Convention 2.1. For the moment, by definable we mean definable with arbitrary parameters from the structure in question. But later (starting in § 3) we will abandon this practice and work with a fixed set of parameters. The reason for this change will be made abundantly clear when it happens.

Notation 2.4 (Coordinate projections).

For each $n\in \unicode[STIX]{x2115}$ , let $[n]$ abbreviate the set $\{1,\ldots ,n\}$ . For any $E\subseteq [n]$ , we write $\operatorname{pr}_{E}(A)$ for the projection of $A$ into the coordinates contained in $E$ . In practice, it is often more convenient to use simple standard descriptions as subscripts. For example, if $E$ is a singleton $\{i\}$ then we shall always write $E$ as $i$ and $\tilde{E}:=[n]\smallsetminus E$ as $\tilde{i}$ ; similarly, if $E=[i]$ , $\{k:i\leqslant k\leqslant j\}$ , $\{k:i<k<j\}$ , $\{\text{all the coordinates in the sort }S\}$ , etc., then we may write $\operatorname{pr}_{{\leqslant}i}$ , $\operatorname{pr}_{[i,j]}$ , $\operatorname{pr}_{(i,j)}$ , $\operatorname{pr}_{S}$ , etc.; in particular, $A_{\text{VF}}$ and $A_{\text{RV}}$ stand for the projections of $A$ into the $\text{VF}$ -sort and $\text{RV}$ -sort coordinates, respectively.

Unless otherwise specified, by writing $a\in A$ we shall mean that $a$ is a finite tuple of elements (or ‘points’) of $A$ , whose length, denoted by $\operatorname{lh}(a)$ , is not always indicated. If $a=(a_{1},\ldots ,a_{n})$ then, for all $1\leqslant i<j\leqslant n$ , following the notational scheme above, $a_{i}$ , $a_{\tilde{i} }$ , $a_{{\leqslant}i}$ , $a_{[i,j]}$ , $a_{[i,j)}$ , etc., are shorthand for the corresponding subtuples of $a$ . We shall write $\{t\}\times A$ , $\{t\}\cup A$ , $A\smallsetminus \{t\}$ , etc., simply as $t\times A$ , $t\cup A$ , $A\smallsetminus t$ , etc., when no confusion can arise.

For $a\in \operatorname{pr}_{\tilde{E}}(A)$ , the fiber $\{b:(b,a)\in A\}\subseteq \operatorname{pr}_{E}(A)$ over $a$ is denoted by $A_{a}$ . Note that, in the discussion below, the distinction between the two sets $A_{a}$ and $A_{a}\times a$ is usually immaterial and hence they may and often shall be tacitly identified. In particular, given a function $f:A\longrightarrow B$ and $b\in B$ , the pullback $f^{-1}(b)$ is sometimes written as $A_{b}$ as well. This is a special case since functions are identified with their graphs. This notational scheme is especially useful when the function $f$ has been clearly understood in the context and hence there is no need to spell it out all the time.

Notation 2.5 (Subsets and substructures).

By a definable set we mean a definable subset in ${\mathcal{R}}$ , and by a subset in ${\mathcal{R}}$ we mean a subset in $R$ , by which we mean a subset of $R^{n}$ for some $n$ , unless indicated otherwise. Similarly for other structures or sets in place of ${\mathcal{R}}$ that have been clearly understood in the context.

Often the ambient total ordering in ${\mathcal{R}}$ induces a total ordering on a definable set $S$ of interest with a distinguished element $e$ . Then it makes sense to speak of the positive and the negative parts of $S$ relative to $e$ , which are denoted by $S^{+}$ and $S^{-}$ , respectively. Also write $S_{e}^{+}$ for $S^{+}\cup e$ , etc. There may also be a natural absolute value map $S\longrightarrow S_{e}^{+}$ , which is always denoted by $|\cdot |$ ; typically $S$ is a sort and $S^{\times }:=S\smallsetminus e$ is equipped with a (multiplicatively written) group structure, in which case the absolute value map is usually given as a component of a (splitting) short exact sequence

$$\begin{eqnarray}\pm 1\longrightarrow S^{\times }\longrightarrow S^{+}\quad \text{or}\quad S^{+}\longrightarrow S^{\times }\longrightarrow \pm 1.\end{eqnarray}$$

Note that $e$ cannot be the identity element of $S^{\times }$ . We will also write $A<e$ to mean that $A\subseteq S$ and $a<e$ for all $a\in A$ , etc. If $\unicode[STIX]{x1D719}(x)$ is a formula then $\unicode[STIX]{x1D719}(x)<e$ denotes the subset of $S$ defined by the formula $\unicode[STIX]{x1D719}(x)\wedge x<e$ .

Substructures of ${\mathcal{R}}$ are written as ${\mathcal{S}}\subseteq {\mathcal{R}}$ . As has been pointed out above, all substructures ${\mathcal{S}}$ of ${\mathcal{R}}$ are actually elementary substructures. If $A\subseteq {\mathcal{R}}^{n}$ is a set definable with parameters coming from ${\mathcal{S}}$ then $A({\mathcal{S}})$ is the subset in ${\mathcal{S}}$ defined by the same formula, that is, $A({\mathcal{S}})=A\cap {\mathcal{S}}^{n}$ . Given a substructure ${\mathcal{S}}\subseteq {\mathcal{R}}$ and a set $A\subseteq {\mathcal{R}}$ , the substructure generated by $A$ over ${\mathcal{S}}$ is denoted by $\langle {\mathcal{S}},A\rangle$ or ${\mathcal{S}}\langle A\rangle$ . Clearly $\langle {\mathcal{S}},A\rangle$ is the definable closure of $A$ over ${\mathcal{S}}$ . Later, we will expand ${\mathcal{R}}$ and introduce more sorts and structures. In that situation we will write ${\mathcal{S}}\langle A\rangle _{T}$ or $\langle {\mathcal{S}},A\rangle _{T}$ to emphasize that this is the ${\mathcal{L}}_{T\text{}}$ -substructure generated by $A$ over the ${\mathcal{L}}_{T\text{}}$ -reduct of ${\mathcal{S}}$ .

Notation 2.6 (Topology).

The default topology on ${\mathcal{R}}$ is of course the order topology and the default topology on ${\mathcal{R}}^{n}$ is the corresponding product topology. Given a subset $S$ in ${\mathcal{R}}$ , we write $\operatorname{cl}(S)$ for its topological closure, $\operatorname{int}(S)$ for its interior, and $\unicode[STIX]{x2202}S:=\operatorname{cl}(S)\setminus S$ for its frontier (not to be confused with the boundary $\operatorname{cl}(S)\setminus \operatorname{int}(S)$ of $S$ , which is also sometimes denoted by $\unicode[STIX]{x2202}S$ ). The same topological discourse applies to a definable set if the ambient total ordering of ${\mathcal{R}}$ induces a total ordering on it.

Convention 2.9. Semantically we shall treat the value group $\unicode[STIX]{x1D6E4}_{0}$ as an imaginary sort. However, syntactically any reference to $\unicode[STIX]{x1D6E4}_{0}$ may be eliminated in the usual way and we can still work with ${\mathcal{L}}_{T\text{RV}}$ -formulas for the same purpose.

Proof. See [Reference van den Dries and LewenbergVdDL95, Theorem 3.10, Corollary 3.13]. ◻

Proof. Let $({\mathcal{R}},\mathfrak{O})$ be a $T_{\text{convex}}$ -model. It is enough to show that there is a canonical $\text{TCVF}$ -model expansion $({\mathcal{R}},\operatorname{RV}_{0}({\mathcal{R}}))$ of $({\mathcal{R}},\mathfrak{O})$ , where ${\mathcal{R}}$ is the $\text{VF}$ -sort, such that any other such expansion $({\mathcal{R}},\operatorname{RV}_{0})$ is isomorphic to it. This canonical expansion is constructed as follows.

Let $\text{RV}({\mathcal{R}})$ be the quotient group ${\mathcal{R}}^{\times }/(1+\mathfrak{M})$ and $\operatorname{rv}:{\mathcal{R}}^{\times }\longrightarrow \text{RV}({\mathcal{R}})$ the quotient map. As in Example 2.10, it is routine to convert the pair $({\mathcal{R}},\operatorname{RV}_{0}({\mathcal{R}}))$ into an ${\mathcal{L}}_{T\text{RV}}$ -structure and check that it satisfies all the axioms in Definition 2.7, where (Ax. 6) is implied by the construction just described above. We shall refer to the obvious bijection between $({\mathcal{R}},\operatorname{RV}_{0}({\mathcal{R}}))$ and $({\mathcal{R}},\operatorname{RV}_{0})$ as the identity map. This map commutes with all the primitives of ${\mathcal{L}}_{T\text{RV}}$ except, possibly, those in the $\Bbbk$ -sort. This is where the syntactical maneuver in Remark 2.3 comes in. Recall that all the functional primitives of ${\mathcal{L}}_{T^{\prime }\text{}}$ define continuous functions in all the models of $T^{\prime }$ , and $T$ is a definitional extension of $T^{\prime }$ . It follows from (Ax. 8) that the identity map indeed induces an ${\mathcal{L}}_{T\text{}}$ -isomorphism between the two $\Bbbk$ -sorts. Thus the two expansions are isomorphic.◻

Proof. By Proposition 2.13, every embedding between two $T_{\text{convex}}$ -models, which is necessarily elementary, expands uniquely to an ${\mathcal{L}}_{T\text{RV}}$ -embedding between two $\text{TCVF}$ -models. This latter embedding is indeed elementary since $\text{TCVF}$ admits quantifier elimination, which will be shown below. It follows that the theory $\text{TCVF}$ is complete. But here we do not really need to go down that route. We can simply observe that, by the proof of Proposition 2.13, $T_{\text{convex}}$ and $\text{TCVF}$ are equivalent in the sense mentioned in Remark 2.3, and hence they are both complete if one of them is.◻

Convention 2.15. From now on, we shall work in the model ${\mathcal{R}}_{\operatorname{rv}}$ of $\text{TCVF}$ , which is the unique ${\mathcal{L}}_{T\text{RV}}$ -expansion of the sufficiently saturated $T_{\text{convex}}$ -model $({\mathcal{R}},\mathfrak{O})$ . We shall write $\text{VF}({\mathcal{R}}_{\operatorname{rv}})$ simply as $\text{VF}$ or ${\mathcal{R}}$ , depending on the context, $\operatorname{RV}_{0}({\mathcal{R}}_{\operatorname{rv}})$ as $\text{RV}_{0}$ , etc. A subset in ${\mathcal{R}}_{\operatorname{rv}}$ may simply be referred to as a set.

When we work in the ${\mathcal{L}}_{T\text{}}$ -reduct ${\mathcal{R}}$ of ${\mathcal{R}}_{\operatorname{rv}}$ instead of ${\mathcal{R}}_{\operatorname{rv}}$ , or just wish to emphasize that a set is definable in ${\mathcal{R}}$ instead of ${\mathcal{R}}_{\operatorname{rv}}$ , the symbol ‘ ${\mathcal{L}}_{T\text{}}$ ’ or ‘ $T$ ’ will be inserted into the corresponding places in the terminology.

Proof. We shall run the usual Shoenfield test for quantifier elimination. To that end, let ${\mathcal{M}}$ be a model of $\text{TCVF}$ , ${\mathcal{S}}$ a substructure of ${\mathcal{M}}$ , and $\unicode[STIX]{x1D70E}:{\mathcal{S}}\longrightarrow {\mathcal{R}}_{\operatorname{rv}}$ an embedding. All we need to do is to extend $\unicode[STIX]{x1D70E}$ to an embedding ${\mathcal{M}}\longrightarrow {\mathcal{R}}_{\operatorname{rv}}$ .

The construction is more or less a variation of that in the proof of [Reference YinYin09, Theorem 3.10]. The strategy is to reduce the situation to Theorem 2.11. In the process of doing so, instead of the dimension inequality of the general theory of valued fields, the Wilkie inequality [Reference van den DriesVdD97, Corollary 5.6] is used (see [Reference van den Dries and LewenbergVdDL95, § 3.2] for the notion of ranks of $T$ -models). Note that, to use this inequality, we need to assume that $T$ is power-bounded.

Let ${\mathcal{S}}_{\ast }=\langle \text{VF}({\mathcal{S}})\rangle$ and $t\in \text{RV}({\mathcal{S}})\smallsetminus \text{RV}({\mathcal{S}}_{\ast })$ . Note that if such a $t$ does not exist then we have ${\mathcal{S}}={\mathcal{S}}_{\ast }$ and its ${\mathcal{L}}_{\text{convex}}$ -reduct is an ${\mathcal{L}}_{\text{convex}}$ -substructure of the ${\mathcal{L}}_{\text{convex}}$ -reduct of ${\mathcal{M}}$ , and hence an embedding as desired can be easily obtained by applying Theorem 2.11 and Proposition 2.13. Let $a\in \text{VF}({\mathcal{M}})$ with $\operatorname{rv}(a)=t$ , and $b\in \text{VF}$ with $\operatorname{rv}(b)=\unicode[STIX]{x1D70E}(t)$ . Observe that, according to $\unicode[STIX]{x1D70E}$ , $a$ and $b$ must make the same cut in $\text{VF}({\mathcal{S}})$ and $\text{VF}(\unicode[STIX]{x1D70E}({\mathcal{S}}))$ , respectively, and hence there is an ${\mathcal{L}}_{T\text{}}$ -isomorphism

$$\begin{eqnarray}\bar{\unicode[STIX]{x1D70E}}:\langle {\mathcal{S}}_{\ast },a\rangle _{T}\longrightarrow \langle \unicode[STIX]{x1D70E}({\mathcal{S}}_{\ast }),b\rangle _{T}\end{eqnarray}$$

with $\bar{\unicode[STIX]{x1D70E}}(a)=b$ and $\bar{\unicode[STIX]{x1D70E}}\restriction \text{VF}({\mathcal{S}})=\unicode[STIX]{x1D70E}\restriction \text{VF}({\mathcal{S}})$ . We shall show that $\bar{\unicode[STIX]{x1D70E}}$ expands to an isomorphism between $\langle {\mathcal{S}}_{\ast },a\rangle$ and $\langle \unicode[STIX]{x1D70E}({\mathcal{S}}_{\ast }),b\rangle$ that is compatible with $\unicode[STIX]{x1D70E}$ .

Case (1). There is an $a_{1}\in \langle {\mathcal{S}}_{\ast },a\rangle _{T}$ such that

$$\begin{eqnarray}|\mathfrak{O}({\mathcal{S}}_{\ast })|<a_{1}<|\text{VF}({\mathcal{S}}_{\ast })\smallsetminus \mathfrak{O}({\mathcal{S}}_{\ast })|.\end{eqnarray}$$

Set $|\unicode[STIX]{x1D6E4}|({\mathcal{S}}_{\ast })=G$ . Since $\mathfrak{O}(\langle {\mathcal{S}}_{\ast },a\rangle )$ is $T$ -convex, by [Reference van den DriesVdD97, Lemma 5.4] and [Reference van den Dries and LewenbergVdDL95, Remark 3.8],

1. – either $a_{1}\in \mathfrak{O}(\langle {\mathcal{S}}_{\ast },a\rangle )$ and $|\unicode[STIX]{x1D6E4}|(\langle {\mathcal{S}}_{\ast },a\rangle )=G$ or

2. – $a_{1}\notin \mathfrak{O}(\langle {\mathcal{S}}_{\ast },a\rangle )$ and $|\unicode[STIX]{x1D6E4}|(\langle {\mathcal{S}}_{\ast },a\rangle )\cong G\oplus \unicode[STIX]{x1D542}$ .

By the Wilkie inequality, if

$$\begin{eqnarray}|\unicode[STIX]{x1D6E4}|(\langle {\mathcal{S}}_{\ast },a\rangle )\cong G\oplus \unicode[STIX]{x1D542}\end{eqnarray}$$

then $\Bbbk (\langle {\mathcal{S}}_{\ast },a\rangle )=\Bbbk ({\mathcal{S}}_{\ast })$ and hence $|\text{vrv}|(t)\notin G$ , which implies $|\text{vrv}|(\unicode[STIX]{x1D70E}(t))\notin \unicode[STIX]{x1D70E}(G)$ ; conversely, if

$$\begin{eqnarray}|\unicode[STIX]{x1D6E4}|(\langle \unicode[STIX]{x1D70E}({\mathcal{S}}_{\ast }),b\rangle )\cong \unicode[STIX]{x1D70E}(G)\oplus \unicode[STIX]{x1D542}\end{eqnarray}$$

then $|\text{vrv}|(t)\notin G$ . Therefore

$$\begin{eqnarray}|\unicode[STIX]{x1D6E4}|(\langle {\mathcal{S}}_{\ast },a\rangle )\cong G\oplus \unicode[STIX]{x1D542}\quad \text{if and only if }|\unicode[STIX]{x1D6E4}|(\langle \unicode[STIX]{x1D70E}({\mathcal{S}}_{\ast }),b\rangle )\cong \unicode[STIX]{x1D70E}(G)\oplus \unicode[STIX]{x1D542},\end{eqnarray}$$

which, by [Reference van den Dries and LewenbergVdDL95, Remark 3.8], is equivalent to saying that $a_{1}\in \mathfrak{O}(\langle {\mathcal{S}}_{\ast },a\rangle )$ if and only if $\bar{\unicode[STIX]{x1D70E}}(a_{1})\in \mathfrak{O}(\langle \{({\mathcal{S}}_{\ast }),b\rangle )$ .

Subcase (1a). $a_{1}\in \mathfrak{O}(\langle {\mathcal{S}}_{\ast },a\rangle )$ . Subcase (1a) of the proof of [Reference van den Dries and LewenbergVdDL95, Theorem 3.10] shows that $\bar{\unicode[STIX]{x1D70E}}$ expands to an ${\mathcal{L}}_{\text{convex}}$ -isomorphism and hence to an ${\mathcal{L}}_{T\text{RV}}$ -isomorphism, which is also denoted by $\bar{\unicode[STIX]{x1D70E}}$ . Since $|\unicode[STIX]{x1D6E4}|(\langle {\mathcal{S}}_{\ast },a\rangle )=G$ , we may assume $t\in \Bbbk ({\mathcal{M}})$ . By the Wilkie inequality, $\Bbbk (\langle {\mathcal{S}}_{\ast },a\rangle )$ is precisely the $T$ -model generated by $t$ over $\Bbbk ({\mathcal{S}}_{\ast })$ . So $\text{RV}(\langle {\mathcal{S}}_{\ast },a\rangle )=\langle \text{RV}({\mathcal{S}}_{\ast }),t\rangle$ and

$$\begin{eqnarray}\bar{\unicode[STIX]{x1D70E}}\restriction \text{RV}(\langle {\mathcal{S}}_{\ast },a\rangle )=\unicode[STIX]{x1D70E}\restriction \text{RV}(\langle {\mathcal{S}}_{\ast },a\rangle ).\end{eqnarray}$$

Subcase (1b). $a_{1}\notin \mathfrak{O}(\langle {\mathcal{S}}_{\ast },a\rangle )$ . As above, Subcase (1b) of the proof of [Reference van den Dries and LewenbergVdDL95, Theorem 3.10] shows that $\bar{\unicode[STIX]{x1D70E}}$ expands to an ${\mathcal{L}}_{T\text{RV}}$ -isomorphism and this time $\Bbbk (\langle {\mathcal{S}}_{\ast },a\rangle )=\Bbbk ({\mathcal{S}}_{\ast })$ . Again it is clear that

$$\begin{eqnarray}\bar{\unicode[STIX]{x1D70E}}\restriction \text{RV}(\langle {\mathcal{S}}_{\ast },a\rangle )=\unicode[STIX]{x1D70E}\restriction \text{RV}(\langle {\mathcal{S}}_{\ast },a\rangle ).\end{eqnarray}$$

Case (2). Case (1) fails. Then there is also no $b_{1}\in \langle \unicode[STIX]{x1D70E}({\mathcal{S}}_{\ast }),b\rangle _{T}$ such that

$$\begin{eqnarray}|\mathfrak{O}(\unicode[STIX]{x1D70E}({\mathcal{S}}_{\ast }))|<b_{1}<|\text{VF}(\unicode[STIX]{x1D70E}({\mathcal{S}}_{\ast }))\smallsetminus \mathfrak{O}(\unicode[STIX]{x1D70E}({\mathcal{S}}_{\ast }))|.\end{eqnarray}$$

Using Case (2) of the proof of [Reference van den Dries and LewenbergVdDL95, Theorem 3.10], compatibility between $\bar{\unicode[STIX]{x1D70E}}$ and $\unicode[STIX]{x1D70E}$ may be deduced as in Case (1) above.

Iterating this procedure, we may assume ${\mathcal{S}}={\mathcal{S}}_{\ast }$ . The theorem follows.◻

Proof. Let ${\mathcal{M}}_{\ast }=\langle \text{VF}({\mathcal{M}})\rangle$ and ${\mathcal{N}}_{\ast }=\langle \text{VF}({\mathcal{N}})\rangle$ . Let $t\notin \text{RV}({\mathcal{M}}_{\ast })$ and $a\in \operatorname{rv}^{-1}(t)$ . Since $\unicode[STIX]{x1D70E}$ is immediate, $a$ makes the same cut in $\text{VF}({\mathcal{M}})$ and $\text{VF}({\mathcal{N}})$ according to $\unicode[STIX]{x1D70E}$ . By the proof of Theorem 2.16, $\unicode[STIX]{x1D70E}$ may be extended to an immediate isomorphism $\langle {\mathcal{M}},a\rangle \longrightarrow \langle {\mathcal{N}},a\rangle$ . Iterating this procedure, we reach a stage where the assertion simply follows from Theorem 2.11.◻

Proof. For the polynomially bounded case, see [Reference van den Dries and SpeisseggerVdDS00, Proposition 9.2] and the remark thereafter. Apparently this is established in full generality (power-bounded) in [Reference TyneTyn03], which is in a repository that is password-protected. ◻

Proof. Without loss of generality, we may assume that ${\mathcal{M}}$ , ${\mathcal{N}}$ are $\text{VF}$ -generated. According to $\unicode[STIX]{x1D70E}$ , $a$ and $b$ must make the same cut respectively in $\text{VF}({\mathcal{M}})$ and $\text{VF}({\mathcal{N}})$ , and hence there is an ${\mathcal{L}}_{T\text{}}$ -isomorphism $\bar{\unicode[STIX]{x1D70E}}:\langle {\mathcal{M}},a\rangle _{T}\longrightarrow \langle {\mathcal{N}},b\rangle _{T}$ with $\bar{\unicode[STIX]{x1D70E}}(a)=b$ that extends $\unicode[STIX]{x1D70E}\restriction \text{VF}({\mathcal{M}})$ . We shall first show that $\bar{\unicode[STIX]{x1D70E}}$ expands to an ${\mathcal{L}}_{T\text{RV}}$ -isomorphism. There are two cases to consider, corresponding to the two cases in the proof of Theorem 2.16.

Case (1). There is an $a^{\prime }\in \langle {\mathcal{M}},a\rangle _{T}$ such that

$$\begin{eqnarray}|\mathfrak{O}({\mathcal{M}})|<a^{\prime }<|\text{VF}({\mathcal{M}})\smallsetminus \mathfrak{O}({\mathcal{M}})|.\end{eqnarray}$$

Let $f$ be a function symbol of ${\mathcal{L}}_{T\text{}}$ and $c\in \text{VF}({\mathcal{M}})$ such that $f(a,c)=a^{\prime }$ . Let $b^{\prime }=\bar{\unicode[STIX]{x1D70E}}(f(a,c))$ . Then we also have

$$\begin{eqnarray}|\mathfrak{O}({\mathcal{N}})|<b^{\prime }<|\text{VF}({\mathcal{N}})\smallsetminus \mathfrak{O}({\mathcal{N}})|.\end{eqnarray}$$

If $a^{\prime }\notin \mathfrak{O}(\langle {\mathcal{M}},a\rangle )$ then $\unicode[STIX]{x1D6E4}(\langle {\mathcal{M}},a\rangle )\neq \unicode[STIX]{x1D6E4}({\mathcal{M}})$ . By the valuation property, there is a $d\in \text{VF}({\mathcal{M}})$ such that $\operatorname{val}(a-d)\notin \unicode[STIX]{x1D6E4}({\mathcal{M}})$ . Then $\operatorname{val}(b-\unicode[STIX]{x1D70E}(d))\notin \unicode[STIX]{x1D6E4}({\mathcal{N}})$ and hence, by the Wilkie inequality, $\mathfrak{O}(\langle {\mathcal{N}},b\rangle )$ is the convex hull of $\mathfrak{O}({\mathcal{N}})$ in $\langle {\mathcal{N}},b\rangle _{T}$ . This implies $b^{\prime }\notin \mathfrak{O}(\langle {\mathcal{N}},b\rangle )$ . By symmetry and [Reference van den Dries and LewenbergVdDL95, Remark 3.8], we see that $a^{\prime }\in \mathfrak{O}(\langle {\mathcal{M}},a\rangle )$ if and only if $b^{\prime }\in \mathfrak{O}(\langle {\mathcal{N}},b\rangle )$ , and hence

$$\begin{eqnarray}\bar{\unicode[STIX]{x1D70E}}(\mathfrak{O}(\langle {\mathcal{M}},a\rangle ))=\mathfrak{O}(\langle {\mathcal{N}},b\rangle ).\end{eqnarray}$$

Case (2). Case (1) fails. We may proceed exactly as in Case (2) of the proof of Theorem 2.16.

This concludes our proof that $\bar{\unicode[STIX]{x1D70E}}$ expands to an ${\mathcal{L}}_{T\text{RV}}$ -isomorphism.

Next, we show that $\bar{\unicode[STIX]{x1D70E}}$ is indeed immediate. If $\text{RV}(\langle {\mathcal{M}},a\rangle )\,=\,\text{RV}({\mathcal{M}})$ then also $\text{RV}(\langle {\mathcal{N}},b\rangle )=\text{RV}({\mathcal{N}})$ , and there is nothing more to be done. So suppose $\text{RV}(\langle {\mathcal{M}},a\rangle )\neq \text{RV}({\mathcal{M}})$ . We claim that there is a $d\in \text{VF}({\mathcal{M}})$ such that $\operatorname{rv}(a-d)\notin \text{RV}({\mathcal{M}})$ . We consider two (mutually exclusive) cases.

Case (1). $\unicode[STIX]{x1D6E4}(\langle {\mathcal{M}},a\rangle )\neq \unicode[STIX]{x1D6E4}({\mathcal{M}})$ . Then the valuation property gives such a $d$ directly.

Case (2). $\Bbbk (\langle {\mathcal{M}},a\rangle )\neq \Bbbk ({\mathcal{M}})$ . Let $a^{\prime }$ be as above. Let $\mathfrak{O}^{\prime }$ be the T-convex subring of $\text{VF}({\mathcal{M}})$ that does not contain $a^{\prime }$ , that is, $\mathfrak{O}^{\prime }$ is the convex hull of $\mathfrak{O}({\mathcal{M}})$ in $\langle {\mathcal{M}},a\rangle _{T}$ . Let $\operatorname{val}^{\prime }$ , $\unicode[STIX]{x1D6E4}^{\prime }(\langle {\mathcal{M}},a\rangle )$ be the corresponding signed valuation map and signed value group. Then the valuation property yields a $d\in \text{VF}({\mathcal{M}})$ such that $\operatorname{val}^{\prime }(a-d)\notin \unicode[STIX]{x1D6E4}^{\prime }({\mathcal{M}})$ . Since

$$\begin{eqnarray}|\unicode[STIX]{x1D6E4}^{\prime }|(\langle {\mathcal{M}},a\rangle )\cong |\unicode[STIX]{x1D6E4}|({\mathcal{M}})\oplus \unicode[STIX]{x1D542},\end{eqnarray}$$

there is a $\unicode[STIX]{x1D6FE}\in |\unicode[STIX]{x1D6E4}|({\mathcal{M}})$ such that (exactly) one of the following two relations holds:

$$\begin{eqnarray}\displaystyle |\mathfrak{O}_{\unicode[STIX]{x1D6FE}}({\mathcal{M}})|<|a-d| & {<} & \displaystyle |\text{VF}({\mathcal{M}})\smallsetminus \mathfrak{O}_{\unicode[STIX]{x1D6FE}}({\mathcal{M}})|,\nonumber\\ \displaystyle |\mathfrak{M}_{\unicode[STIX]{x1D6FE}}({\mathcal{M}})|<|a-d| & {<} & \displaystyle |\text{VF}({\mathcal{M}})\smallsetminus \mathfrak{M}_{\unicode[STIX]{x1D6FE}}({\mathcal{M}})|,\nonumber\end{eqnarray}$$

where

$$\begin{eqnarray}\mathfrak{O}_{\unicode[STIX]{x1D6FE}}=\{c\in \text{VF}:|\text{val}|(c)\geqslant \unicode[STIX]{x1D6FE}\}\quad \text{and}\quad \mathfrak{M}_{\unicode[STIX]{x1D6FE}}=\{c\in \text{VF}:|\text{val}|(c)>\unicode[STIX]{x1D6FE}\}.\end{eqnarray}$$

It is not hard to see that, in either case, $\operatorname{rv}(a-d)\notin \text{RV}({\mathcal{M}})$ .

Since $\operatorname{rv}(a-d)=\operatorname{rv}(b-\unicode[STIX]{x1D70E}(d))=:t$ , by the Wilkie inequality, $\text{RV}(\langle {\mathcal{M}},a\rangle )=\langle \text{RV}({\mathcal{M}}),t\rangle$ and hence $\bar{\unicode[STIX]{x1D70E}}$ must be immediate.◻

Proof. See [Reference van den DriesVdD97, Proposition 5.8]. ◻

Notation 2.24. Recall Convention 2.9. There are two ways of treating an element $\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6E4}$ : as a point, when we study $\unicode[STIX]{x1D6E4}$ as an independent structure (see Convention 2.9); or as a subset of ${\mathcal{R}}_{\operatorname{rv}}$ , when we need to remain in the realm of definable sets in ${\mathcal{R}}_{\operatorname{rv}}$ . The former perspective simplifies the notation but is of course dispensable.

We shall write $\unicode[STIX]{x1D6FE}$ as $\unicode[STIX]{x1D6FE}^{\sharp }$ when we want to emphasize that it is the set $\operatorname{vrv}^{-1}(\unicode[STIX]{x1D6FE})$ in ${\mathcal{R}}_{\operatorname{rv}}$ that is being considered. More generally, if $I$ is a set in $\unicode[STIX]{x1D6E4}$ then we write $I^{\sharp }=\bigcup \{\unicode[STIX]{x1D6FE}^{\sharp }:\unicode[STIX]{x1D6FE}\in I\}$ . Similarly, if $U$ is a set in $\text{RV}$ then $U^{\sharp }$ stands for $\bigcup \{\operatorname{rv}^{-1}(t):t\in U\}$ .

Proof. This is immediate by a straightforward argument combining the exchange principle, Lemma 3.6 below, and compactness.

Alternatively, we may just quote [Reference Macpherson, Marker and SteinhornMMS00, Theorem 4.11]. ◻

Proof. By [Reference Macpherson, Marker and SteinhornMMS00, Theorem 4.11] we may assume $n=k$ . Then, for some $\unicode[STIX]{x1D6FE}\in \operatorname{vrv}(U)$ , $U_{\unicode[STIX]{x1D6FE}}$ contains an open subset of $\text{RV}^{n}$ . The lemma follows.◻

Proof. Over a definable finite partition of $D$ , we may assume that $D\subseteq (\unicode[STIX]{x1D6E4}^{+})^{n}$ and the restriction $\operatorname{pr}_{{\leqslant}k}\restriction D$ is injective. It follows from [Reference van den DriesVdD97, Theorem B] that the induced function $f:D_{{\leqslant}k}\longrightarrow D_{{>}k}$ is piecewise $\unicode[STIX]{x1D542}$ -linear. Thus, for every $\unicode[STIX]{x1D6FE}\in D_{{\leqslant}k}$ and every $t\in \unicode[STIX]{x1D6FE}^{\sharp }$ there is a $t$ -definable point in $f(\unicode[STIX]{x1D6FE})^{\sharp }$ . The assertion follows.◻

Proof. See [Reference van den DriesVdD98, § 8.2.11]. ◻

Notation 2.33 (The definable sort $\operatorname{DC}$ of discs).

At times it will be more convenient to work in the traditional expansion ${\mathcal{R}}_{\operatorname{rv}}^{\text{eq}}$ of ${\mathcal{R}}_{\operatorname{rv}}$ by all definable sorts. However, for our purpose, a much simpler expansion ${\mathcal{R}}_{\operatorname{rv}}^{\bullet }$ suffices. This expansion has only one additional sort $\operatorname{DC}$ that contains, as elements, all the open and closed discs (since each point in $\text{VF}$ may be regarded as a closed disc of valuative radius $\infty$ , for convenience, we may and occasionally do think of $\text{VF}$ as a subset of $\operatorname{DC}$ ). Heuristically, we may think of a disc that is properly contained in an $\text{RV}$ -disc as a ‘thickened’ point of certain stature in $\text{VF}$ . For each $\unicode[STIX]{x1D6FE}\in |\unicode[STIX]{x1D6E4}|$ , there are two additional cross-sort maps $\text{VF}\longrightarrow \operatorname{DC}$ in ${\mathcal{R}}_{\operatorname{rv}}^{\bullet }$ ; one sends $a$ to the open disc, the other to the closed disc, of radius $\unicode[STIX]{x1D6FE}$ that contains $a$ .

The expansion ${\mathcal{R}}_{\operatorname{rv}}^{\bullet }$ can help reduce the technical complexity of our discussion. However, as is the case with the imaginary $\unicode[STIX]{x1D6E4}$ -sort, it is conceptually inessential since, for the purpose of this paper, all allusions to discs as (imaginary) elements may be eliminated in favor of objects already definable in ${\mathcal{R}}_{\operatorname{rv}}$ .

Whether parameters in $\operatorname{DC}$ are used or not shall be indicated explicitly, if it is necessary. Note that it is redundant to include in $\operatorname{DC}$ discs centered at $0$ , since they may be identified with their valuative radii.

For a disc $\mathfrak{a}\subseteq \text{VF}$ , the corresponding imaginary element in $\operatorname{DC}$ is denoted by $\ulcorner \mathfrak{a}\urcorner$ when notational distinction makes the discussion more streamlined; $\ulcorner \mathfrak{a}\urcorner$ may be heuristically thought of as the ‘name’ of $\mathfrak{a}$ . Conversely, a set $D\subseteq \operatorname{DC}$ is often identified with the set $\{\mathfrak{a}:\ulcorner \mathfrak{a}\urcorner \in D\}$ , in which case $\bigcup D$ denotes a subset of $\text{VF}$ .

Notation 2.34. For each $\unicode[STIX]{x1D6FE}\in |\unicode[STIX]{x1D6E4}|$ , let $\mathfrak{M}_{\unicode[STIX]{x1D6FE}}$ and $\mathfrak{O}_{\unicode[STIX]{x1D6FE}}$ be the open and closed discs around $0$ with radius $\unicode[STIX]{x1D6FE}$ , respectively. Assume $\unicode[STIX]{x1D6FE}\geqslant 0$ . Let $\text{RV}_{\unicode[STIX]{x1D6FE}}=\text{VF}^{\times }/(1+\mathfrak{M}_{\unicode[STIX]{x1D6FE}})$ , which is a subset of $\operatorname{DC}$ . It is an abelian group and also inherits an ordering from $\text{VF}^{\times }$ . The canonical map $\text{VF}^{\times }\longrightarrow \text{RV}_{\unicode[STIX]{x1D6FE}}$ is denoted by $\operatorname{rv}_{\unicode[STIX]{x1D6FE}}$ and is augmented by $\operatorname{rv}_{\unicode[STIX]{x1D6FE}}(0)=0$ .

If $\ulcorner \mathfrak{b}\urcorner \in \operatorname{DC}$ , $b\in \mathfrak{b}$ , and $\operatorname{rad}(\mathfrak{b})\leqslant |\text{val}|(b)+\unicode[STIX]{x1D6FE}$ then $\mathfrak{b}$ is a union of discs of the form $\operatorname{rv}_{\unicode[STIX]{x1D6FE}}^{-1}(\ulcorner \mathfrak{a}\urcorner )$ . In this case, we shall abuse the notation slightly and write $\ulcorner \mathfrak{a}\urcorner \in \mathfrak{b}$ , $\mathfrak{b}\subseteq \text{RV}_{\unicode[STIX]{x1D6FE}}$ , etc.

For each $\ulcorner \mathfrak{a}\urcorner \in \text{RV}_{\unicode[STIX]{x1D6FE}}$ , let $\operatorname{Tor}(\ulcorner \mathfrak{a}\urcorner )\subseteq \text{RV}_{\unicode[STIX]{x1D6FE}}$ be the $\ulcorner \mathfrak{a}\urcorner$ -definable subset such that $\operatorname{rv}_{\unicode[STIX]{x1D6FE}}^{-1}(\operatorname{Tor}(\ulcorner \mathfrak{a}\urcorner ))$ forms the smallest closed disc containing $\mathfrak{a}$ . Set

$$\begin{eqnarray}\displaystyle \operatorname{Tor}^{\times }(\ulcorner \mathfrak{a}\urcorner ) & = & \displaystyle \operatorname{Tor}(\ulcorner \mathfrak{a}\urcorner )\smallsetminus \ulcorner \mathfrak{a}\urcorner ,\nonumber\\ \displaystyle \operatorname{Tor}^{+}(\ulcorner \mathfrak{a}\urcorner ) & = & \displaystyle \{t\in \operatorname{Tor}(\ulcorner \mathfrak{a}\urcorner ):t>\ulcorner \mathfrak{a}\urcorner \},\nonumber\\ \displaystyle \operatorname{Tor}^{-}(\ulcorner \mathfrak{a}\urcorner ) & = & \displaystyle \{t\in \operatorname{Tor}(\ulcorner \mathfrak{a}\urcorner ):t<\ulcorner \mathfrak{a}\urcorner \}.\nonumber\end{eqnarray}$$

If $\ulcorner \mathfrak{a}\urcorner =(\ulcorner \mathfrak{a}_{1}\urcorner ,\ldots ,\ulcorner \mathfrak{a}_{n}\urcorner )$ with $\ulcorner \mathfrak{a}_{i}\urcorner \in \text{RV}_{\unicode[STIX]{x1D6FE}_{i}}$ then $\prod _{i}\operatorname{Tor}(\ulcorner \mathfrak{a}_{i}\urcorner )$ is simply written as $\operatorname{Tor}(\ulcorner \mathfrak{a}\urcorner )$ ; similarly for $\operatorname{Tor}^{\times }(\ulcorner \mathfrak{a}\urcorner )$ , $\operatorname{Tor}^{+}(\ulcorner \mathfrak{a}\urcorner )$ , etc.

If $\unicode[STIX]{x1D6FE}=0$ then we may, for all purposes, identify $\operatorname{Tor}^{\times }(\ulcorner \mathfrak{a}\urcorner )$ , $\operatorname{Tor}(\ulcorner \mathfrak{a}\urcorner )$ , etc., with $\operatorname{Tor}^{\times }(\unicode[STIX]{x1D6FC}):=|\text{vrv}|^{-1}(\unicode[STIX]{x1D6FC})\subseteq \text{RV}$ , $\operatorname{Tor}(\unicode[STIX]{x1D6FC}):=\operatorname{Tor}^{\times }(\unicode[STIX]{x1D6FC})\cup \{0\}$ , etc., where $\unicode[STIX]{x1D6FC}=\operatorname{rad}(\mathfrak{a})$ .

Convention 3.1. Since apart from ${\leqslant}$ the language ${\mathcal{L}}_{T\text{}}$ only has function symbols, we may and shall assume that, in any ${\mathcal{L}}_{T\text{RV}}$ -formula, every ${\mathcal{L}}_{T\text{}}$ -term occurs in the scope of an instance of the function symbol $\operatorname{rv}$ . For example, if $f(x)$ , $g(x)$ are ${\mathcal{L}}_{T\text{}}$ -terms then the formula $f(x)<g(x)$ is equivalent to $\operatorname{rv}(f(x)-g(x))<0$ . The ${\mathcal{L}}_{T\text{}}$ -term $f(x)$ in $\operatorname{rv}(f(x))$ shall be referred to as a top ${\mathcal{L}}_{T\text{}}$ -term.

Proof. See [Reference van den DriesVdD97, Proposition 4.2]. ◻

Proof. By compactness, this is immediately reduced to the case $n=1$ . In that case, let $\unicode[STIX]{x1D719}(x,y)$ be a quantifier-free formula that defines $f$ . Let $\unicode[STIX]{x1D70F}_{i}(x,y)$ enumerate the top ${\mathcal{L}}_{T\text{}}$ -terms in $\unicode[STIX]{x1D719}(x,y)$ . For each $a\in \text{VF}$ and each $t_{i}(a,y)$ , let $B_{a,i}\subseteq \text{VF}$ be the characteristic finite subset of the function $t_{i}(a,y)$ given by $o$ -minimal monotonicity (see [Reference van den DriesVdD98, § 3.1]). It is not difficult to see that if $f(a)\notin \bigcup _{i}B_{a,i}$ then there would be a $b\neq f(a)$ such that

$$\begin{eqnarray}\operatorname{rv}(\unicode[STIX]{x1D70F}_{i}(a,b))=\operatorname{rv}(\unicode[STIX]{x1D70F}_{i}(a,f(a)))\end{eqnarray}$$

for all $i$ and hence $\unicode[STIX]{x1D719}(a,b)$ holds, which is impossible since $f$ is a function. The lemma follows.◻

Proof. This is immediate by Lemma 3.3, $o$ -minimal monotonicity, and HNF.◻

Proof. This follows easily from monotonicity. Also, the proof of [Reference YinYin09, Lemma 4.11] still works. ◻

Proof. The first assertion follows directly from Lemma 3.3. It can also be easily seen through an induction on $n$ with the trivial base case $n=0$ . For any $b\in t_{n}^{\sharp }$ , by the inductive hypothesis, we have $a\in \text{VF}(\langle b\rangle )$ . If $a$ were not definable then, by the exchange principle, we would have $b\in \text{VF}(\langle a\rangle )$ and hence $t_{n}^{\sharp }\subseteq \text{VF}(\langle a\rangle )$ , which is impossible. The second assertion is similar, using the exchange principle in the $\text{RV}$ -sort (see Remark 2.27).◻

Proof. We may assume $n=1$ . Then this is immediate by Lemma 3.6 and compactness.◻

Proof. We proceed by induction on $n$ . Let $b\in \mathfrak{a}_{n}$ and $t\in \text{RV}$ such that $|\text{vrv}|(t)=\operatorname{rad}(\mathfrak{a}_{n})$ . Then $a$ is $(\ulcorner \mathfrak{a}_{1}\urcorner ,\ldots ,\ulcorner \mathfrak{a}_{n-1}\urcorner ,t,b)$ -definable. By the inductive hypothesis and Lemma 3.6, we have $a\in \text{VF}(\langle b\rangle )$ . If $a$ were not definable then we would have $b\in \text{VF}(\langle a\rangle )$ and hence $\mathfrak{a}_{n}\subseteq \text{VF}(\langle a\rangle )$ , which is impossible unless $\mathfrak{a}_{n}$ is a definable point in $\text{VF}$ .◻

Proof. If $\mathfrak{b}_{1}$ and $\mathfrak{b}_{2}$ are two proper subdiscs of $\mathfrak{a}$ such that $f\restriction (\mathfrak{a}\smallsetminus \mathfrak{b}_{1})$ and $f\restriction (\mathfrak{a}\smallsetminus \mathfrak{b}_{2})$ are both constant then $\mathfrak{b}_{1}$ and $\mathfrak{b}_{2}$ must be concentric, that is, $\mathfrak{b}_{1}\cap \mathfrak{b}_{2}\neq \emptyset$ , for otherwise $f$ would be constant. Therefore, it is enough to show that $f\restriction (\mathfrak{a}\smallsetminus \mathfrak{b})$ is constant for some proper subdisc $\mathfrak{b}\subseteq \mathfrak{a}$ . To that end, without loss of generality, we may assume that $\mathfrak{a}$ is centered at $0$ . For each $\unicode[STIX]{x1D6FE}\in \operatorname{val}(\mathfrak{a})\subseteq \unicode[STIX]{x1D6E4}$ , by [Reference van den DriesVdD97, Theorem A] and $o$ -minimality, $f(\operatorname{val}^{-1}(\unicode[STIX]{x1D6FE}))$ contains a $\unicode[STIX]{x1D6FE}$ -definable element $t_{\unicode[STIX]{x1D6FE}}\in \Bbbk$ . By weak $o$ -minimality, $f(\operatorname{val}^{-1}(\unicode[STIX]{x1D6FE}))=t_{\unicode[STIX]{x1D6FE}}$ for all but finitely many $\unicode[STIX]{x1D6FE}\in \operatorname{val}(\mathfrak{a})$ . Let $g:\operatorname{val}(\mathfrak{a})\longrightarrow \Bbbk$ be the definable function given by $\unicode[STIX]{x1D6FE}\longrightarrow t_{\unicode[STIX]{x1D6FE}}$ . By Lemma 2.23, the image of $g$ is finite. The assertion follows.◻