Proof of Lemma 2.3.1.

(1) and (2) are straightforward from the definition. (3) also follows directly by preparing X.

(4) Using (3), we can assume that $\#C_q$ does not depend on q. We define $a_q := \frac 1{\#C_q}\sum _{x \in C_q} x$ and $\hat f_q\colon C_q\to \mathrm {RV}_{|n|}, x \mapsto \operatorname {rv}_{|n|}(x - a_q)$ for some n which is a multiple of $\#C_q$ . This implies that $\hat f_q$ is not constant on $C_q$ . We then apply induction to the family consisting of all fibers of $f_q$ , for all q. (We take the final n to be a multiple of all cardinalities $\#C_q$ appearing during this process.)

(5) For $x\in K$ , define

$$ \begin{align*} \mu(x) &= \min\{|x-c|\mid c\in C\},\\ C(x) &= \{c\in C\mid |x-c| = \mu(x)\}, \\ a(x) &= \frac{1}{\#C(x)}\sum_{c\in C(x)}c. \end{align*} $$

The map $a: K\to K$ has finite image, so by (4), we can find an injection $\alpha : \operatorname {im} a\to \mathrm {RV}_{|n|}^k$ . Let $m = \max \{C(x)\mid x\in K\}!$ and define the map f as

$$\begin{align*}f(x) := \begin{cases}(\alpha(a(x)), \operatorname{rv}_{|m|\lambda}(x-a(x))) & \text{ if } |x-a(x)|\geq \mu(x)\lambda|m| \\ (\alpha(a(x)), \operatorname{rv}_{|m|\lambda}(0) ) & \text{ if } |x-a(x)|< \mu(x)\lambda|m|. \end{cases} \end{align*}$$

If x and $x'$ are in the same ball $\lambda $ -next to C, then $C(x)=C(x')$ and so $a(x) = a(x')$ . If x and $x'$ are in the same fibre of f and we are in the first case, then the fact that $\operatorname {rv}_{|m|\lambda }(x-a(x)) = \operatorname {rv}_{|m|\lambda }(x'-a(x'))$ implies that $\operatorname {rv}_{\lambda }(x-c) = \operatorname {rv}_{\lambda }(x'-c)$ for any $c\in C$ . Thus, any fibre of f is contained in a ball $\lambda $ -next to C.

If $\lambda = |n'|$ for some integer $n'$ , then one can take $p = n'\cdot m$ . Then f will be constant on any ball $|p|$ -next to C.

Proof. For each $\lambda $ and each $(\xi , \xi ') \in \mathrm {RV}^k_{|n|} \times \mathrm {RV}_{\lambda }^{\ell }$ , let $C_{\lambda ,\xi ,\xi '}$ be a finite A-definable set $|m_{\lambda ,\xi ,\xi '}|\lambda $ -preparing the fiber $W_{\xi ,\xi '} \subset K$ . Using compactness, we may assume that for varying $\lambda ,\xi ,\xi '$ , there are only finitely many different sets $C_{\lambda ,\xi ,\xi '}$ and integers $m_{\lambda ,\xi ,\xi '}$ . Let C be the union of the $C_{\lambda ,\xi ,\xi '}$ and m be the least common multiple of the $m_{\lambda ,\xi ,\xi '}$ .

Proof. Let $\phi $ be given as in the statement. Whether a pair $(m, \psi )$ works as desired in a model K can be expressed by an ${\mathcal L}$ -sentence, by Lemma 2.3.1. By compactness and $\ell $ -h $^{\mathrm {mix}}$ -minimality, we deduce that there exist finitely many pairs $(m_i, \psi _i)$ which cover all models. We may furthermore assume that the sets $\psi _i(K)$ are finite for each model K. We are done by letting m be the least common multiple of the $m_i$ (so that $|m| \le |m_i|$ for each i) and $\psi $ be the disjunction of the $\psi _i$ .

Proof. (1) The case $k = 1$ follows by applying Proposition 2.3.2. (Each $W_{\xi }$ is contained in the set C one obtains.) For $k \ge 2$ , use induction (adding parameters to the language using Lemma 2.3.1 (1)).

(2) Apply (1) to the graph of f.

(3) C is a fiber of an A-definable subset $W \subset K \times \mathrm {RV}_{\bullet }^k$ , for some k. Apply (1) to W.

Proof. (1) Suppose for contradiction that f has infinitely many infinite fibers. Let $X \subset K$ be the subset of the domain of f where f is locally constant (i.e, of points $x \in K$ such that f is constant on $B_{<\lambda }(x)$ for some $\lambda \in \Gamma _K^{\times }$ ). Since each infinite fiber of f contains a ball (by preparation of the fiber), the restriction of f to X still has infinite image.

Let $C \subset K$ be a finite set $|m|$ -preparing X, for some integer $m \ge 1$ . Then enlarge C and m in such a way that for each $\lambda \in \Gamma _K^{\times }$ , whether f is constant on $B_{<\lambda }(x)$ only depends on the ball $|m|$ -next to C containing x. This is possible by applying Corollary 2.3.4 (and Remark 2.3.3) to the set $W \subset K \times \Gamma _K^{\times }$ of those $(x,\lambda )$ for which f is constant on $B_{<\lambda }(x)$ .

By Corollary 2.3.6 (1), there exists a ball $B_0 \subset X \ |m|$ -next to C such that $f(B_0)$ is still infinite. Fix such a $B_0$ , and also fix a $\lambda _0 \in \Gamma _K^{\times }$ such that f is constant on every open ball of radius $\lambda _0$ contained in $B_0$ .

The family of all those balls can definably be parametrized by a subset $Z \subset \mathrm {RV}_{\lambda _1}$ , for $\lambda _1 = \lambda _0/\operatorname {rad}_{\mathrm {op}} B_0$ . Now f induces a map from Z to K with infinite image, contradicting Corollary 2.3.6 (1).

(2) First, find C and m such that $C \ |m|$ -prepares every infinite fiber of f (using (1) to see that there are only finitely many infinite fibers). Then apply Lemma 2.3.1 (4) to the family of finite fibers of f to obtain an injective map from each finite fiber of f to $\mathrm {RV}^k_{|n|}$ for some n and k. We put all those maps together to one single map $g\colon Y \to \mathrm {RV}^k_{|n|}$ , where Y is the union of all finite fibers of f. Enlarge C and m in such a way that g is constant on each ball $B \subset Y \ |m|$ -next to C (by applying Corollary 2.3.4 to the graph of g). Then, for each ball $B \ |m|$ -next to C, either B is entirely contained in an infinite fiber of f (and hence f is constant on B, as desired) or $B \subset Y$ and g is constant on B. In that case, f is injective on B since $f(x_1) = f(x_2)$ for $x_1, x_2 \in B$ would imply that $x_1$ and $x_2$ lie in the same fiber of f, contradicting that g is injective on each fiber of f.

(3) Use (2) to obtain a finite $\emptyset $ -definable set C and an integer m such that on any ball $|m|$ -next to C, f is either constant or injective. Let $W_0\subset K\times (\Gamma _K^{\times })^2$ consist of those $(x, \lambda , \mu )$ with $\mu \leq 1$ such that for every $y\in f(B_{<\lambda }(x))$ , there are open balls $B', B"$ with $y\in B'\subset f(B_{<\lambda }(x))\subset B"$ and

$$\begin{align*}\operatorname{rad}_{\mathrm{op}} B' \geq \mu \operatorname{rad}_{\mathrm{op}} B". \end{align*}$$

We enlarge C and m, such that C also $|m|$ -prepares this set $W_0$ .

The set C is already as desired, but m will later be enlarged to some $m\cdot m"$ . Note that we already simplified the statement we will need to prove. First, it suffices to consider balls B which are contained in a ball $B_1 \ |m|$ -next to C on which f is injective, and secondly, for each $\lambda \le 1$ , it suffices to find a single $\lambda \cdot |m"|$ -shrinking B of $B_1$ for which the lemma holds (using some n which we still need to specify). Indeed, the fact that $C \ m$ -prepares $W_0$ implies that then the lemma also holds for all translates of B within $B_1$ (using the same n).

Before we can continue, we need to do some preparation on the range side of f. We want to find a finite $\emptyset $ -definable D and a positive integer p such that for every $\lambda \le 1$ and every ball $B \ \lambda |m|$ -next to C, the set $f(B)$ is $\lambda |p|$ -prepared by D. To see that such a D exists, first note that by Lemma 2.3.1 (5), there exists a $\lambda $ -definable map $g_{\lambda }\colon K \to \mathrm {RV}_{|n'|}^k\times \mathrm {RV}_{|m'|\cdot \lambda }$ such that every ball $\lambda |m|$ -next to C is a union of fibers of $g_{\lambda }$ . We may assume (by compactness) that the maps $g_{\lambda }$ form a $\emptyset $ -definable family. Now apply Proposition 2.3.2 to the set

$$\begin{align*}W := \{(f(x), \lambda, g_{\lambda}(x)) \in K \times \Gamma_K^{\times} \times \mathrm{RV}_{|n'|}^k \times \bigcup_{\mu} \mathrm{RV}_{\mu} \mid x \in K, \lambda \le 1\} \end{align*}$$

and let D and p be the result. To see that this works, let $\lambda \le 1$ and $B \ \lambda |m|$ -next to C be given. We have $B = g_{\lambda }^{-1}(\Xi )$ for $\Xi := g_{\lambda }(B)$ , and the image $f(B)$ consists of those $y \in K$ for which the fiber $W_{(y, \lambda )}$ is not disjoint from $\Xi $ . Since this fiber is constant when y runs over a ball $B' \ \lambda |p|$ -next to D, we either have $B' \subset f(B)$ or $B' \cap f(B) = \emptyset $ , as desired.

Now that we constructed D and p (on the range side of f), we construct another set $C'$ and integer $m'$ on the domain side of f. For $\lambda \in \Gamma _K^{\times }$ , $\lambda \leq 1$ , use Lemma 2.3.1 to get a map $h_{\lambda }: K\to \mathrm {RV}_{|p|}^k\times \mathrm {RV}_{\lambda |n"|}$ such that each fibre is contained in a ball $\lambda |p|$ -next to D. We may assume that $h_{\lambda }$ is a $\emptyset $ -definable family of maps, with a parameter $\lambda $ . Let $C'$ be a finite $\emptyset $ -definable set which $\lambda |m'|$ -prepares every map $h_{\lambda }\circ f$ . This can again be done using Proposition 2.3.2.

We need one last ingredient before we can verify that the lemma holds. Since $C'$ is finite, there exists an integer $p' \ge 1$ such that every ball $B_1 \ |m m'|$ -next to C has a $|p'|$ -shrinking $B_1^{\prime } \subset B_1$ disjoint from $C'$ .

We now claim that the lemma holds using C, $|mm'p'|$ and $n = m'p'$ . According to the beginning of the proof, it suffices to check that given a ball $B_1 \ |m|$ -next to C on which f is injective, and given a $\lambda \le 1$ , there exists a $\lambda \cdot |p' m'|$ -shrinking B of $B_1$ for which the claim holds. Choose a $|p'|$ -shrinking $B_1^{\prime }$ of $B_1$ disjoint from $C'$ and choose B to be any $\lambda |m'|$ -shrinking of $B_1^{\prime }$ . To finish the proof, we show that the lemma holds for this B.

On the one hand, since B is a $\lambda \cdot |m'|$ -shrinking of $B_1^{\prime }$ (and $B_1^{\prime }$ is disjoint from $C'$ ), B is contained in a ball $\lambda \cdot |m'|$ -next to $C'$ . By definition of $C'$ , this means that $h_{\lambda } \circ f$ is constant on B, and hence (by definition of $h_{\lambda }$ ), $f(B)$ is contained in a ball $B" \ \lambda |p|$ -next to D.

On the other hand, B is $\lambda |mm'p'|$ -next to C, so that $f(B)$ is $\lambda |pm'p'|$ -prepared by D (by definition of D). Thus, for any $y\in f(B)$ , we obtain that the entire ball $B' \ \lambda |pm'p'|$ -next to D containing y is contained in $f(B)$ . This ball $B'$ is just the (unique) $|m'p'|$ -shrinking of $B"$ containing y; in particular, $\operatorname {rad}_{\mathrm {op}} B' = |m'p'|\operatorname {rad}_{\mathrm {op}} B"$ , as desired.

(4) If the residue field of K has characteristic zero, then this is exactly [Reference Cluckers, Halupczok and Rideau10, Lemma 2.8.4]. So we may assume that we are in mixed characteristic (though the following proof easily also adapts to the equicharacteristic zero case.) The family of radius $\alpha $ open balls in ${\mathcal M}_K$ can be definable parametrized by the set $\Lambda = \operatorname {rv}_{\alpha }({\mathcal M}_K)$ . Namely, for $\xi \in \Lambda $ , let $B_{\xi }\subset {\mathcal M}_K$ be the unique open ball of radius $\alpha $ containing $\operatorname {rv}_{\alpha }^{-1}(\xi )$ . Using 1-h $^{\mathrm {mix}}$ -minimality, we can find a finite set $C\subset K$ and a positive integer m such that $C \ |m|\alpha $ -prepares every set $f(B_{\xi })$ , for every $\xi \in \Lambda $ . In other words, if $\xi \in \Lambda $ , and if $B'$ is an open ball $|m|\alpha $ -next to C then either $B'\subset f(B_{\xi })$ or $B'\cap f(B_{\xi }) = \emptyset $ . Now, if $B'$ has radius strictly larger than $\beta $ , then it follows that $B'\cap f(B_{\xi }) = \emptyset $ . Therefore, $f({\mathcal M}_K)$ is contained in the union of C and all balls $|m|\alpha $ -next to C of radius at most $\beta $ . This union is equal to the finite union of all closed balls of the form $B_{\leq \beta /(|m|\alpha )}(c), c\in C$ . In particular the open ball $B'$ is contained in this finite union of closed balls of radius $\beta /(|m|\alpha )$ . Since we are in mixed characteristic, there exists some positive integer q such that $\operatorname {rad}_{\mathrm {op}} B'\leq \beta /(|q|\alpha )$ . But then $B"$ is an open ball containing $f({\mathcal M}_K)$ of radius at most $\beta / (|pq|\alpha )$ , finishing the proof.

Proof. We may assume that $A=\emptyset $ by Lemma 2.3.1. Using Lemma 2.3.8 (3), we can find a finite $\emptyset $ -definable set $C_0$ and positive integers $m,n$ such that

• ○ f is constant or injective on balls $|m|$ -next to $C_0$ , and

• ○ if B is an open ball contained in a ball $|m|$ -next to $C_0$ , then either $f(B)$ is a singleton or there are open balls $B'\subset f(B)\subset B"$ such that the radii of $B'$ and $B"$ differ by at most $|n|$ .

For an open ball $B = B_{<\alpha }(x)$ contained in a ball $|m|$ -next to C on which f is injective, define $\mu (x, \alpha )$ to be the (convex) set of $\mu \in \Gamma _K^{\times }$ for which $f(B)$ is contained in an open ball of radius $\mu $ and contains an open ball of radius $|n|\mu $ . Note that we have $|n|\mu (x,\alpha )\leq \mu (x,\alpha )$ , in the sense that for every $\nu \in |n|\mu (x,\alpha )$ and every $\nu ' \in \mu (x,\alpha )$ , we have $\nu \leq \nu '$ . (In the following, inequalities between convex subsets of $\Gamma _K$ are always meant in this sense.) Also define

$$\begin{align*}s(x,\alpha) = \{\mu/\alpha \mid \mu \in \mu(x,\alpha)\}. \end{align*}$$

(An element $\nu \in s(x,\alpha )$ is a kind of ‘scaling factor’: the ball $B_{<\alpha }(x)$ is sent into a ball of radius $\alpha \nu $ .)

We will now enlarge C and m to obtain the following claim. As a side remark, note that we will keep n fixed for the entire proof so that the definitions of $\mu (x,\alpha )$ and $s(x, \alpha )$ are unaltered.

To prove Claim 1, let $W\subset K\times (\Gamma _K^{\times })^2$ consist of those $(x, \lambda , \mu )$ such that $f(B_{<\lambda }(x))$ is contained in a ball $B'$ of radius $\mu $ , and contains a ball of radius $|n|\mu $ . Enlarge C and m such that $C \ |m|$ -prepares this set W. Note that item (1) of the claim then holds by preparation. Let B be $|m|$ -next to C. If f is constant on B, then there is nothing to check, so we can assume that f is injective on B. By item (1), $\mu (x,\alpha )$ and $s(x,\alpha )$ are constant when x runs over B and $\alpha \leq \operatorname {rad}_{\mathrm {op}} B$ is fixed, so we write simply $\mu (\alpha )$ and $s(\alpha )$ . It remains to prove item (2) of Claim 1 (after possibly enlarging m once more). Fix $\mu \in \mu (\alpha )$ and take $\alpha \leq \beta \leq \operatorname {rad}_{\mathrm {op}} B$ . Then any ball of radius $\alpha $ inside B has image under f contained in a ball of radius at most $\mu $ . Hence, by a rescaled version of Lemma 2.3.8(4), there exists an integer $p \ge 1$ such that for $x \in B$ , the image $f(B_{<\beta }(x))$ is contained in an open ball of radius $\frac {\mu \cdot \beta }{|p| \cdot \alpha }$ . In particular, we have

$$\begin{align*}|n|\mu(\beta)\leq \frac{\mu(\alpha)\cdot \beta}{|p| \cdot\alpha}. \end{align*}$$

But this means precisely that

$$\begin{align*}s(\beta)\leq s(\alpha)/|np|, \end{align*}$$

which proves Claim 1 (after replacing m by, for example, $mnp$ ).

To prove the lemma, we also need an inequality opposite to the one of Claim 1 (2).

Claim 2. After possibly further enlarging C and m, in Claim 1 (2), we can additionally obtain

$$\begin{align*}s(y,\alpha) \leq s(x,\beta)/|m|. \end{align*}$$

The idea of the proof of Claim 2 is to apply Claim 1 to $f^{-1}$ (which, in reality, only exists piecewise). This is made precise as follows:

Denote by Y the set of $y\in K$ for which $f^{-1}(y)$ is finite. This is a cofinite $\emptyset $ -definable set in K, by Lemmas 2.3.1 and 2.3.8 (1). Use Lemma 2.3.1 (4) to obtain a $\emptyset $ -definable family of injections

$$\begin{align*}h_y: f^{-1}(y)\to \mathrm{RV}_{|n_0|}^k, \end{align*}$$

for $y\in Y$ . For $\eta \in \mathrm {RV}_{|n_0|}^k$ , define

$$ \begin{align*} Y_{\eta} &= \{y\in Y\mid \eta\in \operatorname{im} h_y\}, \\ g_{\eta} &: Y_{\eta} \to K: y\mapsto h_y^{-1}(\eta). \end{align*} $$

Then we have that

$$\begin{align*}\bigcup_{y\in Y}\{y\}\times f^{-1}(y) = \bigsqcup_{\eta\in \mathrm{RV}_{|n_0|}^k}\operatorname{graph}(g_{\eta}). \end{align*}$$

For each $\eta $ , apply Lemma 2.3.8 (3) to $g_{\eta }$ (extended by $0$ outside of $Y_{\eta }$ ) to obtain a finite $\eta $ -definable set $D_{\eta }$ and integers $m_{\eta }, n_{\eta }$ . By compactness, we may take $n' := n_{\eta }$ independent of $\eta $ . Now enlarge $D_{\eta }$ and $m_{\eta }$ using Claim 1, so that (1) and (2) hold for $g_{\eta }$ (where the corresponding $\mu $ and s are defined using $n'$ ). We may, moreover, assume that $D_{\eta } \ |m_{\eta }|$ -prepares $Y_{\eta }$ (enlarging $D_{\eta }$ and $m_{\eta }$ once more, if necessary). After that, apply compactness once more to make $m':=m_{\eta }$ independent of $\eta $ and to turn $(D_{\eta })_{\eta }$ into a $\emptyset $ -definable family. By Corollary 2.3.6, the union $D=\bigcup _{\eta } D_{\eta }$ is a finite $\emptyset $ -definable set.

Using Lemma 2.3.1 (5), take a $\emptyset $ -definable function $\chi : K\to \mathrm {RV}_{|n_1|}^{k'}$ such that every ball $|m'|$ -next to D is a union of fibres of $\chi $ . Then choose $C_2$ and $C_3$ such that, after possibly enlarging m, $C_2 \ |m|$ -prepares the family of sets $f^{-1}(\chi ^{-1}(\eta ))$ for $\eta \in \mathrm {RV}_{|n_1|}^{k'}$ , and $C_3 \ |m|$ -prepares the family of images $g_{\eta }(Y_{\eta })$ . We will now prove that Claim 2 holds after replacing C by $C \cup C_2 \cup C_3$ (and some further enlargement of m).

So suppose that we have open balls $B_1, B$ with $\operatorname {rad}_{\mathrm {op}} B_1 = \alpha \leq \operatorname {rad}_{\mathrm {op}} B = \beta $ which are contained in the same ball $|m|$ -next to C. If f is constant on B, then we are done, so assume that f is injective on B. We may assume that $B_1\subset B$ since $\mu (x, \alpha )$ is independent of x as x runs over B. Let $B'\subset f(B)\subset B"$ be open balls whose radii differ by at most $|n|$ . By definition of $C_2$ , $\chi \circ f$ is constant on B, so (by definition of $\chi $ ), $f(B)$ is contained in a ball $|m'|$ -next to D. Perhaps after shrinking $B"$ , we can assume that also $B"$ is contained in a ball $|m'|$ -next to D. By definition of $C_3$ , there is a (unique) $\eta \in \mathrm {RV}_{|n_0|}^k$ such that $B \subset g_{\eta }(Y_{\eta })$ ; this implies that $f|_B$ and $g_{\eta }|_{f(B)}$ are mutually inverse bijections between B and $f(B)$ . Using that we applied Lemma 2.3.8 to $g_{\eta }$ , take open balls $\tilde {B}'\subset g_{\eta }(B')\subset \overline {B}'$ whose radii differ by at most $|n'|$ and do the same with open balls $\tilde {B}"\subset g_{\eta }(B")\subset \overline {B}"$ . Note that we have a chain of inclusions

$$\begin{align*}\tilde{B}' \subset g_{\eta}(B') \subset g_{\eta}(f(B)) = B \subset g_{\eta}(B") \subset \overline{B}". \end{align*}$$

Choose similar balls corresponding to $B_1$ : $B_1^{\prime }, B_1^{\prime \prime }, \tilde {B}_1^{\prime }, \dots $ . We may certainly assume that $\operatorname {rad}_{\mathrm {op}} B_1^{\prime } \le \operatorname {rad}_{\mathrm {op}} B"$ . Therefore, our application of Claim 1 to $g_{\eta }$ yields that

$$\begin{align*}\frac{\operatorname{rad}_{\mathrm{op}} \overline{B}"}{\operatorname{rad}_{\mathrm{op}} B"} \le \frac{\operatorname{rad}_{\mathrm{op}} \overline{B}^{\prime}_1}{|m'|\operatorname{rad}_{\mathrm{op}} B^{\prime}_1}. \end{align*}$$

Combining this with $\operatorname {rad}_{\mathrm {op}} B \le \operatorname {rad}_{\mathrm {op}}\overline {B}"$ , $|n'|\operatorname {rad}_{\mathrm {op}} \overline {B}^{\prime }_1 \le \operatorname {rad}_{\mathrm {op}}\tilde {B}^{\prime }_1 \le \operatorname {rad}_{\mathrm {op}} B_1$ and $|n|\operatorname {rad}_{\mathrm {op}} B^{\prime \prime }_1 \le \operatorname {rad}_{\mathrm {op}} B^{\prime }_1$ , we obtain

$$\begin{align*}\frac{\operatorname{rad}_{\mathrm{op}} B}{\operatorname{rad}_{\mathrm{op}} B"} \le \frac{\operatorname{rad}_{\mathrm{op}} \overline{B}"}{\operatorname{rad}_{\mathrm{op}} B"} \le \frac{\operatorname{rad}_{\mathrm{op}} \overline{B}^{\prime}_1}{|m'|\operatorname{rad}_{\mathrm{op}} B^{\prime}_1} \le \frac{\operatorname{rad}_{\mathrm{op}} B_1}{|n'm'|\operatorname{rad}_{\mathrm{op}} B^{\prime}_1} \le \frac{\operatorname{rad}_{\mathrm{op}} B_1}{|nn'm'|\operatorname{rad}_{\mathrm{op}} B^{\prime\prime}_1}. \end{align*}$$

Since $\frac {\operatorname {rad}_{\mathrm {op}} B^{\prime \prime }_1}{\operatorname {rad}_{\mathrm {op}} B_1} \in s(\alpha )$ and $\frac {\operatorname {rad}_{\mathrm {op}} B"}{\operatorname {rad}_{\mathrm {op}} B} \in s(\beta )$ , we deduce

$$\begin{align*}s(\alpha) \le s(\beta)/|n^3n'm'| \end{align*}$$

(where the additional factor $n^2$ takes into account the length of the convex sets $s(\alpha )$ and $s(\beta )$ ). This finishes the proof of Claim 2 (where we take m to be a multiple of $n^3n'm'$ ).

We are now ready to prove the lemma itself. Take $x,y$ in the same ball $B \ |m|$ -next to C. Denote by $\beta $ the open radius of B. Let $\mu \in \mu (x, \beta )$ , so that $f(B)$ is contained in a ball of radius $\mu $ and contains a ball of radius $|n|\mu $ . We will show that we can take $\mu _B = \mu /\beta $ .

If we choose $\alpha> |x-y|$ , then x and y are contained in an open ball of radius $\alpha $ . If we, moreover, choose $\mu '\in \mu (x,\alpha )$ , then $f(x), f(y)$ are contained in an open ball of radius $\mu '$ . Thus,

$$\begin{align*}|f(x)-f(y)|/\alpha < \mu'/\alpha \leq \frac{1}{|m|} \mu /\beta. \end{align*}$$

Since this holds for any $\alpha> |x-y|$ , this gives that

$$\begin{align*}|f(x)-f(y)| \leq \frac{1}{|m|} \frac{\mu}{\beta} |x-y|. \end{align*}$$

For the other inequality, set $\alpha = |x-y|$ and denote by $B'$ the open ball of radius $\alpha $ around x. By the injectivity of f on B, $f(y)$ is not in $f(B')$ . But $f(B')$ contains an open ball of radius $|n|\mu '$ (for $\mu ' \in \mu (x, \alpha )$ , as before) and so

$$\begin{align*}|f(x)-f(y)|/\alpha \geq |n|\mu'/\alpha \geq |nm|\mu/\beta. \end{align*}$$

Thus,

$$\begin{align*}|f(x)-f(y)| \geq |mn|\frac{\mu}{\beta}|x-y|. \end{align*}$$

This proves the Lemma.

Proof. This is direct from Item (1) from Lemma 2.3.8 and Lemma 2.3.9.

Proof of Lemma 2.4.1.

(1) We induct on k, so first assume that $k=1$ . Using $\exists ^{\infty }$ -elimination, we may assume that the cardinality of $C_{\xi }$ is constant, say, equal to m. Let $\sigma _1, \dots , \sigma _m$ be the elementary symmetric polynomials in m variables, considered as functions on m-element-subsets of K. Then the map $K^k \to K, x \mapsto \sigma _i(C_{\operatorname {rv}_{\lambda }(x)})$ is locally constant everywhere except possibly at $0$ , so by (T2), it has finite image. Since $\sigma _1(C_{\xi }), \dots , \sigma _m(C_{\xi })$ together determine $C_{\xi }$ , there are only finitely many different $C_{\xi }$ .

For arbitrary k, consider $C_{\xi }$ as a definable family $C_{\xi _1, \xi _2}$ with $\xi _1\in \mathrm {RV}_{\lambda }$ and $\xi _2\in \mathrm {RV}_{\lambda }^{k-1}$ . Then by induction on k, the union $\cup _{\xi } C_{\xi } = \cup _{\xi _1\in \mathrm {RV}_{\lambda }}\cup _{\xi _2\in \mathrm {RV}_{\lambda }^k} C_{\xi _1, \xi _2}$ is finite.

(2) Using (T1 $^{\mathrm {mix}}$ ), we find an A-definable C. We consider C as an $(A\cap K)$ -definable family $C_{\xi }$ , parametrized by $\xi \in \mathrm {RV}_{|n|}^k$ . Using (1), the union $C' = \cup _{\xi \in \mathrm {RV}_{|n|}^k} C_{\xi }$ is $(A\cap K)$ -definable, finite, and f satisfies (T1 $^{\mathrm {mix}}$ ) with respect to $C'$ .

Proof. Using Lemma 2.3.1 we may as well assume that $A=\emptyset $ . Take a finite $\emptyset $ -definable set C and an integer m such that (T $1^{\mathrm {mix}}$ ) holds for f with respect to C and m. Let $\chi : K\to \mathrm {RV}_{|p|}^k$ be a $\emptyset $ -definable map coming from Lemma 2.3.1(5) for $C,m$ . So every ball $|m|$ -next to C is a union of fibres of $\chi $ and every ball $|n'|$ -next to C is contained in a fibre of $\chi $ . Let D be a finite $\emptyset $ -definable set $|n|$ -preparing the family $(f(\chi ^{-1}(\eta )))_{\eta \in \mathrm {RV}_{|p|}^k}$ ; here, we use Corollary 2.3.4. Let $\psi : K\to \mathrm {RV}_{|p'|}^{k'}$ be a $\emptyset $ -definable map such that every ball $|n|$ -next to D is a union of fibres of $\psi $ and every ball $|n"|$ -next to D is contained in a fibre of $\psi $ . Finally, use Corollary 2.3.4 again to $|q|$ -prepare the family $(f^{-1}(\psi ^{-1}(\eta )))_{\eta \in \mathrm {RV}_{|p'|}^{k'}}$ with a finite $\emptyset $ -definable set $C_0$ .

We claim that $C'=C\cup C_0$ suffices. So let $B'$ be a ball $|q|$ -next to $C'$ and let B be the ball $|m|$ -next to C containing $B'$ . We can assume that $\mu _B\neq 0$ , for else f is constant on B. Fix any open ball $B"$ in $B'$ and let $x,y\in B"$ . Then

$$\begin{align*}|f(x)-f(y)|\leq \frac{\mu_B}{|m|}|x-y| < \frac{\mu_B}{|m|}\operatorname{rad}_{\mathrm{op}} B". \end{align*}$$

Therefore, if we denote by $B_1^{\prime \prime }$ the open ball of radius $\frac {\mu _B}{|m|}\operatorname {rad}_{\mathrm {op}} B"$ around $f(x)$ , then $f(B")$ is contained in $B_1^{\prime \prime }$ . By definition of $C_0$ , $f(B')$ is contained in a ball $B_1$ $|n|$ -next to D. However, $f(B)$ is $|n|$ -prepared by D. Since $B_1\cap f(B)\neq \emptyset $ , we have

$$\begin{align*}f(B")\subset f(B')\subset B_1\subset f(B). \end{align*}$$

Using property (T $1^{\mathrm {mix}}$ ), we see that

$$\begin{align*}\operatorname{rad}_{\mathrm{op}} B_1\geq |m|\mu_B\operatorname{rad}_{\mathrm{op}} B'. \end{align*}$$

Now take $z\in K$ such that $|f(x)-z| < |m|\mu _B\operatorname {rad}_{\mathrm {op}} B"$ . Then we have $z\in B_1\subset f(B)$ , and so there is some $x'\in B$ with $f(x')=z$ . Applying (T $1^{\mathrm {mix}}$ ) one more time yields that

$$\begin{align*}|x-x'|\leq \frac{1}{|m|\mu_B} |f(x)-f(x')| < \operatorname{rad}_{\mathrm{op}} B". \end{align*}$$

We conclude that $f(B")$ contains the open ball of radius $|m|\mu _B\operatorname {rad}_{\mathrm {op}} B"$ around $f(x)$ .

Notation 2.5.1. The notion of balls $\lambda $ -next to a finite set $C \subset K$ now has different meanings for $|\cdot |$ and for $|\cdot |_{\mathrm {ecc}}$ , with notation from Definition 2.1.3. To make clear which of the valuations we mean, we either write $|1|$ -next or $|1|_{\mathrm {ecc}}$ -next (instead of just $1$ -next).

Proof of Lemma 2.5.3.

It suffices to prove that for any $a \in C$ , all realizations of $p:=\mathrm {tp}_{{\mathcal L}}(a/\emptyset )$ lie in C; indeed, by $\aleph _0$ -saturation, this then implies that p is algebraic (using that C is finite) and hence isolated by some formula $\phi _p(x)$ . Therefore, C is defined by the disjunction of finitely many such $\phi _p(x)$ .

So now suppose for contradiction that there exist $a \in C$ and $a' \in K \setminus C$ which have the same ${\mathcal L}$ -type over $\emptyset $ . Then by our homogeneity assumption, there exists an ${\mathcal L}$ -automorphism of K sending a to $a'$ (and hence not fixing C setwise). Such an automorphism also preserves ${\mathcal O}_{{\mathrm {ecc}}}$ and hence is an ${\mathcal L}_{\mathrm {ecc}}$ -automorphism, but this contradicts C being ${\mathcal L}_{\mathrm {ecc}}$ -definable.

Proof. By Remark 2.5.4, we may consider a model K of ${\mathcal T}$ which is sufficiently saturated and sufficiently homogeneous (as in Lemma 2.5.3). Consider an integer n, $\lambda \in \Gamma _K^{\times }$ , $A\subset K$ , $\xi \in \mathrm {RV}_{\lambda }$ , $\eta \in \mathrm {RV}_{|n|}^k$ for some integers $k,n>0$ , and an ${\mathcal L}(A\cup \{\eta ,\xi \})$ -definable set $X\subset K$ . We have to show that X can be $|m|\cdot \lambda $ -prepared by some finite ${\mathcal L}(A)$ -definable set C for some integer $m>0$ .

Let $\lambda _{\mathrm {ecc}}$ be the image of $\lambda $ in $\Gamma _{K,{\mathrm {ecc}}}$ . Since $B_{<\lambda _{\mathrm {ecc}}}(1) \subset B_{<\lambda }(1)$ , we have a canonical surjection $\mathrm {RV}_{\lambda _{\mathrm {ecc}}} \to \mathrm {RV}_{\lambda }$ . Similarly, there is a canonical surjection $\mathrm {RV}_{{\mathrm {ecc}}}\to \mathrm {RV}_{|n|}$ . We fix any preimage $(\xi _{\mathrm {ecc}}, \eta _{\mathrm {ecc}}) \in \mathrm {RV}_{\lambda _{\mathrm {ecc}}} \times \mathrm {RV}_{\mathrm {ecc}}^k$ of $(\xi ,\eta )$ , so that X is ${\mathcal L}(A \cup \{\xi _{\mathrm {ecc}}, \eta _{\mathrm {ecc}}\})$ -definable. By $1$ -h-minimality for the ${\mathcal L}_{\mathrm {ecc}}$ -structure on K, there exists a finite ${\mathcal L}_{\mathrm {ecc}}(A)$ -definable set C such that for every pair $x, x'$ in the same ball $\lambda _{\mathrm {ecc}}$ -next to C, we have $x \in X \iff x' \in X$ . By Lemma 2.5.3, C is already ${\mathcal L}(A)$ -definable; we claim that it is as desired. Suppose for contradiction that there exists no m as in the corollary; that is, for every integer $m \ge 1$ , there exists a pair of points $(x, x') \in K^2$ which lie in the same ball $\lambda \cdot |m|$ -next to C such that $x \in X$ but $x' \notin X$ . By $\aleph _0$ -saturation (in the language ${\mathcal L}$ ), we find a single pair $(x, x') \in K^2$ of points with $x \in X$ but $x' \notin X$ and which lie in the same ball $\lambda \cdot |m|$ -next to C for every $m \ge 1$ . The latter implies that x and $x'$ lie in the same ball $\lambda _{\mathrm {ecc}}$ -next to C (by Remark 2.5.2), so we get a contradiction to our choice of C.

Proof. (i) $\Rightarrow $ (iii): Let f be as in (iii). We have to check that for every ${\mathcal L}(A\cup B)$ -definable set $X\subset K$ , $c_1 \in X$ if and only if $c_2\in X$ . By (i), there exists a finite $C \subset A$ and an integer m such that X is $mI$ -prepared by C. Since f is an ${\mathcal L}_I(A\cup B)$ -isomorphism and B contains $\operatorname {rv}_{mI}(\langle A,c_1\rangle )$ , for all $a\in C$ and all $r\geq 1$ , we have

$$\begin{align*}\operatorname{rv}_{mI}(c_2-a) = \operatorname{rv}_{mI}(f(c_1)-a) = f(\operatorname{rv}_{mI}(c_1-a)) = \operatorname{rv}_{mI}(c_1-a). \end{align*}$$

Since X is I-prepared by C, it follows that $c_1 \in X$ if and only if $c_2\in X$ .

(iii) $\Rightarrow $ (ii): Let f be as in (ii). Since f is ${\mathcal L}$ -elementary if and only its restriction to every finite domain is, we may assume $B_i$ small. Using the assumption that ${\left .{f}\right |{}_{AB_1}}$ is ${\mathcal L}$ -elementary, we can extend $({\left .{f}\right |{}_{AB_1}})^{-1}$ ${\mathcal L}$ -elementarily to some g defined at $c_2$ . Let $c^{\prime }_1 := g(c_2)$ . Then $g\circ f\colon \{c_1\} \to \{c^{\prime }_1\}$ is a partial ${\mathcal L}_I(A\cup B_1)$ -isomorphism. Since $\bigcup _n\operatorname {rv}_{nI}(\langle A, c_1\rangle )\subset B_1$ , it follows by (iii) that $g\circ f$ is an elementary ${\mathcal L}$ -isomorphism. As g is also ${\mathcal L}$ -elementary, so is f.

(ii) $\Rightarrow $ (i): Let X be as in (i), and let $B \subset \bigcup _n\mathrm {RV}_{nI}$ be a finite subset such that X is ${\mathcal L}(A \cup B)$ -definable. Let ${\mathcal L}_I^{\prime }$ denote the expansion of L _I by the full L(A)-induced structure on ⋃ _n RV _nI .

Consider any $c_1, c_2 \in K$ which have the same qf- ${\mathcal L}_I^{\prime }$ -type over $A \cup B$ . Then the map $f\colon c_1 \to c_2$ is an ${\mathcal L}_I^{\prime }(A \cup B)$ -isomorphism and extends to $f\colon A B_1 c_1 \to A_2 B_2 c_2$ , where $B_i := B \cup \bigcup _n \operatorname {rv}_{nI}(\langle A,c_i\rangle )$ . By definition of ${\mathcal L}_I^{\prime }$ , the restriction $f|_{AB_1}$ is ${\mathcal L}$ -elementary, so by (ii), the map f is ${\mathcal L}$ -elementary. Moreover, since f is the identity on $A \cup B$ , this implies that $c_1$ and $c_2$ have the same ${\mathcal L}$ -type over $A \cup B$ .

We just proved that the ${\mathcal L}(A \cup B)$ -type of any element $c \in K$ is implied by its qf- ${\mathcal L}_I^{\prime }(A \cup B)$ -type. By a classical compactness argument (cf. the proof of [Reference Tent and Ziegler22, Theorem 3.2.5]), it follows that any ${\mathcal L}(A\cup B)$ -formula in one valued field variable is equivalent to a quantifier free ${\mathcal L}_I^{\prime }(A\cup B)$ -formula. In particular, this applies to our set X. Since ${\mathcal L}_I^{\prime }(B)$ is an $\bigcup _n\mathrm {RV}_{nI}$ -expansion of ${\mathcal L}_I$ , X is indeed mI-prepared by some finite C ⊂ A.

Proof. Let $A_1, A_2, f, c_1$ be given. We may assume that $\mathrm {acl}_K A_i = A_i$ . Set $B_1 := \bigcup _n\operatorname {rv}_{nI}(\langle A_1, c_1\rangle _{\mathbb Q})$ . We claim that we can extend f by setting $f(c_1) := c_2$ for any $c_2\in K$ realizing $f_*\mathrm {qftp}_{{\mathcal L}_I}(c_1/A_1B_1)$ . To see this, we use 2.6.2 (i) $\Rightarrow $ (ii) (i.e., we need to verify that $f|_{A_1B_1}$ is a partial elementary ${\mathcal L}$ -automorphism (this is clear by assumption) and that f is a partial ${\mathcal L}_I$ -isomorphism). This follows from the restrictions $f|_{A_1B_1c_1}$ and $f|_{\bigcup _n\mathrm {RV}_{nI}}$ being partial ${\mathcal L}_I$ -isomorphisms, since the only ${\mathcal L}_I$ -interaction between $A_1c_1$ and $\bigcup _n\mathrm {RV}_{nI}$ is via $B_1$ .

Proof. (ii) $\Rightarrow $ (i) is trivial, so we assume (i) and prove (ii). Let ${\mathcal L}' \supset {\mathcal L}$ be an $\bigcup _n\mathrm {RV}_{nI}$ -expansion and suppose that $X \subset K$ is ${\mathcal L}'(A)$ -definable, for some $A \subset K$ . We need to find a finite ${\mathcal L}(A)$ -definable $C \subset K$ $mI$ -preparing X for some $m \ge 1$ . We may replace K by a sufficiently saturated elementary extension. We may also assume that $A = \mathrm {acl}_{{\mathcal L},K}(A)$ (since every finite $\mathrm {acl}_{{\mathcal L},K}(A)$ -definable set is contained in a finite A-definable set). Now we are in the setting of Lemma 2.6.2. Applying the lemma for ${\mathcal L}'$ shows that it suffices to verify that every partial ${\mathcal L}_I(\tilde A)$ -isomorphism $f\colon \{c_1\} \to \{c_2\}$ is an elementary ${\mathcal L}'(\tilde A)$ -isomorphism, for $\tilde A :=A \cup \bigcup _n\mathrm {RV}_{nI}$ , and applying the lemma for ${\mathcal L}$ shows that such an f is an elementary ${\mathcal L}(\tilde A)$ -isomorphism. Thus, we can finish the proof by showing (more generally) that any elementary ${\mathcal L}(\tilde A)$ -isomorphism $f\colon B_1 \to B_2$ is already an elementary ${\mathcal L}'(\tilde A)$ -isomorphism, for small sets $B_1, B_2 \subset K$ . We show that such f preserve every ${\mathcal L}'(\tilde A)$ -formula, by induction over the structure of the formula. For quantifier free formulas, this is clear. To deal with an existence quantifier, use Lemma 2.6.3 to extend f to a realization of the existential quantifier.

Proof. (i) follows easily from Proposition 2.6.4 (applied in sufficiently saturated models $K \models {\mathcal T}$ ), using $I = {\mathcal M}_K$ in the case $\ell = 0$ , and using $I = B_{<\lambda }(0)$ for every $\lambda \le 1$ in the case $\ell = \omega $ . In the latter case, we need to first add $\lambda $ to the language and then get rid of it again using that $\mathrm {RV}$ -unions preserve finiteness (Corollary 2.3.6 (1)). The full argument is as follows: to $\lambda $ -prepare a set $X \subset K$ which is $A \cup \mathrm {RV}_{\lambda }$ -definable (where $A \subset K$ ) in the ⋃ _n RV _n -expansion of K, apply Proposition 2.6.4 with $I = B_{<\lambda }(0)$ in the language ${\mathcal L}(\lambda )$ to get a finite ${\mathcal L}(A,\lambda )$ -definable set $C' = \phi (K,\lambda )$ $\lambda $ -preparing X. By Corollary 2.3.6 (1), the ${\mathcal L}(A)$ -definable set $C = \bigcup _{\lambda '} \phi (K,\lambda ')$ is still finite (and it $\lambda $ -prepares X since it contains $C'$ ).

For (ii), let ${\mathcal L}'$ be an $(\bigcup _n\mathrm {RV}_{|n|})$ -expansion of the language ${\mathcal L}$ , and let K be a model of the expansion of ${\mathcal T}$ . First of all, by part (i) (and Lemma 2.4.1 (2)) we know that $\operatorname {Th}_{{\mathcal L}'}(K)$ is still $0^{\mathrm {mix}}$ -h-minimal. Let $f: K\to K$ be ${\mathcal L}'(A)$ -definable for some $A\subset K\cup \mathrm {RV}_n$ . By Proposition 2.6.4, the ${\mathcal L}'(A \cup \{x\})$ -definable set $\{f(x)\}$ can be prepared by a finite ${\mathcal L}(A \cup \{x\})$ -definable set $C_x$ ; note that this implies $f(x) \in C_x$ . We may assume (using compactness) that $C_x$ is definable uniformly in x. By Lemma 2.3.1, we find an ${\mathcal L}(A)$ -definable family of injective maps $g_x: C_x\to \mathrm {RV}_m^k$ . We define $h: K\to \mathrm {RV}_m^k: x\mapsto g_x(f(x))$ and

$$\begin{align*}\tilde{f}: K\times \mathrm{RV}_m^k\to K: (x, \zeta)\mapsto \begin{cases} g_x^{-1}(\zeta) & \text{ if } \zeta\in g_x(C_x), \\ 0 & \text{ else.} \end{cases} \end{align*}$$

Note that $f(x) = \tilde {f}(x, h(x))$ for every $x\in K$ and that h is ${\mathcal L}'(A)$ -definable while $\tilde {f}$ is ${\mathcal L}(A)$ -definable. By Lemma 2.3.1, we can find an ${\mathcal L}(A)$ -definable family of finite sets $C_{\xi }$ , for $\xi \in \mathrm {RV}_m^k$ such that the map $x\mapsto \tilde {f}(x, \xi )$ satisfies (T1 $^{\mathrm {mix}}$ ) with integer p and (T2) with respect to $C_{\xi }$ . Let $C_1$ be the union of all of these $C_{\xi }$ , which is still finite by Lemma 2.4.1. By $0^{\mathrm {mix}}$ -h-minimality, there exists a finite ${\mathcal L}'(A)$ -definable set $C_2\subset K$ and a positive integer $p'$ such that h is constant on the balls $|p'|$ -next to $C_2$ . Now take $C = C_1\cup C_2$ and let $q = pp'$ . Then C is a finite ${\mathcal L}'(A)$ -definable set for which f satisfies properties (T1 $^{\mathrm {mix}}$ ) and (T2) with respect to C (with integer q).

Proof of Theorem 2.2.8.

1. (1) implies (2). This is Corollary 2.5.5.

2. (2) implies (3). This is Corollary 2.3.10.

3. (3) implies (4). Let $|\cdot |_c$ be a nontrivial equicharacteristic zero coarsening of the valuation $|\cdot |$ .

The coarsened valuation ring ${\mathcal O}_{K,c}$ is a pullback of some subset of $\mathrm {RV}$ . Hence, by Item (ii) of Proposition 2.6.5, the theory of K in ${\mathcal L}' = {\mathcal L}\cup \{{\mathcal O}_{K,c}\}$ still satisfies (T1,T2). We work in this language ${\mathcal L}'$ .

We will use [Reference Cluckers, Halupczok and Rideau10, Theorem 2.9.1], which says that the theory of K in ${\mathcal L}_c$ with respect to the norm $|\cdot |_c$ is $1$ -h-minimal if for every set $A\subset K\cup \mathrm {RV}_c$ and every A-definable function $f: K\to K$ , the following hold:

1. (1) there exists some finite A-definable set C such that for every ball B which is $|1|_c$ -next to C, there is a $\mu _B\in \Gamma _{K,c}$ such that for $x,y\in B$ ,

   $$\begin{align*}|f(x)-f(y)|_c = \mu_B |x-y|_c, \text{ and} \end{align*}$$

2. (2) the set $\{y\in K\mid f^{-1}(y) \text { is infinite}\}$ is finite.

Let us say that a finite set C $\mathfrak {m}_c$ -prepares a function f if (1) holds for f and C.

So let $f: K\to K$ be A-definable for some $A\subset K\cup \mathrm {RV}_{c}$ . By adding parameters from K to the language, we may as well assume that $A\subset \mathrm {RV}_c$ . We will find a finite $\emptyset $ -definable C which $\mathfrak {m}_c$ -prepares f. We induct on $r=\# A$ . The case $r = 0$ is clear. Indeed, in that case, there even exists a finite ${\mathcal L}$ -definable set C such that f satisfies (T1) with respect to $C,m$ . In particular, C will $\mathfrak {m}_c$ -prepare f.