_{q,\xi}$ for some $q\in\mathbb{Q}$ and
$\xi\in\overline{M}\setminus\\{+\infty,-\infty\\}$.

In the second case, there are restrictions on $\xi$. The set

$\\{s\in M:(y\leq qx+s)\in p(x,y)\\}=\\{s\in M:s\leq\xi\\}$

must be definable. By o-minimality, $\xi$ must be one of
$+\infty,-\infty,m,m^{+},$ or $m^{-}$ for some $m\in M$. By Remark B.5, we
conclude that $p$ is one of the following:

  * •

$p_{\xi}$ for some $\xi\in\operatorname{Cut}(\mathbb{Q})$

  * •

$r_{q,m}$ or $r_{q,m^{+}}$ or $r_{q,m^{-}}$ for some $q\in\mathbb{Q}$ and
$m\in M$.

Let $M^{\prime}$ be an $\aleph_{1}$-saturated elementary extension of $M$. By
Proposition B.7,

$\displaystyle p^{M^{\prime}}_{\xi}$ $\displaystyle\text{ is the unique heir
of }p^{M}_{\xi}$ $\displaystyle r^{M^{\prime}}_{q,m}$ $\displaystyle\text{ is
the unique heir of }r^{M}_{q,m}$ $\displaystyle r^{M^{\prime}}_{q,m^{+}}$
$\displaystyle\text{ is the unique heir of }r^{M}_{q,m^{+}}$ $\displaystyle
r^{M^{\prime}}_{q,m^{-}}$ $\displaystyle\text{ is the unique heir of
}r^{M}_{q,m^{-}}.$

By Lemma B.1.2, we may replace $M$ with $M^{\prime}$, and therefore assume
that $M$ is $\aleph_{1}$-saturated.

Now we apply Lemma B.1.3 to rule out the case of $p_{\xi}$. We break into
cases depending on $\xi$:

  * •

Suppose $p=p_{+\infty}$. Define

$\displaystyle V_{q,t}$ $\displaystyle=\\{(x,y):x>t\text{ and }y>qx\\}$
$\displaystyle\text{ for }q\in\mathbb{Q},\leavevmode\nobreak\ t\in M$
$\displaystyle\mathcal{V}_{q}$ $\displaystyle=\\{V_{q,t}:t\in M\\}$
$\displaystyle\text{ for }q\in\mathbb{Q}.$

Each family $\mathcal{V}_{q}$ is a definable downward directed family. One can
verify that $\mathcal{V}_{q}$ is finer than $\mathcal{V}_{q^{\prime}}$ if and
only if $q\geq q^{\prime}$. Moreover, the union
$\bigcup_{n=1}^{\infty}\mathcal{V}_{n}$ generates $p_{+\infty}$. By Lemma
B.1.3, $p_{+\infty}$ does not have a definable basis.

  * •

The case of $p=p_{-\infty}$ is handled similarly.

  * •

Suppose $p=p_{\alpha}$ for some irrational
$\alpha\in\mathbb{R}\setminus\mathbb{Q}$. Take two sequences of rational
numbers

$\displaystyle q_{1}$ $\displaystyle>q_{2}>q_{3}>\cdots$ $\displaystyle
q_{1}^{\prime}$ $\displaystyle<q_{2}^{\prime}<q_{3}^{\prime}<\cdots$

with
$\lim\limits_{i\to\infty}q_{i}=\lim\limits_{i\to\infty}q^{\prime}_{i}=\alpha$.
For each $i$, let

$\displaystyle V_{i,t}$ $\displaystyle=\\{(x,y):x>t\text{ and
}q^{\prime}_{i}x<y<q_{i}x\\}$ $\displaystyle\mathcal{V}_{i}$
$\displaystyle=\\{V_{i,t}:t\in M\\}.$

Again, $\mathcal{V}_{i}$ is finer than $\mathcal{V}_{j}$ if and only if $i\geq
j$, and $\bigcup_{i=1}^{\infty}\mathcal{V}_{i}$ generates $p_{\alpha}$.
Therefore $p_{\alpha}$ does not have a definable basis, by Lemma B.1.3.

  * •

Suppose $p=p_{q^{-}}$ for some rational $q\in\mathbb{Q}$. Take a sequence of
rational numbers

$q_{1}^{\prime}<q_{2}^{\prime}<q_{3}^{\prime}<\cdots$

with $\lim\limits_{i\to\infty}q^{\prime}_{i}=q$. For each $i$, let

$\displaystyle V_{i,s,t}$ $\displaystyle=\\{(x,y):x>t\text{ and
}q^{\prime}_{i}x<y<qx+s\\}$ $\displaystyle\mathcal{V}_{i}$
$\displaystyle=\\{V_{i,s,t}:s,t\in M\\}.$

As before, $\mathcal{V}_{i}$ is finer than $\mathcal{V}_{j}$ if and only if
$i\geq j$, and $\bigcup_{i=1}^{\infty}\mathcal{V}_{i}$ generates $p_{q^{-}}$.
So $p_{q^{-}}$ does not have a definable basis.

  * •

The case of $p=p_{q^{+}}$ is handled similarly.

Therefore, we have ruled out all cases where $p=p_{\xi}$, and it must be that
$p=r_{q,\xi}$, where $\xi$ is $m,m^{+},$ or $m^{-}$ for some $m\in M$. ∎

###  B.3 The counterexample

Define

$\displaystyle\mathbb{R}^{+}=\\{x\in\mathbb{R}:x>0\\}.$
$\displaystyle\mathbb{R}^{-}=\\{x\in\mathbb{R}:x<0\\}.$

######  Proposition B.9.

  1. 1.

In the structure $(\mathbb{R}^{+},\cdot,\leq)$, let $p(x,y)$ be a 2-type
containing $x>t$ for all $t\in\mathbb{R}^{+}$. Suppose $p(x,y)$ has a
definable basis. Then there is some $q\in\mathbb{Q}$ and $c\in\mathbb{R}^{+}$
such that $p(x,y)$ is generated by one of the following three families:

$\displaystyle\\{x>t:t\in\mathbb{R}^{+}\\}$ $\displaystyle\cup\\{y=x^{q}\cdot
c\\}$ $\displaystyle\\{x>t:t\in\mathbb{R}^{+}\\}$
$\displaystyle\cup\\{x^{q}\cdot c<y<x^{q}\cdot c\cdot\mu$
$\displaystyle:\mu>1\\}$ $\displaystyle\\{x>t:t\in\mathbb{R}^{+}\\}$
$\displaystyle\cup\\{x^{q}\cdot c\cdot\mu^{-1}<y<x^{q}\cdot c$
$\displaystyle:\mu>1\\}$

  2. 2.

In the structure $(\mathbb{R},\cdot,\leq)$, let $p(x,y)$ be a 2-type
containing $y>0$ and $x>t$ for all $t\in\mathbb{R}^{+}$. Suppose $p(x,y)$ has
a definable basis. Then there is some $q\in\mathbb{Q}$ and
$c\in\mathbb{R}^{+}$ such that $p(x,y)$ is generated by one of the following
three families:

$\displaystyle\\{x>t:t\in\mathbb{R}^{+}\\}$ $\displaystyle\cup\\{y=x^{q}\cdot
c\\}$ $\displaystyle\\{x>t:t\in\mathbb{R}^{+}\\}$
$\displaystyle\cup\\{x^{q}\cdot c<y<x^{q}\cdot c\cdot\mu$
$\displaystyle:\mu>1\\}$ $\displaystyle\\{x>t:t\in\mathbb{R}^{+}\\}$
$\displaystyle\cup\\{x^{q}\cdot c\cdot\mu^{-1}<y<x^{q}\cdot c$
$\displaystyle:\mu>1\\}$

  3. 3.

In the structure $(\mathbb{R},\cdot,\leq)$, let $p(x,y)$ be a 2-type
containing $y<0$ and $x>t$ for all $t\in\mathbb{R}^{+}$. Suppose $p(x,y)$ has
a definable basis. Then there is some $q\in\mathbb{Q}$ and
$c\in\mathbb{R}^{-}$ such that $p(x,y)$ is generated by one of the following
three families:

$\displaystyle\\{x>t:t\in\mathbb{R}^{+}\\}$ $\displaystyle\cup\\{y=x^{q}\cdot
c\\}$ $\displaystyle\\{x>t:t\in\mathbb{R}^{+}\\}$
$\displaystyle\cup\\{x^{q}\cdot c>y>x^{q}\cdot c\cdot\mu$
$\displaystyle:\mu>1\\}$ $\displaystyle\\{x>t:t\in\mathbb{R}^{+}\\}$
$\displaystyle\cup\\{x^{q}\cdot c\cdot\mu^{-1}>y>x^{q}\cdot c$
$\displaystyle:\mu>1\\}$

###### Proof.

  1. 1.

Apply Proposition B.8 to the densely ordered abelian group
$(\mathbb{R}^{+},\cdot,\leq)$.

  2. 2.

This follows from Part 1. The two structures $(\mathbb{R},\cdot,\leq)$ and
$(\mathbb{R}^{+},\cdot,\leq)$ are bi-interpretable, and we can interpret
$(\mathbb{R}^{+},\cdot,\leq)$ as the definable set $\mathbb{R}^{+}$ in
$(\mathbb{R},\cdot,\leq)$.

  3. 3.

This follows by pushing forward Part 2 along the definable bijection

$\displaystyle\mathbb{R}^{+}\times\mathbb{R}^{+}$
$\displaystyle\to\mathbb{R}^{+}\times\mathbb{R}^{-}$ $\displaystyle(x,y)$
$\displaystyle\mapsto(x,-y).\qed$

Let $C$ be the 4-ary predicate on $\mathbb{R}$ defined by

$C(x,y,z,w)\iff\left(x<y<0<z<w\text{ and }\frac{x}{y}=\frac{w}{z}\right).$

######  Lemma B.10.

The two structures $(\mathbb{R},\leq,\cdot)$ and $(\mathbb{R},\leq,C)$ have
the same definable sets.777However, $(\mathbb{R},\leq,C)$ has fewer
$\emptyset$-definable sets than $(\mathbb{R},\leq,\cdot)$. This will be seen
implicitly in the proof of Lemma B.12 below.

###### Proof.

We claim that the following four structures all have the same definable sets:

$\displaystyle M_{1}$ $\displaystyle=(\mathbb{R},\leq,\cdot)$ $\displaystyle
M_{2}$ $\displaystyle=(\mathbb{R},\leq,\cdot,1,-1,C)$ $\displaystyle M_{3}$
$\displaystyle=(\mathbb{R},\leq,1,-1,C)$ $\displaystyle M_{4}$
$\displaystyle=(\mathbb{R},\leq,C).$

First of all, $M_{2}$ is a definitional expansion of $M_{1}$, so they have the
same definable sets. It is an easy exercise that the relation $x\cdot y=z$ is
definable in $M_{3}$. Therefore $M_{2}$ is a definitional expansion of
$M_{3}$. Finally, $M_{3}$ is an expansion of $M_{4}$ by naming parameters. ∎

Because the definable sets in $(\mathbb{R},\leq,C)$ are the same as in
$(\mathbb{R},\leq,\cdot)$, the following makes sense, and follows immediately
from Proposition B.9.3:

######  Proposition B.11.

In the structure $(\mathbb{R},\leq,C)$, let $p(x,y)$ be a 2-type containing
$y<0$ and $x>t$ for all $t\in\mathbb{R}^{+}$. Suppose $p(x,y)$ has a definable
basis. Then there is some $q\in\mathbb{Q}$ and $c\in\mathbb{R}^{-}$ such that
$p(x,y)$ is generated by one of the following three families:

$\displaystyle\\{x>t:t\in\mathbb{R}^{+}\\}$ $\displaystyle\cup\\{y=x^{q}\cdot
c\\}$ $\displaystyle\\{x>t:t\in\mathbb{R}^{+}\\}$
$\displaystyle\cup\\{x^{q}\cdot c>y>x^{q}\cdot c\cdot\mu$
$\displaystyle:\mu>1\\}$ $\displaystyle\\{x>t:t\in\mathbb{R}^{+}\\}$
$\displaystyle\cup\\{x^{q}\cdot c\cdot\mu^{-1}>y>x^{q}\cdot c$
$\displaystyle:\mu>1\\}$

######  Lemma B.12.

None of the three types in Proposition B.11 is $\emptyset$-definable in the
structure $(\mathbb{R},\leq,C)$.

###### Proof.

Let $\sigma:\mathbb{R}\to\mathbb{R}$ be the function

$\sigma(x)=\begin{cases}2x&x<0\\\ x&x\geq 0.\end{cases}$

Then $\sigma$ is an automorphism of $(\mathbb{R},\leq,C)$: it clearly
preserves $\leq$, and it preserves $C$ because if $x<y<0<z<w$, then

$\frac{x}{y}=\frac{w}{z}\iff\frac{2x}{2y}=\frac{w}{z}\iff\frac{\sigma(x)}{\sigma(y)}=\frac{\sigma(w)}{\sigma(z)}.$

Suppose $p(x,y)$ is the type generated by one of the following families, for
some $q\in\mathbb{Q}$ and $c\in\mathbb{R}^{-}$:

$\displaystyle\\{x>t:t\in\mathbb{R}^{+}\\}$ $\displaystyle\cup\\{y=x^{q}\cdot
c\\}$ $\displaystyle\\{x>t:t\in\mathbb{R}^{+}\\}$
$\displaystyle\cup\\{x^{q}\cdot c>y>x^{q}\cdot c\cdot\mu$
$\displaystyle:\mu>1\\}$ $\displaystyle\\{x>t:t\in\mathbb{R}^{+}\\}$
$\displaystyle\cup\\{x^{q}\cdot c\cdot\mu^{-1}>y>x^{q}\cdot c$
$\displaystyle:\mu>1\\}$

Then the pushforward along $\sigma$ is, respectively, the type generated by
one of these families:

$\displaystyle\\{x>t:t\in\mathbb{R}^{+}\\}$ $\displaystyle\cup\\{y=2x^{q}\cdot
c\\}$ $\displaystyle\\{x>t:t\in\mathbb{R}^{+}\\}$
$\displaystyle\cup\\{2x^{q}\cdot c>y>2x^{q}\cdot c\cdot\mu$
$\displaystyle:\mu>1\\}$ $\displaystyle\\{x>t:t\in\mathbb{R}^{+}\\}$
$\displaystyle\cup\\{2x^{q}\cdot c\cdot\mu^{-1}>y>2x^{q}\cdot c$
$\displaystyle:\mu>1\\}$

The new type $\sigma_{*}p$ is never equal to $p$, because $c\neq 2c$.
Therefore $p$ is not fixed by some automorphism over $\emptyset$, and $p$ is
not $\emptyset$-definable. ∎

######  Theorem B.13.

The structure $(\mathbb{R},\leq,C)$ is o-minimal. Let $\mathcal{S}$ be the
definable family of sets $\\{(x,y):y<0,\leavevmode\nobreak\ x>t\\}$ for
$t\in\mathbb{R}$. Then $\mathcal{S}$ is a $\emptyset$-definable downward
directed family. If $\mathcal{S}^{\prime}$ is a definable downward directed
family finer than $\mathcal{S}$ that generates a 2-type $p(x,y)$, then
$\mathcal{S}^{\prime}$ and $p$ are _not_ $\emptyset$-definable.

###### Proof.

The structure $(\mathbb{R},\leq,C)$ is o-minimal because, for example, it is a
reduct of RCF. The definable family $\mathcal{S}$ is clearly downward
directed. Because $\mathcal{S}^{\prime}$ is finer than $\mathcal{S}$, the type
$p(x,y)$ contains the formulas $y<0$ and $x>t$ for all $t\in\mathbb{R}$.
Therefore $p(x,y)$ must be one of the three types listed in Proposition B.11.
By Lemma B.12, $p(x,y)$ is not $\emptyset$-definable. If
$\mathcal{S}^{\prime}$ were $\emptyset$-definable, then $p$ would be as well.
∎

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