Proof of Lemma 2.4.1

We verify Definition 2.3.3: Let $K^{\prime }\models \operatorname {Th}_{{\mathcal L}(A)}(K)$ , and let $X\subset K^{\prime }$ be ${\mathcal L}(A\cup A^{\prime }\cup \mathrm {RV}_{K^{\prime }}\cup A^{\prime \prime })$ -definable for some $A^{\prime } \subset K^{\prime }$ and some $A^{\prime \prime } \subset \mathrm {RV}_{K^{\prime },\lambda }$ satisfying $\#A^{\prime \prime } \le \ell $ , with $\lambda \le 1$ in $\Gamma _{K^{\prime }}$ . Choose $\tilde A \subset \mathrm {RV}_{K^{\prime }}$ such that every element of $A \cap \mathrm {RV}_{K^{\prime }}^{\mathrm {eq}}$ is ${\mathcal L}(\tilde A)$ -definable. Then X is ${\mathcal L}((A \cap K) \cup \tilde {A}\cup A^{\prime }\cup \mathrm {RV}_{K^{\prime }}\cup A^{\prime \prime })$ -definable, so by $\ell $ -h-minimality of $\operatorname {Th}_{{\mathcal L}}(K^{\prime })$ , X can be $\lambda $ -prepared by a finite ${\mathcal L}((A \cap K) \cup A^{\prime })$ -definable set C. In particular, C is ${\mathcal L}(A\cup A^{\prime })$ -definable, as desired.

Proof. If $\phi (x, q)$ defines $X_{q}$ and $\psi (z, q)$ defines $C_{q}$ , then the above set of pairs $(q,\lambda )$ is defined by the following ${\mathcal L}$ -formula:

$$ \begin{align*} \forall x,x^{\prime}\colon \big(\underbrace{\forall z\colon (\psi(z, q) \to \operatorname{rv}_{\lambda}(x - z) = \operatorname{rv}_{\lambda}(x^{\prime} - z))}_{\text{i.e., }(\operatorname{rv}_{\lambda}(x-c))_{c\in C_{q}} = (\operatorname{rv}_{\lambda}(x^{\prime}-c))_{c\in C_{q}}} \to (\phi(x, q) \leftrightarrow \phi(x^{\prime}, q))\big). \end{align*} $$

Proof. We need to show that if K satisfies (2.3.1), then so does any other model $K^{\prime }$ of $\operatorname {Th}(K)$ .

Suppose for contradiction that $K^{\prime }$ is a model not satisfying (2.3.1): that is, there exist a $\lambda \le 1$ in $\Gamma ^{\times }_{K^{\prime }}$ , tuples $a \in (K^{\prime })^{n}$ , $\zeta \in \mathrm {RV}_{K^{\prime }}^{n^{\prime }}$ , $\xi \in \mathrm {RV}_{K^{\prime },\lambda }^{n^{\prime \prime }}$ with $n, n^{\prime }$ arbitrary and $n^{\prime \prime } \le \ell $ , and an $(a, \zeta , \xi )$ -definable set $X = \phi (K^{\prime }, a, \zeta , \xi ) \subset K^{\prime }$ such that no finite non-empty a-definable set $C \subset K \ \lambda $ -prepares X.

For fixed $\phi $ , the non-existence of C can be expressed by an infinite conjunction of ${\mathcal L}$ -formulas in $(\lambda , a, \zeta , \xi )$ . Indeed, for every formula $\psi (z, a)$ that could potentially define C and for every integer $k \ge 1$ , there is (by Lemma 2.4.2) an ${\mathcal L}$ -formula $\chi _{\psi }(\lambda , a, \zeta , \xi )$ expressing ‘ $\psi (K^{\prime }, a)$ has cardinality k and $\psi (K^{\prime }, a)$ does not $\lambda $ -prepare $\phi (K^{\prime }, a, \zeta , \xi )$ .’

The fact that this partial type $\{\chi _{\psi } \mid \psi \text { as above}\}$ is realized in $K^{\prime }$ implies that it is also realized in K, so that Condition (2.3.1) fails in K.

Proof of Lemma 2.4.4

‘ $\Rightarrow $ ’: Suppose for contradiction that B contains two elements $x, x^{\prime }$ with $\mathrm {tp}(x/A \cup \mathrm {RV}) \ne \mathrm {tp}(x^{\prime }/A \cup \mathrm {RV})$ . This means that there exists an $(A \cup \mathrm {RV})$ -definable set X containing x but not $x^{\prime }$ . By $0$ -h-minimality, there exists a finite A-definable set $C \ 1$ -preparing X. In particular, $C \subset \mathrm {acl}_{K}(A)$ and hence $C \cap B = \emptyset $ . However, by Example 2.1.9, this implies that B is either contained in X or disjoint from X, contradicting the properties of x and $x^{\prime }$ .

‘ $\Leftarrow $ ’: Let X be $(A \cup \mathrm {RV})$ -definable for some parameter set $A \subset K$ that we may assume to be finite, and suppose that no finite A-definable $C \subset K \ 1$ -prepares X. This means (using Example 2.1.9 again) that for every finite A-definable $C \subset K$ , there exists an open ball $B \subset K$ that is disjoint from C and such that neither $B \subset X$ nor $B \subset K \setminus X$ . Taking all those conditions on B together (for all finite A-definable C), we obtain a (finitely satisfiable) type in an imaginary variable running over the open balls in K. A realization of this type is a ball B that is disjoint from $\mathrm {acl}_{K}(A)$ on the one hand but, on the other hand, contains elements x and $x^{\prime }$ that satisfy $x \in X$ and $x^{\prime } \notin X$ and hence have different types over $A \cup \mathrm {RV}$ .

Assumption 2.5.1. For every $K^{\prime } \equiv K$ , every definable (with any parameters) subset of $K^{\prime }$ can be $1$ -prepared by a finite set C. (We do not impose definability conditions on C.)

Proof. If X is infinite, it is not contained in the finite set C preparing it, which implies that it contains a ball. The definability of $Q^{\prime }$ then follows since the condition that a set contains an open ball can be expressed by a formula. The existence of the bound N then follows by compactness: If no bound would exist, then in a sufficiently saturated model, we would find a $q \in Q^{\prime }$ with $\#X_{q}> N$ for every N.

Proof of Lemma 2.5.3

Using Lemma 2.5.2, we can bound the cardinality $\#C_{q}$ and then assume that it is constant. We do an induction over $\#C_{q}$ .

If $C_{q}$ is always a singleton or empty, we can define $f_{q}$ to always be constant. Otherwise, the lemma is obtained by repeatedly taking averages of the elements of $C_{q}$ and subtracting. More precisely, setting $a_{q} := \frac 1{\#C_{q}}\sum _{x \in C_{q}} x$ , we get that the map $\hat f_{q}\colon C_{q}\to \mathrm {RV}, x \mapsto \operatorname {rv}(x - a_{q})$ is not constant on $C_{q}$ . Therefore, each fiber $\hat f^{-1}_{q}(\xi )$ of $\hat f$ (for $\xi \in \hat f_{q}(C_{q})$ ) has cardinality less than $C_{q}$ , so by induction, we obtain a definable family of injective maps $g_{q,\xi }\colon \hat f^{-1}_{q}(\xi ) \to \mathrm {RV}^{k}$ . Now set $f_{q}(x) := (\hat f_{q}(x), g_{q,\hat f_{q}(x)}(x))$ .

Proof. Given $x \in K$ , let $\mu (x) := \min \{|x-c| \mid c \in C\}$ be the minimal distance to elements of C, let $C(x) := \{c \in C \mid |x-c| = \mu (x)\}$ be the set $c \in C$ realizing that distance, and let $a(x) := \frac 1{\#C(x)}\sum _{c \in C(x)} c$ be the average of those elements. Note that the map $a\colon K \to K$ has a finite image. Using Lemma 2.5.3, we find an injective map $\alpha $ from the image of a to $\mathrm {RV}^{k}$ . If $\lambda < 1$ , we define

$$ \begin{align*} f(x) := (\alpha(a(x)), \operatorname{rv}_{\lambda}(x - a(x))). \end{align*} $$

In the case $\lambda = 1$ , to obtain the more precise statement, we define

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

It is now just a computation to check that the lemma holds. For the first part, suppose that $f(x_{1}) = f(x_{2}) = (\zeta , \xi ) \in \mathrm {RV}^{k} \times \mathrm {RV}_{\lambda }$ for some $x_{1}, x_{2} \in K$ ; our aim is to show that either $x_{1} = x_{2} \in C$ or they both lie in the same ball $\lambda $ -next to C. Set $a_{0} := a(x_{1}) = a(x_{2})$ .

If $\xi = 0$ , then either $x_{i} = a_{0}$ or $|x_{i} - a_{0}| < \mu (x_{i})$ . The latter is only possible if $a_{0} \notin C$ , and it implies that $x_{i}$ and $a_{0}$ lie in the same ball $1$ -next to C. This shows: Either $x_{1} = x_{2} \in C$ or $\lambda = 1$ , and both lie in the same ball $1$ -next to C.

If $\xi \ne 0$ , then the fact that $|x_{i} - a(x)| \le \mu (x_{i})$ shows that $\operatorname {rv}_{\lambda }(x_{i} - a(x))$ determines $\operatorname {rv}_{\lambda }(x_{i} - c)$ for each $c \in C$ , so we are done with the first part of the lemma.

For the second part, pick $x_{1}, x_{2}$ in the same ball $\lambda $ -next to C. Then one obtains $\mu (x_{1}) = \mu (x_{2}) =: \mu _{0}$ , $C(x_{1}) = C(x_{2}) =: C_{0}$ and $a(x_{1}) = a(x_{2}) =: a_{0}$ . If $|x_{i} - a_{0}| = \mu $ , then $\operatorname {rv}(x_{i} - a_{0})$ is determined by $\operatorname {rv}(x_{i} - c)$ for any $c \in C_{0}$ , so $f(x_{1}) = f(x_{2})$ . Otherwise, if $|x_{i} - a_{0}| < \mu $ , then $f(x_{i}) = (\alpha (a_{0}), 0)$ for $i = 1,2$ by definition.

Proof. For each $\lambda $ and each $\xi \in \mathrm {RV}^{k}\times \mathrm {RV}_{\lambda }^{\ell }$ , let $C_{\xi }$ be a finite A-definable set $\lambda $ -preparing $W_{\xi }$ . By a usual compactness argument (see Remark 2.6.3 below), we may suppose that $C := \bigcup _{\xi } C_{\xi }$ is finite and A-definable. It prepares each $W_{\xi }$ for each $\xi $ and hence also W.

Proof of Corollary 2.6.6

Clear.

Proof. We suppose that $W\subset \mathrm {RV}^{k} \times \mathrm {RV}_{\bullet }^{k^{\prime }} \times K$ and proceed by induction on $k^{\prime }$ . If $k^{\prime } \le 1$ , we obtain both claims of the corollary from Proposition 2.6.2 applied with $\ell = k^{\prime }$ , namely: We find a finite set C such that, for all $\lambda $ and $\xi \in \mathrm {RV}^k\times \mathrm {RV}_\lambda ^{k'}$ , $W_{\xi }$ is $\lambda $ -prepared by C. Since $W_{\xi }$ is finite, we have $W_\xi \subset C$ , and hence $\bigcup _\xi W_\xi \subset C$ is finite.

The case of $k^{\prime }>1$ now follows by induction on $k^{\prime }$ and the case $k^{\prime }=1$ .

Proof. Apply Corollary 2.6.7 to the graph of f.

Proof. Add constants for A to the language. We have $C = W_ {\xi _{0}}$ for some $\emptyset $ -definable $W \subset K \times \mathrm {RV}^{k}$ and some $\xi _{0} \in \mathrm {RV}^{k}$ . We may assume that all fibers $W_{\xi }$ have cardinality at most the cardinality of C. Let $C^{\prime }$ be their union, which is finite by Corollary 2.6.7.

Proof. We may assume that A is finite; we do an induction on the cardinality of A.

Let $A = \hat A \cup \{a\}$ . Then we have an $\hat A$ -definable set $Y \subset K \times \mathrm {RV}^{n}$ such that X is equal to the fiber $Y_{a} \subset \mathrm {RV}^{n}$ . By applying Corollary 2.6.6 to Y, we find a finite $\hat A$ -definable set $C \subset K$ such that either $a \in C$ or, for every $a^{\prime } \in K$ in the same ball $1$ -next to C as a, we have $Y_{a^{\prime }} = Y_{a}$ . Using Lemma 2.5.4, we find an $\hat A$ -definable map $f\colon K \to \mathrm {RV}^{k}$ (for some k) whose fibers are exactly the elements of C and the balls $1$ -next to C. In particular, the set $X = Y_{a}$ is definable using $\hat A$ and $f(a)$ as parameters. Thus we have $X = Z_{f(a)}$ for some $\hat A$ -definable set $Z \subset \mathrm {RV}^{k} \times \mathrm {RV}^{n}$ . By induction, Z is $\mathrm {dcl}_{\mathrm {RV}}(\hat A)$ -definable, so X is $\mathrm {dcl}_{\mathrm {RV}}(\hat A)\cup \{f(a)\}$ -definable and hence $\mathrm {dcl}_{\mathrm {RV}}(A)$ -definable.

Proof. First we suppose that $Q \subset \Gamma ^{\times }_{K}$ and that each $B_{q}$ is an open ball of radius q. By Corollary 2.6.6 (and Remark 2.6.9), there exists a finite set $C \ 1$ -preparing the family of balls $B_{q}$ . Now one checks that at least one of the following two situations occurs. First: For each $q\in Q$ , the intersection of C with $B_{q}$ is non-empty. Second: The set Q has a minimum $q_{0}\in Q$ . (Indeed, suppose that the intersection of C with $B_{q_{0}}$ is empty for some $q_{0}\in Q$ ; then Q contains no $q < q_{0}$ , since $B_{q}$ would not be $1$ -prepared by C.) In both situations, the lemma follows.

Finally, we reduce the general case to the case that $Q\subset \Gamma ^{\times }_{K}$ and that each $B_{q}$ is an open ball of radius q. To this end, for $\gamma \in \Gamma ^{\times }_{K}$ , let $B^{\prime }_{\gamma }$ be the (necessarily unique) open ball of radius $\gamma $ containing some $B_{q}$ ( $q \in Q$ ) if such a ball exists, and let it be the empty set otherwise. Then it is clear that the non-empty $B^{\prime }_{\gamma }$ form a nested definable family of open balls. Moreover, the intersection of the non-empty $B^{\prime }_{\gamma }$ equals the intersection of the $B_{q}$ (since each $B_{q}$ is equal to the intersection of all open balls containing $B_{q}$ ).

Proof of Theorem 2.7.2

Let $P\in {\mathcal O}_{K}[X]$ be a polynomial such that $P(0)\in {\mathcal M}_{K}$ and $P^{\prime }(0)\in {\mathcal O}^{\times }_{K}$ ; we need to prove that P has a root in ${\mathcal M}_{K}$ . (Note that the uniqueness of such a root then follows automatically.) The idea is to use ‘Newton approximation’ as in the usual proof of Hensel’s lemma for complete discretely valued fields, but where complete and discretely valued is replaced by definably spherically complete.

To make this formal, we suppose that P has no root in ${\mathcal M}_{K}$ and we set $B_{x} := B_{\le |P(x)|}(x)$ for $x \in {\mathcal M}_{K}$ . Note that the $B_{x}$ form a definable family of balls, parameterized by $x \in {\mathcal M}_{K}$ . We will prove that (a) all these balls form a chain under inclusion and (b) that an element in the intersection of all those balls (which is non-empty by Lemma 2.7.1) is, after all, a root of P.

(a) Let $x_{1}, x_{2} \in {\mathcal M}_{K}$ be given, and set $\varepsilon := x_{2} - x_{1}$ . To see that the balls $B_{x_{1}}$ and $B_{x_{2}}$ are not disjoint, we verify that $|\epsilon | \le \max \{|P(x_{1})|, |P(x_{2})|\}$ . Taylor expanding P around $x_{1}$ yields

(2.7.1) $$ \begin{align} |P(x_{1}+\varepsilon) - P(x_{1}) - \varepsilon P^{\prime}(x_{1})| \le |\varepsilon^{2}|, \end{align} $$

which, together with $|P^{\prime }(x_{1})| = 1$ , implies $|\epsilon | \le \max \{|P(x_{1})|, |P(x_{1} + \epsilon )|\}$ .

(b) Let $x_{1}$ be in the intersection $\bigcap _{x \in {\mathcal M}_{K}} B_{x}$ , and suppose that $P(x_{1}) \ne 0$ . Then equation (2.7.1) with $\varepsilon := - \frac {P(x_{1})}{P^{\prime }(x_{1})}$ implies $|P(x_{1} + \varepsilon )| \le |\varepsilon ^{2}| < |\varepsilon |$ and hence $x_{1} \notin B_{x_{1} + \varepsilon }$ , contradicting our choice of $x_{1}$ .

Proof. We may assume f to be $\emptyset $ -definable (say, after adding enough parameters from K to the language). Suppose f takes infinitely many values y infinitely many times. Then for each such y, $f^{-1}(y)$ contains a ball (by Lemma 2.5.2). Thus, letting $X \subset K$ be the set of points where f is locally constant, $f(X)$ is still infinite.

Let $W\subset K\times \Gamma ^{\times }_{K}$ consist of those $(x,\lambda )$ such that f is constant on $B_{<\lambda }(x)$ . This set W is $\emptyset $ -definable, so we find a finite set $C \ 1$ -preparing W (by Corollary 2.6.6). By enlarging C, we may moreover assume that C also $1$ -prepares X.

From the fact that $f(X)$ is infinite, we can deduce that there exists a ball $B_{0} \subset X \ 1$ -next to C such that $f(B_{0})$ is still infinite. Indeed, letting $g\colon K \to \mathrm {RV}^{k}$ be a map whose fibers are the singletons in C and the balls $1$ -next to C (using Lemma 2.5.4), if $f(g^{-1}(\xi ))$ would be finite for every $\xi \in g(X)$ , then so would be $f(X)$ (by Corollary 2.6.7).

Choose $x \in B_{0}$ and $\lambda _{0} \in \Gamma ^{\times }_{K}$ such that f is constant on $B_{<\lambda _{0}}(x)$ . Since $C \ 1$ -prepares W, f is constant on $B_{<\lambda _{0}}(x^{\prime })$ for every $x^{\prime } \in B_{0}$ .

Set $\lambda _{1} := \lambda _{0}/\operatorname {rad}_{\mathrm {op}}(B_{0})$ . Then the family $F_{1}$ of open balls of radius $\lambda _{0}$ contained in $B_{0}$ can be definably parametrized by a subset of $\mathrm {RV}_{\lambda _{1}}$ (using some parameters). Indeed, if we fix $c \in K$ such that $B_{0}$ is $1$ -next to c, then each member of $F_{1}$ is of the form $c + \operatorname {rv}_{\lambda _{1}}^{-1}(\xi )$ for some $\xi \in \mathrm {RV}_{\lambda _{1}}$ . We define $F_{2}$ to be the family of $f(B)$ , for B in $F_{1}$ . Then each family member of $F_{2}$ is a singleton, yet their union is infinite, contradicting Corollary 2.6.7.

Proof. First, consider the set $Y_{\infty }$ of $y \in K$ such that $f^{-1}(y)$ is infinite. By Lemma 2.8.1, this set $Y_{\infty }$ is finite, so we can find a finite $\emptyset $ -definable set $C \ 1$ -preparing $f^{-1}(y)$ for each $y \in Y_{\infty }$ . Indeed, for each $y \in Y_{\infty }$ , one finds a finite y-definable set $C_{y} \ 1$ -preparing $f^{-1}(y)$ (uniformly in y), and one lets C be the union of the sets $C_{y}$ .

For each $y \in K \setminus Y_{\infty }$ , the set $f^{-1}(y)$ is finite, so by Lemma 2.5.3, there exists a $\emptyset $ -definable family of injective functions $g_{y}\colon f^{-1}(y) \to \mathrm {RV}^{k}$ for some $k \ge 1$ . For convenience, we set $g_{y}(x) := 0$ if $y \in Y_{\infty }$ and $x \in f^{-1}(y)$ , so that we can define a function $h\colon K \to \mathrm {RV}^{k}$ by $h(x) := g_{f(x)}(x)$ . We then enlarge our above set C (using Corollary 2.6.6) so that it also $1$ -prepares (the graph of) h.

We claim that this set C is as desired, so let B be a ball $1$ -next to C. If $f(B) \cap Y_{\infty } \ne \emptyset $ , then B is contained in one of the sets $f^{-1}(y)$ , for $y \in Y_{\infty }$ , and hence f is constant on B. Otherwise, we use that h is constant on B to deduce that f is injective on B. Indeed, $f(x_{1}) = f(x_{2}) = y$ for some $y \in K \setminus Y_{\infty }$ implies $h(x_{1}) = g_{y}(x_{1}) = g_{y}(x_{2}) = h(x_{2})$ , so injectivity of h on $f^{-1}(y)$ implies $x_{1} = x_{2}$ .

Proof. Define $W \subset K \times \Gamma ^{\times }_{K}$ to consist of those $(x, \lambda )$ for which the open ball $B_{<\lambda }(x)$ ‘has a good image’: that is, $f(B_{<\lambda }(x))$ is a singleton or an open ball. By Corollary 2.6.6, we find a finite $\emptyset $ -definable subset $C_{0}$ of K such that for any ball $B \ 1$ -next to $C_{0}$ , the fiber $W_{x} \subset \Gamma ^{\times }_{K}$ does not depend on $x \in B$ .

Fix a ball $B_{0} \ 1$ -next to $C_{0}$ . We first prove that any open ball B strictly contained in $B_{0}$ has a good image. Suppose otherwise, namely that $B_{1}$ is an open ball strictly contained in $B_{0}$ with bad image. Then the fact that $W_{x}$ does not depend on $x \in B_{0}$ implies that every translate of $B_{1}$ contained in $B_{0}$ also has a bad image. We can find an infinite definable (with parameters) family $F_{1}$ consisting of such translates of $B_{1}$ and parameterized by a subset of $\mathrm {RV}$ . Indeed, the sets $x_{0} + \operatorname {rv}^{-1}(\xi )$ form such a family for a suitable $x_{0} \in K$ and when $\xi $ runs over a suitable subset of $\mathrm {RV}$ .

Consider the family $F_{2}$ of the sets $f(B)$ for B in $F_{1}$ , use Corollary 2.6.6 to find a finite set $D \subset K \ 1$ -preparing the family $F_{2}$ , and let $g\colon K \to \mathrm {RV}^{k}$ be a function whose fibers are the balls $1$ -next to D and the individual points of D (as obtained using Lemma 2.5.4). Since none of the balls B in $F_{1}$ are good, none of the sets $f(B)$ in $F_{2}$ are exactly a point or an open ball, so each $f(B)$ consists of several fibers of g; in other words, $g \circ f\colon K \to \mathrm {RV}^{k}$ is non-constant on every B in $F_{1}$ .

Now we get a contradiction by applying Corollary 2.6.6 to the graph of $g \circ f$ . Indeed, any set $C^{\prime } \ 1$ -preparing that graph would have to contain at least one point in each ball B from $F_{1}$ (since $g \circ f$ is not constant on B), so $C^{\prime }$ cannot be finite. This finishes the proof that balls strictly contained in $B_{0}$ have a good image.

The only problematic open balls left (i.e., which are disjoint from $C_{0}$ and might have bad image) are the ones $1$ -next to $C_{0}$ . To get hold of those, we run a similar argument as above: We let $F_{1}$ be the family of balls $1$ -next to $C_{0}$ ; we find a finite set $D \ 1$ -preparing $f(B)$ for each B in $F_{1}$ ; and we define $g\colon K \to \mathrm {RV}^{k}$ as before, so that in particular if $g \circ f$ is constant on a ball $B \ 1$ -next to $C_{0}$ , then that ball B has good image.

Now we find a finite set $C^{\prime } \ 1$ -preparing the graph of $g \circ f$ and we set $C := C_{0} \cup C^{\prime }$ . In this way, among the balls $1$ -next to $C_{0}$ , all those that have bad image are not disjoint from C. Note that since this time, $F_{1}$ is $\emptyset $ -definable, and hence so are D, g and $C^{\prime }$ .

Proof. (1) The family of all open balls of radius $\alpha $ contained in B can be (definably) parametrized by a definable set $\Lambda \subset \mathrm {RV}_{\alpha }$ , namely, set $\Lambda := \operatorname {rv}_{\alpha }(B)$ , and for $\xi \in \Lambda $ , let $B_{\xi }$ be the open ball of radius $\alpha $ containing $\operatorname {rv}_{\alpha }^{-1}(\xi )$ (which exists, since $\operatorname {rad}_{\mathrm {op}}(\operatorname {rv}_{\alpha }^{-1}(\xi )) \le \alpha $ ). By Proposition 2.6.2, there exists a finite set C such that for each $\xi \in \Lambda $ , the set $f(B_{\xi })$ is $\alpha $ -prepared by C. This implies that for every open ball $B^{\prime }$ that is $\alpha $ -next to C and every $\xi \in \Lambda $ , one has

$$ \begin{align*}\text{either } B^{\prime}\subset f(B_{\xi}), \quad\text{or } B^{\prime}\cap f(B_{\xi})=\emptyset. \end{align*} $$

Hence, if the radius of the open ball $B^{\prime }$ is larger than $\beta $ , then $B^{\prime }\cap f(B_{\xi })$ is empty for all $\xi \in \Lambda $ . (Indeed, by assumption, $f(B_{\xi })$ is contained in an open ball of radius $\beta $ , so $f(B_{\xi })$ cannot contain the larger ball $B^{\prime }$ .)

Thus, we find that $f(B)$ is contained in the union of C with those open balls $\alpha $ -next to C that have a radius at most $\beta $ . This union equals the union over $c\in C$ of the closed balls $B_{\le \beta /\alpha }(c)$ . This proves (1).

(2) If the value group is dense, then (2) follows from (1) using that the largest open balls contained in a finite union of closed balls of radius $\beta /\alpha $ have radius $\beta /\alpha $ .

If the value group is discrete (see just above Remark 2.1.5), we apply Part (1) to $g(x) := f(\varpi x)\colon {\mathcal O}_{K} \to K$ , where $\varpi \in K$ is a uniformizing element. Since g sends open balls of radius $|\varpi |^{-1} \cdot \alpha $ to open balls of radius $\beta $ , we obtain that $f({\mathcal M}_{K}) = g({\mathcal O}_{K})$ is contained in a finite union of closed balls of radius $|\varpi | \cdot \beta /\alpha $ , which is the same as a finite union of open balls of radius $\beta /\alpha $ . Now the claim follows from the assumption that $f({\mathcal M}_{K})$ is itself an open ball.

Proof. Choose $C \ \emptyset $ -definable such that for every ball $B \ 1$ -next to C, we have:

• • f is constant or injective on B (using Lemma 2.8.2);

• • $f(B^{\prime })$ is a point or an open ball for every open ball $B^{\prime } \subset B$ (using Lemma 2.8.3).

Moreover, we assume (using Corollary 2.6.6) that $C \ 1$ -prepares the graph of the function $r\colon K \times \Gamma ^{\times }_{K} \to \Gamma _{K}$ defined by

$$ \begin{align*}r(x,\lambda) = \begin{cases} \operatorname{rad}_{\mathrm{op}}(f(B_{<\lambda}(x))) & \text{if } f(B_{<\lambda}(x)) \text{ is an open ball}\\ 0 & \text{otherwise}. \end{cases} \end{align*} $$

We claim that this C is as desired, so fix a ball $B \ 1$ -next to C for the remainder of the proof. Note that for x running over B, the function $r(x,\cdot )$ is independent of x, so from now on we write $r(\lambda )$ instead of $r(x, \lambda )$ .

If f is constant on B, there is nothing to do, so we may assume that f is injective on B. We claim that the lemma holds with $\mu _{B} := \operatorname {rad}_{\mathrm {op}}(f(B))/\operatorname {rad}_{\mathrm {op}}(B)$ . (Note that indeed, $f(B)$ is an open ball.)

(1) Fix $\alpha \in \Gamma _{K}$ with $0 < \alpha < \operatorname {rad}_{\mathrm {op}}(B)$ . Any open ball $B^{\prime } \subset B$ of radius $\alpha $ is sent to an open ball $f(B^{\prime })$ of radius $r(\alpha )$ . By applying Lemma 2.8.4 (to a suitably rescaled function), we deduce that the radius of the open ball $f(B)$ is at most $\operatorname {rad}_{\mathrm {op}}(B)\cdot r(\alpha )/\alpha $ ; this implies $r(\alpha )/\alpha \ge \mu _{B}$ .

To get the other inequality, namely $r(\alpha )/\alpha \le \mu _{B}$ , we apply the same argument to the inverse function $f^{-1}\colon f(B) \to B$ . This inverse is well-defined since f is injective on B, so it remains to verify that $f^{-1}$ sends open balls $B^{\prime \prime } \subset f(B)$ of radius $r(\alpha )$ to open balls of radius $\alpha $ . Indeed, choose any $x \in f^{-1}(B^{\prime \prime })$ ; then $f(B_{< \alpha }(x)) = B^{\prime \prime }$ , and hence $f^{-1}(B^{\prime \prime }) = B_{< \alpha }(x)$ .

(2) For every $x_{1}, x_{2} \in B$ , (1) implies $|f(x_{1}) - f(x_{2})| \le \mu _{B}\cdot |x_{1} - x_{2}|$ . Indeed, applying (1) to a ball of the form $B_{<\alpha }(x_{1})$ with $|x_{1} - x_{2}| < \alpha \le \operatorname {rad}_{\mathrm {op}}(B)$ yields $|f(x_{1}) - f(x_{2})| < \mu _{B}\cdot \alpha $ for every $\alpha> |x_{1} - x_{2}|$ , and hence $|f(x_{1}) - f(x_{2})| \le \mu _{B}\cdot |x_{1} - x_{2}|$ . The same argument with f replaced by $f^{-1}\colon f(B) \to B$ yields the other inequality.

Proof of Proposition 2.8.6

First, enlarge $C_{0}$ to a set $C_{1}$ using Lemma 2.8.5 so that on each ball $B \ 1$ -next to $C_{1}$ , f sends open balls to open balls and the quotient

(2.8.1) $$ \begin{align} |f(x_{1}) - f(x_{2})|/|x_{1} - x_{2}| \end{align} $$

is constant. Then let D be a set containing $f(C_{1})$ and preparing the family $f(B)$ , where B runs over the balls $1$ -next to $C_{1}$ . (This family is parametrized by $\mathrm {RV}$ -variables, by Lemma 2.5.4, so Corollary 2.6.6 applies.) Now denote by $X \subset K$ the locus where f is not locally constant, and let C be $\big (X\cap f^{-1}(D)\big ) \cup C_{1}$ . Then clearly C is finite. We claim that these C and D are as desired.

Note that we have $C \supset C_{1}$ , so the only thing that is not clear from the construction is that balls $\lambda $ -next to C are sent to elements of D or balls $\lambda $ -next to D. We first deal with the case $\lambda = 1$ .

Let B be a ball $1$ -next to C, and let $B_{1}$ be the ball $1$ -next to $C_{1}$ containing B. If $f(B_{1})$ is a singleton, then it is an element of D, and we are done. If $f(B_{1})$ is not a singleton, then $B_{1}\subset X$ by the constancy of equation (2.8.1) on $B_{1}$ . If furthermore $f(B_{1}) \cap D = \emptyset $ , then $B_{1} \cap C = \emptyset $ , so $B = B_{1}$ and $f(B_{1})$ is a ball $1$ -next to D. Finally, suppose that $f(B_{1}) \cap D \ne \emptyset $ . Then $B_{1} \cap C \ne \emptyset $ , so B is $1$ -next to some $c \in C \cap B_{1}$ . Then using that equation (2.8.1) is constant on $B_{1}$ , we obtain that $f(B)$ is a ball $1$ -next to $f(c) \in D$ .

Now suppose that $\lambda < 1$ . Any ball $B \ \lambda $ -next to C is contained in a ball $B^{\prime } \ 1$ -next to C. If f is constant on $B^{\prime }$ , we are done; otherwise, $f(B^{\prime })$ is a ball $1$ -next to D, and we deduce that $f(B)$ is $\lambda $ -next to D, using once more that equation (2.8.1) is constant on $B^{\prime }$ .

Proof. We first prove both claims for $k = 1$ .

(1) The composition $f \circ \operatorname {rv}\colon K \to K$ is locally constant everywhere except possibly at $0$ , so the claim follows from (T2).

(2) By Lemma 2.5.2, the cardinality of $C_{\xi }$ is bounded independently of $\xi $ , so we may assume that the cardinality is constant: say, equal to m. Let $\sigma _{1}, \dots , \sigma _{m} \in {\mathbb Z}[x_{1}, \dots , x_{m}]$ be the elementary symmetric functions in m variables, which we consider functions on the set of m-element subsets of K. By (1), for each i, the function $f_{i}(\xi ) := \sigma _{i}(C_{\xi })$ has a finite image. Since $\sigma _{1}(C), \dots , \sigma _{m}(C)$ together determine C, there are only finitely many different sets $C_{\xi }$ , which implies the claim.

Now we deduce (2) for arbitrary k by induction: Given a definable family $C_{\xi , \xi ^{\prime }} \subset K$ , for $\xi \in \mathrm {RV}$ and $\xi ^{\prime } \in \mathrm {RV}^{k}$ , we first obtain that $D_{\xi ^{\prime }} := \bigcup _{\xi \in \mathrm {RV}} C_{\xi , \xi ^{\prime }}$ is finite for every $\xi ^{\prime }$ and then that the entire union is finite.

Finally, we obtain (1) for arbitrary k by applying (2) to the family of singletons $C_{\xi } := \{f(\xi )\}$ .

Proof. Using (T1), we first find an A-definable set C. We consider it a member of an $(A\cap K)$ -definable family of sets $C_{\xi }$ , parametrized by $\xi \in \mathrm {RV}^{k}$ . By Lemma 2.9.4, the union $C^{\prime } := \bigcup _{\xi \in \mathrm {RV}^{k}} C$ is still finite. Moreover, it is $(A\cap K)$ -definable, and since it contains C, it satisfies the requirements of (T1).

For the in-particular part, apply this improved (T1) to the characteristic function of an A-definable set $X \subset K$ (where $A \subset K \cup \mathrm {RV}$ ), to find a finite $(A \cap K)$ -definable $C \ 1$ -preparing X.

Proof. Let C be an $(A\cap K)$ -definable set satisfying (T1). By Lemma 2.5.4, the family of balls $B \ 1$ -next to C can be parametrized using $\mathrm {RV}$ -parameters, so using Corollary 2.6.6, we can prepare the family of $f(B)$ using a finite $(A\cap K)$ -definable set D. Similarly, we find a finite $(A\cap K)$ -definable set $C_{0}$ preparing the family $f^{-1}(B_{1})$ , where $B_{1}$ runs over the balls $1$ -next to D. We claim that the set $C^{\prime } := C \cup C_{0}$ does the job.

So let $B^{\prime } \subset K$ be $1$ -next to $C^{\prime }$ , let B be the ball $1$ -next to C containing $B^{\prime }$ , and let $\mu _{B} \in \Gamma _{K}$ be as in (T1): that is, such that we have

(2.9.1) $$ \begin{align} |f(x_{1}) - f(x_{2})| = \mu_{B} \cdot |x_{1} - x_{2}| \end{align} $$

for all $x_{1}, x_{2} \in B$ . We suppose that $\mu _{B} \ne 0$ (otherwise $f(B^{\prime })$ is a singleton) and we have to show that for every open ball $B^{\prime \prime } \subset B^{\prime }$ , $f(B^{\prime \prime })$ is an open ball of radius $\operatorname {rad}_{\mathrm {op}}(B^{\prime \prime })\cdot \mu _{B}$ .

Let $B^{\prime \prime }_{1}$ be the smallest ball containing $f(B^{\prime \prime })$ . Using equation (2.9.1), one obtains that $B^{\prime \prime }_{1}$ is an open ball with radius $\operatorname {rad}_{\mathrm {op}} B^{\prime \prime }_{1} = \operatorname {rad}_{\mathrm {op}}(B^{\prime \prime })\cdot \mu _{B}$ , so it remains to show that $f(B^{\prime \prime })$ is equal to the entire ball $B^{\prime \prime }_{1}$ .

By our definition of $C_{0}$ , $f(B^{\prime })$ (and hence also $B^{\prime \prime }_{1}$ ) is contained in a ball $B_{1}$ that is $1$ -next to D, and by definition of D (and since $B_{1}$ is certainly not disjoint from $f(B)$ ), $B_{1}$ is contained in $f(B)$ . In particular, the entire ball $B^{\prime \prime }_{1}$ is contained in $f(B)$ . However, using equation (2.9.1) again, we obtain that no element of $B \setminus B^{\prime \prime }$ can be sent into $B^{\prime \prime }_{1}$ , so we deduce $f(B^{\prime \prime }) = B^{\prime \prime }_{1}$ , as desired.