\section{Introduction}


In this note we will study the topological properties of sets definable in densely ordered Abelian groups satisfying an extra model-theoretic hypothesis (having ``burden 2'') which in some sense limits the combinatorial complexity of combinations of instances of formulas (the precise definition will be recalled below). Typical examples of such groups are the structures $\mc{R}_1 = \langle \R; <, +, \Q \rangle$, the additive group of real numbers endowed with a unary predicate for the set $\Q$ of rationals, and $\mc{R}_2 = \langle \R; <, +, \Z\rangle$, the same group but with a predicate for the integers. In fact, both of the structures $\mc{R}_1$ and $\mc{R}_2$ are of \emph{dp-rank 2}, which is equivalent to having burden 2 and being NIP.\footnote{Note that it can be tricky to prove precise upper bounds on the burden or the dp-rank of a structure. See \cite{DG} for a detailed calculation of the dp-rank of the structures mentioned here.}

Recall that in a divisible ordered Abelian group (or ``OAG'') of dp-rank 1, no infinite discrete subset of the domain can be definable  (see \cite{Goodrick_dpmin}), nor can any dense and codense subset be definable (by a result of Simon \cite{S}). However, the examples $\mc{R}_1$ and $\mc{R}_2$ above show that neither of these results hold for OAGs of dp-rank 2. One of the new results of this article is that in a divisible OAG of dp-rank 2, there cannot be \emph{both} a definable infinite discrete set and a definable dense-codense set (see Theorem~\ref{sec2main} below).

The goal of this article is to understand the topological properties of unary definable sets in a densely ordered OAG $\mathcal{R} = \langle R; < , +, \ldots \rangle$ with burden at most $2$. Suppose that $X \subseteq R$ is definable in such a structure. The case when $X$ is open may be considered the ``nicest'' situation since the topological structure around any point in $X$ is as simple as possible. A densely ordered OAG in which every infinite definable $X \subseteq R$ has interior is called a \emph{viscerally ordered structure}, and we already analyzed such structures in \cite{viscerality}, giving a cell decomposition theorem and developing dimension theory for definable sets.

 If $X \subseteq R$ is not open, three other typical cases are:

\begin{enumerate}

\item $X$ is \emph{discrete}, that is, around any point $a \in X$, there is an open interval $I$ such that $I \cap X = \{a\}$;

\item $X$ is \emph{dense and codense} in some interval $I$, so that in any subinterval $J \subseteq I$ there are points in both $X$ and $R \setminus X$; or

\item $X$ is \emph{Cantor-like}, that is, $X$ is closed, nowhere dense, and has no isolated points.

\end{enumerate}

We conjecture that in a divisible OAG of dp-rank $2$, if there is an infinite definable set $X$ satisfying one of the three conditions above, then there cannot be any other infinite definable set satisfying either of the other two (giving a basic trichotomy). Though we cannot quite prove this, in this section we will show that the existence of an infinite definable discrete set precludes the existence of other definable sets satisfying either (2) or (3) above (Theorem~\ref{sec2main}). In the case when $\mathcal{R}$ has burden $2$ and is definably complete, there can never be a definable Cantor-like set (Theorem~\ref{no_cantor_dc} below). Finally, in any OAG with burden $2$, if there is both an infinite discrete set $D$ definable in such a structure and also a definable $X$ which is dense and codense in some interval, then the Independence Property can be witnessed by translations of $X$ (Theorem~\ref{IP_translations}).

After proving the above general results in Section 2, the final section (Section 3) is devoted to the study of a concrete example of a divisible OAG of burden 2 in which both an infinite discrete set and a dense codense set are definable.

Note that the present work focuses on the topological properties of definable sets. Given an OAG of burden $2$ in which an infinite discrete set $D$ is definable, it turns out that $D$ must have a very simple ``arithmetical'' structure, similar to the subsets of $\Z$ definable in Presburger arithmetic. For much more on this topic, see our previous article \cite{DG} or our recent preprint \cite{discrete_burden_2}, which can be seen as a companion to the present work.


\subsection{Notation and basic definitions}

Mostly we will follow standard notational conventions from model theory (as in \cite{Guide_NIP}, for example). ``Formulas'' and ``models'' are as in first-order logic and overlined variables ($\overline{x}, \overline{a}, \ldots$) denote finite tuples. Unlike some authors, a plain variable such as $x$ (not $\overline{x}$) is always a single variable, and we rarely work in $T^{eq}$. Also note that ``definable'' for us always means ``definable over some set of parameters.''


The abbreviation OAG stands for Ordered Abelian Group, which is a structure $\langle G; +, < \rangle$ consisting of an Abelian group $\langle G; +\rangle$ endowed with a total ordering $<$ which is translation invariant ($x < y$ implies that $x+z < y + z$). Some OAGs are \emph{discretely ordered} and have a least positive element (such as $\langle \Z; +, ,\rangle$), but in the present article all OAGs will be \emph{densely ordered}. Note that if an OAG is \emph{divisible} -- that is, for every $x \in G$ and every positive integer $n$, there is a $y \in G$ such that $ny = x$ -- then it is densely ordered, but there are examples of OAGs of finite dp-rank which are densely ordered and not divisible. See, for instance, \cite{Guide_NIP} or \cite{pure_OAGs} for more examples.

All topological properties of sets mentioned in this article (``open,'' ``dense,'' and so on) refer to the order topology generated by all open intervals.

At some points in Section 2 it will be useful to work in the Dedekind completion $\overline{R}$ of an OAG $\mathcal{R}$, for which the following notion from \cite{MMS} will be useful.

\begin{definition}
\label{ded_sort}
Suppose that $\mathcal{R} = \langle R; <, \ldots \rangle$ is a linearly ordered structure and $\{X_{\overline{a}} \, : \, \overline{a} \in Z\}$ is a definable family of subsets $X_{\overline{a}}$ of $R$, with $Z \subseteq R^n$ definable over $\emptyset$ and $X_{\overline{a}} = \{b \in R \, : \, \mathcal{R} \models \varphi(b; \overline{a}) \}$ for some formula $\varphi(x; \overline{y})$. Consider the definable equivalence relation $\sim$ on $n$-tuples from $Z$ given by

$$\overline{a}_1 \sim \overline{a}_2 \Leftrightarrow \sup(X_{\overline{a}_1}) = \sup(X_{\overline{a}_2}).$$

(We formally define $\sup(X_{\overline{a}}) = - \infty$ or $\sup(X_{\overline{a}}) = + \infty$ when $X_{\overline{a}} = \emptyset$, or when $X_{\overline{a}}$ is unbounded, respectively.) Then $Z / \sim$, which is a sort in $\mathcal{R}^{eq}$, is a \emph{sort in $\overline{R}$.} We naturally identify such a sort with a subset of the Dedekind completion $\overline{R}$ of $\mathcal{R}$ with the induced ordering.

Thinking of sorts in $\overline{R}$ as sorts in $\mathcal{R}^{eq}$, we can naturally talk about definable subsets of sorts in $\overline{R}$, and of functions to and from such sorts.
\end{definition}

\begin{definition}
If $\mathcal{R} = \langle R; <, +, \ldots, \rangle$ is an OAG, then $\mathcal{R}$ is \emph{definably complete} if for every nonempty definable subset $X \subseteq R$ which has an upper bound, $\sup(X) \in R$.
\end{definition}

The next observation (\cite[Proposition 2.2]{ivp}) will occasionally be useful.



\begin{fact}
\label{DC_divisible}
If $\mathcal{R}$ is a densely ordered OAG which is definably complete, then $\mathcal{R}$ is divisible.
\end{fact}

The less commonly used definitions we need are those of \emph{burden} and \emph{dp-rank}. These notions originally go back to Shelah \cite{strong_dep}, but we will use the versions as given by Adler \cite{adler_strong_dep}. We recall them briefly here. Below, ``$T$'' always denotes some complete theory.

\begin{definition}
An \emph{ict-pattern of depth $\kappa$} is a sequence $\{ \varphi_i(\overline{x}; \overline{y}) \, : \, i < \kappa \}$ of formulas and a sequence $\{\overline{a}_{i,j} \, : \, i < \kappa, j < \omega \}$ of tuples from some model $\mathcal{M} \models T$ such that for every function $\eta \, : \, \kappa \rightarrow \omega$, the partial type

\begin{equation}
\{\varphi_i(\overline{x}; \overline{a}_{i, j})^{\textup{if } j = \eta(i)} \, : \, i < \kappa, j < \omega \}
\end{equation}
is consistent, where the exponent ``$\textup{if } j = \eta(i)$'' means that the formula is negated if $j \neq \eta(i)$. If $p(\overline{x})$ is a partial type, an ict-pattern as above is \emph{in $p(\overline{x})$} if every partial type as in (1) is consistent with $p(\overline{x})$.

The partial type $p(\overline{x})$ has \emph{dp-rank less than $\kappa$} if there is \textbf{no} ict-pattern of depth $\kappa$ in $p(\overline{x})$. If the least $\kappa$ such that the dp-rank of $p(\overline{x})$ is less than $\kappa$ is a successor cardinal, say $\kappa = \lambda^+$, then we say that the dp-rank of $p(\overline{x})$ is $\lambda$.

The dp-rank of the theory $T$ is the dp-rank of the partial type $x=x$ (in a single free variable $x$), and $T$ is \emph{dp-minimal} if its dp-rank is $1$.
\end{definition}


\begin{definition}
An \emph{inp-pattern of depth $\kappa$} is a sequence $\{ \varphi_i(\overline{x}; \overline{y}) \, : \, i < \kappa \}$ of formulas, a sequence $\{k_i \, : \, i < \kappa\}$ of positive integers, and a sequence $\{\overline{a}_{i,j} \, : \, i < \kappa, j < \omega \}$ of tuples from some model $\mathcal{M} \models T$ such that:

$\bullet$ For each $i < \kappa$, the ``$i$-th row'' $$\{\varphi_i(\overline{x}; \overline{a}_{i, j}) \, : \, j < \omega \}$$ is $k_i$-inconsistent; and

$\bullet$ For each function $\eta \, : \, \kappa \rightarrow \omega$, the partial type

\begin{equation}
\{ \varphi_i(\overline{x}; \overline{a}_{i, \eta(i)}) \, : \, i < \kappa\}
\end{equation}

is consistent.

If $p(\overline{x})$ is a partial type, an inp-pattern as above is \emph{in $p(\overline{x})$} if every partial type as in (2) is consistent with $p(\overline{x})$.

The partial type $p(\overline{x})$ has \emph{burden less than $\kappa$} if there is \textbf{no} inp-pattern of depth $\kappa$ in $p(\overline{x})$. If the least $\kappa$ such that the burden of $p(\overline{x})$ is less than $\kappa$ is a successor cardinal, say $\kappa = \lambda^+$, then we say that the burden of $p(\overline{x})$ is $\lambda$.

The burden of the theory $T$ is the burden of the partial type $x=x$ (in a single free variable $x$), and $T$ is \emph{inp-minimal} if its burden is $1$.


\end{definition}

\begin{fact}
(Adler, \cite{adler_strong_dep}) The dp-rank of a theory is less than some cardinal $\kappa$ if and only if $T$ is NIP. In case $T$ is NIP, the dp-rank of $T$ is equal to the burden of $T$. In particular, $T$ is dp-minimal if, and only if, $T$ is both NIP and inp-minimal.
\end{fact}

To mention some related work, fields of finite dp-rank have been recently classified by Johnson, who also showed that any valued field with finite dp-rank is Henselian \cite{Johnson_dp_finite}. Interesting examples of finite burden structures which are not finite dp-rank include pseudo real-closed fields \cite{samy_prcf}. For more background on these concepts and how they relate to NIP, see the introduction to our companion article \cite{discrete_burden_2} or the survey \cite{goodrick_survey}.



\section{Topologoical properties of definable sets in burden-2 OAGs}


In this section we will prove the general topological results for OAGs of burden 2 mentioned in the introduction.

Throughout this section, we will work under the following assumptions, unless otherwise stated: 


\begin{itemize}

\item $\mathcal{R} = \langle R; <, +, \ldots\rangle$ is a densely-ordered OAG with complete theory $T$; 

\medskip

\item $\mathcal{R}$ is sufficiently saturated (generally just $|T|^+$-saturated will be enough);

\medskip

\item The burden of $\mathcal{R}$ is at most $2$.

\end{itemize}

Note that in this section, we do not generally assume that $\mathcal{R}$ is definably complete (unless we explicitly say so).




\begin{lem}
\label{rank_1_interval}
Suppose that $I \subseteq R$ is an interval such that $I$ has burden $1$. Then if $X \subseteq I$ is definable and nowhere dense, then $X$ is finite.
\end{lem}

\begin{proof}
This is essentially the same result as Lemma~3.3(1) of \cite{Goodrick_dpmin}, except that we allow $I$ to be any interval instead of the whole universe $R$.  The same proof as in \cite{Goodrick_dpmin} goes through, working within $I$.
\end{proof}

\begin{lem}
\label{rank_2_intervals}
If there is an infinite discrete set definable in $\mathcal{R}$ then there is $\epsilon>0$ so that $(0,\epsilon)$ has burden $1$.
\end{lem}

\begin{proof}
Suppose to the contrary that $D \subseteq R$ is infinite, discrete, and definable, and that for every $\epsilon > 0$ in $R$ there is an inp-pattern of depth $2$ consistent with $(0, \epsilon)$.

Now select an increasing sequence of elements $\{a_i \, : \, i \in \omega\} \subseteq D$ and an $\epsilon > 0$ such that for every $i$ we have $(a_i - 2 \epsilon, a_i + 2 \epsilon) \cap D = \{a_i\}$. We can now construct an inp-pattern of depth $3$ by attaching translated copies of the inp-pattern within $(0, \epsilon)$ onto each point of $a_i$ and adding a third row consisting of pairwise disjoint intervals, leading to a contradiction. 

More precisely, if $\varphi_0(x, \overline{b}_{0,j})$ and $\varphi_1(x, \overline{b}_{1,j})$ witness an inp-pattern of depth $2$ consistent with $(0, \epsilon)$, then for each $\ell \in \{0,1\}$ let $\varphi'_\ell(x, \overline{b}_{\ell,j})$ be the formula expressing ``there is a unique point $a \in D$ such that $a < x < a + \epsilon$, and for this unique point $a$, the formula $\varphi_\ell(x-a, \overline{b}_{\ell,j})$ holds.'' These will be the first two rows of our inp-pattern, and the third row will consist of the pairwise disjoint intervals $I_i = (a_i - \epsilon, a_i + \epsilon)$. The inconsistency of each row is easy to check, and if $c_{i,j} \in (0, \epsilon)$ satisfies the formula $\varphi_0(x, \overline{b}_{0,i}) \wedge \varphi_1(x, \overline{b}_{1,j})$, then for any $k \in \omega$ the point $a_k + c_{i,j}$ satisfies $\varphi'_0(x, \overline{b}_{0,i}) \wedge \varphi'_1(x, \overline{b}_{1,j})$ and lies within the interval $I_k$.
\end{proof}

This lemma has an immediate and useful corollary:

\begin{cor}  
\label{no_Cantor}
If there is $X \subseteq R$ definable, infinite, and discrete then there is no definable Cantor-like set in $\mathcal{R}$.
\end{cor}

\begin{proof}  
By the previous Lemma, there is $\epsilon > 0$ such that $(0,\epsilon)$ has burden $1$.  Suppose for contradiction that there is $Y$ definable and Cantor-like.  Translating $Y$ as necessary, we may assume that $Y \cap (0,\epsilon) \neq \emptyset$; and intersecting $Y$ with a closed subinterval of $(0, \epsilon)$, we may further assume that there is a Cantor-like definable subset of $(0, \epsilon)$. This contradicts Lemma~\ref{rank_1_interval}.
\end{proof}

Using the above Corollary, we can rule out the existence of any Cantor-like definable set in the case when $\mathcal{R}$ additionally satisfies definable completeness:

\begin{thm}
\label{no_cantor_dc} If $\mathcal{R}$ is a densely ordered, definably complete OAG with burden at most $2$, then there is no definable Cantor-like set in $\mathcal{R}$.
\end{thm}

\begin{proof}
Assume otherwise, and let $X \subseteq R$ be a definable Cantor-like set. Then $R \setminus X$ is an infinite union of pairwise disjoint open convex sets, each of which is an interval by definable completeness; we call these the \emph{complementary intervals}. All but at most two complementary intervals are bounded, and if $I = (a,b)$ is a bounded complementary interval, then we can use the fact that $\mathcal{R}$ is divisible (Fact~\ref{DC_divisible} above) to define its midpoint $I_m = \frac{a+b}{2}$. The collection of all such midpoints $I_m$ is itself definable and comprises and infinite discrete set, yielding a contradiction to Corollary~\ref{no_Cantor}.
\end{proof}
 

Now we will focus on the case when $\mathcal{R}$ has dp-rank 2 (which, recall, is equivalent to NIP plus having burden $2$). In this case, we will show that if $R$ is divisible and defines an infinite discrete set, then it cannot also define a set which is dense and codense in some infinite interval.

To this end, we will use the \emph{Shelah expansion} $\mathcal{R}^{Sh}$ of the structure $\mathcal{R}$. This concept, and the facts below, are due to Shelah \cite{Dependent_Shelah}, but we will follow the notation and presentation of Simon \cite{Guide_NIP}.

\begin{definition}
\label{Sh_expansion}
Suppose that $\mathcal{R} \prec \mathcal{U}$ and $\mathcal{U}$ is $|R|^+$ -saturated. The \emph{Shelah expansion} $\mathcal{R}^{Sh}$ of $\mathcal{R}$ is the expansion of $\mathcal{R}$ with the following new predicates: for every partitioned formula $\varphi(\overline{x}; \overline{y})$ and every finite tuple $\overline{b} \in U^{|\overline{y}|}$, define a predicate $S_{\varphi(\overline{x}; \overline{b})} (\overline{x})$ on $R^{|\overline{x}|}$ such that $$\mathcal{R}^{Sh} \models S_{\varphi(\overline{x}; \overline{b})} (\overline{a}) \Leftrightarrow \mathcal{U} \models \varphi(\overline{a}; \overline{b}).$$

\end{definition}

The subsets of $R^n$ defined by the new basic predicates $ S_{\varphi(\overline{x}; \overline{b})}$ as above are called \emph{externally definable sets}. An important example for our purposes is that if $C \subseteq R$ is convex, then using $|R|^+$-saturation we may find $a, b \in U$ such that $(a,b) \cap R = C$, and thus $C$ is externally definable.

The next fact summarizes the important basic properties of Shelah expansions.

\begin{fact}
\label{Sh_expansion_properties}
(\cite{Dependent_Shelah}, and see also \cite{Guide_NIP})
\begin{enumerate}
\item The subsets of $R^n$ which are definable in the Shelah expansion $\mathcal{R}^{Sh}$ are independent of the choice of the saturated extension $\mathcal{U}$ in Definition~\ref{Sh_expansion}, and hence we may talk about ``the'' Shelah expansion.
\item The structure $\mathcal{R}^{Sh}$ admits elimination of quantifiers.
\item The structure $\mathcal{R}^{Sh}$ is NIP.
\item If $X \subseteq R^n$ is type-definable in $\mathcal{R}$, then the dp-rank of $X$ as calculated in $\mathcal{R}$ is equal to the dp-rank of $X$ as calculated in $\mathcal{R}^{Sh}$.
\end{enumerate}
\end{fact}

\begin{proof}
While complete proofs of (1), (2), and (3) can be found in \cite{Guide_NIP}, we take the opportunity to explain how (4) follows from (2). Note that a fact similar to (4) has been claimed by Onshuus and Usvyatsov (see \cite{onsh_usv}) but only in the special case when the theory is dp-minimal.

Let $d_1$ be the dp-rank of $X$ as calculated in the original structure $\mathcal{R}$, and let $d_2$ be the dp-rank of $X$ as calculated in the Shelah expansion $\mathcal{R}^{Sh}$ (each of which may be a cardinal or $\infty$). On the one hand, if $\kappa$ is any cardinal and $d_1 \geq \kappa$, then there is an ict-pattern of depth $\kappa$ in $X$ with parameters from some elementary extension of $\mathcal{R}$, and the same array of formulas is an ict-pattern of depth $\kappa$ in the expanded language of $\mathcal{R}^{Sh}$. Therefore $d_1 \leq d_2$. 

On the other hand, suppose that there is an ict-pattern 

\begin{equation}
\{\varphi_i(\overline{x}; \overline{a}_{i,j}) \, : \, i < \kappa, \, j < \omega \}
\end{equation}
of depth $\kappa$ consistent with $X$ in some elementary extension $\mathcal{R}'$ of $\mathcal{R}^{Sh}$. By quantifier elimination, we may assume that
 $$\varphi_i(\overline{x}; \overline{y}) = S_{\psi_i(\overline{x}; \overline{y}; \overline{b}_i)}(\overline{x}; \overline{y})$$
  where $\overline{b}_i$ is a tuple of parameters from the $|R|^+$-saturated model $\mathcal{U} \succ \mathcal{R}$ used to define $\mathcal{R}^{Sh}$.

Working in $\mc{R}^{Sh}$, for any $m,n \in \mathbb{N}$, any $l_1, \dots, l_i, \dots l_m \in \kappa$, and any formula $\theta(\overline{x})$ in $X$, there are parameters $\ob{c}_{i,j} \in R$ with $1 \leq i \leq  m$ and $1 \leq j \leq n$ so that 
for any $\eta: \{1, \dots, m\} \to \{1, \dots, n\}$ there is  $\ob{d}_{\eta} \in \theta(R)$ so that 
$$\mc{R}^{Sh} \models  S_{\psi_{l_i}(\overline{x}; \overline{y}; \overline{b}_{l_i})}(\overline{d}_{\eta}; \overline{c}_{i,j}) \text{ if and only if } \eta(i)=j.$$
But then
\[\mc{U} \models \psi_{l_i}(\ob{d}_{\eta}, \ob{c}_{i,j}, \ob{b}_{l_i}) \text{ if and only if } \eta(i)=j.\]
Thus by compactness, in $\mc{U}$ the formulas $\psi_i(\ob{x}, \ob{y}, \ob{b}_i)$ form an ict-pattern with $\kappa$ rows consistent with $X$ and hence $d_1 \geq d_2$.
 


\end{proof}

The fact below was proved by Simon \cite{S} in the case when the entire structure is a dp-minimal divisible OAG, but the same proof can be relativized to convex definable subgroups $G$ to yield:

\begin{fact}
\label{Simon_interior}
(\cite{S}, Theorem 3.6) If $G \subseteq R$ is a type-definable convex divisible subgroup of $R$, $\textup{dp-rk}(G) = 1$, and $X \subseteq G$ is infinite and definable, then $X$ has nonempty interior.
\end{fact}

Now we state our main result on divisible OAGs of dp-rank $2$.

\begin{thm}\label{sec2main}  If $\mathcal{R}$ is an expansion of a divisible ordered Abelian group of dp-rank at most $2$ and there is an infinite definable discrete set in $\mathcal{R}$ then there is no $X$ definable which is dense and codense in an interval $I$ of $\mathcal{R}$ nor is there a definable Cantor-like set.
\end{thm}

\begin{proof}
Say $X \subseteq R$ is and $\mathcal{R}$-definable. By Corollary~\ref{no_Cantor}, we know that $X$ is not Cantor-like. It only remains to consider the case when $X$ is dense in some interval $I$.

First pick $\epsilon \in R$ as in the conclusion of Lemma~\ref{rank_2_intervals} so that $(-\epsilon, \epsilon)$ has dp-rank $1$. Now let $G$ be the subset of $R$ defined as

$$G = \bigcap_{n \in \N} (-\frac{\epsilon}{n+1}, \frac{\epsilon}{n+1} )$$
and note that $G$ is a convex subgroup of $R$. By $\omega$-saturation, $G$ is nonzero. Also, $G$ is definable in $\mathcal{R}^{Sh}$, so by Fact~\ref{Sh_expansion_properties}, it has dp-rank $1$ as calculated in the Shelah expansion.

Translating and truncating $I$ as necessary, we may assume that $I \subseteq G$. By Fact~\ref{Simon_interior}, $X$ cannot be codense in $I$, as we wanted.

\end{proof}


We have considerably more precise results if we also assume that $\mathcal{R}$ is definably complete.

\begin{cor}\label{discrete_interior_dichotomy}
Suppose that $\mathcal{R}$ is an expansion of a definably complete divisible OAG of dp-rank at most $2$ in which an infinite discrete set is definable. Then for any definable $X \subseteq R$, either $X$ is discrete or $X$ has nonempty interior.
\end{cor}

\begin{proof}
Suppose that $X \subseteq R$ is definable and has empty interior. By Theorem~\ref{sec2main}, $X$ is nowhere dense. By Corollary~2.13 of \cite{DG}, it follows that $X$ is discrete.
\end{proof}



\begin{cor}\label{opencore}
Suppose that $\mathcal{R}$ is a definably complete expansion of a divisible ordered Abelian group with dp-rank at most $2$.  If there is $X \subseteq R$ definable which is dense and codense in some interval then any model of $T=\textup{Th}{(\mc{R})}$ has o-minimal open core.
\end{cor}

\begin{proof}
We claim that the theory of $\mathcal{R}$ has uniform finiteness. If not, then there would be an infinite discrete definable subset of $R$ contradicting the previous theorem.

Therefore by~\cite[Theorem A]{DMS}, $\mathcal{R}$ has o-minimal open core.
\end{proof}

We note that Theorem \ref{sec2main} no longer holds once the dp-rank of $\mathcal{R}$ exceeds $2$: for example, in \cite{DG} we have shown that the structure $\langle \R; +, <, \Q, \Z\rangle$ has dp-rank $3$.

For the remainder of this section, we return to the general situation when $\mc{R}$ is only assumed to have burden $2$ (rather than dp-rank $2$). Once again adapting arguments of Simon from \cite{S} allows us to obtain information on dense-codense definable sets in the case when there is also an infinite definable discrete set.

We begin with a definition.

\begin{definition}
\label{bounded_away_zero}
Let $X \subseteq R$ be definable and $f:X \to \ob{R}$ be a definable function from $X$ to some sort in the Dedekind completion $\ob{R}$ of $\mathcal{R}$ (see Definition~\ref{ded_sort} above) with $f(x)>0$ for all $x \in X$.   We say that \emph{$f$ is  bounded away from $0$} if whenever $I$ is an interval so that $I \cap X$ is infinite there is $\epsilon>0$ and a subinterval $J \subseteq I$ with $J \cap X$ infinite so that $f(a)>\epsilon$ for all $a \in J$.  

We say "functions on $X$ are bounded away from $0$" to mean that all definable $f: X \to \ob{R}$ with $f$ postive on $X$ are bounded away from $0$.
\end{definition}


\begin{lem}
\label{bdd_away_from_zero_bd_1}
If $X \subseteq R$ has burden $1$, then definable functions on $X$ are bounded away from $0$.
\end{lem}

\begin{proof}
This can be shown by adapting the proof of Lemma~3.19 \cite{Goodrick_dpmin}. Namely, if definable functions on $X$ are not bounded away from $0$, then pick any collection of pairwise-disjoint intervals $\langle I_i \, : \, i \in \omega \rangle$ such that each set $I_i \cap X$ is infinite, and then we can select elements $\epsilon_0 > \epsilon_1 > \ldots > 0$ from $R$ such that for every $(i,j) \in \omega^2$,

$$X \cap I_i \cap \{x \in R \, : \, \epsilon_j > f(x) > \epsilon_{j+1} \} \neq \emptyset.$$

Thus we have an inp-pattern in $X$ of depth $2$: one row consists of formulas expressing $x \in I_i$, and the other consists of formulas expressing $ \epsilon_j > f(x) > \epsilon_{j+1}$.

\end{proof}


The next lemma generalizes a key fact about definable functions in the dp-minimal case from \cite{Goodrick_dpmin}.

\begin{lem}
\label{bdd_away_from_zero_discrete}
If there is an infinite discrete set definable in $\mathcal{R}$, then functions on $R$ are bounded away from $0$.

\end{lem}

\begin{proof}
Suppose, to the contrary, that $f : R \rightarrow \overline{R}$ is definable, $f(x) > 0$ for every $x \in R$, and $f$ is not bounded away from $0$. Then there is a nonempty interval $I$ such that for any $\epsilon > 0$ and any subinterval $J$ of $I$, there is $a \in J$ such that $f(a) > \epsilon$. Therefore, $f$ is not bounded away from $0$ on any subinterval of $I$. By Lemma~\ref{bdd_away_from_zero_bd_1}, any subinterval of $I$ has burden $2$. By  Lemma~\ref{rank_2_intervals}, there is some interval $(0, \epsilon)$ of burden $1$, and shrinking $\epsilon$ as necessary, we may assume that $\epsilon$ is less than the diameter of $I$. Since burden is translation-invariant, we conclude that there is a subinterval of $I$ of burden $1$, a contradiction.


\end{proof}

We record an immediate consequence of Lemma \ref{bdd_away_from_zero_discrete}.


\begin{cor}\label{away_dense} Suppose there is an infinite discrete set definable in $\mathcal{R}$.  Suppose that $I$ is an interval and $X \subseteq I$ is definable, dense, and codense in $I$.  Then functions on $X$ are bounded away from $0$.
\end{cor}

\begin{proof}
If $g \, : \, X \rightarrow \overline{R}$ is definable and has positive values, pick any $a > 0$ and extend $g$ to a definable function $f \, : \, R \rightarrow \overline{R}$ by the rule that $f(x) = a$ when $x \notin X$. Now apply the previous Lemma to conclude that $f$, and hence $g$, is bounded away from $0$.
\end{proof}

\begin{definition}  For a definable set $X$ we write $a \sim_{X, \delta} b$ or simply $a \sim_\delta b$ for the equivalence relation on $R$ defined by $$ \forall \, \epsilon \in (-\delta, \delta) \left[ a + \epsilon \in X \Leftrightarrow b + \epsilon \in X \right],$$ and $a \sim_X b$ (or simply $a \sim b$) means that for some $\delta > 0$, we have $a \sim_\delta b$. 
\end{definition}

First we note the following easy but useful fact:

\begin{lem}
\label{translation_fact}
If $X$ is definable, $a, b \in R$, $a \sim_{X,\delta} b$, and $|\epsilon| < \delta$, then  $a + \epsilon \sim_{X,\delta - |\epsilon|} b + \epsilon$. In particular, $a + \epsilon \sim_X b + \epsilon$.
\end{lem}

\begin{lem}
\label{single_class_reduction}
Let $X$ be definable.  Fix any $a \in X$ and let $\widetilde{X} = \{b \in X \, : \, b \sim_X a\}$. Then $|\widetilde{X} / \sim_{\widetilde{X}} | = 1$.
\end{lem}

\begin{proof}
Let $b \in \widetilde{X}$, and we will show that $b \sim_{\widetilde{X}} a$. Choose $\delta > 0$ such that $b \sim_{X, \delta} a$ and suppose that $\epsilon \in (-\delta, \delta)$. By Lemma~\ref{translation_fact}, we have $ b + \epsilon \sim_X a + \epsilon$, so in particular $b + \epsilon \sim_X a$ if and only if $a + \epsilon \sim_X a$. In other words, $b + \epsilon \in \widetilde{X}$ if and only if $a + \epsilon \in \widetilde{X}$, so $b \sim_{\widetilde{X},\delta} a$.
\end{proof}

Now we look for type-definable subgroups related to definable subsets of $R$. The proof of the next lemma is similar to that of Theorem~3.6 from \cite{S}.


\begin{lem}
\label{single_class_group}
Suppose that:
\begin{enumerate}
\item $X \subseteq R$ is definable;
\item $I$ is an open interval such that $X \cap I$ is infinite; 
\item functions on $X \cap I$ are bounded away from $0$;
\item $X \cap I$ is not discrete; and
\item $|(X \cap I) / \sim| = 1$.
\end{enumerate}

Then for any $a \in X \cap I$, there is a nonzero type-definable convex subgroup $C$ of $(R, +)$ such that $(X - a) \cap C$ is also a subgroup of $(R, +)$.



\end{lem}

\begin{proof}
Fix $a \in X \cap I$ and define $f: X \rightarrow \overline{R}$ to be the function $$f(x) = \sup \{ \delta \in R \, : \, x \sim_\delta a \}.$$ 
Note that $f(x)>0$ on $I$.  Since functions on $X \cap I$ are bounded away from zero, we may pick an $\epsilon > 0$ and an open subinterval  $J \subseteq I$ such that $X \cap J$ is infinite and $f(x) > \epsilon$ for every $x \in J \cap X$.

\begin{claim}
There is a positive element $\epsilon_0 < \epsilon$ such that $f(x) > \epsilon_0$ for every $x \in (a - \epsilon_0, a+ \epsilon_0)$.
\end{claim}

\begin{proof}
 First, we may assume that $f(a) > \epsilon$ (by replacing $\epsilon$ by a positive element less than $f(a)$ if necessary). Pick some $b \in X \cap J$ and some $\epsilon' > 0$ such that $a \sim_{\epsilon'} b$, $(b - \epsilon', b + \epsilon') \subseteq J$, and $\epsilon' \leq \epsilon$. Then if $\epsilon''$ is any positive element such that $2 \epsilon'' < \epsilon'$, for every $z \in (-\epsilon'', \epsilon'')$, we have $a + z \sim_{\epsilon''} b + z$ (since $a \sim_{\epsilon'} b$ and using Lemma~\ref{translation_fact}), but also  $b + z \sim_{\epsilon'} a$ (since $b + z \in J$), and thus $a+z \sim_{\epsilon''} a$. Therefore, given that $f(a) > \epsilon > \epsilon''$, it follows that for any $a + z \in (a - \epsilon'', a + \epsilon'')$, we have $f(a+z) \geq \epsilon''$. Then if $0 < \epsilon_0 < \epsilon''$ and $a + z \in (a - \epsilon_0, a + \epsilon_0)$, we have $f(a + z) > \epsilon_0$, as desired.
 \end{proof}

Pick $\epsilon_0$ as in the Claim above and small enough that $(a - \epsilon_0, a + \epsilon_0) \subseteq I$. Replace $I$ by the subinterval $(a - \epsilon_0, a + \epsilon_0)$ and note that all the hypotheses of the Lemma still hold (since (4) and (5) imply that $X \cap I$ has no isolated points, hence $X \cap (a - \epsilon_0, a + \epsilon_0)$ is infinite, and the rest are trivial).

Define
 $$C = \bigcap_{n \in \N}\{x \in R : -\epsilon_0 < nx < \epsilon_0\}$$
 and note that $C$ is closed under addition by the triangle quality. Thus $C$ is a convex type-definable subgroup of $(R, +)$. Let $H = (X - a) \cap C$ and suppose that $g, h \in H$. Then $g + h \in C$. Furthermore, $a + g \sim_{\epsilon_0} a$, so using Lemma~\ref{translation_fact} and the fact that $|h| < \epsilon_0$, we obtain $$a + g + h \sim a + h ,$$ but $a+h \in X$ since $h \in X - a$, so $g + h \in X - a$. This shows that $H$ is closed under addition. For closure under negation, if $g \in (X-a) \cap C$, then it is immediate that $-g \in C$, while $a + g \sim_{\epsilon_0} a$ and $|g| < \epsilon_0$ imply (using Lemma~\ref{translation_fact} again) that $$ a + g -g   \sim a - g$$ $$\Rightarrow a \sim a-g,$$ and therefore $a-g \in X$, hence $-g \in X - a$.

\end{proof}

Finally we come to our generalization of Theorem~\ref{sec2main} in the case when $\mathcal{R}$ has burden $2$ instead of dp-rank $2$.  The example in the subsequent section demonstrates the necessity of  the extra hypothesis that $X / \sim$ is finite.


\begin{thm}
\label{dense_finite_twiddle}
If $\mathcal{R}$ is a divisible OAG of burden $\leq 2$ in which an infinite discrete set is definable, then there cannot be a definable set $X \subseteq R$ which is dense and codense in some nonempty interval and such that $X / \sim $ is finite.
\end{thm}

\begin{proof}


Suppose towards a contradiction there is such a set $X$.  Since  $X / \sim $ is finite, there is some $a \in X$ and some nonempty interval $I$ in which $[a]_\sim$ is dense and codense. Let $Z = [a]_\sim \cap I$, and by Lemma~\ref{single_class_reduction}, we have that $| Z / \sim_Z | = 1$.

Since an infinite discrete set is definable in $\mathcal{R}$, by Corollary~\ref{away_dense}, functions on $Z$ are bounded away from zero. Fix any $a \in Z \cap I$, and we may apply Lemma~\ref{single_class_group} to obtain a nonzero convex subgroup $C$ of $\langle R; + \rangle$ such that $H := C \cap (Z -a)$ is a subgroup. By our assumptions on $Z$, $H$ is dense and codense in $C$. Pick any $g \in C \setminus H$.


\begin{claim}
The set $\{\frac{g}{n} \, : \, n \in \N \setminus \{0\} \}$ contains representatives of infinitely many cosets of $H$.
\end{claim}

\begin{proof}
On the one hand, if the image of $g$ in $R / H$ has infinite order, then for any distinct $n, m \in \N \setminus \{0\}$, we have $mg - ng \notin H$, and hence $\frac{g}{n} - \frac{g}{m} = \frac{mg - ng}{nm} \notin H$, so we are done. Otherwise, let $k$ be the least positive natural number such that $k g \in H$, and note that if $\textup{GCD}(k,m) =1$, then $mg \notin H$. Thus whenever $1 \leq i < j$, $$\frac{g}{k^j} - \frac{g}{k^i} = \frac{(1 - k^{j-i} ) g}{k^j},$$ but $\textup{GCD}(1 - k^{j-i} , k) = 1$, so this difference cannot be in $H$.
\end{proof}

By the Claim, the convex set $C$ intersects infinitely many cosets of the group $H$. But now we can easily build a depth-2 inp-pattern within any interval $J \subseteq C$, as follows: one row consists of translates of the definable set $Z$ which correspond to distinct cosets of $H$ within $C$ and hence are $2$-inconsistent and dense throughout the interval $J$, while the second row consists of pairwise disjoint subintervals of $J$. This contradicts Lemma~\ref{rank_2_intervals} and finishes the proof.
\end{proof}



Thus we have established that for divisible $\mathcal{R}$ we can not simultaneously have an infinite definable discrete set and a definable dense codense set $X$ with $X/\sim$ finite.
Next we consider the case in which $X \subseteq R$ is definable and $X / \sim$ is infinite. First we need a preliminary lemma:

\begin{lem}
\label{dividing_acc_point}
If there is an infinite definable discrete set in $\mathcal{R}$, then there is no definable $X \subseteq R$ such that $X$ divides over some elementary submodel $\mathcal{R}_0$ of $\mathcal{R}$ and $0$ is an accumulation point of $X$.

\end{lem}

\begin{proof}
Otherwise, say there is such a set $X = X_{\overline{a}}$ defined over parameters $\overline{a}$. Then within any open interval $I$ around $0$, no matter how small, we can construct an inp-pattern of depth $2$ as follows: first, since $X_{\overline{a}}$ divides, pick a set of $\mathcal{R}_0$-conjugates $\{X_{\overline{a}_i} \, : \, i \in \omega\}$ of $X_{\overline{a}}$ which are $k$-inconsistent for some $k$, and this will be the first row of the inp-pattern. For the second row, we can pick pairwise disjoint subintervals of $I$ which each intersect every set $X_{\overline{a_i}}$ (since $0$ is an accumulation point of $X_{\overline{a}}$ and so too of all of its conjugates). But this contradicts Lemma~\ref{rank_2_intervals}.
\end{proof}

The next proposition is similar to the well-known fact that in an NIP theory, global types which do not divide over a small submodel are invariant over said submodel (see Chapter 5 of \cite{Guide_NIP}, or Proposition 2.1 of \cite{HruPil}). However, we require a version of this which assumes only that the formula we are working with is NIP.
  
\begin{prop}
\label{dividing_invariant}
Suppose $\varphi(\overline{x}; \overline{y})$ is an NIP formula in any theory $T$, $\mathcal{N} \prec \mathcal{M}$ are models of $T$ such that $\mathcal{M}$ is $|N|^+$- saturated, and $p(\overline{x})$ is a complete $\varphi(\overline{x}; \overline{y})$-$M$-type\footnote{That is, a maximal consistent collection of instances $\varphi(\overline{x}; \overline{b})$ of $\varphi(\overline{x}; \overline{y})$ with $\overline{b}$ from $M$.} which does not divide over $N$. Then $p(\overline{x})$ is $N$-invariant.
\end{prop}


\begin{proof}
The same proof as in \cite{HruPil} goes through. Namely, suppose that $\overline{a}_0, \overline{a}_1 \in M$ and $\textup{tp}(\overline{a}_0 / N) = \textup{tp}(\overline{a}_1 / N)$, and assume towards a contradiction that $\varphi(\overline{x}; \overline{a}_0) \wedge \neg \varphi(\overline{x}; \overline{a}_1) \in p(\overline{x})$. As $N$ is a model, there is some $\overline{b}$ such that both $\overline{a}_0, \overline{b}$ and $\overline{a}_1, \overline{b}$ can be extended to infinite $N$-indiscernible sequences (see Fact~1.12~(ii) of \cite{CLPZ}), and by saturation we may assume $\overline{b} \in M$. So replacing either $\overline{a}_0$ or $\overline{a}_1$ by $\overline{b}$ as needed, there is an infinite $N$-indiscernible sequence $\{ \overline{a}_i \, : \, i < \omega \}$. Since $p(\overline{x})$ does not divide over $N$, the partial type $\{\varphi(\overline{x}; \overline{a}_{2i}) \wedge \neg \varphi(\overline{x}; \overline{a}_{2i+1}) \, : \, i < \omega \}$ must be consistent. As any Boolean combination of NIP formulas is NIP, $\varphi(\overline{x}; \overline{y}_1) \wedge \neg \varphi(\overline{x}; \overline{y}_2)$ is NIP, but by the previous sentence this formula has infinite alternation rank, which is a contradiction.
\end{proof}

\begin{prop}
\label{no_inf_twiddle}
Suppose that there is an infinite discrete set definable in $\mathcal{R}$, $X \subseteq R$ is definable, and the formula $\varphi(x; y)$ expressing that $x \in X - y$ is NIP.  Then $X / \sim$ is finite.
\end{prop}

\begin{proof}
Suppose that $X$ is definable over the $\omega$-saturated model $\mathcal{R}_0$. Let $\mathcal{R}_1$ be an $|R_0|^+$-saturated elementary extension of $\mathcal{R}_0$.  If $X / \sim$ is infinite, then it has unboundedly many equivalence classes (since $\sim$ is a definable equivalence relation) and thus we can find $a, b \in X(R_1)$ such that $\textup{tp}(a/ \mathcal{R}_0) = \textup{tp}(b / \mathcal{R}_0)$ and $a$ is not in the same $\sim$-class as $b$. Hence $0$ is an accumulation point of the set $Z := (X - a) \Delta (X - b)$, and by Lemma~\ref{dividing_acc_point}, $Z$ does not divide over $R_0$. By $\omega$-saturation, $Z$ does not fork over $R_0$ (see, for instance, Proposition~5.14 of \cite{Guide_NIP}). Therefore $Z$ has an extension to a complete $\varphi(x;y)$-type $p(x)$ over $R_1$ which does not fork over $R_0$. But $p(x)$ cannot be $R_0$-invariant since it contains $Z$, so by Proposition~\ref{dividing_invariant}, we conclude that $\varphi(x; y)$ cannot be NIP. 

\end{proof}

Putting this all together, we obtain:

\begin{thm}
\label{IP_translations}
Suppose that $\mathcal{R}$ is a divisible OAG of burden $2$ in which an infinite discrete set is definable and also there is a definable $X \subseteq R$ which is dense and codense in some nonempty interval. Then the formula $\varphi(x; y)$ expressing that $x \in X - y$ has the independence property.

\end{thm}

\begin{proof}
By Theorem~\ref{dense_finite_twiddle}, $X / \sim$ is infinite. Hence by Proposition~\ref{no_inf_twiddle}, we conclude that $\varphi(x;y)$ has the independence property.
\end{proof}


\begin{quest}
Are there divisible OAGs of burden $2$ (or dp-rank $2$) in which both a Cantor-like set $C$ and a set $X$ which is dense and codense on an infinite interval are definable?
\end{quest}



\section{A theory with burden $2$ with both an infinite definable discrete set and a dense-codense set}


In this section, we construct an example of a complete theory $T$ with the following properties:  \begin{itemize}
\item $T$ expands the theory divisible ordered Abelian groups;
\item $T$ is definably complete;
\item $T$ has burden 2; and
\item if $\mc{M} \models T$ then there are 
   both an infinite definable discrete subset of $M$ and a definable dense and codense subset of $M$. 

\end{itemize}
   
This establishes that the hypotheses of Theorem \ref{IP_translations} can in fact be satisfied. In order to accomplish this we need to develop some background facts that slightly generalize some of the work in \cite{cp}.

Recall that in \cite{cp} the authors show that given a theory $T$ in a language $\mc{L}$ which eliminates $\exists^{\infty}$ 
and a distinguished unary predicate $S$ from $\mc{L}$ if we expand $\mc{L}$ to $\mc{L}(P)$ by adding a new unary predicate $P$ then 
the $\mc{L}(P)$-theory $$T_{S}= T \cup \{\forall x \left[ P(x) \rightarrow S(x) \right]\}$$ has a model companion $T_G$.  $T_G$ has a simple axiomatization given in \cite[Theorem 2.4]{cp}.

Here we point out the the results from \cite[section 2]{cp} all hold if we slightly weaken the assumption that $T$ eliminates $\exists^{\infty}$ to the assumption that  $T$ eliminates $\exists^{\infty}$ relative to $S$.

\begin{definition}  For a theory $T$ and a distinguished unary predicate $S$ we say that {\em $T$ eliminates $\exists^{\infty}$ relative to $S$} if for any formula $\varphi(x, \ob{y})$ so that $T \models \forall x \left[\varphi(x, \ob{y}) \rightarrow S(x)\right]$ there is $n \in \mathbb{N}$ so that if $\mc{M} \models T$ and $\ob{a} \in M^{|\ob{y}|}$ and
 $|\varphi(M, \ob{a}) |> n$ then $\varphi(M, \ob{a})$ is infinite.
 \end{definition}
 
 It is straightforward to see that if we assume that $T$ eliminates $\exists^{\infty}$ relative to $S$ then we can form a model companion $T_G$ of $T_S$  just as in \cite{cp} and Theorem 2.4, Proposition 2.5, and Corollaries 2.6 from \cite{cp} still hold in this case.
 
Let $T$ be the theory of the structure $\langle \R; +, <, \mathbb{Z}\rangle$ formulated in the language
$\mc{L}=\{+, <, 0, 1, \floor{\;}, S, \lambda\}_{\lambda \in \Q}$ where $S$ is interpreted as the interval $(0,1)$, the $\lambda$ are simply multiplication by $\lambda \in \Q$ and 
 $\floor{x}=\max\{y \in \mathbb{Z} : y\leq x\}$.  By results from the appendix of \cite{ivp} the $\mc{L}$-theory $T$ eliminates quantifiers and is universally axiomatizable.  We record some further facts about $T$ which we will use repeatedly without further mention.
 
 \begin{fact}\label{zfact}
 \begin{enumerate}  \item $T$ has dp-rank 2;
 \item If $\mc{M} \models T$ and $X \subseteq M$ is definable then $X$ is either discrete or has interior;
 \item If $\mc{M} \models T$ and $X \subseteq (0,1)$ is definable then $X$ is a finite union of points and intervals, so $T$ eliminates $\exists^{\infty}$ relative to $S$;
 \item If $\mc{M} \models T$, $X \subseteq M$ is definable and discrete and $f: M \to M$ is definable then $f[X]$ is discrete;
 \item If $\mc{M} \models T$, $U \subseteq M$ is an open interval and $f: U \to M$ is definable then there is an open subinterval $V \subseteq U$ so that $f$ is equal to a linear function of the form 
 $x \mapsto \lambda x +a$ on $V$.
 \end{enumerate}
\end{fact}

\begin{proof} See the note after the proof of Theorem 3.1 in \cite{DG} for (1).  As $T$ is clearly definably complete (2) follows from Corollary \ref{discrete_interior_dichotomy}.  (3) is  a special case of \cite[Lemma 3.3(1)]{DG}.  (4) follows from \cite[Corollary 2.17]{DG}.  (5) follows from \cite[Lemma 3.2(2)]{DG} and the fact that definable functions in models of $T$ are given piecewise by terms.
\end{proof}

As $T$ eliminates $\exists^{\infty}$ relative to $S$ we can form the theory $T_G$.  For convenience we fix a completion $T^*_G$ of $T_G$ by specifying that if $\mc{M} \models T_G$ then $P(M) \cap \textup{dcl}{(\emptyset)}=\emptyset$ (see \cite[Corollary 2.6]{cp}). 

\begin{lem} $T_G^*$ has quantifier elimination and is definably complete.
\end{lem}

\begin{proof} As $T$ is universally axiomatizable and model complete, functions definable in models of $T$ are given piecewise by terms.  Also $\textup{acl}=\textup{dcl}$ in models of $T$.  Thus it follows from \cite[Corollary 2.6]{cp} that $T_G^*$ has quantifier elimination.

  By a simple inductive construction and using the axiomatization for $T_G$ in \cite[Theorem 2.4]{cp} there is $G \subseteq \R$ so that \[\langle \R; +, <, 0, 1, \floor{\;}, S, G, \lambda\rangle_{\lambda \in \Q}\] is a model of $T_G^*$.  Thus as $T^*_G$ is complete it is definably complete.
\end{proof}


 The axioms for $T_G$ imply that if $\mc{M} \models T_G^*$  then $P(M)$ must be dense and codense in $(0,1)$.  Thus to establish that $T_G^*$ is our desired example we must show that $T_G^*$ is of burden $2$.  Ideally we would simply like to reference \cite[Theorem 7.3]{chernikov} but as $T$ does not satisfy exchange for algebraic closure it is not clear that this result is applicable.  We provide an ad hoc proof. We need some preliminary lemmas and observations.  The following is an immediate consquence of the axiomatization for $T_G$ given in \cite{cp}.

\begin{fact}\label{genfact} Let $\mc{M} \models T_G$ and let $U \subseteq M$ be an open interval. Furthermore suppose that $f_i: U \to (0,1)$ for $1 \leq i \leq n$ and $g_j: U \to (0,1)$  for $1 \leq j \leq m$ are definable functions which are continuous, non-constant, and monotone.  If there is $V \subseteq U$ an open interval so that $f_i(x) \not= g_j(x)$ for all $x \in V$ and all $i,j$ then there is $x \in V$ so that 
$f_i(x) \in P(M)$ for all $i$ and $g_j(x) \notin P(M)$ for all $j$.
\end{fact}

\begin{lem}\label{loc_const}  Let $\mc{M} \models T_G^*$ and $f: M \to (0,1)$ be definable.  Then there are only finitely many $a \in (0,1)$ so that $f^{-1}(a)$ has non-empty interior.
\end{lem}

\begin{proof} Let 
\[X=\{b \in M : f \text{ is constant in a neighborhood of } b\}.\]  Either $X$ is empty in which case we are done or it is a non-empty open set and hence a union of open intervals by definable completeness.  Also by definable completeness if $I$ is any one of these intervals $f$ is constant on $I$.  Let $X_0$ be the set of midpoints of these intervals.  $X_0$ is definable and discrete. Thus $f[X_0]$ is discrete and as $f[X_0] \subseteq (0,1)$ it must be finite.  This establishes the lemma.  

\end{proof}

\begin{lem}\label{good_form} Let $\mc{M} \models T_G^*$ and let $X \subseteq M$ be definable.  $X$ is a finite union of sets definable by formulas of the form:
\[\psi(x) \wedge \bigwedge_{i \in I}P(t_i(x)) \wedge \bigwedge_{j \in J}\neg P(s_j(x))\]
where $\psi(x)$ is an $\mc{L}$-formula possibly with parameters, $I$ and $J$ are finite sets, and the $t_i$'s and $s_j$'s are terms possibly with parameters so that:
\begin{enumerate}
\item $\psi(M)$ is either discrete or open;
\item If $\mc{M} \models \psi(a)$ then $t_{i_1}(a) \not= t_{i_2}(a)$ for $i_1 \not= i_2$, $s_{j_1}(a) \not= s_{j_2}(a)$ for $j_2 \not= j_2$ and $t_i(a) \not= s_j(a)$ for all $i,j$;
\item If $\mc{M} \models \psi(a)$ then $t_i(a) \in (0,1)$ and $s_j(a) \in (0,1)$ for all $i,j$;
\item If $I$ is an open interval with $I \subseteq \psi(M)$ then no $t_i$ and no $s_j$ is constant on $I$.
\end{enumerate}
\end{lem}

\begin{proof} By quantifier elimination for $T_G$ we reduce to the case where $X$ is defined by a formula of the form:
\[\psi_0(x) \wedge \bigwedge_{i \in I_0}P(t_i^0(x)) \wedge \bigwedge_{j \in J_0}\neg P(s^0_j(x))\] where $\psi_0$ is an $\mc{L}$-formula, $I_0$ and $J_0$ are finite sets, and the $t^0_i$'s and $s^0_j$'s are terms.  The proof is a straightforward induction on $|I_0+J_0|$ where (1) follows as any definable set in $\mc{M}$ either is discrete or has interior, (2) is immediate, (3) is immediate as $P(M) \subseteq (0,1)$, and (4) follows by Lemma \ref{loc_const} and the fact that $T$ eliminates $\exists^{\infty}$ relative to $(0,1)$.

\end{proof}

\begin{lem}\label{disc_def}  If $\mc{M} \models T_G^*$ and $X \subseteq M$ is definable and discrete then $X$ is definable in 
$\mc{M} \restriction \mc{L}$.
\end{lem}

\begin{proof} Without loss of generality we may assume that $X$ is defined by a formula $\varphi(x)$ of the form given in Lemma \ref{good_form}, say	
\[\varphi(x) = \psi(x) \wedge \bigwedge_{i \in I}P(t_i(x)) \wedge \bigwedge_{j \in J}\neg P(s_j(x)).\]


 If $\psi(M)$ has interior we can find an open interval $U \subseteq \psi(M)$ so that each $t_i$ and $s_j$ is equal to a linear term of the form $\lambda x + a$ on $U$.  By assumption none of the $s_i$ and $t_i$ are equal on $U$ and none of them are constant.  Thus by Fact \ref{genfact}  it must be the case that $\varphi(M)$ is dense in $U$ and thus this case is impossible.
 
If $\psi(M)$ is discrete the image of $\psi(M)$ under any of the $t_i$ or $s_j$ must be discrete.   As these images are subsets of $(0,1)$ they must be finite.  It follows that $X$ is definable in $\mc{M} \restriction \mc{L}$.

\end{proof}

Although we only need the following lemma in the specific case of models of $T$ we provide a statement and proof in much greater generality. 

\begin{lem}\label{disc_rank}  Suppose that $\mc{N}$ is an expansion of a densely ordered group of finite burden $n$.  If $X \subseteq N$ is definable, infinite, and discrete, then the burden of $X$ is less than or equal to $n-1$.
\end{lem}

\begin{proof}  Without loss of generality we assume that $\mc{N}$ is sufficiently saturated.  Suppose the result fails.  Thus $X$ has burden $n$.  We can find $\delta>0$ so that 
\[X'=\{x \in X : (x-\delta, x+\delta) \cap X =\{x\}\}\] also has burden $n$.  Let the formulas $\varphi_i(x, \ob{y})$ for $1 \leq i \leq n$ and parameters $\ob{a}_{i,j}$ for $1 \leq i \leq n$ and $j \in \omega$ witness that $X'$ has burden $n$.
We can find $0<\varepsilon_j^0<\varepsilon_j^1<\delta$ for $j \in \omega$ so that $\varepsilon^1_j<\varepsilon^0_{j+1}$ for all $j \in \omega$.  Let $\varphi_{n+1}(x,y_1y_2)$ be the formula
\[ \exists z \in X'(z<x \wedge \forall w(w \in X' \rightarrow w \leq z \vee w>x) \wedge y_1<x-z<y_2)\]
and let $\varphi'_i(x,\ob{y})$ for $1 \leq i \leq n$ be the formula
\[\exists z (z \in X' \wedge \forall w(w \in X' \rightarrow w \leq z \vee w>x) \wedge \varphi_i(x, \ob{y})).\]
It now follows that the formulas $\varphi'_1(x, \ob{y}) \dots \varphi'_n(x, \ob{y})$ and $\varphi_{n+1}(x,y_1y_2)$ with respective sequences of parameters $\ob{a}_{1,j}, \dots \ob{a}_{n,j}$ and $\epsilon^0_j,\epsilon^1_j$ witness that the theory of $\mc{N}$ has burden at least $n+1$, a contradiction.
\end{proof}

\begin{prop}  $T_G^*$ has burden $2$.
\end{prop}

\begin{proof} Fix $\mc{M}$ a sufficiently saturated model of $T$ and  suppose the result fails.  Let $\varphi_k(x, \ob{y})$ for $k \in \{1, 2, 3\}$ together with mutually indiscernible parameters $\ob{a}_{k,l}$ with $k \in \{1, 2, 3\}$ and $l \in \R$ be an inp-pattern with three rows.

If any of the $\varphi_k(x, \ob{a}_{k,l})$, say $k=1$, defines a discrete set $X$ then by Lemma \ref{disc_def} $X$ is definable in $\mc{M} \restriction \mc{L}$.  As $\mc{M} \restriction \mc{L}$ has dp-rank $2$,  $X$ must have dp-rank $1$ by Lemma \ref{disc_rank}.  But then $\varphi_j(x, \ob{y})$ for $j \in \{2,3\}$ and $\ob{a}_{j_l}$ for $j \in \{2,3\}$ and $l \in \R$ witness that $X$ has burden two, a contradiction.  

Consider $\varphi_1(x, \ob{y})$.  By  \cite[Lemma 7.1]{chernikov} we may assume  that this formula is of the form given by Lemma \ref{good_form}, say
\[\varphi(x, \ob{y}) = \psi_1(x, \ob{y}) \wedge \bigwedge_{i \in I}P(t_i(x, \ob{y})) \wedge \bigwedge_{j \in J}\neg P(s_j(x, \ob{y})).\]
Suppose that $\{\psi_1(x, \ob{a}_{1,l}) : l \in \R\}$ is consistent. 
 Then we may find an open interval $V=(c,d)$ so that $V$ is a subset of $\psi_1(M, \ob{a}_{1,l})$ for all $l \in \R$.  By shrinking $V$ further we may also assume  that all the terms $t_i(x, \ob{a}_{1,l})$ and $s_j(x, \ob{a}_{1,l})$ are given by linear functions of the form $\lambda x +a$ on $V$.
   As $\mc{M} \restriction \mc{L}$ is of dp-rank 2, by \cite
   [Thm 4.16]{Guide_NIP} we may find an open interval $W \subset \R$
  so that $\{\ob{a}_{1,j} : j \in W\}$ is indiscernible over $cd$ as a sequence in $\mc{M} \restriction \mc{L}$.  For notational convenience assume that $W=\R$. 
  
 Fix $\kappa \in \omega$ such that $\{\varphi_1(x, \ob{a}_{1,l}) : l \in \R\}$ is $\kappa$-inconsistent.  By Fact \ref{genfact} for $T_G$ and the assumptions on the formula $\varphi_1(x, \ob{y})$ from Lemma \ref{good_form}, this is only possible if for some $i \in I$, some $j \in J$, and some $l_1, l_2 \in \R$,
  we have that $t_i(x, \ob{a}_{1, l_1})=s_j(x, \ob{a}_{1, l_2})$ for all $x$ in some subinterval $V_0 \subseteq V$.   As these functions are linear over $\Q$, it follows that  $t_i(x, \ob{a}_{1, l_1})=s_j(x, \ob{a}_{1, l_2})$ on all of $V$.  Then by indiscernibility of $\ob{a}_{1,l}$ over $cd$ (as an $\mc{M} \restriction \mc{L}$-sequence) it follows that $t_i(x, \ob{a}_{1,0})=s_j(x, \ob{a}_{1,0})$ on $V$.  But this violates the properties $\varphi_1$ guaranteed by Lemma \ref{good_form}.
  
  Thus it must be the case that $\{\psi_1(x, \ob{a}_{1,l}) : l \in \R\}$ is inconsistent.  But the same holds for 
  $\{\psi_k(x, \ob{a}_{k,l}) : l \in \R\}$ for $k \in \{2,3\}$.  Thus $\psi_k(x, \ob{y})$ for $k \in \{1,2,3\}$ and $\ob{a}_{k,l}$ for $k \in \{1,2,3\}$ and $l \in \R$ is an inp-pattern with three rows in $\mc{M}\restriction \mc{L}$ which is impossible as $\mc{M}\restriction \mc{L}$ has dp-rank $2$

\end{proof}

This example demonstrates that the results in Section 2 of this paper are in some sense sharp.  In particular notice that as indicated in Theorem \ref{IP_translations} the formula $\tau(x,y):=x \in P-y$ has the independence property by the axioms for $T_G$.



