Fact 2.2.

1. 1. If X is a compact definable topological space, then X is definably compact.

2. 2. If $X, Y$ are definably compact, then $X \times Y$ is definably compact.

3. 3. If $f : X \to Y$ is definable and continuous, and X is definably compact, then the image $f(X) \subseteq Y$ is definably compact.

4. 4. If X is a Hausdorff definable topological space and $Y \subseteq X$ is definably compact, then Y is closed.

5. 5. If X is definably compact and $Y \subseteq X$ is closed and definable, then Y is definably compact.

6. 6. If X is a definable topological space and $Y_1, Y_2 \subseteq X$ are definably compact, then $Y_1 \cup Y_2$ is definably compact.

Proof For $t \in M {\setminus} \{0\}$ , let $O_t$ be the n-dimensional ball ${\cal B}(0,v(t))^n$ . Each $O_t$ is clopen in $M^n$ . Therefore $\{X {\setminus} O_t : t \in M {\setminus} \{0\}\}$ is a downwards-directed definable family of closed subsets of X, with empty intersection. By definable compactness, there is some t such that $X {\setminus} O_t = \emptyset $ , or equivalently, $X \subseteq O_t$ . Then X is bounded.

Closedness follows similarly, or by Fact 2.2(4).

Proof Equivalently, if $\{Y_t\}$ is a downwards-directed definable family of non-empty, closed, bounded sets, then $\bigcap _t Y_t \ne \emptyset $ . This claim can be expressed as a countable conjunction of $\mathcal {L}$ -sentences. (We need infinitely many sentences because there is no bound on the complexity of the definable family $\{Y_t\}$ .) As a countable conjunction of $\mathcal {L}$ -sentences, the claim holds in M if and only if it holds in ${\mathbb Q}_p$ . Therefore, we may assume that $M = {\mathbb Q}_p$ . In this case, the set X will be compact, and hence definably compact by Fact 2.2(1).

Proof For any $\bar {x} = (x_1,\ldots ,x_n) \in M^n$ and $\gamma \in \Gamma _M$ , let ${\cal B}(\bar {x},\gamma )$ denote the ball of valuative radius $\gamma $ around $\bar {x}$ , i.e., $\prod _{i = 1}^n {\cal B}(x_i,\gamma )$ .

Let $W_{\gamma }$ be the set of $x \in U$ such that ${\cal B}(x,\gamma ) \subseteq U$ and $\bar {0} \in {\cal B}(x,-\gamma )$ . We claim that the family $W_{\gamma }$ is a $\Gamma $ -exhaustion.

First of all, for all $x'$ sufficiently close to x, we have ${\cal B}(x,\gamma ) = {\cal B}(x',\gamma )$ and ${\cal B}(x,-\gamma ) = {\cal B}(x',-\gamma )$ , and so $x \in W_\gamma \iff x' \in W_\gamma $ . Therefore $W_\gamma $ is clopen. Additionally,

$$ \begin{align*} x \in W_\gamma \implies \bar{0} \in {\cal B}(x,-\gamma) \iff x \in {\cal B}(\bar{0},-\gamma). \end{align*} $$

Therefore $W_\gamma $ is bounded. By Lemma 2.5, $W_\gamma $ is definably compact.

If $\gamma ' \ge \gamma $ , then ${\cal B}(x,\gamma ') \subseteq {\cal B}(x,\gamma )$ and ${\cal B}(x,-\gamma ') \supseteq {\cal B}(x,-\gamma )$ . Therefore

$$ \begin{align*} x \in W_\gamma \implies x \in W_{\gamma'}, \end{align*} $$

and the family $\{W_\gamma \}$ is monotone.

Lastly, if $x \in U$ , then for sufficiently large $\gamma $ , we have ${\cal B}(x,\gamma ) \subseteq U$ , because U is open. Also, $\bar {0} \in {\cal B}(x,-\gamma )$ for sufficiently large $\gamma $ . Thus $x \in W_\gamma $ for all sufficiently large $\gamma $ . This shows $U = \bigcup _\gamma W_\gamma $ .

Proof Cover X with finitely many open sets $U_i$ homeomorphic to open subsets of $M^n$ . For each i, let $\{W_{i,\gamma }\}_{\gamma \in \Gamma _M}$ be a $\Gamma $ -exhaustion of $U_i$ . Let $V_\gamma = \bigcup _i W_{i,\gamma }$ . Then the family $\{V_\gamma \}$ is a $\Gamma $ -exhaustion of X.

Proof

1. 1. If $Y \subseteq W_\gamma $ , then Y is contained in the definably compact set $W_\gamma $ . Conversely, suppose Y is bounded, witnessed by a definably compact set $Z \subseteq X$ with $Y \subseteq Z$ . The filtered intersection

   $$ \begin{align*} \bigcap_\gamma (Z {\setminus} W_\gamma) \end{align*} $$
   is empty, so there is some $\gamma $ such that $W_\gamma \supseteq Z \supseteq Y$ .

2. 2. If Y is definably compact, then Y is closed (Fact 2.2(4)), and Y is bounded because $Y \subseteq Y$ . Conversely, suppose that Y is closed and bounded. Then Y is a definable closed subset of a definably compact set, so Y is definably compact by Fact 2.2(5).

3. 3. If $\overline {Y}$ is definably compact, then Y is bounded because $Y \subseteq \overline {Y}$ . Conversely, suppose that Y is bounded. Then $Y \subseteq Z$ for some definably compact set $Z \subseteq X$ . The closure $\overline {Y}$ is a definable closed subset of Z, so $\overline {Y}$ is definably compact by Fact 2.2(5).

Proof An exercise using the fact that any definable function $M \to M^n$ is continuous off a finite set.

Proof (1) $\Rightarrow $ (2): The set of cluster points is the intersection

$$ \begin{align*} \bigcap_{\gamma \in \Gamma_M} \overline{f({\cal B}(0,\gamma) \cap D)}. \end{align*} $$

This is non-empty by definable compactness of Y.

(2) $\Rightarrow $ (3) is trivial, and (3) $\Rightarrow $ (4) follows by rescaling.

(4) $\Rightarrow $ (5): By definable Skolem functions, there is some definable function $f : M {\setminus} \{0\} \to Y$ such that $f(x) \in Z_{v(x)}$ for all $x \in M {\setminus} \{0\}$ . By Lemma 2.14, there is some $\delta \in \Gamma _M$ such that f is continuous on ${\cal B}(0,\delta ) {\setminus} \{0\}$ . By (4), f has a cluster point $a \in Y$ . Then $a \in \bigcap _{\gamma } Z_\gamma $ . Otherwise, take $\gamma $ large enough that $a \notin Z_\gamma $ . Because a is a cluster point and $Z_\gamma $ is closed in Y, there is some $x \ne 0$ such that $v(x) \ge \gamma $ and $f(x) \notin Z_\gamma $ . By choice of f, $f(x) \in Z_{v(x)} \subseteq Z_\gamma $ , a contradiction.

(5) $\Rightarrow $ (1): We first claim that Y is closed. Take $p \in \overline {Y}$ . Because X is a definable manifold, we can identify a neighborhood of p in X with the closed ball $R(M)^n$ in $M^n$ . For $\gamma \ge 0$ , let $B_\gamma $ be the closed ball of radius $\gamma $ around p. For $\gamma \le 0$ let $B_\gamma = B_0$ . Then $B_\gamma \cap Y$ is a non-empty closed subset of Y for any $\gamma $ , because $p \in \overline {Y}$ . By (4), the intersection $\bigcap _{\gamma } (B_\gamma \cap Y)$ is non-empty, and so $p \in Y$ . Therefore Y is closed.

Similarly, Y is bounded. Take a $\Gamma $ -exhaustion $\{U_\gamma \}_{\gamma \in \Gamma _M}$ of the definable manifold X. If Y is unbounded, then $Y {\setminus} U_\gamma $ is a closed non-empty subset of Y for each $\gamma $ . Applying (5) to the family of sets $Y {\setminus} U_\gamma $ , we see that $Y \not \subseteq \bigcup _{\gamma } U_\gamma = X$ , a contradiction. Therefore Y is closed and bounded. By Proposition 2.10(2), Y is definably compact.

Fact 2.20. If $a'\in N\backslash M$ is infinitesimally close to $ a\in M$ over M, then there is a coset $C\subseteq N\backslash \{0\}$ of $\bigcap _{n\geq 1}P_n(N)$ such that $\operatorname {\mathrm {tp}}(a'/M)$ is determined by the partial type

$$ \begin{align*} \{v(x-a)>\gamma \mid \gamma\in \Gamma_M\}\cup \{x-a\in C\}, \end{align*} $$

and $\operatorname {\mathrm {tp}}(a'/M)$ is definable over M.

Proof We may assume $M = {\mathbb Q}_p$ , in which case the lemma is an easy exercise using spherical completeness of ${\mathbb Q}_p$ .

Proof Take $N \succ M$ containing a realization b of p. Because p is one-dimensional, there is some singleton $c \in N$ such that $\operatorname {\mathrm {dcl}}(Mb) = \operatorname {\mathrm {dcl}}(Mc)$ . (In fact, we can take c to be a coordinate of the tuple b.) Replacing c with $1/c$ if necessary, we may assume that $v(c) \ge 0$ . Then $\operatorname {\mathrm {tp}}(c/M)$ is definable and one-dimensional. Let $\mathcal {C}$ be the family of M-definable balls which contain c. Then $\mathcal {C}$ is definable, because $\operatorname {\mathrm {tp}}(c/M)$ is definable. Moreover, $\mathcal {C}$ satisfies the four conditions of Lemma 2.21:

1. 1. $\mathcal {C}$ is non-empty, because it contains the ball $R(M)$ of radius 0.

2. 2. $\mathcal {C}$ is a chain, because any two balls which intersect are comparable, and $\mathcal {C}$ cannot contain two disjoint balls.

3. 3. $\mathcal {C}$ is upwards-closed, trivially.

4. 4. $\mathcal {C}$ has no least element. Otherwise, if B were the smallest M-definable ball containing c, then we could write B as a disjoint union of smaller balls $B = B_1 \cup \cdots \cup B_p$ , and one of the $B_i$ would belong to $\mathcal {C}$ .

By Lemma 2.21, $\mathcal {C}$ is the class of balls around some point d. So there is some $d \in M$ such that c is contained in every M-definable ball around d. Therefore, c is infinitesimally close to d over M. Take $a = c - d$ .

Proof Let N be an $\aleph _1$ -saturated elementary extension of M, and let $\mathbb M$ be a monster model extending N. Let $p^N$ be the heir of p over N. We first show that $p^N$ specializes to a point in $Y(N)$ . Take $c \in Y(\mathbb M)$ realizing $p^N$ . By Lemma 2.22, we can write c as $g(a)$ for some N-definable function $g : \mathbb M \to Y(\mathbb M)$ and some $a \in \mathbb M$ infinitesimally close to 0 over N. Because N is $\aleph _1$ -saturated, there is some $u \in N$ such that $a/u \in P_n(\mathbb M)$ for all n. Replacing a with $a/u$ , we may assume that $a \in P_n(\mathbb M)$ for all n. For each n, let $S_n \subseteq Y(N)$ be the definable set of cluster points of $g \restriction P_{n}(N)$ . Each $S_n$ is closed, and non-empty by Lemma 2.15(2). The intersection $\bigcap _n S_n$ is filtered, and therefore non-empty by $\aleph _1$ -saturation. Take $b \in \bigcap _n S_n$ . Let $\Sigma (x)$ be the partial type saying that x is infinitesimally close to 0, $g(x)$ is infinitesimally close to b, and $x \in P_n$ for all n. Then $\Sigma (x)$ is finitely satisfiable, by choice of b. Take $a' \in \mathbb M$ realizing $\Sigma (x)$ . By Fact 2.20, $\operatorname {\mathrm {tp}}(a'/N) = \operatorname {\mathrm {tp}}(a/N)$ . Therefore a satisfies $\Sigma (x)$ , and so $g(a)$ is infinitesimally close to b. It follows that $p^N(x)$ specializes to b.

Let Z be the set of $b \in Y(N)$ such that $p^N$ specializes to b. The set Z is M-definable, because $p^N$ is definable over M. The above argument shows $|Z|> 0$ . On the other hand, $|Z| \le 1$ because $Y(N)$ is Hausdorff. Therefore Z is a singleton $\{b\}$ , and the element b lies in $Y(M)$ . Then p specializes to b.

Proof One direction is Lemma 2.23. Conversely, suppose every one-dimensional definable type in Y specializes to a point. We claim that Y is definably compact. We use criterion (3) of Lemma 2.15. Let $f : R(M) {\setminus} \{0\} \to Y$ be a definable continuous function. Take a monster model $\mathbb M \succ M$ and a non-zero $a \in \mathbb M$ infinitesimally close to 0 over M. Let $b = f(a)$ . By Fact 2.20, $\operatorname {\mathrm {tp}}(a/M)$ is definable. Therefore $\operatorname {\mathrm {tp}}(b/M)$ is one-dimensional and definable. Then $\operatorname {\mathrm {tp}}(b/M)$ specializes to a point $c \in Y(M)$ . We claim that c is a cluster point of f. For any M-definable neighborhoods $U_1 \ni 0$ and $U_2 \ni c$ , we have $(a,f(a)) = (a,b) \in U_1 \times U_2$ . As $M \prec \mathbb M$ , there must be some $(a',f(a')) \in U_1(M) \times U_2(M)$ . This shows that c is a cluster point of f.

Proof By Lemma 2.22, we have $b = f(a)$ for some M-definable function $f : \mathbb M \to X$ and some $a \in \mathbb M$ infinitesimally close to 0 over M. By Lemma 2.14, f is continuous on ${\cal B}(0,\gamma _0)$ for some sufficiently large $\gamma _0 \in \Gamma _M$ ; note that $v(a)> \gamma _0$ . We claim that a is infinitesimally close to 0 over N. Otherwise, there is some $\gamma \in \Gamma _N$ such that $v(a) < \gamma $ . Let A be the definable set of $x \in \mathbb M$ such that $\gamma _0 < v(x) < \gamma $ ; note that $a \in A$ . The set A is definably compact and N-definable. Also, f is N-definable and continuous on A. Therefore, the image $f(A)$ is N-definable, and definably compact. By Proposition 2.10, there is some $t \in \Gamma _N$ such that $f(A) \subseteq O_t$ . Then $b = f(a) \in f(A) \subseteq O_t(\mathbb M)$ , contradicting the assumptions.

This shows that a is infinitesimally close to 0 over N. By Fact 2.20, $\operatorname {\mathrm {tp}}(a/N)$ is the heir of $\operatorname {\mathrm {tp}}(a/M)$ , implying that $\operatorname {\mathrm {tp}}(b/N) = \operatorname {\mathrm {tp}}(f(a)/N)$ is the heir of $\operatorname {\mathrm {tp}}(f(a)/M) = p$ .

Proof By Proposition 2.8, the group G admits a $\Gamma $ -exhaustion $\{W_t : t \in \Gamma _M\}$ . Replacing $\{W_t\}$ with $\{W_{t + \gamma }\}$ , we may assume that $W_0$ is non-empty. Replacing $\{W_t\}$ with $\{a \cdot W_t\}$ , we may assume that $\operatorname {\mathrm {id}}_G \in W_0$ .

Because G is a definable manifold, there is some definable neighborhood basis $\{N_t : t \in \Gamma _M,\ t < 0\}$ such that each $N_t$ is clopen, and $N_t$ depends monotonically on t. Define

$$ \begin{align*} B_t = \begin{cases} W_t & t \ge 0, \\ W_0 \cap N_t & t < 0. \end{cases} \end{align*} $$

Then $\{B_t : t \in \Gamma _M\}$ is a definable neighborhood basis and a $\Gamma $ -exhaustion. Lastly, define $O_t = B_t \cap B_t^{-1}$ . Then $\{O_t : t \in \Gamma _M\}$ has all the desired properties.

Proof (1) is by continuity. For (2), note that the set $O_t \cdot O_t$ is an image of the definably compact space $O_t \times O_t$ under the definable continuous map $(x,y) \mapsto x \cdot y$ . Therefore $O_t \cdot O_t$ is definably compact. Then $t''$ exists by Proposition 2.10.

Proof Define $S_\delta = \{(a,b) \in O_\delta \times O_t : b^{-1} a b \notin O_\epsilon \}$ . Suppose for the sake of contradiction that $S_\delta \ne \emptyset $ for all $\delta $ . The family $S_\delta $ is definable, and depends monotonically on $\delta $ . Each set $S_\delta $ is closed, because $O_\epsilon , O_\delta $ , and $O_t$ are clopen. By definable compactness of $O_{\delta } \times O_t$ , the intersection $\bigcap _\delta S_\delta $ is non-empty. Therefore there are $a, b \in G$ such that

1. 1. $a \in O_\delta $ for all $\delta $ .

2. 2. $b \in O_t$ .

3. 3. $b^{-1} a b \notin O_\epsilon $ .

The first point implies $a = \operatorname {\mathrm {id}}_G$ , which then implies $b^{-1} a b = \operatorname {\mathrm {id}}_G \in O_\epsilon $ , a contradiction.

Proof Let $(F,\{Y_i\},\{X_i\},I)$ be a bad gap configuration. Replacing M with the Shelah expansion $M^{Sh}$ , we may assume that the bad gap configuration is definable. Passing to an elementary extension and naming parameters, we may assume that M is $\aleph _1$ -saturated and the bad gap configuration is $\emptyset $ -definable.

Note that $FY_i = Y_iF$ is a subgroup of G, and that the index of $Y_i$ if $FY_i$ is finite, no more than $|F|$ . Let $D_i$ be the definable set $X_i \cap (I \diamond X_{i-1} {\setminus} F Y_{i-1})$ . By assumption, $D_i/X_{i-1}$ is infinite.

Claim 1. Suppose $a_i, a^{\prime }_i \in D_i$ for $i = 1, \ldots , n$ , and suppose

(1) $$ \begin{align} Y_n a_n a_{n-1} \cdots a_1 = Y_n a^{\prime}_n a^{\prime}_{n-1} \cdots a^{\prime}_1. \end{align} $$

Then $a_n F Y_{n-1} = a^{\prime }_n F Y_{n-1}$ . If moreover $a_n Y_{n-1} = a^{\prime }_n Y_{n-1}$ , then

(2) $$ \begin{align} Y_{n-1} a_{n-1} \cdots a_1 = Y_{n-1} a^{\prime}_{n-1} \cdots a^{\prime}_1. \end{align} $$

Proof Note that $a_i, a_i' \in X_i$ . Equation (1) implies that

(3) $$ \begin{align} a^{\prime}_n a^{\prime}_{n-1} \cdots a^{\prime}_1 = \epsilon a_n a_{n-1} \cdots a_1 = a_n \epsilon^{a_n} a_{n-1} \cdots a_1, \end{align} $$

for some $\epsilon \in Y_n$ . Then $\epsilon ^{a_n} \in Y_n^{X_n} \subseteq Y_{n-1} \subseteq X_{n-1}$ . For $i < n$ , we have $a_i, a^{\prime }_i \in X_i \subseteq X_{n-1}$ . Therefore (3) implies that $a^{\prime }_n X_{n-1} = a_n X_{n-1}$ . Both $a^{\prime }_n$ and $a_n$ are in $I \diamond X_{n-1} {\setminus} F Y_{n-1}$ , so by Remark 4.4 we have $a^{\prime }_n F Y_{n-1} = a_n F Y_{n-1}$ as desired. Now suppose that $a_n Y_{n-1} = a^{\prime }_n Y_{n-1}$ . Then $a^{\prime }_n = a_n \delta $ for some $\delta \in Y_{n-1}$ . Then Equation (3) implies

$$ \begin{align*} a_n \epsilon^{a_n} a_{n-1} \cdots a_1 &= a_n \delta a^{\prime}_{n-1} a^{\prime}_{n-2} \cdots a^{\prime}_1, \\ \epsilon^{a_n} a_{n-1} \cdots a_1 &= \delta a^{\prime}_{n-1} a^{\prime}_{n-2} \cdots a^{\prime}_1. \end{align*} $$

Both $\epsilon ^{a_n}$ and $\delta $ are in $Y_{n-1}$ , so Equation (2) holds.

For any $n \in \mathbb N$ , we claim that $\operatorname {\mathrm {dp-rk}}(G) \ge n$ . By assumption, the interpretable set $D_i/X_{i-1}$ is infinite. The interpretable set $D_i/Y_{i-1}$ is even bigger, because $Y_{i-1} \subseteq X_{i-1}$ . By the properties of dp-rank in Section 3, $\prod _{i = 1}^n D_i/Y_{i-1}$ has dp-rank at least n. Take a tuple $\bar {b} \in \prod _{i = 1}^n D_i/Y_{i-1}$ such that $\operatorname {\mathrm {dp-rk}}(\bar {b}/\emptyset ) \ge n$ . Each $b_i$ is a coset $a_iY_{i-1}$ for some $a_i \in D_i$ . Let $c = a_na_{n-1} \cdots a_1 \in G$ .

Proof Let S be the set of $(a^{\prime }_1,\ldots ,a^{\prime }_n) \in \prod _j D_j$ such that

• • $a^{\prime }_n a^{\prime }_{n-1} \cdots a^{\prime }_1 = c$ .

• • $a^{\prime }_j Y_{j-1} = b_j = a_j Y_{j-1}$ , for $j> i$ .

Then $(a_1,\ldots ,a_n) \in S$ and S is definable over $c, b_{i+1}, \ldots , b_n$ . If $(a^{\prime }_1,\ldots ,a^{\prime }_n) \in S$ , then

$$ \begin{align*} Y_n a^{\prime}_n a^{\prime}_{n-1} \cdots a^{\prime}_1 = Y_n c = Y_n a_n a_{n-1} \cdots a_1. \end{align*} $$

By Claim 1 applied $(n-i+1)$ times, we see that $a_i F Y_{i-1} = a^{\prime }_i F Y_{i-1}$ . We have shown

$$ \begin{align*} \{a^{\prime}_i F Y_{i-1} : (a^{\prime}_1,\ldots,a^{\prime}_n) \in S\} = \{a_i F Y_{i-1}\}. \end{align*} $$

It follows that $a_i F Y_{i-1}$ is definable over $c, b_{i+1}, \ldots , b_n$ . The fibers of the map $G/Y_{i-1} \to G/(FY_{i-1})$ are finite, and so $a_i Y_{i-1} = b_i$ is algebraic over $c, b_{i+1}, \ldots , b_n$ .

By Claim 2 and induction, $\bar {b} \in \operatorname {\mathrm {acl}}(c)$ . Therefore

$$ \begin{align*}n \le \operatorname{\mathrm{dp-rk}}(\bar{b}/\emptyset) \le \operatorname{\mathrm{dp-rk}}(c/\emptyset) \le \operatorname{\mathrm{dp-rk}}(G).\end{align*} $$

As n was arbitrary, G has infinite dp-rank, a contradiction.

Proof For definable sets in pCF, dp-rank agrees with dimension. In particular, dp-rank is finite. Therefore Lemma 4.7 applies to G.

Proof Take $t_0 \in \Gamma _{\mathbb M}$ such that $S \subseteq O_{t_0}$ . By Proposition 2.29, we can build an ascending sequence

$$ \begin{align*} t_0 < t_1 < t_2 < \cdots \end{align*} $$

in $\Gamma _{\mathbb M}$ such that $O_{t_i} \cdot O_{t_i} \subseteq O_{t_{i+1}}$ for each i. By saturation, we can also find some $t_\omega> t_i$ for all finite i. Set $X = \bigcup _{i < \omega } O_{t_i}$ . The set X is externally definable (Remark 4.5). The set X is bounded, because $X \subseteq O_{t_\omega }$ . We have $X = X^{-1}$ because $O_{t_i} = O_{t_i}^{-1}$ for each i. Lastly, X is closed under the group operation by choice of the $t_i$ ’s.

Proof We claim that $I/X$ is infinite. Otherwise, I is contained in a finite union of cosets: $I \subseteq \bigcup _{i = 1}^n a_i X$ . Take $t \in \Gamma _M$ such that $X \subseteq O_t$ . Then I is a subset of the definably compact set $\bigcup _{i = 1}^n a_i O_t$ , so I is bounded, a contradiction.

Now take $a_1, a_2, a_3, \ldots \in I$ such that the cosets $a_iX$ are pairwise distinct. By saturation, there is some $t \in \Gamma $ such that $\{a_1,a_2,\ldots \} \subseteq O_t$ . Then $\{a_1,a_2,\ldots \}$ and X are bounded. By Lemma 4.9, there is an externally definable bounded subgroup $X'$ containing $\{a_1,a_2,\ldots \} \cup X$ . Then $(X' \cap I)/X$ is infinite, witnessed by the $a_iX$ .

Proof Suppose not.

Let K be the normal subgroup witnessing near-abelianity. By Lemma 4.9, there is a bounded externally definable subgroup $X_0 \supseteq A \cup K$ . By Lemma 4.10 we can recursively build an increasing chain of bounded externally definable subgroups

$$ \begin{align*} X_0 \subseteq X_1 \subseteq X_2 \subseteq \cdots \end{align*} $$

such that

• • $(X_1 \cap I)/X_0$ is infinite.

• • For $n> 1$ , $(X_n \cap (I \diamond X_{n-1} {\setminus} X_0))/X_{n-1}$ is infinite. This is possible because $I \diamond X_{n-1} {\setminus} X_0$ is unbounded by the claim.

Let $Y_i = X_0$ for all i, and let $F = \{\operatorname {\mathrm {id}}_G\}$ . Note $X_0$ is normal, because it contains K which contains $[G,G]$ . We have constructed a bad gap configuration in G, contradicting Lemma 4.8.

Proof Take $t_0 \in \Gamma _M$ such that $S \supseteq O_{t_0}$ . By Proposition 2.29 and Lemma 2.30, there is a descending sequence

$$ \begin{align*} t_0> t_1 > t_2 > \cdots \end{align*} $$

in $\Gamma _{\mathbb M}$ such that $O_{t_{i+1}} \cdot O_{t_{i+1}} \subseteq O_{t_i}$ and also $O_{t_{i+1}}^B \subseteq O_{t_i}$ . Take $X = \bigcap _{i = 1}^\infty O_{t_i}$ . Then X is an externally definable subgroup with $X^B \subseteq X$ . We can take some $t_\omega $ less than all the $t_i$ ’s, and then $O_{t_\omega } \subseteq X$ . Therefore X has interior, and is an open subgroup.

Proof Suppose not.

Proof Take $t, t'$ such that

$$ \begin{gather*} O_{t'} \subseteq A \text{ and } B \subseteq O_t, \\ O_t {\setminus} (F \cdot O_{t'}) \supseteq B {\setminus} (F \cdot A), \\ I \diamond O_t {\setminus} (F \cdot O_{t'}) \subseteq I \diamond B {\setminus} (F \cdot A). \\[-36pt]\end{gather*} $$

Take any bounded open externally definable subgroup $X_0 \subseteq G$ . By Lemma 4.12 there is an externally definable open subgroup $Y_0 \subseteq X_0$ such that $Y_0^F \subseteq Y_0$ . Recursively build chains

$$ \begin{align*} X_0 \subseteq X_1 \subseteq \cdots, \\ Y_0 \supseteq Y_1 \supseteq \cdots, \end{align*} $$

where

• • $X_i$ is a bounded externally definable subgroup, chosen large enough to ensure that $(X_i \cap (I \diamond X_{i-1} {\setminus} F Y_{i-1}))/X_{i-1}$ is infinite (Lemma 4.10).

• • $Y_i$ is an open externally definable subgroup with $Y_i^F = Y_i$ , chosen small enough that $Y_i^{X_i} \subseteq Y_{i-1}$ (Lemma 4.12).

This gives a bad gap configuration in G, contradicting Lemma 4.8.

Proof We may replace M with a monster model, and then apply Lemma 4.11.

Proof We may replace M with a monster model, and then apply Lemma 4.13.

Notation 5.1.

1. (1) If $\phi (x)$ and $\psi (x)$ are G-formulas then $\phi \cdot \psi $ denotes the G-formula

   $$ \begin{align*} (\phi\cdot \psi)(x):= \exists u\exists v(\phi(u)\wedge \psi(v)\wedge x=u\cdot v). \end{align*} $$
   Thus $(\phi \cdot \psi )(M) = \phi (M) \cdot \psi (M)$ .

2. (2) More generally, if $q(x)$ and $r(x)$ are partial G-types then $q\cdot r$ denotes the G-type

   $$ \begin{align*} (q\cdot r)(x):= \{\phi\cdot \psi(x) \mid q(x)\vdash \phi(x),~ r(x)\vdash \psi(x)\}. \end{align*} $$
   Thus $(q \cdot r)(N) = q(N) \cdot r(N)$ for an $|M|^+$ -saturated elementary extension $N \succ M$ .

3. (3) If $g \in G(M)$ and $\phi (x)$ is a G-formula, then $g \cdot \phi $ denotes the G-formula

   $$ \begin{align*} (g \cdot \phi)(x) := \exists u (\phi(u) \wedge x = g \cdot u). \end{align*} $$
   Thus $(g \cdot \phi )(M) = g \cdot \phi (M)$ .

4. (4) If $g \in G(M)$ and $p(x)$ is a partial G-type then $g \cdot p$ denotes the G-type

   $$ \begin{align*} (g \cdot p)(x) := \{g \cdot \phi(x) \mid p(x) \vdash \phi(x)\}. \end{align*} $$
   Thus $(g \cdot p)(N) = g \cdot p(N)$ for an $|M|^+$ -saturated $N \succ M$ .

Fact 5.6. $\operatorname {\mathrm {stab}}_\phi (\Sigma )$ is a definable subgroup of G if $\Sigma $ is definable.

Fact 5.7. For every partial type $\Sigma $ over M,

$$ \begin{align*} \operatorname{\mathrm{stab}}(\Sigma)=\bigcap_{ \phi\in \mathcal{L}}\operatorname{\mathrm{stab}}_\phi(\Sigma). \end{align*} $$

In particular, if $\Sigma $ is definable then $\operatorname {\mathrm {stab}}(\Sigma )$ is an intersection of definable subgroups.

Proof Indeed, $\operatorname {\mathrm {stab}}_\phi (\Sigma )(M)$ is defined by the formula

$$ \begin{align*} \forall y \forall g : ((d_\Sigma z) \phi(g \cdot z; y)) \leftrightarrow ((d_\Sigma z) \phi(x \cdot g \cdot z; y)), \end{align*} $$

and $\operatorname {\mathrm {stab}}_\phi (\Sigma ^N)$ is defined by the same formula, because $\Sigma ^N$ and $\Sigma $ have the same definition schema.

Fact 5.10 ([Reference Peterzil and Starchenko15, Corollary 2.5 and Claim 2.15])

1. 1. If $N \succ M$ , then $\mu (N)$ is a subgroup of $G(N)$ normalized by $G(M)$ .

2. 2. For any definable $q \in S_G(M)$ , the partial type $\mu \cdot q$ is definable.

Fact 5.11 ([Reference Peterzil and Starchenko15, Remark 2.16])

If p is a definable type over M and $N \succ M$ , then the product of the canonical extensions is equal to the canonical extension of the product $:$

$$ \begin{align*} \mu^N \cdot p^N = (\mu \cdot p)^N. \end{align*} $$

Proof Clause (1) is by Claim 2.22 in [Reference Peterzil and Starchenko15].

For (2), we must show

$$ \begin{align*} {\operatorname{\mathrm{st}}_N^{\mathbb M}}(p^N(\mathbb M)\beta^{-1}) = \bigcap_{\psi\in p^N}{\operatorname{\mathrm{st}}_N^{\mathbb M}}(\psi(\mathbb M)\beta^{-1}). \end{align*} $$

The $\subseteq $ direction is clear. For $\supseteq $ , suppose that $g \in \bigcap _{\psi \in p^N}{\operatorname {\mathrm {st}}_N^{\mathbb M}}(\psi (\mathbb M)\beta ^{-1})$ . Then for any $\psi \in p^N$ , there is $h_\psi \in \psi (\mathbb M)$ such that $h_\psi \cdot \beta ^{-1} \cdot g^{-1}$ satisfies $\mu ^N$ . By compactness there is $h \in p^N(\mathbb M)$ such that $h \cdot \beta ^{-1} \cdot g^{-1}$ satisfies $\mu ^N$ . Then $g \in {\operatorname {\mathrm {st}}_N^{\mathbb M}}(p^N(\mathbb M)\beta ^{-1})$ .