Proof. First notice that, since $\operatorname {dist}(b,A)=\min \{\operatorname {dist}(b,a):a\in A\}$ , there must be an element $c\in A$ such that $\operatorname {dist}(b,A)=\operatorname {dist}(b,c)$ . If there are two elements $c_1$ and $c_2$ in A such that $\operatorname {dist}(b,c_1)=\operatorname {dist}(b,c_2)=\operatorname {dist}(b,A)<\infty $ , then we would have $b\in P(c_1,c_2)$ . Hence, since $A=\operatorname {conv}(A)$ , we have $b\in A$ . Therefore, $\operatorname {dist}(b,A)=0$ , and so $c_1=c_2=b$ .

Proof. For part (1), let c be the unique element in $P(a_1,a_2)$ such that $\operatorname {dist}(b,c)=\operatorname {dist}(b,P(a_1,a_2))$ . By the triangle inequality, $d=\operatorname {dist}(a_1,a_2)\leq \operatorname {dist}(b,a_1)+\operatorname {dist}(b,a_2)=k_1+k_2$ . Also, by uniqueness of paths in G we have $P(b,a_1)=P(b,c)\cup P(c,a_1)$ and $P(b,a_2)=P(b,c)\cup P(c,a_2)$ , so $\operatorname {dist}(b,a_1)+\operatorname {dist}(b,a_2)=2\operatorname {dist}(b,c)+d$ , from which it follows that the distance from b to the path $P(a_1,a_2)$ satisfies $\operatorname {dist}(b,P(a_1,a_2))=\frac {1}{2}(k_1+k_2-d)$ .

We now proceed to prove part (2). Since $k_1\leq k_2+d$ , we have that $k_1-\ell \leq d.$ Hence, there is a unique element c in $P(a_1,a_2)$ which is at distance $k_1-\ell $ from $a_1$ . Note that the distance from c to $a_2$ is equal to

$$ \begin{align*} \operatorname{dist}(c,a_2)&=\operatorname{dist}(a_1,a_2)-\operatorname{dist}(a_1,c)=d-(k_1-\ell)=k_2+d-k_1-k_2+\ell=k_2-\ell. \end{align*} $$

Let b be an arbitrary vertex in G. If $\operatorname {dist}(b,c)=\ell $ and $P(b,c)\cap P(a_1,a_2)=\{c\}$ , then we have $P(b,a_1)=P(b,c)\cup P(c,a_1)$ and $P(b,a_2)=P(b,c)\cup P(c,a_2)$ , which implies that $\operatorname {dist}(b,a_1)=k_1.$ Similarly, we have $\operatorname {dist}(b,a_2)=k_2$ , so $G\models D_{k_1}(b,a_1)\wedge D_{k_2}(b,a_2)$ .

Now, suppose that $\operatorname {dist}(b,a_1)=k_1$ and $\operatorname {dist}(b,a_2)=k_2$ . Hence, we can pick $c'$ to be an element in $P(a_1,a_2)$ such that $\operatorname {dist}(b,c')=\ell '$ is minimal. Then we have $P(b,c')\cap P(a_1,a_2)=\{c'\}$ , because if there were another element in $P(b,c')$ that lies in $P(a_1,a_2)$ it would contradict the minimality of $\operatorname {dist}(b,c')$ . Since paths are unique, we have that $P(b,a_1)=P(b,c')\cup P(c',a_1)$ and $P(b,a_2)=P(b,c')\cup P(c',a_2)$ , so $\operatorname {dist}(a_1,c')=k_1-\ell '$ and $\operatorname {dist}(c',a_2)=k_2-\ell '$ . Hence,

$$\begin{align*}d=\operatorname{dist}(a_1,a_2)=\operatorname{dist}(a_1,c')+\operatorname{dist}(c',a_2)=(k_1-\ell')+(k_2-\ell'),\end{align*}$$

and so we obtain $\ell '=\ell $ , and since $c'$ is an element in $P(a_1,a_2)$ at distance $k_1-\ell $ from $a_1$ and $k_2-\ell $ from $a_2$ , we can conclude that $c=c'$ .

Proof. As in the proof of Lemma 2.6, we have that $k_1-\ell \leq d_{12}$ , so we can take c to be the unique element in the path $P(a_1,a_2)\subseteq \operatorname {conv}(A)$ at distance $k_1-\ell $ from $a_1$ . We will now show that the equivalence holds.

First, suppose that $b\in G$ satisfies $G\models \bigwedge _{i=1}^n D_{k_i}(b,a_i)$ . In particular, $G\models D_{k_1}(b,a_1)\wedge D_{k_2}(b,a_2)$ , so by Lemma 2.6 we have that $\operatorname {dist}(b,c)=\ell $ and $P(b,c)\cap P(a_1,a_2)=\{c\}$ . Suppose now that d is the unique element in $\operatorname {conv}(A)$ satisfying $\operatorname {dist}(b,d)=\operatorname {dist}(b,\operatorname {conv}(A))$ . Since d belongs to the convex closure of A, $d\in P(a_i,a_j)$ for some $a_i,a_j\in A$ . Therefore, by Lemma 2.6, we have that $\operatorname {dist}(b,d)=\frac {1}{2}(k_i+k_j-d_{ij})=\ell _{ij}\leq \ell $ . By the minimality of $\ell $ , we conclude that $\ell =\ell _{12}=\ell _{ij}$ , so $c=d$ and $P(b,c)\cap \operatorname {conv}(A)=\{c\}$ .

Conversely, suppose that $\operatorname {dist}(b,c)=\ell $ and $P(b,c)\cap \operatorname {conv}(A)=\{c\}$ . These conditions imply that $\operatorname {dist}(b,\operatorname {conv}(A))=\ell $ and c is the closest element in $\operatorname {conv}(A)$ to b. By Lemma 2.6, since $k_1\leq k_2+d_{12}$ and $\ell =\ell _{12}=\frac {1}{2}(k_1+k_2-d_{12})$ , we already have that $G\models D_{k_1}(b,a_1)\wedge D_{k_2}(b,a_2)$ . For $i\geq 3$ , since $G\models D_{d_{1i}}(a_i,a_1)\wedge D_{d_{2i}}(a_i,a_2)$ , we know by Lemma 2.6 that there is a unique element $e_i$ be the unique element in $P(a_1,a_2)$ such that $\operatorname {dist}(a_i,e_i)=\operatorname {dist}(a_i,P(a_1,a_2))=\frac {1}{2}(d_{1i}+d_{2i}-d_{12})$ . We have either $e_i\in P(a_1,c)\setminus \{c\}$ or $e_i\in P(c,a_2)$ , and we will consider these two cases separately.

If $e_i\in P(a_1,c)\setminus \{c\}$ , then $\operatorname {dist}(a_1,e_i)<\operatorname {dist}(a_1,c)$ , from which we have

$$ \begin{align*} d_{1i}-\frac{1}{2}(d_{1i}+d_{2i}-d_{12})&<k_1-\ell=k_1-\frac{1}{2}(k_1+k_2-d_{12}),\\ d_{1i}-d_{2i}+d_{12}&<k_1-k_2+d_{12},\\ d_{1i}+k_2&<k_1+d_{2i}. \end{align*} $$

So, condition ii(a) holds and since the path from b and $a_i$ is given by

$$\begin{align*}P(b,a_i)=P(b,c)\cup P(c,e_i) \cup P(e_i,a_i),\end{align*}$$

we have

$$ \begin{align*} \operatorname{dist}(b,a_i)&=\operatorname{dist}(b,c)+\operatorname{dist}(c,e_i)+\operatorname{dist}(e_i,a_i)\\ &=\operatorname{dist}(b,c)+[\operatorname{dist}(a_1,c)-\operatorname{dist}(a_1,e_i)]+\operatorname{dist}(e_i,a_i)\\ &=\operatorname{dist}(b,c)+[\operatorname{dist}(a_1,c)-(\operatorname{dist}(a_1,a_i)-\operatorname{dist}(a_i,e_i))]+\operatorname{dist}(e_i,a_i)\\ &=\operatorname{dist}(b,c)+\operatorname{dist}(a_1,c)-\operatorname{dist}(a_1,a_i)+2\operatorname{dist}(a_i,e_i)\\ &=\ell+(k_1-\ell)-d_{1i}+(d_{1i}+d_{2i}-d_{12})\\ &=k_1+d_{2i}-d_{12}=k_i. \end{align*} $$

If $e_i\in P(c,a_2)$ , by a symmetric argument we obtain $k_1 + d_{2i} \leq k_2 + d_{1i}$ and $\operatorname {dist}(b,a_i) = k_i$ .

Therefore, we conclude that $G\models D_{k_i}(b,a_i)$ for every $i=1,2,\ldots ,n$ , which finishes the proof.

Proof. Since $\deg _{\operatorname {conv}(A)}(c)=m$ , there are by definition different elements $e_1,\ldots ,e_m\in \operatorname {conv}(A)$ all adjacent to c, and so we can pick elements $a_{s_1},\ldots ,a_{s_m}\in A$ such that $e_i\in P(c,a_{s_i})$ .

By Lemma 2.6, we have $c\in P(a_{s_1},a_{s_2})$ and $\ell =\ell _{s_1,s_2}$ . Moreover, by Remark 2.8 we know that condition (ii) of Proposition 2.7 holds, and it guarantees that for every ${i\leq m}$ we have either $\ell _{s_i,s_1}=\ell $ or $\ell _{s_i,s_2}=\ell $ . Since $c\in P(a_{s_i},a_{s_1})\cap P(a_{s_i},a_{s_2})$ , we have $\operatorname {dist}(c,a_{s_1})+\operatorname {dist}(c,a_{s_i})=d_{s_1,s_i}$ and $\operatorname {dist}(c,a_{s_2})+\operatorname {dist}(c,a_{s_i})=d_{s_2,s_i}$ . Therefore, we have the equalities $k_{s_1}-\ell +k_{s_i}-\ell =d_{s_1,s_i}$ and $k_{s_2}-\ell +k_{s_i}-\ell =d_{s_2,s_i}$ , which implies by definition that $\ell =\ell _{s_1,s_i}=\ell _{s_2,s_i}$ . This proves part (i) of the Proposition.

For part (ii), suppose that $t\neq s_1,\ldots ,s_m$ and $a_t\neq c$ . Since $G\models D_{k_t}(b,a_t)$ , the distance between c and $a_t$ is finite and so the path $P(a_t,c)$ includes an element $e_i$ . If we choose $s_2^t$ to be $s_i$ and let $s_1^t$ be an arbitrary in $\{s_1,\ldots ,s_m\}$ different from $s_i$ , we will have by Lemma 2.6 that $\ell _{t,s_1^t}=\operatorname {dist}(b,P(a_{s_1^t},a_t))=\ell $ and $\ell _{t,s_2^t}=\operatorname {dist}(b,P(a_{s_2^t},a_t))\geq \operatorname {dist}(b,e_i)=\ell +1$ . This finishes the proof.

Proof. First, suppose that every vertex in G has degree k. For fixed vertices $(u,f),(v,g)\in L[G]$ , and in order to have $(u,f)R(v,g)$ , it is necessary that $uEv$ , and once this occurs there is only one possibility for g, namely being the same function as f but changing its value exactly in the edge $e=\{u,v\}$ . Therefore, any $(u,f)\in L[G]$ is connected precisely with k other vertices of the form $(v,g)$ in $L[G]$ .

Now, suppose that $x_1=(v_1,f_1),\ldots ,x_m=(v_m,f_m)$ are pairwise distinct vertices in $L[G]$ that form a cycle, and put $x_{m+1}=(v_{m+1},f_{m+1})=(v_1,f_1)$ . Let $s<t$ in $\{1,\ldots ,m+1\}$ such that $v_s=v_t$ , and $t-s$ is minimal. Note that $t-s\geq 3$ as ${t=s+1}$ would imply an edge between a vertex and itself in G, and $t=s+2$ would correspond to a path in $L[G]$ of the form $x_sRx_{s+1}Rx_s$ , which is not possible in a cycle. Hence, $t\geq s+3$ , and we may suppose without loss of generality that $x_s=x_1$ . Putting $\ell =t-s$ , we then have a cycle $v_1, v_2, \ldots , v_\ell , v_{\ell +1}=v_1$ in G, so $\ell \geq n$ .

On the other hand, considering the functions $f_1,\ldots ,f_m$ , notice that $f_{i+1}$ is obtained from $f_i$ by changing exactly one of the values. So, $f_{\ell +1}$ is obtained from $f_1$ by performing changes in $\ell $ different entries, and to complete the cycle in $L[G]$ the function $f_{\ell +1}$ would need to be restored to $f_1$ . That is, it would be necessary to change at least change these $\ell $ different entries back to their original value. Therefore, $m\geq 2n$ , that is, the length of the minimal cycle in $L[G]$ is at least $2n$ .

To show that the length of the minimal cycle is exactly $2n$ , we can first consider a cycle given by $v_1,\ldots ,v_n,v_{n+1}=v_1$ in G. Let us enumerate the edges $e_i=\{v_i,v_{i+1}\}$ for $i=1,\ldots ,n$ . This produces the following cycle in $L[G]$ of length exactly $2n$ :

Proof. Given that the vertex set of $L[G]$ is $V(G)\times \{0,1\}^{|E(G)|}$ , it is clear that $|L[G]|=|G|\cdot 2^{|E(G)|}$ . On the other hand, by the Hand-shaking lemma, we have that

$$\begin{align*}2|E(G)|=\sum_{v\in V(G)}\deg(v)=|V(G)|\cdot k,\end{align*}$$

so $|E(G)|=|V(G)|\cdot \frac {k}{2}$ , and the result follows.

Proof. For $1\leq k<\omega $ , let us put $\ell _k=\lceil \log _2(g(k))\rceil $ and define the graphs $H_{k,0}',\ldots ,H_{k,\ell _k}'=H_k$ recursively. First, let $H_{k,0}$ be the complete graph $K_{d(k)+1}$ , which is a $d(k)$ -regular graph whose minimal cycle has length $3$ . Now, supposing $H_{k,j}'$ has been given and $j<\ell _k$ , we take $H_{k,j+1}'=L[H_{k,j}']$ . Finally, we put $H_k:=H^{\prime }_{k,\ell _k}$ , and we have the following:

1. (1) $|H_k|\geq |K_{k+1}|=k+1$ , so $\displaystyle {\lim _{k\to \infty } |H_k|=\infty }$ .

2. (2) Since $H_{k,0}'$ is a $d(k)$ -regular graph, we have by Proposition 3.4 that its $\ell _k$ -th lift $H_k=H^{\prime }_{k,\ell _k}$ is also $d(k)$ -regular.

3. (3) Since the minimal cycle in $H_{k,0}'$ has length 3, again by Proposition 3.4 we have that the minimal cycle in $H_k$ has length $3\cdot 2^{\ell _k} \geq 3\cdot g(k)$ .

Thus, the family of graphs $\{H_k:1\leq k<\omega \}$ satisfies the required conditions.

Proof. Let us consider the functions $d_1,d_2,g:\mathbb {N}\to \mathbb {N}$ defined by $d_1(k)=r$ and $d_2(k)=g(k)=k$ for every $k\in \mathbb {N}$ . Let $\mathcal {C}_{d_1,g}=\{G_k:k<\omega \}$ and ${\mathcal {C}_{d_2,g}=\{H_k:k<\omega \}}$ be families of finite graphs satisfying the conditions provided by Proposition 3.6, and consider the ultraproducts $G=\prod _{\mathcal {U}}G_k$ and $H=\prod _{\mathcal {U}}H_k$ with respect to a non-principal ultrafilter $\mathcal {U}$ on $\omega $ . We will now show that $G\models \mathcal {T}_r$ , and $H\models \mathcal {T}_\infty $ .

Property (1) in Proposition 3.6 implies that the vertex sets of $G, H$ are infinite, and property (3) implies that neither $G_k$ nor $H_k$ contain cycles of length $n<g(k)=k$ . So, the collection of indices $\{k<\omega :G_k,H_k\models \tau _n\}$ contains $\mathbb {N}^{\geq n}$ and so belongs to the ultrafilter $\mathcal {U}$ . Therefore, by Łoś’ theorem, both G and H are acyclic graphs. Finally, note that every graph $G_k$ is r-regular and so G is r-regular, and since every graph $H_k$ is k-regular, we have that for any fixed $n<\omega H_k\models \sigma _n$ whenever $n\leq k$ . Thus, by Łoś Theorem, every vertex of H has infinite degree.

Proof. Since $\mathcal {T}_\infty $ has quantifier elimination, every definable set $X \subseteq M^1$ is determined by a formula $\varphi (x,\overline {a})$ that is equivalent to a disjunction of formulas $\tau _j(x,\overline {a})$ , where each $\tau _j(x,\overline {a})$ is a conjunction of basic formulas. By increasing the number of conjunctions if necessary, we can assume that the sets defined by the formulas $\tau _i(x,\overline {a})$ are all disjoint. Consequently, we have $|X|=\sum _{i=1}^{k}|\tau _i(\mathcal {M},\overline {a})|$ and thus it suffices to prove the statement for sets defined by a single conjunction of basic formulas.

We will first show the result for a set X defined by a formula of the form $\varphi (x, \overline {a}) \equiv \bigwedge _{j = 1}^{n} D_{k_i}(x,a_i)$ . If the definable set X is empty, then we can take $p_X(t_1,t_2)=0$ . So, we can assume that $X\neq \emptyset $ , which implies that the set $\displaystyle {A=\operatorname {conv}(\{a_1,\ldots ,a_n\})}$ is finite. By Proposition 2.7, there is a non-negative integer $\ell $ and a unique element $c\in A$ such that for every $b\in \mathcal {M}$ , $b\in E$ if and only if $\operatorname {dist}(b,c)=\ell $ and $P(b,c)\cap A=\{c\}$ . Hence, since c has precisely $\beta -\operatorname {deg}_A(c)$ neighbours that do not belong to A and each of these neighbours adds $(\beta -1)^{\ell -1}$ new elements at distance $\ell $ from c, we have $|X|=(\beta -\deg _A(c))(\beta -1)^{\ell -1}$ . Therefore, we can choose

$$\begin{align*}p_X(t_1,t_2)=(t_2-\deg_A(c))(t_2-1)^{\ell-1}.\end{align*}$$

Next, we will analyze the case when X is defined by a formula of the form

$$\begin{align*}\varphi(x,\overline{a}) \equiv \bigwedge_{i=1}^{n_1} D_{k_i}(x,a_i) \wedge \bigwedge_{j=1}^{n_2} \neg D_{t_j}(x,a_j'). \end{align*}$$

Let us fix the formula $\displaystyle {\theta (x,\overline {a})\equiv \bigwedge _{i=1}^{n_1}D_{k_i}(x,a_i)}$ . By the previous case, for every subset $S\subseteq \{1,\ldots ,n_2\}$ there is a polynomial $q_S(t_1,t_2)$ such that

$$\begin{align*}\left|\theta(\mathcal{M},\overline{a})\wedge \bigwedge_{j\in S}D_{t_j}(x,a_j')\right|=q_S(\alpha,\beta).\end{align*}$$

Therefore, by the inclusion–exclusion principle, we have that

$$ \begin{align*} |\varphi(\mathcal{M},\overline{a})|&=\bigg|\theta(\mathcal{M},\overline{a}) \bigg| - \bigg| \theta(\mathcal{M},\overline{a}) \wedge \bigvee_{j=1}^{n_2} D_{t_j}(x,a_j') \bigg|\\ &=\bigg|\theta(\mathcal{M},\overline{a}) \bigg| - \bigg| \bigvee_{j=1}^{n_2}[\,\,\theta(\mathcal{M},\overline{a}) \wedge D_{t_j}(x,a_j)\,\,] \bigg|\\ &=\bigg|\theta(\mathcal{M},\overline{a}) \bigg| - \left(\sum_{r=1}^{n_2}(-1)^r\cdot \sum_{S\subseteq \{1,\ldots,n_2\}, |S|=r}\bigg|\theta(\mathcal{M},\overline{a})\wedge \bigwedge_{j\in S}D_{t_j}(x,a_j')\bigg|\right)\\ &=q_{\emptyset}(\alpha,\beta)-\sum_{r=1}^{n_2}(-1)^r\cdot \left( \sum_{S\subseteq \{1,\ldots,n_2\}, |S|=r}q_S(\alpha,\beta)\right), \end{align*} $$

obtaining the polynomial

$$\begin{align*}p_X(t_1,t_2)=q_{\emptyset}(t_1,t_2)-\sum_{r=1}^{n_2}(-1)^r\cdot \left(\sum_{S\subseteq \{1,\ldots,n_2\}, |S|=r}q_S(t_1,t_2)\right).\end{align*}$$

Finally, for a formula of the form $\varphi (x,\overline {a})\,:=\, \displaystyle {\bigwedge _{j=1}^n \neg D_{k_j}(x,a_j)}$ , we can apply the same argument replacing the formula $\theta (x,\overline {a})$ by $x=x$ . Hence, we have

$$\begin{align*}|\varphi(\mathcal{M},\overline{a})|=\left|\mathcal{M}\setminus \bigcup_{j=1}^n D_{k_j}(x,a_j)\right|=\alpha-\sum_{r=0}^n(-1)^r\cdot \left(\sum_{S\subseteq \{1,\ldots,n\}, |S|=r}q_S(\alpha,\beta)\right),\end{align*}$$

thus obtaining the polynomial

$$\begin{align*}p_X(t_1,t_2)=t_1-\sum_{r=0}^n(-1)^r\cdot \left(\sum_{S\subseteq \{1,\ldots,n\}, |S|=r}q_S(t_1,t_2)\right).\end{align*}$$

This completes the proof.

Proof. We will first show the result for a formula $\varphi (x,\overline {y})\,:=\, \bigwedge _{i = 1}^n D_{k_i} (x,y_i)$ . If we take an arbitrary tuple $\overline {a} \in M^{|\overline {y}|}$ we know by Lemma 2.6 that, if $\operatorname {dist}(a_{j_1},a_{j_2})> k_{j_1} + k_{j_2}$ for any $1 \leq j_1 < j_2 \leq n$ , then $\varphi (\mathcal {M},\overline {a}) = \emptyset $ . Therefore, if $\varphi (\mathcal {M},\overline {a}) \neq \emptyset $ then the configuration of distances $\overline {d}=\langle d_{ij}:=\operatorname {dist}(a_i,a_j)\rangle _{i,j} $ must satisfy $d_{ij} \leq k_i+k_j$ for every $i,j$ . This guarantees that the possible configurations for which $\varphi (x,\overline {a})$ defines a non-empty set are finite. In fact, the number of this configurations is bounded by the product $\displaystyle {\prod _{1\leq i < j \leq n} (k_i+k_j+1).}$

Moreover, Proposition 2.7 and Remark 2.8 establish conditions under which, given a configuration of finite distances $\overline {d}$ and non-negative integers $k_1,k_2,\dots , k_n$ , the set defined by $\bigwedge _{i = 1}^n D_{k_i} (x,a_i)$ is empty, and furthermore, these conditions are first-order definable. So, given integers $k_1,\ldots ,k_n$ , we can denote by J the set of possible configurations of distances of the form $\overline {d}=(d_{ij})_{ij}$ that satisfy the conditions (i),(ii) described in Proposition 2.7. Notice that J is a finite set, and consider the formula

$$ \begin{align*}\theta(\overline{y}) \equiv \bigvee_{\overline{d}=(d_{ij})_{ij} \in J} \ \bigwedge_{i < j \leq n} D_{d_{ij}}(y_i,y_j).\end{align*} $$

Hence, a tuple $\overline {a}$ satisfies $\theta (\overline {y})$ if and only if the set defined by $\displaystyle {\bigwedge _{i=1}^n D_{k_i}(x,a_i)}$ is not empty.

In Theorem 3.11 we showed that for each tuple $\overline {a} \in M^{|\overline {y}|}$ , the polynomials depend both on the distance $\ell $ from the vertex $c\in A=\operatorname {conv}(a_1,\ldots ,a_n)$ to each element on $ \varphi (\mathcal {M},\overline {a})$ and the degree $\operatorname {deg}_A(c)$ . As in the previous case, there are only finitely many possible values for $\ell $ and $\operatorname {deg}_A(c)$ . In fact, $\ell \leq \min \{k_1,\dots ,k_n\}$ and $\operatorname {deg}_A(c) \leq n$ .

Let us fix a tuple $\overline {a} \in M^n$ and suppose that $\ell = 0$ . This means that the set defined by $\varphi (x,\overline {a})$ contains only one element in A. By Proposition 2.7 it also implies that there are indices $i,j\leq n$ such that $\ell _{ij} = 0$ . Hence, we can define the formula and the corresponding polynomial for this case as

$$ \begin{align*} & \psi_0(\overline{y}) \,:=\, \theta(\overline{y}) \wedge \bigvee_{i<j} D_{k_i+k_j}(y_i,y_j), & &p_0(t_1,t_2) = 1. \end{align*} $$

Now suppose that $\ell> 0$ and $\operatorname {deg}_A(c) = 1$ . In this case $c=a_s$ for some s, $k_s = \min \{k_1, \dots , k_n\}$ and for every $i\leq n$ , $\operatorname {dist}(a_i,a_s) = k_i - k_s $ . In this way, we can define the formula and the respective polynomial as

$$ \begin{align*} & \psi_{\ell,1}(\overline{y}) \,:=\, \theta(\overline{y}) \wedge \bigwedge_{i \neq s} D_{k_i-k_s}(y_i,y_j) & & p_1(t_1,t_2) = (t_2-1)^{k_s}. \end{align*} $$

For the general case when $\ell> 0$ and $\operatorname {deg}(c) = m> 1$ , we have to define a formula that stating that $\operatorname {deg}(c) = m> 1$ . In this case, there are m vertices $a_{s_1}, a_{s_2}, \dots a_{s_m}$ such that $\ell _{s_i,s_j}$ is equal to $\ell $ for every $i,j \leq m$ . Proposition 2.9 guarantees that the vertices $a_f$ with $f \notin \{s_1,\dots ,s_m\}$ there exists a pair of vertices $a_{s_1^t}$ and $a_{s_2^t}$ such that either $\ell = \ell _{f,s_i}$ or $\ell = \ell _{f,s_j}$ but not both, since otherwise $\operatorname {deg}(c)$ would be grater than m. Therefore, we can take the defining formula to be

$$ \begin{align*} &\psi_{\ell,m}(\overline{y})\,:=\, \theta(\overline{y}) \wedge \bigvee_{s_1 < \dots < s_m} \left( \bigwedge_{s_i<s_j} D_{k_{s_i}+k_{s_j}-2\ell}(y_i,y_j) \wedge \right. \\ &\bigwedge_{i \neq s_1,\dots,s_m} \left( \bigvee_{s_i < s_j}(D_{k_i+k_{s_i}-2\ell} (y_i,y_{k_1}) \vee D_{k_i+k_{s_j}-2\ell}(y_i,y_{k_2}) )\right. \\ &\left.\left.\vphantom{\bigwedge_{i \neq s_1,\dots,s_m}}\wedge \lnot (D_{k_i+k_{d_i}-2\ell} (y_i,y_{s_i}) \wedge D_{k_i+k_{s_j}-2\ell}(y_i,y_{s_j}) ) \right) \right), \end{align*} $$

and the respective polynomial is $p_{\ell ,m}(t_1,t_2) = (t_2-m)(t_2-1)^{\ell -1}$ .

The only case that is left to analyze occurs when $\varphi (x,\overline {a})$ defines an empty set. By the construction of the formula $\theta (\overline {y})$ , we can take $\psi (\overline {y})$ as $\lnot \theta (\overline {y})$ and the respective polynomial as $p_0(t_1,t_2) = 0$ . As we saw in Theorem 3.11, this concludes the case when $\displaystyle {\varphi (x,\overline {y})\,:=\, \bigwedge _{i = 1}^n D_{s_i} (x,y_i).}$

On the other hand, if $\displaystyle {\varphi (x,\overline {y}) \,:=\, \bigwedge _{i=1}^{n_1} D_{k_i}(x,y_i) \wedge \bigwedge _{j=1}^{n_2} \lnot D_{t_j}(x,y_j)}$ with $n_2> 0$ , recall that the proof of Theorem 3.11 provides us an explicit construction of the polynomials that we are looking for in function of polynomials in the previous case. By this construction, the maximum number of polynomials can be bounded above by the product

$$ \begin{align*} &\prod_{1\leq i < j \leq n_1} (k_i+k_j+1) \cdot \prod_{r = 1}^{n_2} \left(\prod_{1\leq i < j \leq n_1} (k_i+k_j+1) \cdot \prod_{i=1}^{n_1} (k_i+t_r+1)\right) \cdot\\ & \prod_{1 \leq r_1 < r_2 \leq n_2} \left( \prod_{1\leq i < j \leq n_1} (k_i+k_j+1) \cdot \prod_{i=1}^{n_1} (k_i+t_{r_1}+1) \cdot \prod_{i=1}^{n_1} (k_i+t_{r_2}+1)\right) \cdots \\ & \prod_{1\leq i < j \leq n_1} (k_i+k_j+1) \cdot \prod_{r=1}^{n_2} \left( \prod_{i=1}^{n_1} (k_i+t_{r}+1)\right) < \infty.\\[-44pt] \end{align*} $$

Fact 4.5 (Projection Lemma, Lemma 2.3.1 in [Reference Wolf18]).

Let $\mathcal {C}$ be a class of $\mathcal {L}$ -structures such that the definition of an R-mec holds in $\mathcal {C}$ for all formulas $\varphi (x,\overline {y})$ with a single object variable x. Then $\mathcal {C}$ is an $R'$ -mec in $\mathcal {L}$ , where $R'$ is generated under addition and multiplication by the functions in R.

Proof. For a given formula $\varphi (x,\overline {y})$ , consider the finite set of polynomials

$$\begin{align*}E_\varphi=\{p_1^\varphi(t_1,t_2),\ldots,p_{k_\varphi}^{\varphi}(t_1,t_2)\}\subseteq \mathbb{Z}[t_1,t_2]\end{align*}$$

and the formulas $\psi _1(\overline {y}),\ldots ,\psi _{k_\varphi }(\overline {y})$ given by Theorem 3.12. By quantifier elimination in $\mathcal {T}_\infty $ (Proposition A.1), there is a quantifier-free $\mathcal {L}'$ -formula $\theta (x,\overline {y})$ such that $\mathcal {T}_\infty \models \forall x,\overline {y}\left (\varphi (x,\overline {y})\leftrightarrow \theta (x,\overline {y})\right ).$ Moreover, since $\mathcal {T}_\infty $ is axiomatized by the collection of sentences $\{\sigma _n,\tau _n:n<\omega \}$ , by compactness there is some $k_0\in \mathbb {N}$ such that $\{\sigma _n,\tau _n:n\leq k_0\}\models \forall x,\overline {y}\left (\varphi (x,\overline {y})\leftrightarrow \theta (x,\overline {y})\right ).$

We will consider now a further expansion of the language $\mathcal {L}'$ to a two-sorted language $\mathcal {L}^+$ defined as follows:

• • In $\mathcal {L}^+$ there are two sorts: one sort $\mathbb {K}$ with the language $\mathcal {L}'$ , and another sort $\mathbb {D}$ carrying the language of ordered rings $\mathcal {L}_{or}=\{+,-,\cdot ,<\}$ together with two constant symbols, $\{A,B\}$ .

• • For every formula $\varphi (\overline {x},\overline {y})$ in the language $\mathcal {L}^{\prime }_g$ , we add a function symbol $f_\varphi :\mathbb {K}^{|\overline {y}|}\to \mathbb {D}$ .

Every finite structure $H_k$ in $\mathcal {C}_{d,g}$ induces an $\mathcal {L}^+$ -structure $H_k^+$ in a canonical way. Namely, we can consider the pair $(H_k,\mathbb {R})$ with the usual interpretation of the languages $\mathcal {L}'$ and $\mathcal {L}_{or}$ in each sort, together with the interpretations $A^{H_k}:=|H_k|$ and $B^{H_k}:=d(k)$ , the degree of regularity of the graph $H_k$ . Also, we can interpret the functions $f_{\varphi (\overline {x},\overline {y})}: H_k^{|\overline {y}|}\to \mathbb {R}$ as $f_\varphi (\overline {a}):=|\varphi (H_k^{|\overline {x}|},\overline {a})|$ .

Therefore, in the language $\mathcal {L}^+$ , the statement saying that the collection of polynomials $p_1(t_1,t_2)$ , $p_2(t_1,t_2),\ldots ,p_k(t_1,t_2)$ , together with the formulas $\psi _1(\overline {y}),\ldots ,\psi _k(\overline {y})$ satisfy the conclusion of Theorem 4.6 for the formula $\varphi (x,\overline {y})$ can be written in a first-order fashion using the following $\mathcal {L}^+$ -sentence:

$$ \begin{align*} &\Phi(\varphi,p_1,\ldots,p_k,\psi_1,\ldots,\psi_k):\\ &\forall \overline{y}\left(\bigvee_{i=1}^k \psi_i(\overline{y})\right)\wedge \forall \overline{y}\left(\bigwedge_{1\leq i<j\leq k}\neg\left(\psi_i(\overline{y})\wedge \psi_j(\overline{y})\right)\right)\\&\quad\wedge\forall \overline{y}\left(\bigwedge_{1\leq i\leq k}[f_\varphi(\overline{y})=p_i(A,B)\leftrightarrow \psi_i(\overline{y})]\right). \end{align*} $$

Note that there must be an integer N such that for every $k\geq N$ we have

$$\begin{align*}H_k^+\models \Phi(\varphi,p_1,\ldots,p_k,\psi_1,\ldots,\psi_k),\end{align*}$$

since otherwise there would be a subsequence of graphs of graphs $\{H_{k_i}:i<\omega \}\subseteq \mathcal {C}_{d,g}$ such that $H_{k_i}^+\models \neg \Phi (\varphi ,p_1,\ldots ,p_k,\psi _1,\ldots ,\psi _k)$ , and every ultraproduct with respect to an ultrafilter $\mathcal {U}$ containing the set $\{k_i:i<\omega \}$ would contradict Theorem 3.12.

So, by taking $C=\max \{|H_k|:k\leq N\}$ , completing the collection of possible polynomials with $j=0,1,\ldots ,C$ and using the formulas $\psi _j'\,:=\, \exists ^{=j}x\varphi (x,\overline {y})$ , we obtained the desired result.

Proof. Note that Theorem 4.6 establishes precisely the polynomial exact conditions for formulas in one object variable $\varphi (x,\overline {y})$ . This, together with the Projection Lemma, yields the desired result.

Proof. See Appendix, Proposition A.8, and Theorem A.11.

Proof. We will prove this result by induction on k. For $k=1$ , suppose that $\varphi (x;\overline {a})$ is a formula defining a non-empty set $X\subseteq \mathcal {M}^1$ , with $\overline {a}=a_1,\ldots ,a_s$ . Using quantifier elimination and normal prenex form, we may assume that $\varphi (x;\overline {a})$ is equivalent to a disjunction of conjunctions of basic formulas. Moreover, by adding formulas in the conjunctions we may assume without loss of generality that the disjunctives are pairwise inconsistent.

Since the Morley rank of a finite union of definable sets is equal to the maximum of the Morley ranks of its parts, we may assume that $\varphi (x;\overline {a})$ is equivalent to a formula of the form

$$\begin{align*}\bigwedge_{i\in I}D_{k_i}(x,a_i)\wedge \bigwedge_{j\in J}\neg D_{\ell_j}(x,a_j)\end{align*}$$

for some sets $I,J\subseteq \{1,\ldots ,s\}$ and non-negative integers $\{k_i:i\in I\}$ and $\{\ell _j:j\in J\}$ . Let b be a generic element in X, that is, an element $b\in X$ such that $\operatorname {RM}(X)=\operatorname {RM}(b/\overline {a})$ .

If $I\neq \emptyset $ , we know by Proposition 2.7 that there is an element $c\in \operatorname {conv}(\{a_i:i\in I\})$ and $\ell \in \mathbb {N}$ such that for any element $b'$ , $\mathcal {M}\models \bigwedge _{i\in I}D_{k_i}(b',a_i)$ if and only if $\operatorname {dist}(b',c)=\ell $ and $P(b',c)\cap \operatorname {conv}(\{a_i:i\in I\})=\{c\}$ . By Theorem 5.1 we obtain the equalities

$$\begin{align*}\operatorname{RM}(X)=\operatorname{RM}(b/\overline{a})=D(b/\overline{a})=\ell=(\omega\cdot 0)\oplus \ell,\end{align*}$$

and calculating the size of X we obtain $|X|=(\beta -\operatorname {deg}_{\operatorname {conv}(\overline {a})}(c))\cdot (\beta -1)^{\ell -1}=p_X(\alpha ,\beta )$ , showing that the leading monomial of $p_X(t_1,t_2)$ is $t_2^{\ell }$ , as desired.

Suppose now that $I=\emptyset $ . Hence $\varphi (x;\overline {a})$ is equivalent to a conjunction of the form $\displaystyle {\bigwedge _{j\in J}\neg D_{\ell _j}(x,a_j)}$ , which defines the set $\displaystyle {\bigcup _{j\in J}(D_{\ell _j}(\mathcal {M};a_j)^c)=\mathcal {M}\setminus \left (\bigcup _{j\in J}D_{\ell _j}(\mathcal {M};a_j)\right )}$ . Using inclusion–exclusion, we obtain that the size $|X|$ has the form $\alpha -q(\beta )$ for some polynomial $q(t_2)\in \mathbb {Z}[t_2]$ , showing that the leading monomial of $p_X(t_1,t_2)$ is $t_1^1$ . On the other hand, since $\operatorname {RM}\left (\bigwedge _{j\in J}D_{\ell _j}(x;a_j)\right )$ is finite, we have $\operatorname {RM}(X)=\operatorname {RM}(\mathcal {M})=\omega =(\omega \cdot 1)\oplus 0$ . This completes the proof of the case $k=1$ .

Suppose now the result holds for every definable subset of $\mathcal {M}^k$ . Let X be a non-empty definable subset of $\mathcal {M}^{k+1}$ defined by a formula of the form $\varphi (\overline {x},z;\overline {a})$ with $|\overline {x}|=k$ . Applying Theorem 3.12 to the formula with one object-variable defined by $\theta (z;\overline {x},\overline {a}):=\varphi (\overline {x},z;\overline {a})$ , we know that there are finitely many formulas $\psi _1(\overline {x};\overline {a}),\ldots ,\psi _r(\overline {x};\overline {a})$ forming a partition of $\mathcal {M}^k$ and finitely many polynomials $q_1,\ldots ,q_r\in \mathbb {Z}[t_1,t_2]$ such that for every $\overline {b}'\in \mathcal {M}^k$ and $i\leq r$ , $\mathcal {M}\models \psi _i(\overline {b}',\overline {a})$ if and only if $|\varphi (\mathcal {M};\overline {b}',\overline {a})|=q_i(\alpha ,\beta )$ .

For $i=1,\ldots ,r$ , let us define the set $Z_i$ by the formula $\varphi (\overline {x},z;\overline {a})\wedge \psi _i(\overline {x};\overline {a})$ , and the set $Y_i\subseteq \mathcal {M}^k$ by $\psi _i(\overline {x};\overline {a})$ . Hence, we have

$$\begin{align*}|X|=\displaystyle{\sum_{i=1}^r|Z_i|=\sum_{i=1}^r q_i(\alpha,\beta)\cdot |Y_i|=\sum_{i=1}^r q_i(\alpha,\beta)\cdot p_{Y_i}(\alpha,\beta)},\end{align*}$$

and so the leading-degree of $p_X(t_1,t_2)$ is equal to the maximum of the leading-degrees of the products $q_i(\alpha ,\beta )\cdot p_{Y_i}(\alpha ,\beta )=p_{Z_i}(\alpha ,\beta )$ . Suppose now that $(\overline {b}_i,c_i)$ be a generic element for the set $Z_i$ , which in particular implies that $\overline {b}_i$ satisfies $\psi _i(\overline {x};\overline {a})$ .

By induction hypothesis we know that, for each $i\leq r$ , $\operatorname {RM}(Y_i)=(\omega \cdot m_i)\oplus n_i$ if and only if the leading monomial of $p_{Y_i}(t_1,t_2)$ has the form $A_it_1^{m_i}t_2^{n_i}$ . By the case $k=1$ , we have two cases for $q_i(t_1,t_2)$ :

• • If $\operatorname {dist}(c_i/\operatorname {conv}(\overline {a}\overline {b}_i))=\infty $ , the leading monomial of $q_i(t_1,t_2)$ has the form $B_it_1$ , and so the leading monomial of $p_{Z_i}(t_1,t_2)$ would have the form $A_it_1^{m_i}t_2^{n_i}\cdot B_it_1=A_iB_i\cdot t_1^{m_i+1}t_2^{n_i}$ . Also, by induction hypothesis and Theorem 5.1, the Morley rank of $Z_i$ is given by

  $$ \begin{align*} \operatorname{RM}(Z_i)&=\operatorname{RM}(\overline{b}_i,c_i/\overline{a})=\operatorname{RM}(\overline{b}_i/\overline{a})\oplus d(c_i/\overline{b}_i\overline{a})\\ &=(\omega\cdot m_i)\oplus n_i \oplus \omega=(\omega\cdot (m_i+1))\oplus n_i. \end{align*} $$

• • If $\operatorname {dist}(c_i/\operatorname {conv}(\overline {a}\overline {b}_i))=\ell <\infty $ , the leading monomial of $q_i(t_1,t_2)$ has the form $C_it_2^{\ell }$ , and so the leading monomial of $p_{Z_i}(t_1,t_2)$ is $A_it_1^{m_i}t_2^{n_i}\cdot C_it_2^{\ell }=A_iC_i\cdot t_1^{m_i}t_2^{n_i+\ell }$ . Again, by induction hypothesis and Theorem 5.1, the Morley rank of $Z_i$ is given by

  $$ \begin{align*} \operatorname{RM}(Z_i)&=\operatorname{RM}(\overline{b}_i,c_i/\overline{a})=\operatorname{RM}(\overline{b}_i/\overline{a})\oplus d(c_i/\overline{b}_i\overline{a})\\ &=(\omega\cdot m_i)\oplus n_i \oplus \ell=(\omega\cdot m_i)\oplus (n_i\oplus \ell). \end{align*} $$

Therefore, each set $Z_i$ satisfies the conclusion of the theorem, and since $\operatorname {RM}(X)$ is the maximum of $\{\operatorname {RM}(Z_1),\ldots ,\operatorname {RM}(Z_r)\}$ and the leading degree of $p_X(t_1,t_2)$ corresponds to the maximum (in the lexicographical order) of the leading degrees of the polynomials $p_{Z_1}(t_1,t_2),\ldots ,p_{Z_r}(t_1,t_2)$ the conclusion also holds for the set X.

Proof. Let $\mathcal {M},\mathcal {N}$ be $\mathcal {L}'$ -structures that are both models of $\mathcal {T}_\infty $ (resp. $\mathcal {T}_r$ ) and let $\mathcal {A}$ be a finitely generated common $\mathcal {L}'$ -substructure. Since $\mathcal {L}'$ is a relational language, $\mathcal {A}$ is finite. By Theorem 8.4.1 in [Reference Hodges11], it suffices to show that for every $b\in \mathcal {M}$ there is an elementary extension $\mathcal {N}'$ of $\mathcal {N}$ and a partial $\mathcal {L}'$ -embedding $f:\mathcal {A}\cup \{b\} \to \mathcal {N}'$ such that $f\upharpoonright _{\mathcal {A}}=\operatorname {id}_{\mathcal {A}}$ . There are three cases to consider:

1. (1) Suppose that $b\in \operatorname {conv}_{\mathcal {M}}(\mathcal {A})$ . Then b lies in a path between two elements $a_1,a_2$ of $\mathcal {A}$ , then there are integers $k_1,k_2<\omega $ such that $\mathcal {M}\models D_{k_1}(b,a_1)\wedge D_{k_2}(b,a_{2})\wedge D_{k_1+k_2}(a_1,a_2)$ . Since $\mathcal {A}$ is an $\mathcal {L}'$ -substructure of both $\mathcal {M}, \mathcal {N}$ , we must have $N\models D_{k_1+k_2}(a_1,a_2)$ and there is a unique element $b'$ in the path $P(a_1,a_2)$ at distance $k_1$ from $a_1$ and $k_2$ from $a_2$ . Thus, we can take $f=\operatorname {id}_{\mathcal {A}}\cup \{(b,b')\}$ . In this case, for an arbitrary $a\in \mathcal {A}$ we have

   $$\begin{align*}\mathcal{M}\models D_\ell(b,a)\Leftrightarrow \mathcal{A}\models D_{k_1+\ell}(a_1,a)\vee D_{k_2+\ell}(a_2,a) \Leftrightarrow \mathcal{N}\models D_{k_1+\ell}(b',a),\end{align*}$$
   and so f is a partial $\mathcal {L}'$ -embedding.

2. (2) Suppose $b\not \in \operatorname {conv}_{\mathcal {M}}(\mathcal {A})$ and $\operatorname {dist}(b,\mathcal {A})$ is finite. Since $\operatorname {conv}(\mathcal {A})\setminus \mathcal {A}$ is finite, we have by case (1) that there is an $\mathcal {L}'$ -embedding $f_1:\operatorname {conv}_{\mathcal {M}}(\mathcal {A})\to \mathcal {N}$ extending $\operatorname {id}_{\mathcal {A}}$ . Now, since $\operatorname {dist}(b,\mathcal {A})$ is finite, there is an element $c\in \operatorname {conv}(\mathcal {A})$ such that $\ell =\operatorname {dist}(b,c)=\operatorname {dist}(b,\operatorname {conv}(\mathcal {A}))$ . This implies in particular that for every $a\in \mathcal {A}$ , we have $\operatorname {dist}(a,b)=\operatorname {dist}_{\mathcal {M}}(a,c)+\ell $ . Note also that the only neighbour of c in the path $P(c,b)$ does not belong to $\operatorname {conv}_{\mathcal {M}}(\mathcal {A})$ .

   Since both $\mathcal {M}$ and $\mathcal {N}$ are models of $\mathcal {T}_\infty $ (resp. of $\mathcal {T}_r$ ), there is a neighbour $d'\in \mathcal {N}$ of $f(c)$ which does not belong to $f(\operatorname {conv}(\mathcal {A}))$ . Let $b'$ be an element in $\mathcal {N}$ such that $\operatorname {dist}_{\mathcal {N}}(b',f_1(c))=\ell $ and $\operatorname {dist}_{\mathcal {N}}(b',d')=\ell -1$ . Thus, since $f_1$ is an $\mathcal {L}'$ -embedding, we have for every $a\in \mathcal {A}$ that $\operatorname {dist}_{\mathcal {N}}(b',a)=\operatorname {dist}_{\mathcal {N}}(a,f_1(c))+\ell =\operatorname {dist}_{\mathcal {M}}(a,c)+\ell =\operatorname {dist}_{\mathcal {M}}(b,a)$ . Therefore, the map $f=f_1\cup \{(b,b')\}$ is an $\mathcal {L}'$ -embedding from $\operatorname {conv}(\mathcal {A})\cup \{b\}$ to $\mathcal {N}$ .

3. (3) Suppose now that b is not at finite distance from any element in $\mathcal {A}$ , and consider the type

   $$\begin{align*}q(x)=\{\neg D_k(x,a):k<\omega,\ a\in\mathcal{A}\}.\end{align*}$$

   Since $\mathcal {A}$ is finite and $\mathcal {N}$ is a model of $\mathcal {T}_\infty $ (resp. $\mathcal {T}_r$ ), this type is finitely satisfiable in $\mathcal {N}$ , and so there is an elementary extension $\mathcal {N}'$ of $\mathcal {N}$ with an element $b'$ realizing $q(x)$ . Thus, by putting $f(b)=b'$ , we obtain a partial $\mathcal {L}'$ -embedding $f=\operatorname {id}_{\mathcal {A}}\cup \{(b,c)\}:A\cup \{b\}\to \mathcal {N}'$ .

Thus, we conclude that $\mathcal {T}_\infty $ has quantifier elimination in the language $\mathcal {L}'=\{D_k:k<\omega \}$ . The same proof also shows that the theory $\mathcal {T}_r$ eliminates quantifiers in the language $\mathcal {L}'$ .

Proof. Note that for any $\mathcal {M}\models \mathcal {T}_r$ and $a\in M$ , the formula $D_k(x,a)$ defines a set of size $r(r-1)^{k-1}$ , so it is finite. By quantifier elimination in $\mathcal {L}'$ , this is enough to show that $\mathcal {T}_r$ is strongly minimal. This also shows that the unique generic type over a set A is completely determined by the partial type $\pi (x)=\{\neg D_k(x,a):a\in A, k<\omega \}.$

Suppose now that A is an arbitrary subset of a model $\mathcal {M}$ of $\mathcal {T}_r$ that is $|A|$ -saturated, and let $b\in M$ an arbitrary element. We have two options:

• • If $\mathcal {M}\models D_k(b,a)$ for some $k<\omega $ and $a\in A$ , then $b\in \operatorname {acl}(a)\subseteq \operatorname {acl}(A)$ because b lies in the set defined by the formula $D_k(x,a)$ which has $r(r-1)^{k-1}$ elements.

• • If $\mathcal {M}\models \neg D_k(b,a)$ for every $k<\omega $ and $a\in A$ , then $b\models \pi (x)$ , and so $b\not \in \operatorname {acl}(A)$ .

Therefore, we have shown that $\operatorname {acl}_M(a)=\{b\in M: M\models D_k(b,a) \text {for some} k<\omega \}$ and that $\operatorname {acl}(A)=\bigcup _{a\in A}\operatorname {acl}(a)$ . This implies that the pregeometry in models of $\mathcal {T}_r$ is trivial.

Proof. Let $A=\{a_n:n<\omega \}$ be an enumeration of a countable subset of $M\models \mathcal {T}_\infty $ . By quantifier elimination, every type $q(x)\in S_1(A)$ is completely determined by the formulas of the form $D_k(x,a_n)$ that belong to $q(x)$ , with $k,n<\omega $ . We have three cases to consider for the type $q(x)$ :

1. (1) There are two elements $a_{m_1},a_{m_2}$ such that $D_{k_1}(x,a_{m_1}),D_{k_2}(x,a_{m_2})\in q(x)$ and also $\mathcal {M}\models D_{k_1+k_2}(a_{m_1},a_{m_2})$ . In this case, there is a unique realization b of $q(x)$ that lies in the path between $a_{m_1}$ and $a_{m_2}$ , and so $q(x)$ is completely determined by the tuple $(m_1,m_2,\operatorname {dist}(a_{m_1},a_{m_2}))\in \mathbb {N}^3$ . There are at most $\aleph _0$ such types.

2. (2) There is an element $a_{m_1}\in A$ such that $D_\ell (x,a_{m_1})\in \operatorname {tp}(b/A)$ and $\operatorname {dist}(b,a_{m_1})\leq \operatorname {dist}(b,a_n)$ for every $n<\omega $ . In this case, the formulas $D_k(x,a_n)$ that belong to $q(x)$ are precisely those corresponding to the pairs $(k,n)$ where $k=\operatorname {dist}(a_{m_1},a_n)+\ell $ . So, the type $q(x)$ is completely determined by the pair $(\ell ,m_1)$ . There are at most $\aleph _0$ such types.

3. (3) If there is no pair $(k,n)\in \mathbb {N}^2$ such that $D_k(x,a_n)\in q(x)$ , then $q(x)$ is determined by the partial type $\pi (x)=\{\neg D_k(x,a_n): n,k<\omega \}$ . There is only one type with this condition.

This shows that $|S_1(A)|\leq \aleph _0+\aleph _0+1=\aleph _0$ , and since A is an arbitrary countable set we conclude that $\mathcal {T}_\infty $ is $\omega $ -stable. Moreover, since each vertex in $\mathcal {M}$ has infinite degree, the only algebraic type is given by case (1), and so every algebraic element lies in the unique path between two elements from A. This shows that $\operatorname {acl}(A)=\operatorname {conv}(A)$ .

Fact A.6. Let $p(\overline {x})$ be a complete type with parameters over A. Then $\operatorname {SU}(p(\overline {x})\leq \operatorname {RM}(p(\overline {x}))$ .

Proof. By quantifier elimination, $\operatorname {tp}(b/A)$ is completely determined by the formulas of the form $D_k(x,a)$ that hold for b, and so the Morley rank of $\operatorname {tp}(b/A)$ is equal to the Morley rank of a formula of the form

$$\begin{align*}\theta(x)=\bigwedge_{i=1}^m D_{k_i}(x,a_i)\wedge \bigwedge_{j=1}^n \neg D_{\ell_j}(x,a^{\prime}_j)\in\operatorname{tp}(b/A)\end{align*}$$

for some elements $a_1,\ldots ,a_m,a_1',\ldots ,a_n'\in A$ .

It follows by induction that the formula $D_k(x,a)$ has SU-rank at least k, which also implies that $\operatorname {SU}(\neg D_k(x,a)) \geq \omega $ . Now we will show that the formula $D_k(x,a)$ has Morley rank at most k. For a contradiction, let m be minimal such that $RM(D_m(x,a)) \geq m+1$ and let S be a definable subset of $D_m(x,a)$ with parameters $a,a_1,\ldots ,a_n$ such that $\operatorname {RM}(S)=m$ . Let $A=\operatorname {conv}(a,a_1,\ldots ,a_n)$ , and let $b\in S$ such that $\operatorname {RM}(S)=\operatorname {RM}(\operatorname {tp}(b/A))=m$ . By Proposition 2.7, we know that there is a unique element $c\in A$ and an integer $\ell $ such that $\operatorname {dist}(b,A)=\operatorname {dist}(b,c)=\ell $ and $P(b,c)\cap A=\{c\}$ . By construction we have $\ell \leq m$ , and given that c is definable over $a,a_1,\ldots ,a_m$ and the formula $D_\ell (x,c)$ belongs to $\operatorname {tp}(b/A)$ , we have by minimality of m that $m\leq \ell $ . From this we conclude that $c=a$ , and so the definable set S is contained in the set defined by a formula of the form

$$\begin{align*}D_m(x,a)\wedge \bigwedge_{d\in \mathcal{N}(a)\cap A}\neg D_{m-1}(x,d),\end{align*}$$

where $\mathcal {N}(a)$ is the set of vertices in M that are adjacent to a. With these conditions, we will have that

$$ \begin{align*} m+1&=\operatorname{RM}(D_m(x,a))=\max\{\operatorname{RM}(S),\operatorname{RM}(D_m(x,a)\wedge \neg S)\}\\ &=\max\left\{\operatorname{RM}(S),\max_{d\in\mathcal{N}(a)\cap A}\{\operatorname{RM}\left(D_m(x,a)\wedge D_{m-1}(x,d)\right)\}\right\}\leq m, \end{align*} $$

thus obtaining a contradiction.

Let now b be an element in M and let A be a finite algebraically closed subset of M. By Fact A.6 we already have $\operatorname {SU}(b/A)\leq \operatorname {RM}(b/A)$ , so it suffices to show that either $\operatorname {dist}(b,A)=k< \omega $ and we have $\operatorname {RM}(b/A)\leq k\leq \operatorname {SU}(b/A)$ , or $\operatorname {dist}(b,A)=\infty $ and $\operatorname {RM}(b/A)\leq \omega \leq \operatorname {SU}(b/A)$ . If $\operatorname {dist}(b,A)=\infty $ , then $\operatorname {tp}(b/A)=\{\neg D_k(x,a):a\in A, k<\omega \}$ , which implies that $\operatorname {SU}(b/A)\geq \omega $ because each finite conjunction of these formulas has SU-rank at least $\omega $ . So we obtain

$$\begin{align*}\omega\leq \operatorname{SU}(b/A)\leq \operatorname{RM}(b/A)\leq \operatorname{RM}(\neg D_\ell(x,a))\leq \omega.\end{align*}$$

On the other hand, if $\operatorname {dist}(b/A)=k<\omega $ , we have $\operatorname {RM}(b/A)\leq \operatorname {RM}(D_k(x,a))\leq k\leq \operatorname {SU}(b/A).$

Proof. First, notice that $c\not \in \operatorname {conv}(A\overline {b})=\operatorname {acl}(A\overline {b})$ , so we must have $\operatorname {RM}(\overline {b},c/A)>\operatorname {RM}(\overline {b}/A)$ , so $\operatorname {RM}(\overline {b},c/A)\geq \alpha \oplus 1$ .

Suppose now that $\varphi (\overline {x})$ is a minimal formula for $\operatorname {tp}(\overline {b}/A)$ , that is, a formula with parameters from A such that $\operatorname {RM}(\overline {b}/A)=\operatorname {RM}(\varphi (\overline {x}))=\alpha $ . Since $\varphi (\overline {x})$ is a formula with parameters in a finite subset $A_0\subseteq A$ , we may assume that $\varphi (\overline {x})$ implies all the formulas of the form $D_r(x_i,x_j)$ or $D_s(x_i,a)$ (with $a\in A_0$ ) that belong to $\operatorname {tp}(b_1,\ldots ,b_n/A_0)$ .

Since $D(c/A\overline {b})=1$ , there is an element $e\in \operatorname {conv}(A\overline {b})$ such that $\operatorname {dist}(e,c)=1$ . Moreover, since the convex closure is the union of the paths between vertices at finite distance, we may assume that there are elements $e_1,e_2$ from $A\overline {b}$ such that $e\in P(e_1,e_2)$ . Without loss of generality, we may suppose that $e_1=b_1$ and $e_2=a\in A$ as all the other cases can be analyzed in a similar fashion. Finally, let us fix the distances $\operatorname {dist}(b_1,e)=k_1$ and $\operatorname {dist}(e,a)=k_2$ .

Consider the formula $\zeta (\overline {x},y):=\varphi (\overline {x})\wedge D_{k_1+1}(y,x_1)\wedge D_{k_2+1}(y,a)$ . Note that $\mathcal {M}\models \zeta (\overline {b},c)$ , so to prove $\operatorname {RM}(\overline {b},c/A)\leq \alpha \oplus 1$ it is enough to show that $\operatorname {RM}(\zeta (\overline {x},y))\leq \alpha \oplus 1$ .

For a contradiction, suppose that $\operatorname {RM}(\zeta (\overline {x},y))\geq \alpha \oplus 2$ . By definition of the Morley rank, there is an A-definable set $S(\overline {x},y)$ implying $\zeta (\overline {x},y)$ such that $\operatorname {RM}(S(\overline {x},y))=\alpha \oplus 1$ . Let $\overline {b}',c'$ be a realization of $S(\overline {x},y)$ such that $\operatorname {RM}(\overline {b}',c')=\alpha \oplus 1$ .

By quantifier elimination and the fact that the Morley rank of a finite disjunction is equal to the maximum of the Morley rank of its disjunctives, we may assume that $S(\overline {x},y)$ is equivalent to the conjunction of $\zeta (\overline {x},y)$ with formulas of the form $D_{r_i}(y,x_i), D_{s_i}(y,a_i), \neg D_{p_j}(y,x_{j})$ or $\neg D_{q_j}(y,a_j)$ for some non-negative integers $r_i,s_i,p_j,q_j$ and vertices $a_i,a_j$ from A.

Let $e'$ be the unique element in $P(b_1',a)$ at distance $1$ from $c'$ . If $\overline {b}',c'\models D_{r_i}(y,x_i)$ , then $\operatorname {dist}(c',b_i')=r_i$ and we have either $\operatorname {dist}(e',b_i')=r_i+1$ or $\operatorname {dist}(e',b_i')=r_i-1$ . However, $\operatorname {dist}(e',b_i')=r_i+1$ would imply that $c'\in P(e',b_i')\subseteq P(b_1',b_i')$ , which is impossible because that would imply $c'\in \operatorname {acl}(A\overline {b}')$ and $\operatorname {RM}(\overline {b}',c'/A)=\operatorname {RM}(\overline {b}'/A)=\alpha $ . Hence, $D_{r_i-1}(e',b_i')$ holds, and there is a path with length less than $r_i$ connecting $b_i'$ with $P(b_1',a)$ .

Let $h'$ be the element in $P(b_1',a)$ that is closest to $b_i'$ and put $t=\operatorname {dist}(b_i',P(b_1',a))=\operatorname {dist}(b_i',h')$ , $r=\operatorname {dist}(h',e')$ . Notice that $r+t=\operatorname {dist}(b_i',e')=r_i-1$ , and we have two cases:

• • If $h'\in P(b_1',e')$ , we have $\operatorname {dist}(b_1',b_i')=t+k_1-r$ and $\operatorname {dist}(b_i',a)=t+k_2+r$ .

• • If $h'\in P(e',a)$ , we have $\operatorname {dist}(b_i',b_1')=t+k_1+r$ and $\operatorname {dist}(b_i',a)=t+k_2-r$ .

Therefore, by checking all the $k_1+k_2=\operatorname {dist}(b_1,a)$ possibilities we have that $D_{r_i}(y,x_j)$ is implied by the disjunction

$$\begin{align*}\bigvee_{r+t=r_i-1}(D_{t+k_1-r}(x_1,x_i)\wedge D_{k_2+r+t}(x_i,a))\vee ( D_{t+r+k_1}(x_i,x_1)\wedge D_{t+k_2-r}(x_i,a)),\end{align*}$$

which itself is implied by the formula $\zeta (\overline {x},y)$ . So, $\overline {b},c\models D_{r_i}(y,x_i)$ . A similar case occurs if $\overline {b}',c'$ satisfies a formula of the form $D_{s_i}(y,a)$ for some $a\in A_0$ .

Suppose now that $\overline {b}',c'\models \neg D_{p_j}(y,x_j)$ . If $\overline {b},c\models D_{p_j}(y,x_j)$ then $\operatorname {dist}(c,b_j)=p_j$ , and as in the previous paragraph this implies that either $c\in P(b_1,b_j)$ (which is not possible since $c\not \in \operatorname {acl}(A\overline {b})$ ) or the formula $D_{p_j}(y,x_j)$ is implied by $\zeta (\overline {x},y)$ , which is clearly a contradiction because $\overline {b}',c'\models \zeta (\overline {x},y)$ . Therefore, $\overline {b},c$ also satisfies the formula $\neg D_{p_j}(y,x_j)$ . A similar situation will occur for a formula of the form $\neg D_{q_j}(y,a_j)$ for some element $q_j\in A_0$ .

We have shown that $\zeta (\overline {x},y)$ implies each of the formulas in the conjunction defining $S(\overline {x},y)$ , so $\zeta (\mathcal {M}^{n+1})=S(\mathcal {M}^{n+1})$ , which is a contradiction because $\operatorname {RM}(\zeta (\overline {x},y))\geq \alpha \oplus 2$ and $\operatorname {RM}(S(\overline {x},y)=\alpha \oplus 1$ . This contradiction shows that $\operatorname {RM}(\zeta (\overline {x},y))\leq \alpha \oplus 1$ and so, $\operatorname {RM}(\overline {b},c)\leq \alpha \oplus 1$ .

Proof. We will show the result by induction on n, and notice that the case $n=1$ is precisely Proposition A.8. Suppose the result holds for every tuple $\overline {b}$ of length n, and so there is an ordinal $\alpha $ such that

$$\begin{align*}\alpha=\operatorname{SU}(\overline{b}/A)=\operatorname{RM}(\overline{b}/A)=\bigoplus_{i=1}^n D(b_i/A\overline{b}_{<i}).\end{align*}$$

For the inductive step, it suffices to show that given an arbitrary element $c\in M$ we have $\operatorname {SU}(\overline {b},c/A)=\operatorname {RM}(\overline {b},c/A)=\alpha \oplus D(c/A\overline {b}).$ By Fact A.6 we know already that $\operatorname {SU}(\overline {b},c/A)\leq \operatorname {RM}(\overline {b},c/A)$ .

Suppose now that $D(c/A\overline {b})=m$ for some $m\in \mathbb {N}$ . Hence, there is an element ${e\in \operatorname {conv}(A\overline {b})}$ and elements $c_0=e,c_1,\ldots ,c_m=c$ that form a path from e to c of length m. Note that for every $i=1,\ldots ,m$ we have $c_{i-1}\in \operatorname {acl}(A,\overline {b},c_i)$ and $D(c_i/A\overline {b}c_{<i})=1$ . So, by Lemma A.10, we obtain

$$ \begin{align*} \operatorname{RM}(\overline{b},c/A)&=\operatorname{RM}(\overline{b},c_1,\ldots,c_{m-1},c/A)=\operatorname{RM}(\overline{b},c_1,\ldots,c_{m-1}/A)\oplus 1\\ &=\operatorname{RM}(\overline{b},c_1,\ldots,c_{m-2}/A)\oplus 2=\cdots = \operatorname{RM}(\overline{b},c_1/A)\oplus (m-1)\\ &=\operatorname{RM}(\overline{b}/A)\oplus m=\alpha\oplus m=\alpha\oplus D(c/A\overline{b}). \end{align*} $$

Suppose now that $D(c/A\overline {b})=\omega $ . By quantifier elimination $\operatorname {tp}(\overline {b},c/A)$ is completely characterized by $\operatorname {tp}(\overline {b}/A)\cup \{\neg D_k(y,x_i)\wedge \neg D_\ell (y,a): a\in A; k,\ell <\omega , 1\leq i\leq n\}$ . Let $\varphi (\overline {x})$ be a minimal formula for $\operatorname {tp}(\overline {b}/A)$ defined over a finite set of parameters $A_0\subseteq A$ , and assume that the formula is a conjunction including all instances of possible distances between elements in the tuple $\overline {b}$ and elements in $A_0$ , when they are finite.

Towards a contradiction, assume that $\operatorname {RM}(\overline {b},c/A)\geq \alpha \oplus (\omega +1)$ . Note that the formula $\theta (\overline {x},y):=\varphi (\overline {x})$ in $\operatorname {tp}(\overline {b},c/A)$ must have Morley rank at least $\alpha \oplus (\omega +1)$ . By definition of Morley rank, there must be a formula $S(\overline {x},y)$ implying $\varphi (\overline {x})$ that has Morley rank $\alpha \oplus \omega $ . Hence $S(\overline {x},y)\not \in \operatorname {tp}(\overline {b},c/A)$ , and so it must imply a formula of the form $\varphi (\overline {x})\wedge D_k(y,x_i)$ or $\varphi (\overline {x})\wedge D_\ell (y,a)$ . In either case, if $\overline {b}',c'$ is a generic tuple for $S(\overline {x},y)$ , $\operatorname {RM}(\overline {b}'/A)\leq \operatorname {RM}(\varphi (\overline {x}))=\alpha $ and $D(c'/A\overline {b}')$ would be finite. So, by the previous case, $\operatorname {RM}(S(\overline {x},y))=\operatorname {RM}(\overline {b}',c'/A)\leq \operatorname {RM}(\overline {b}'/A)\oplus D(c'/A\overline {b}')\leq \alpha \oplus D(c'/A\overline {b}')<\alpha \oplus \omega $ , which is a contradiction. This proves that when $D(c/A\overline {b})=\omega $ we have $\operatorname {RM}(\overline {b},c/A)\leq \alpha \oplus \omega $ .

Proof. Suppose first that there is a path between two elements $a\in A$ and $b\in B$ that does not pass through any vertex in C. In particular, $a\not \in C$ . If ${1\leq \operatorname {dist}(a,C)<\infty }$ , then there is an element $e\in P(a,b)\setminus C$ that lies in the unique path between an element $c\in C$ and b. Thus, $\operatorname {RM}(a/BC)=\operatorname {dist}(a,e)<\operatorname {dist}(a,C)=\operatorname {RM}(a/C)$ . On the other hand, if $\operatorname {dist}(a,C)=\infty $ , then $\operatorname {RM}(a/BC)\leq \operatorname {dist}(a,b)<\omega =\operatorname {RM}(a/C)$ . In any case, we conclude that , which implies .

Now, suppose that every path between a vertex in A and a vertex in B passes through C. By finite character, to conclude that it suffices to show that whenever $a_1,\ldots ,a_n$ is a finite tuple of elements in A, then . We prove this by induction on n. For $n=1$ , given an arbitrary element $a_1\in A$ , we know by the assumption that the distance between $a_1$ and $BC$ is equal to the distance between $a_1$ and C. By Theorem A.11, this implies $\operatorname {RM}(a_1/BC)=\operatorname {RM}(a_1/C)$ , and so for every element $a_1\in A$ .

Suppose now that $a_1,\ldots ,a_n,a_{n+1}\in A$ . By induction hypothesis we have that , and by symmetry this implies . Since every path between $a_{n+1}$ and B passes through $Ca_1,\ldots ,a_n$ (because it passes through C), we have by the base case that , and by symmetry, . Transitivity implies , and symmetry again yields , as desired.