2.1 The ordered field ${{\mathbf {No}}}$

With the lexicographically ordered full binary tree $({\mathbf {No}}, <, < _s)$ of surreal numbers now at hand, the s-hierarchical ordered field of surreal numbers may be introduced as follows, where in the definitions of $+$ , $-$ and $\cdot $ , the set-theoretic brackets that enclose the sets of “left-sided options” and “right-sided options” are omitted in accordance with custom.

1. Definition of x+y

   $$ \begin{align*} x+y\ = \ \{x^L+y,\, x+y^L\, |\, x^R+y,\, x+y^R\}. \end{align*} $$

2. Definition of −x

   $$ \begin{align*} -x\ = \ \{-x^R\, |\, -x^L\}. \end{align*} $$

3. Definition of xy

   $$ \begin{align*}xy\ &= \ \{x^Ly +xy^L-x^Ly^L,\, x^Ry+xy^R-x^Ry^R\, | \\&\qquad x^Ly+xy^R -x^Ly^R,\, x^Ry+xy^L-x^Ry^L\}. \end{align*} $$

2.2 Conway names

Let ${\mathbb D}$ be the set of all surreal numbers having finite tree-rank, and

$$ \begin{align*} {\mathbb R}\ = \ {\mathbb D}\cup \big \{ \{ L\, | \, R \}: (L, R)\text{ is a Dedekind gap in }{\mathbb D} \big \}, \end{align*} $$

whereby a Dedekind gap $(L, R)$ in $\mathbb {D}$ we mean a pair of nonempty subsets $L, R$ of ${\mathbb D}$ where $L<R$ , $L \cup R ={\mathbb D}$ , L has no greatest member and R has no least member.

The following result regarding the structure of ${\mathbb R}$ is essentially due to Conway [Reference Conway12, pages 12, 23–25].

Proposition 2.3. ${\mathbb R}$ (with $ +, -, \cdot $ and $ < $ defined à la ${{\mathbf {No}}}$ ) is isomorphic to the ordered field of real numbers defined in any of the more familiar ways, ${\mathbb D}$ being ${{\mathbf {No}}}$ ’s ring of dyadic rationals (i.e., rationals of the form $m/2^n $ where m and n are integers);

$$ \begin{align*} n\ = \ \{ 0,\dots,n - 1 \, | \, \varnothing \}\ \text{ and }\ - n\ = \ \{ \varnothing \, |\, - (n - 1),\dots,0\} \end{align*} $$

for each positive integer n, $0 = \{ \varnothing \, | \, \varnothing \}$ , and the remainder of the dyadics are the arithmetic means of their left and right predecessors of greatest tree-rank; e.g., $1/2 = \{ 0 \, | \, 1 \}$ .

${\mathbb R}$ is the unique Dedekind complete initial subfield of ${\mathbf {No}}$ . Henceforth, all references to the reals are understood to be references to ${\mathbb R}$ .

A striking s-hierarchical feature of ${\mathbf {No}}$ is that every surreal number can be assigned a canonical “proper name” that is a reflection of its characteristic s-hierarchical properties. These Conway names or normal forms are expressed as formal sums of the form

$$ \begin{align*} \sum_{\alpha < \beta } r_{\alpha}\omega ^{y_{\alpha} }, \end{align*} $$

where $\beta $ is an ordinal, $(y_{\alpha })_{\alpha < \beta } $ is a strictly decreasing sequence of surreals, and $(r_{\alpha })_{\alpha < \beta } $ is a sequence of nonzero real numbers, the Conway name of an ordinal being just its Cantor normal form, it being understood that $0$ is the empty sum indexed over $\alpha <\beta =0$ [Reference Conway12, pages 31–33] and [Reference Ehrlich26, Sections 3.1 and 5].

Every nonzero surreal x is the sum of three components, each of which can be succinctly characterized in terms of the Conway name of x: the purely infinite component of x, whose terms solely have positive exponents; the real component of x, whose sole term (if it is not the empty sum) has exponent $0$ ; and the infinitesimal component of x, whose terms solely have negative exponents. Notice that $0$ , being the empty sum, may be regarded as purely infinite.

The surreal numbers having Conway names of the form $\omega ^y$ are called leaders since they denote the simplest positive members of the various Archimedean classes of ${\mathbf {No}}$ . More formally, they may be inductively defined by the formula

$$ \begin{align*} \omega^{y}\ = \ \Big\{ 0,\, n\omega^{y^L}\, \Big|\, \frac{1}{2^n}\omega^{y^R} \Big\}, \end{align*} $$

where n ranges over the positive integers and $y^L$ and $y^R$ range over the elements of $L_{s(y)}$ and $R_{s(y)}$ , respectively. For use in the proof of Lemma 7.3, we need the following lemma:

Proof. By replacing x with $-x$ if need be, we may assume that $r_{\sigma }$ is positive, so $\varepsilon = 1$ . It suffices to find a representation $\{L\, |\,R\}$ of $z+\omega ^{y_{\sigma }}$ with $L<\{x\}<R$ , for the result then follows by Proposition 2.1. As above, we have

$$ \begin{align*} \omega^{y_{\sigma}}\ = \ \Big\{ 0,\, n\omega^{y_{\sigma}^L}\, \Big|\, \frac{1}{2^n}\omega^{y_{\sigma}^R} \Big\}, \end{align*} $$

where n ranges over the positive integers and $y_{\sigma }^L$ and $y_{\sigma }^R$ range over the elements of $L_{s(y_{\sigma })}$ and $R_{s(y_{\sigma })}$ , respectively. We also have

$$ \begin{align*} z\ = \ \Big\{ \sum_{\alpha < \sigma_0 } r_{\alpha}\omega ^{y_{\alpha} }+ q_L \omega^{y_{\sigma_0}} \, \Big|\, \sum_{\alpha < \sigma_0 } r_{\alpha}\omega ^{y_{\alpha} }+ q_R \omega^{y_{\sigma_0}} \Big\}, \end{align*} $$

where $\sigma _0$ ranges over ordinals less than $\sigma $ and $q_L$ and $q_R$ range over rational numbers with $q_L<r_{\sigma _0}<q_R$ [Reference Ehrlich26, Theorem 15]. The formula for addition yields

$$ \begin{align*} z+\omega^{y_{\sigma}}\ &= \ \Big\{\sum_{\alpha < \sigma_0 } r_{\alpha}\omega ^{y_{\alpha} }+ q_L \omega^{y_{\sigma_0}}+\omega^{y_{\sigma}},\, z+n\omega^{y_{\sigma}^L}\, \Big|\, \\&\qquad \sum_{\alpha < \sigma_0 } r_{\alpha}\omega ^{y_{\alpha} }+ q_R \omega^{y_{\sigma_0}}+\omega^{y_{\sigma}},\, z+\frac{1}{2^n}\omega^{y_{\sigma}^R}\Big\}, \end{align*} $$

where n, $y_{\sigma }^L$ , $y_{\sigma }^R$ , $\sigma _0$ , $q_L$ , and $q_R$ are as above. It is straightforward to verify that x lies between the left and right options in the above representation, so $z+\omega ^{y_{\sigma }}\leqslant _s x$ .⊣

We note that Lemma 2.1 is a consequence of [Reference Gonshor36, Theorem 5.12], though translating Gonshor’s statement for our purposes requires a bit of background.

2.3 Infinite sums

There is a notion of convergence in ${\mathbf {No}}$ for sequences and series of surreals that can be conveniently expressed using normal forms written as above with dummy terms. Let $x\in {\mathbf {No}}$ and for each $y\in {\mathbf {No}}$ , let $r_{y}(x)$ be the coefficient of $\omega ^y$ in the normal form of x, it being understood that $r_{y}(x)=0$ , if $\omega ^y$ does not occur. Also let $(x_n)_{n \in {\mathbb N}}$ be a sequence of surreals written in normal form. Following Conway (and adopting notation of Siegel [Reference Siegel49, page 432]), we say that $(x_n)_{n\in {\mathbb N}}$ converges to x if

$$ \begin{align*} r_y (x)\ = \ \lim _{n\rightarrow \infty }r_y (x_{n}),\ \text{ for all } y\in{\mathbf{No}}, \end{align*} $$

and we also write

$$ \begin{align*} x\ = \ \sum_{n=0}^{\infty }x_n \end{align*} $$

to mean the partial sums of the series converge to x.

Among the convergent sequences and series of surreals are those whose mode of convergence is quite distinctive: for each $y\in {\mathbf {No}}$ , there is an $m\in {\mathbb N}$ such that $r_{y}(x_n)=r_{y}(x_m)$ for all $n\geqslant m$ . Thus, for each $y\in \bf {No}$ ,

$$ \begin{align*} r_y (x)\ = \ \lim _{n\rightarrow \infty }r_y (x_{n})\ = \ r_{y}(x_m), \end{align*} $$

where m depends on y.

Conway [Reference Conway12, page 40], calls this mode of convergence absolute convergence, and he (also see [Reference Siegel49, pages 432–434]) proves the following result:

Proposition 2.4. Let f be a formal power series with real coefficients, i.e., let

$$ \begin{align*} f (x) \ = \ \sum_{n=0}^{\infty }r_nx^n. \end{align*} $$

Then $f (\varepsilon )$ is absolutely convergent for all infinitesimals $\varepsilon $ in ${\mathbf {No}}$ .

Conway proof make use of a classical combinatorial result of Neumann [Reference Neumann45, pages 206–209],[Reference Siegel49, Lemma 3.2], [Reference Alling1, pages 260–266]); and, indeed, as Conway indicates, his proof is a straightforward adaptation to ${\mathbf {No}}$ of a classical proof of Neumann [Reference Neumann45, page 210] and [Reference Alling1, page 267] about Hahn fields (see below) and division rings of formal power series more generally.

2.4 Distinguished ordered binary subtrees of ${\mathbf {No}}$

Henceforth, the class of ${\mathbf {No}}$ ’s leaders will be denoted ‘ $Lead_{{\mathbf {No}}}$ ’ or ‘ $\omega ^{{\mathbf {No}}}$ ’ and the class of ${\mathbf {No}}$ ’s purely infinite numbers will be denoted ‘ ${\mathbf {No}}_{\operatorname {PI}}$ ’. $\mathbf {Oz}$ is the canonical integer part of ${\mathbf {No}}$ consisting of the surreals whose Conway names have no negative exponents and whose coefficient for any term whose exponent is $0$ is an integer [Reference Conway12, page 45]. $Lead_{{\mathbf {No}}}$ , ${\mathbf {No}}_{\operatorname {PI}}$ and $\mathbf {Oz}$ all have ordered tree structures inherited from $({\mathbf {No}}, <, < _s)$ . In the subsequent discussion we will appeal to the following results about these substructures of $({\mathbf {No}}, <, < _s)$ , that come from [Reference Ehrlich28, page 3, Note 2], [Reference Ehrlich26, page 1245, Theorem 11] and [Reference Bagayoko and van der Hoeven5], respectively.

3.1 s-hierarchical ordered exponential fields

Following [Reference Ehrlich26, Definition 3], ${(A,+, \cdot , <,<_{s},0,1) }$ is said to be an s-hierarchical ordered field if (i) $(A,+, \cdot , <,0, 1)$ is an ordered field; (ii) $(A,<,<_{s})$ is a lexicographically ordered binary tree; and (iii) for all $x,y\in A$

$$ \begin{align*} x+y\ = \ \{ x^{L}+y,\, x+y^{L} \, | \, x^{R}+y,\, x+y^{R}\} \end{align*} $$

and

$$ \begin{align*} xy\ &= \ \{x^{L}y+xy^{L}-x^{L}y^{L},\, x^{R}y+xy^{R}-x^{R}y^{R} \,| \\&\qquad \,x^{L}y+xy^{R}-x^{L}y^{R},\, x^{R}y+xy^{L}-x^{R}y^{L}\}, \end{align*} $$

where $x^L$ , $x^R$ , $y^L$ and $y^R$ are understood to range over the members of $L_{s(x)}$ , $R_{s(x)}$ , $L_{s(y)}$ and $R_{s(y)}$ , respectively.

Extending this idea, we will say $(A,+, \cdot , \exp , <,<_{s},0,1) $ is an s -hierarchical ordered exponential field if (i) $(A,+, \cdot , \exp , <,0,1) $ is an ordered exponential field; (ii) $(A,+, \cdot , <,<_{s},0,1) $ is an s-hierarchical ordered field; and (iii) for all $x \in A$

$$ \begin{align*} \exp (x)\,{=}\, \Big\{ 0,\, (\exp x^L) [x-x^L]_{n}, \, (\exp x^R) [x-x^R]_{2n+1}\, \Big|\, \frac{\exp x^L}{[x^L-x]_{2n+1}},\, \frac{\exp x^R}{[x^R-x]_{n}}\Big\}, \end{align*} $$

where $x^L$ and $x^R$ range over $L_{s(x)}$ and $R_{s(x)}$ respectively, n ranges over ${\mathbb N}$ , and $[y]_{n} =:1+y+\frac {y^2}{2!}+\cdots +\frac {y^n}{n!}$ for all surreal y, it being understood that expressions containing terms of the form $[y]_{2n+1}$ are to be included only when $[y]_{2n+1}>0$ .

In virtue of the definitions of Conway’s field operations and the definition of the Kruskal–Gonshor surreal exponential function, No is of course an s-hierarchical ordered exponential field. Moreover, extending the argument for initial subfields of No from [Reference Ehrlich26, page 1236], it is evident that

As such, much as the recursive unfolding of the s-hierarchical ordered field of surreal numbers generates a recursive unfolding of all the initial subfields of the surreals, the recursive unfolding of the s-hierarchical ordered exponential field $(\mathbf {No}, \exp )$ gives rise to a recursive unfolding of all the initial exponential subfields of $(\mathbf {No}, \exp )$ , the characterization of which is one of the central focuses of the paper.

Following [Reference Ehrlich26, Definition 7], a mapping $f: A \rightarrow A'$ between two lexicographically ordered binary trees is said to be s -hierarchical if for all $x = \{L\, |\, R\} \in A$ , $f(x) = \{f (L)\, |\, f (R)\}$ . If, in addition, A and $A'$ are s-hierarchical structures (ordered fields; ordered groups; ordered vector spaces; etc.) and the mapping is an embedding (of ordered fields; ordered groups; ordered vector spaces; etc) it is said to be an s -hierarchical embedding.

If $f: A \rightarrow A'$ is an s-hierarchical mapping, then f is the unique initial ordered tree embedding of A into $A'$ [Reference Ehrlich26, Lemma 1]. Moreover, an inductive argument shows that an s-hierarchical mapping of s-hierarchical ordered fields is an s-hierarchical embedding [Reference Ehrlich26, Lemma 2]. By combining the inductive argument for s-hierarchical ordered fields with a similar inductive argument for $\exp x$ , in which the induction hypothesis is that the desired property holds for all the $\exp x^L$ ’s and $\exp x^R$ ’s, one may establish for s-hierarchical ordered exponential fields the analog of the just-said result for s-hierarchical ordered fields. Collecting these results together, we have:

Let A be an s-hierarchical ordered exponential field. Extending the concepts of universality and maximality for s-hierarchical ordered fields [Reference Ehrlich26, page 1239] to s-hierarchical ordered exponential fields, A will be said to be universal if for each s-hierarchical ordered exponential field B, there is an s-hierarchical embedding $f: B \rightarrow A$ , and A will be said to be maximal if there is no s-hierarchical ordered exponential field that properly contains A as an initial exponential subfield.

In virtue of Proposition 3.5 and the fact that there is an s-hierarchial mapping of every lexicographically ordered binary tree into $(\mathbf {{No}},<,<_{s})$ (see the proof of [Reference Ehrlich26, Theorem 5]), we also have:

4.1 Hahn groups

Let ${\mathbb R}((t^S))_{\mathbf {On}}$ be the ordered group of power series (defined á la Hahn [Reference Hahn37]) consisting of all formal power series of the form $\sum _{\alpha < \beta } r_{\alpha } t^{s_{\alpha }}$ where $(s_{\alpha })_{\alpha < \beta \in \mathbf {On}}$ is a possibly empty descending sequence of elements of an ordered class S and $r_{\alpha } \in {\mathbb R} \setminus \{0\}$ for each $\alpha < \beta $ . When S is a set, then ${\mathbb R}((t^S))_{\mathbf {On}}$ is a set as well, and it is often simply written ${\mathbb R}((t^S))$ . When S is a proper class, then ${\mathbb R}((t^S))_{\mathbf {On}}$ is also a proper class. We call S the value class of ${\mathbb R}((t^S))_{\mathbf {On}}$ . In the literature, the appellation “Hahn group” is usually reserved for those structures ${\mathbb R}((t^S))_{\mathbf {On}}={\mathbb R}((t^S))$ , where S is a set. However, we refer to ${\mathbb R}((t^S))_{\mathbf {On}}$ as a Hahn group whether S is a set or a proper class.Footnote ²

For a subclass $B \subseteq S$ , we let $t^B$ denote the class $\{t^b:b \in B\}$ . A subgroup $\Gamma $ of ${\mathbb R}((t^S))_{\mathbf {On}}$ is said to be cross sectional if $\{t^s : s \in S \} \subseteq \Gamma $ .

Let $y = \sum _{\alpha < \beta } r_{\alpha } t^{s_{\alpha }}$ be an element of ${\mathbb R}((t^S))_{\mathbf {On}}$ . We let $\operatorname {supp}(y): = \{s_{\alpha }:\alpha <\beta \}\subseteq S$ denote the support of y, so $\operatorname {supp}(y)$ is empty if and only if $y = 0$ . We also set $c(y):= s_0 = \max \operatorname {supp}(y)$ , so $c(y)$ is the unique element of S with $y \asymp t^{c(y)}$ . An element $x \in {\mathbb R}((t^S))_{\mathbf {On}}$ is said to be a truncation of y if $x = \sum _{\alpha < \sigma } r_{\alpha } t^{s_{\alpha }}$ for some $\sigma \leqslant \beta $ . Given $s \in S$ , let $\sigma \leqslant \beta $ be the least ordinal such that $s_{\sigma }\leqslant s$ , and let $y_{>s}$ be the truncation $\sum _{\alpha < \sigma } r_{\alpha } t^{s_{\alpha }}$ . Then $y-y_{>s} \preccurlyeq t^s$ , and $y - y_{>s} \asymp t^s$ if and only if $s \in \operatorname {supp}(y)$ . A subgroup $\Gamma $ of ${\mathbb R}((t^S))_{\mathbf {On}}$ is said to be truncation closed if every truncation of every member of $\Gamma $ is itself a member of $\Gamma $ . Let $\Gamma $ be a truncation closed subgroup of ${\mathbb R}((t^S))_{\mathbf {On}}$ . We define the $\Gamma $ -truncation of y, denoted $y_{\Gamma }$ , as follows:

1. 1. if $\sum _{\alpha \leqslant \sigma } r_{\alpha } t^{s_{\alpha }}$ belongs to $\Gamma $ for each $\sigma < \beta $ , then we let $y_{\Gamma } := y$ and

2. 2. otherwise, let $\beta _0<\beta $ be least such that $\sum _{\alpha \leqslant \beta _0} r_{\alpha } t^{s_{\alpha }}$ does not belong to $\Gamma $ . Then we let $y_{\Gamma }:= \sum _{\alpha < \beta _0} r_{\alpha } t^{s_{\alpha }}$ .

Note that $y_{\Gamma }$ may or may not belong to $\Gamma $ . Of course, $y_{\Gamma } \in \Gamma $ if $y \in \Gamma $ . If the second case holds and $\beta _0$ is a successor ordinal, then $y_{\Gamma } \in \Gamma $ . If $\Gamma = {\mathbb R}((t^{S'}))_{\mathbf {On}}$ for some subclass $S' \subseteq S$ , then $y_{\Gamma }$ always belongs to $\Gamma $ .

Proof. Suppose that $y_{\Gamma } \neq y^{\prime }_{\Gamma }$ and let $\sum _{\alpha <\beta } r_{\alpha } t^{s_{\alpha }}$ be the greatest common truncation of y and $y'$ in $\Gamma $ . Without loss of generality, we may assume that $\sum _{\alpha <\beta } r_{\alpha } t^{s_{\alpha }}$ is not equal to $y_{\Gamma }$ , so there is $r_{\beta } \in {\mathbb R}\setminus \{0\}$ and $s_{\beta } \in S$ such that $z: = \sum _{\alpha \leqslant \beta } r_{\alpha } t^{s_{\alpha }}\in \Gamma $ is a truncation of y but not $y'$ . Note that $r_{\beta } t^{s_{\beta }} = z-\sum _{\alpha <\beta } r_{\alpha } t^{s_{\alpha }}$ belongs to $\Gamma $ . We have

$$ \begin{align*} y- z\ \prec\ r_{\beta} t^{s_{\beta}},\qquad y'- z\ \succcurlyeq\ r_{\beta} t^{s_{\beta}}. \end{align*} $$

If $y-z$ and $y'-z$ have opposite signs, then z is between y and $y'$ . Suppose that $y-z$ and $y'-z$ are both positive. Take $n>0$ such that $y'-z> \frac {r_{\beta }}{n} t^{s_{\beta }}$ . Then $y<z+\frac {r_{\beta }}{n} t^{s_{\beta }}<y'$ . Since z and $r_{\beta } t^{s_{\beta }}$ belong to $\Gamma $ and since $\Gamma $ is divisible, we see that $z+\frac {r_{\beta }}{n} t^{s_{\beta }}$ belongs to $\Gamma $ , so y and $y'$ do not realize the same cut over $\Gamma $ . The case that $y-z$ and $y'-z$ are both negative is similar.⊣

4.2 Hahn spaces

Let $\boldsymbol {k}$ be an Archimedean ordered field, that is, an ordered field with an Archimedean underlying ordered additive group. We uniquely identify $\boldsymbol {k}$ with a subfield of ${\mathbb R}$ . Via this identification, we view ${\mathbb R}((t^S))_{\mathbf {On}}$ as an ordered $\boldsymbol {k}$ -vector space when convenient (such as in Lemma 4.2 below).

See [Reference Aschenbrenner, van den Dries and van der Hoeven2, Section 2.3] for more information on Hahn spaces. Any ordered ${\mathbb R}$ -vector space, including the Hahn group ${\mathbb R}((t^S))_{\mathbf {On}}$ , is a Hahn space over ${\mathbb R}$ .

4.3 Initial subgroups and subspaces of ${\mathbf {No}}$

For $y \in {\mathbf {No}}$ with $y \neq 0$ , we let $c(y)$ denote the unique element of ${\mathbf {No}}$ such that $y \asymp \omega ^{c(y)}$ . This agrees with our Hahn group definition above, where we identify ${\mathbf {No}}$ with the Hahn group ${\mathbb R}((\omega ^{\mathbf {No}}))_{\mathbf {On}}$ . This also agrees with the map c defined by Gonshor [Reference Gonshor36]. Let $\Gamma $ be an initial divisible subgroup of ${\mathbf {No}}$ . The value class of $\Gamma $ is the class

$$ \begin{align*} S\ : = \ \{s \in {\mathbf{No}}: \omega^s \in \Gamma\}. \end{align*} $$

Then S is an initial subclass of ${\mathbf {No}}$ and $\Gamma $ is a truncation closed, cross sectional subgroup of ${\mathbb R}((\omega ^S))_{\mathbf {On}}$ [Reference Ehrlich and Kaplan31, Theorem 5.1]. Thus, each $\gamma \in {\mathbf {No}}$ has a $\Gamma $ -truncation $\gamma _{\Gamma }$ .

Let $\boldsymbol {k}$ be an Archimedean ordered field. Our identification of $\boldsymbol {k}$ with a subfield of ${\mathbb R}$ allows us to view $\boldsymbol {k}$ as an initial subfield of ${\mathbf {No}}$ . In fact, this identification $\boldsymbol {k} \hookrightarrow {\mathbb R}\subseteq {\mathbf {No}}$ is the unique initial ordered field embedding $\boldsymbol {k}\to {\mathbf {No}}$ [Reference Ehrlich26, Theorem 8]. We view ${\mathbf {No}}$ as a $\boldsymbol {k}$ -vector space via this embedding. By Proposition 1.3, we get that every ordered $\boldsymbol {k}$ -vector space $\Gamma $ admits an initial $\boldsymbol {k}$ -linear embedding into ${\mathbf {No}}$ . For future use, we record the induction step in the proof of this result [Reference Ehrlich26, pages 1241–1242]:

6.1 Subfields parametrized by subspaces of $\Gamma $

Let $\boldsymbol {k}$ be an Archimedean ordered field. If $\log (t^{\Gamma })$ is a $\boldsymbol {k}$ -subspace of ${\mathbb T}$ , then we view $\Gamma $ as an ordered $\boldsymbol {k}$ -vector space, where $r\gamma $ is the unique element of $\Gamma $ with $\log t^{r\gamma } = r\log t^{\gamma }$ for $r \in \boldsymbol {k}$ and $\gamma \in \Gamma $ . We will say that $\Gamma $ is an ordered $\boldsymbol {k}$ -vector space to mean that $\log (t^{\Gamma })$ is a $\boldsymbol {k}$ -subspace of ${\mathbb T}$ , where the operation of scalar multiplication on $\Gamma $ is assumed to be the operation described above. For the remainder of this section, we assume that $\Gamma $ is an ordered $\boldsymbol {k}$ -vector space. It follows easily from Lemma 6.2 that then $\log ({\mathbb T}^{>0})$ is a $\boldsymbol {k}$ -subspace of ${\mathbb T}$ . For $r \in \boldsymbol {k}$ and $y \in {\mathbb T}^{>0}$ , we let $y^r$ be the unique element of ${\mathbb T}^{>0}$ with $\log y^r \,{=}\, r\log y$ . Let $\Delta $ be a $\boldsymbol {k}$ -subspace of $\Gamma $ and set ${\mathbb T}_{\Delta }:={\mathbb R}((t^{\Delta }))_{\mathbf {On}} \subseteq {\mathbb T}$ . Note that the ${\mathbb T}_{\Delta }$ -truncation of y lies in ${\mathbb T}_{\Delta }$ for each $y \in {\mathbb T}$ .

Proof. Suppose that $\Delta $ is a logarithmic $\boldsymbol {k}$ -subspace of $\Gamma $ , let $x \in {\mathbb T}_{\Delta }^{>0}$ and take $\delta \in \Delta $ , $r \in {\mathbb R}^{>0}$ , and $\varepsilon \in {\mathbb T}_{\Delta }$ with $\varepsilon \prec 1$ such that $x = rt^{\delta }(1+\varepsilon )$ . Then

$$ \begin{align*} \log x \ = \ \log t^{\delta} + \ln r + \log(1+\varepsilon). \end{align*} $$

We have $\log t^{\delta } \in {\mathbb T}_{\Delta }$ by assumption and $\ln r \in {\mathbb R}\subseteq {\mathbb T}_{\Delta }$ . We also have

$$ \begin{align*} \log(1+\varepsilon) \ = \ \sum_{n=1}^{\infty } (-1)^{n-1} \varepsilon^n/n\ \in\ {\mathbb T}_{\Delta}, \end{align*} $$

since this sum exists in ${\mathbb T}$ and consists of elements from ${\mathbb T}_{\Delta }$ .

Now suppose that $\Delta $ is an exponential $\boldsymbol {k}$ -subspace of $\Gamma $ and let $y \in \log ({\mathbb T}^{>0})\cap {\mathbb T}_{\Delta }$ . We need to show that $y\in \log ({\mathbb T}_{\Delta }^{>0})$ . Take $r \in {\mathbb R}$ and $\varepsilon \prec 1$ with $y = y_{>0}+r+\varepsilon $ . We have $e^r \in {\mathbb R}^{>0} \subseteq {\mathbb T}_{\Delta }^{>0}$ , and we have

$$ \begin{align*} \exp \varepsilon \ = \ \sum_{n =0 }^{\infty }\varepsilon^n/n!\ \in\ {\mathbb T}_{\Delta}^{>0}, \end{align*} $$

since this sum exists in ${\mathbb T}^{>0}$ and consists of elements from ${\mathbb T}_{\Delta }$ . Thus, r and $\varepsilon $ belong to $\log ({\mathbb T}_{\Delta }^{>0})$ . Since $\log ({\mathbb T}^{>0})\cap {\mathbb T}_{\Delta }$ is a subgroup of ${\mathbb T}_{\Delta }$ containing y, r, and $\varepsilon $ , it follows that $y_{>0} \in \log ({\mathbb T}^{>0})\cap {\mathbb T}_{\Delta }$ . Since $y_{>0}$ is purely infinite, Lemma 6.3 gives that $y_{>0} \in \log (t^{\Gamma }) \cap {\mathbb T}_{\Delta }$ . Thus, $y_{>0} \in \log (t^{\Delta }) \subseteq \log ({\mathbb T}_{\Delta }^{>0})$ , as $\Delta $ is an exponential $\boldsymbol {k}$ -subspace of $\Gamma $ . We conclude that $y \in \log ({\mathbb T}_{\Delta }^{>0})$ .⊣

By Lemma 6.5, the subfield ${\mathbb T}_{\Delta }$ of ${\mathbb T}$ is a logarithmic Hahn field whenever $\Delta $ is a logarithmic $\boldsymbol {k}$ -subspace of $\Gamma $ . If ${\mathbb T}$ is a transserial Hahn field, then so is ${\mathbb T}_{\Delta }$ .

As the reader will recall, a family of structures $(A_{\alpha })_{\alpha <\beta }$ where $\beta \leqslant \mathbf {On}$ is said to be a continuous chain if $A_{\sigma }\subseteq A_{\alpha }$ for all $\sigma <\alpha <\beta $ and if $A_{\alpha } = \bigcup _{\sigma <\alpha }A_{\sigma }$ for each infinite limit ordinal $\alpha <\beta $ .

Proof. We construct a continuous chain $(\Delta _{\alpha })_{\alpha < \mathbf {On}}$ of logarithmic $\boldsymbol {k}$ -subspaces of $\Gamma $ by setting

$$ \begin{align*} \Delta_0\ &: = \ \Delta,\qquad \Delta_{\alpha+1}\ : = \ \big\{\gamma \in \Gamma: \log t^{\gamma} \in {\mathbb T}_{\Delta_{\alpha}}\big\},\qquad\\ \Delta_{\alpha}\ &: = \ \bigcup_{\beta<\alpha}\Delta_{\beta} \text{ when } \alpha>0 \text{ is a limit ordinal.} \end{align*} $$

Note that by induction on $\alpha $ , $\Delta _{\alpha +1}$ is indeed a logarithmic $\boldsymbol {k}$ -subspace of $\Gamma $ containing $\Delta _{\alpha }$ for each ordinal $\alpha $ . Set $\Delta ^E:= \bigcup _{\alpha <\mathbf {On}}\Delta _{\alpha }$ . We claim that $\Delta ^E$ is an exponential $\boldsymbol {k}$ -subspace of $\Gamma $ . To see this, let $\gamma \in \Gamma $ with $\log t^{\gamma } \in {\mathbb T}_{\Delta ^E}$ . Since $\operatorname {supp}(\log t^{\gamma })$ is a set, it must be contained in some $\Delta _{\alpha }$ , so $\log t^{\gamma } \in {\mathbb T}_{\Delta _{\alpha }}$ and $\gamma \in \Delta _{\alpha +1}\subseteq \Delta ^E$ . To see that $\Delta ^E$ is minimal, note that for each $\alpha $ , any exponential $\boldsymbol {k}$ -subspace of $\Gamma $ containing $\Delta _{\beta }$ for each $\beta <\alpha $ must contain $\Delta _{\alpha }$ . By induction, we see that any exponential $\boldsymbol {k}$ -subspace of $\Gamma $ containing $\Delta _0= \Delta $ must contain $\Delta ^E$ .

Now suppose that ${\mathbb T}_{\Delta }$ satisfies axiom (T4). We will prove by induction on $\alpha <\mathbf {On}$ that ${\mathbb T}_{\Delta _{\alpha }}$ also satisfies (T4). This is clear if $\alpha $ is a limit ordinal. Suppose that ${\mathbb T}_{\Delta _{\alpha }}$ satisfies (T4) and let $\gamma _0,\gamma _1,\ldots \in \Delta _{\alpha +1}$ be as in the statement of (T4). Then $\log t^{\gamma _0} \in {\mathbb T}_{\Delta _{\alpha }}$ by definition, so $\gamma _1 \in \Delta _{\alpha }$ . This gives $\gamma _n \in \Delta _{\alpha }$ for all $n>0$ . Since ${\mathbb T}_{\Delta _{\alpha }}$ satisfies (T4), we have

$$ \begin{align*} \big|\log t^{\gamma_n} - (\log t^{\gamma_n})_{>\gamma_{n+1}}\big|\ = \ t^{\gamma_{n+1}} \end{align*} $$

for all n greater than some $n_0$ . Thus, ${\mathbb T}_{\Delta _{\alpha +1}}$ satisfies (T4) as well.⊣

At various points, we will need to impose additional assumptions on $\Gamma $ , namely that $\log (t^{\Gamma })$ is truncation closed and that $t^{\Gamma ^{>0}} \subseteq \log (t^{\Gamma })$ . We already made use of the latter property (with $\Delta $ in place of $\Gamma $ ) in Lemma 6.7. The following lemma shows that if $\Gamma $ enjoys these properties, then we can find a logarithmic $\boldsymbol {k}$ -subspace $\Delta \subseteq \Gamma $ which enjoys the same properties while still being a set. It is worth mentioning that any exponential $\boldsymbol {k}$ -subspace $\Delta \subseteq \Gamma $ inherits these properties from $\Gamma $ , but that if $\Gamma $ is a proper class, then it may not be possible to find a nontrivial exponential $\boldsymbol {k}$ -subspace of $\Gamma $ whose universe is a set. This is a consequence of negative results of van der Hoeven [Reference van der Hoeven38] and Kuhlmann–Kuhlmann–Shelah [Reference Kuhlmann, Kuhlmann and Shelah39] regarding exponentials on Hahn fields.

Proof. Let $\Delta _0$ be the $\boldsymbol {k}$ -subspace of $\Gamma $ generated by A and for each n, let $\Delta _{n+1}$ be the $\boldsymbol {k}$ -subspace of $\Gamma $ generated by $\Delta _n$ and $\bigcup _{\delta \in \Delta _n} \operatorname {supp}(\log t^{\delta })$ . Let $\Delta _{\omega }: = \bigcup _n \Delta _n$ . We note that $\Delta _{\omega }$ is a $\boldsymbol {k}$ -subspace containing A whose universe is a set, and we claim that $\Delta _{\omega }$ is a logarithmic $\boldsymbol {k}$ -subspace of $\Gamma $ . Indeed, let $\delta \in \Delta _{\omega }$ and take n with $\delta \in \Delta _n$ . Then $\log t^{\delta } \in {\mathbb T}_{\Delta _{n+1}}\subseteq {\mathbb T}_{\Delta _{\omega }}$ . If we only require $\Delta $ to be a logarithmic $\boldsymbol {k}$ -subspace of $\Gamma $ , then we may take $\Delta = \Delta _{\omega }$ , but for the “moreover” part of the lemma, we have to do a bit more work.

Suppose $\log (t^{\Gamma })$ is truncation closed and $t^{\Gamma ^{>0}} \subseteq \log (t^{\Gamma })$ . Given $\Delta _{\omega }$ as above, we define an increasing chain $(\Delta _{\omega ,n})_{n \in {\mathbb N}}$ of $\boldsymbol {k}$ -subspaces of $\Gamma $ as follows:

1. 1. Let $\Delta _{\omega ,0}$ be the $\boldsymbol {k}$ -subspace of $\Gamma $ generated by the set

   $$ \begin{align*} B_0\ : = \ \{\gamma \in \Gamma: \log t^{\gamma}\text{ is a truncation of }\log t^{\delta} \text{ for some }\delta \in \Delta_{\omega}\}. \end{align*} $$

2. 2. For each n, let $\Delta _{\omega ,n+1}$ be the $\boldsymbol {k}$ -subspace of $\Gamma $ generated by the set

   $$ \begin{align*} B_{n+1}\ : = \ \Delta_{\omega,n}\cup \{\gamma \in \Gamma: \log t^{\gamma} \in t^{\Delta_{\omega,n}^{>0}}\}.\end{align*} $$

Note that $\Delta _{\omega }$ is contained in $B_0 \subseteq \Delta _{\omega ,0}$ , since $\log t^{\delta }$ is a truncation of itself for $\delta \in \Delta _{\omega }$ . We claim that each $\Delta _{\omega ,n}$ is a logarithmic $\boldsymbol {k}$ -subspace of $\Gamma $ . Since $\log (t^{\Delta _{\omega ,n}})$ is equal to the $\boldsymbol {k}$ -subspace of ${\mathbb T}$ generated by $\log (t^{B_n})$ , it suffices to show that $\log (t^{B_n}) \subseteq {\mathbb T}_{\Delta _{\omega ,n}}$ for each n. This holds for $B_0$ since each element in $\log (t^{B_0})$ is a truncation of some element in $\log (t^{\Delta _{\omega }})\subseteq {\mathbb T}_{\Delta _{\omega }}$ , and so $\log (t^{B_0}) \subseteq {\mathbb T}_{\Delta _{\omega }}$ . Note that $t^{\Delta _{\omega ,n}^{>0}} \subseteq \log (t^{B_{n+1}})$ by our assumption that $t^{\Gamma ^{>0}} \subseteq \log (t^{\Gamma })$ . Thus, we have the equality

$$ \begin{align*} \log(t^{B_{n+1}}) \ = \ t^{\Delta_{\omega,n}^{>0}}\cup \log(t^{\Delta_{\omega,n}}).\end{align*} $$

Assuming that $\Delta _{\omega ,n}$ is a logarithmic $\boldsymbol {k}$ -subspace of $\Gamma $ , this gives $\log (t^{B_{n+1}}) \subseteq {\mathbb T}_{\Delta _{\omega ,n}}$ , so $\Delta _{\omega ,n+1}$ is a logarithmic $\boldsymbol {k}$ -subspace of $\Gamma $ as well. Next, we claim that $\log (t^{\Delta _{\omega ,n}})$ is truncation closed for each n. First, let $\gamma ,\delta \in \Delta _{\omega ,0}$ be given. We will show that $(\log t^{\gamma })_{>\delta } \in \log (t^{\Delta _{\omega ,0}})$ . Take $\gamma _1,\ldots ,\gamma _m \in B_0$ and $r_1,\ldots ,r_m \in \boldsymbol {k}$ with $\gamma = r_1\gamma _1+\cdots +r_m\gamma _m$ , so

$$ \begin{align*} (\log t^{\gamma})_{>\delta}\ = \ \big(r_1\log t^{\gamma_1}+\cdots+r_m \log t^{\gamma_m}\big)_{>\delta} \ = \ r_1(\log t^{\gamma_1})_{>\delta}+\cdots+r_m (\log t^{\gamma_m})_{>\delta}. \end{align*} $$

Since $\log (t^{B_0})$ is truncation closed by construction, each $(\log t^{\gamma _i})_{>\delta }$ is in $\log (t^{B_0})$ . Thus, $(\log t^{\gamma })_{>\delta } \in \log (t^{\Delta _{\omega ,0}})$ , so $\log (t^{\Delta _{\omega ,0}})$ is truncation closed. Now, let n be given and assume that $\log (t^{\Delta _{\omega ,n}})$ is truncation closed. We will show that $\log (t^{\Delta _{\omega ,n+1}})$ is truncation closed. As with $\Delta _{\omega ,0}$ , it suffices to show that $\log (t^{B_{n+1}})$ is truncation closed, and this holds since $\log (t^{B_{n+1}}) = t^{\Delta _{\omega ,n}^{>0}}\cup \log (t^{\Delta _{\omega ,n}})$ is the union of truncation closed sets. Now we let $\Delta := \bigcup _n \Delta _{\omega ,n}$ , so $\Delta $ is a logarithmic $\boldsymbol {k}$ -subspace of $\Gamma $ and $\log (t^{\Delta })$ is truncation closed. To see that $t^{\Delta ^{>0}} \subseteq \log (t^{\Delta })$ , let $\delta \in \Delta ^{>0}$ and take n with $\delta \in \Delta _{\omega ,n}^{>0}$ . Then $t^{\delta } \in \log (t^{\Delta _{\omega ,n+1}}) \subseteq \log (t^{\Delta })$ . It remains to note that $\Delta $ is a set.⊣

6.2 $\Delta $ -paths

We continue to assume that $\Gamma $ is an ordered $\boldsymbol {k}$ -vector space. We fix a proper logarithmic $\boldsymbol {k}$ -subspace $\Delta \subseteq \Gamma $ . A $\Delta $ -path in ${\mathbb T}$ is a sequence $(y_n)_{n\in {\mathbb N}}$ in ${\mathbb T}$ such that:

1. 1. each $y_n$ is a positive infinite element;

2. 2. $c(y_n) \not \in \Delta $ for each n; and

3. 3. $y_{n+1} = \big |\log y_n - (\log y_n)_{{\mathbb T}_{\Delta }}\big |$ for each n.

Note that if $(y_n)_{n\in {\mathbb N}}$ is a $\Delta $ -path, then so is any of its tails $(y_n)_{n\geqslant n_0}$ . The definition of a $\Delta $ -path is motivated by the Main Lemma in [Reference Ressayre, Clote and Krajicek46, page 286].

Proof. Take $n\geqslant 0$ , $\delta \in \Delta $ , and $r_0,\ldots ,r_n\in \boldsymbol {k}$ with

$$ \begin{align*} r_0c(y_0) + \cdots+r_nc(y_n) + \delta \ = \ 0. \end{align*} $$

Then $y_0^{r_0}\cdots y_n^{r_n}t^{\delta } =u$ for some $u \in K^{>0}$ with $u \asymp 1$ . We have

$$ \begin{align*} r_0\log y_0 + \cdots+r_n \log y_n+ \log t^{\delta} \ = \ \log u. \end{align*} $$

For each m, we have $\log y_m = (\log y_m)_{{\mathbb T}_{\Delta }}+\varepsilon _m y_{m+1}$ where $\varepsilon _m = \pm 1$ . Set

$$ \begin{align*} a\ : = \ r_0 (\log y_0)_{{\mathbb T}_{\Delta}}+ \cdots+ r_n(\log y_n)_{{\mathbb T}_{\Delta}}+\log t^{\delta}\ \in\ {\mathbb T}_{\Delta}, \end{align*} $$

so $r_0\log y_0 + \cdots +r_n \log y_n+ \log t^{\delta }=r_0\varepsilon _0y_1+\cdots + r_n \varepsilon _n y_{n+1} +a$ . Since $u \asymp 1$ , we have $\log u \preccurlyeq 1$ , so

$$ \begin{align*} r_0\varepsilon_0y_1+\cdots + r_n \varepsilon_n y_{n+1} +a \ = \ \log u\ \preccurlyeq\ 1. \end{align*} $$

Since $y_m \succ \log y_m \succcurlyeq y_{m+1} \succ 1$ by Lemma 6.1 and since $y_m \not \asymp a$ for each m, this is only possible if each $r_m = 0$ .⊣

A $\Delta $ -path $(y_n)_{n\in {\mathbb N}}$ is said to be $\Delta $ -atomic if $y_n = t^{c(y_n)}$ for each n. Finding a $\Delta $ -atomic $\Delta $ -path seems difficult in general, but if ${\mathbb T}$ is a transserial Hahn field, then any $\Delta $ -path has a $\Delta $ -atomic tail.

Proof. Since

$$ \begin{align*} t^{c(y_{n+1})}\ \asymp\ y_{n+1}\ = \ \big|\log y_n- (\log y_n)_{{\mathbb T}_{\Delta}}\big|, \end{align*} $$

we have $c(y_{n+1}) \in \operatorname {supp}(\log y_n)$ for each n, so (T4) gives $n_0 \in {\mathbb N}$ such that

$$ \begin{align*} t^{c(y_{n+1})}\ = \ \big|\log y_n- (\log y_n)_{{\mathbb T}_{\Delta}}\big| \ = \ y_{n+1} \end{align*} $$

for all $n\geqslant n_0$ . Thus, $(y_n)_{n\geqslant n_0}$ is $\Delta $ -atomic⊣.

Given a $\Delta $ -path $(y_n)_{n\in {\mathbb N}}$ , we let $\Delta _{(y_n)}$ denote the $\boldsymbol {k}$ -linear span of $\Delta \cup \big \{c(y_0),c(y_1),\ldots \big \}$ in $\Gamma $ . Here is a sort of converse to Lemma 6.11:

Proof. First, let $\gamma \in \Delta _{(y_n)}$ . We need to show that $\log t^{\gamma } \in {\mathbb T}_{\Delta _{(y_n)}}$ . Take $\delta \in \Delta $ and $r_0,\ldots ,r_n \in \boldsymbol {k}$ with $\gamma = \delta + r_0c(y_0)+\cdots + r_n c(y_n)$ . Then

$$ \begin{align*} \log t^{\gamma} \,{=} \,\log t^{\delta} + r_0 \log t^{c(y_0)} + \cdots + r_n \log t^{c(y_n)}\,{=}\, \log t^{\delta} + r_0 \log y_0 + \cdots + r_n \log y_n, \end{align*} $$

so it suffices to show that $\log y_n \in {\mathbb T}_{\Delta _{(y_n)}}$ for each n. We have

$$ \begin{align*} \log y_n\ = \ (\log y_n)_{{\mathbb T}_{\Delta}}\pm y_{n+1}\ = \ (\log y_n)_{{\mathbb T}_{\Delta}}\pm t^{c(y_{n+1})} \ \in \ {\mathbb T}_{\Delta_{(y_n)}} \end{align*} $$

for each n, as desired.

Now, let $\gamma _0,\gamma _1,\ldots \in \Delta _{(y_n)}$ be as in the statement of (T4), and take $\delta \in \Delta $ and $r_0,\ldots ,r_k \in \boldsymbol {k}$ with $\gamma _0 = \delta +r_0c(y_0) + \cdots +r_kc(y_k)$ . Then

$$ \begin{align*} \log t^{\gamma_0} \ &= \ \log t^{\delta}+r_0\log y_0+\cdots+r_k \log y_k\ \\& = \ \log t^{\delta}+r_0\big((\log y_0)_{{\mathbb T}_{\Delta}}\pm y_1\big) + \cdots+r_k\big((\log y_k)_{{\mathbb T}_{\Delta}}\pm y_{k+1}\big). \end{align*} $$

Thus, $\gamma _1$ is in the support of $\log t^{\delta }$ or some $(\log y_m)_{{\mathbb T}_{\Delta }}$ or some $y_{m+1}$ , where $m \leqslant k$ . In the first two cases, we would have $\gamma _n \in \Delta $ for all $n>0$ , and we could conclude that

$$ \begin{align*} \big|\log t^{\gamma_n} - (\log t^{\gamma_n})_{>\gamma_{n+1}}\big|\ = \ t^{\gamma_{n+1}} \end{align*} $$

for all n greater than some $n_0$ , since ${\mathbb T}_{\Delta }$ satisfies (T4). Thus, we may assume that $\gamma _1 \in \operatorname {supp} y_{m+1}$ , so $t^{\gamma _1} = y_{m+1}$ , since $(y_n)_{n\in {\mathbb N}}$ is $\Delta $ -atomic. Thus, $\gamma _2$ is in the support of

$$ \begin{align*} \log y_{m+1} \ = \ (\log y_{m+1})_{{\mathbb T}_{\Delta}}\pm y_{m+2}. \end{align*} $$

Again, if $\gamma _2$ is in the support of $(\log y_{m+1})_{{\mathbb T}_{\Delta }}$ , then we are done, so we may assume that $t^{\gamma _2} = y_{m+2}$ . Continuing in this way, we see that either $\gamma _n \in \Delta $ eventually (in which case we are done), or $t^{\gamma _n} = y_{m+n}$ for all $n>0$ . In this second case, we note that since $\log y_{m+n}=(\log y_{m+n})_{{\mathbb T}_{\Delta }}\pm y_{m+n+1} =(\log y_{m+n})_{{\mathbb T}_{\Delta }}\pm t^{\gamma _{n+1}}$ , we have $(\log y_{m+n})_{>\gamma _{n+1}} = (\log y_{m+n})_{{\mathbb T}_{\Delta }}$ . Thus

$$ \begin{align*} \big|\log t^{\gamma_n} - (\log t^{\gamma_n})_{>\gamma_{n+1}}\big|\ = \ \big|\log y_{m+n} - (\log y_{m+n})_{>\gamma_{n+1}}\big|\ = \ y_{m+n+1} \ = \ t^{\gamma_{n+1}} \end{align*} $$

for all $n>0$ . We conclude that ${\mathbb T}_{\Delta _{(y_n)}}$ satisfies (T4).⊣

With some additional assumptions on $\Delta $ , we can find $\Delta $ -paths easily:

Lemma 6.13. Suppose that $\Delta $ is an exponential $\boldsymbol {k}$ -subspace of $\Gamma $ and that $\log (t^{\Gamma })$ is truncation closed. Let $y \in {\mathbb T}$ be positive and infinite with $c(y)\not \in \Delta $ . Define the sequence $(y_n)_{n\in {\mathbb N}}$ by

$$ \begin{align*} y_0\ : = \ y,\qquad y_{n+1}\ : = \ \big|\log y_n - (\log y_n)_{{\mathbb T}_{\Delta}}\big|. \end{align*} $$

Then $(y_n)_{n\in {\mathbb N}}$ is well-defined (that is, each $y_n$ is nonzero) and $(y_n)_{n\in {\mathbb N}}$ is a $\Delta $ -path.

Proof. Let $n \geqslant 0$ be given and suppose that $y_n$ is a positive infinite element and that $c(y_n) \not \in \Delta $ . We need to show that $y_{n+1}$ is positive and infinite and that $c(y_{n+1}) \not \in \Delta $ . First, we claim that $\log y_n\not \in {\mathbb T}_{\Delta }$ . Since $\Delta $ is an exponential $\boldsymbol {k}$ -subspace of $\Gamma $ , if $\log y_n$ were in ${\mathbb T}_{\Delta }$ , then $\log y_n$ would be in $\log ({\mathbb T}_{\Delta }^{>0})$ , giving $y_n \in {\mathbb T}_{\Delta }^{>0}$ , a contradiction. Since $(\log y_n)_{{\mathbb T}_{\Delta }}$ is in ${\mathbb T}_{\Delta }$ , it follows that $y_{n+1}$ is nonzero. Since ${\mathbb T}_{\Delta }$ is a Hahn space over ${\mathbb R}$ , Lemma 4.2 gives that

$$ \begin{align*} y_{n+1} \ = \ \big|\log y_n - (\log y_n)_{{\mathbb T}_{\Delta}}\big| \ \not\asymp\ t^{\delta} \end{align*} $$

for any $\delta \in \Delta $ , which gives $c(y_{n+1}) \not \in \Delta $ . To see that $y_{n+1}$ is infinite, suppose towards contradiction that

$$ \begin{align*} y_{n+1} \ = \ \big|\log y_n - (\log y_n)_{{\mathbb T}_{\Delta}}\big| \ \preccurlyeq\ 1. \end{align*} $$

Since $\log (t^{\Gamma })$ is truncation closed, $\log ({\mathbb T}^{>0})$ is as well by Lemma 6.4, so $(\log y_n)_{{\mathbb T}_{\Delta }} \in \log ({\mathbb T}^{>0})$ . Since $(\log y_n)_{{\mathbb T}_{\Delta }} \in {\mathbb T}_{\Delta }$ and $\Delta $ is an exponential $\boldsymbol {k}$ -subspace of $\Gamma $ , we have $(\log y_n)_{{\mathbb T}_{\Delta }} \in \log ({\mathbb T}_{\Delta }^{>0})$ . Take $z \in {\mathbb T}_{\Delta }^{>0}$ with $(\log y_n)_{{\mathbb T}_{\Delta }} = \log z$ . Then

$$ \begin{align*} \log y_n - \log z\ = \ \log(y_n/z) \ \preccurlyeq\ 1. \end{align*} $$

Lemma 6.2 gives $y_n/ z \asymp 1$ , so $y_n \asymp z \in {\mathbb T}_{\Delta }$ , a contradiction.⊣

7.1 Extending logarithmic subspaces of ${\mathbf {No}}$

In this subsection, let $\boldsymbol {k}$ be an Archimedean ordered field. Note that $\log \omega ^{r\delta } =r\log \omega ^{\delta }$ for each $r \in \boldsymbol {k}$ and each $\delta \in \Delta $ (this is an easy consequence of Proposition 3.3). Let $\Delta $ be an initial logarithmic $\boldsymbol {k}$ -subspace of ${\mathbf {No}}$ . Then ${\mathbf {No}}_{\Delta } = {\mathbb R}((\omega ^{\Delta }))_{\mathbf {On}}$ is an initial logarithmic Hahn subfield of ${\mathbf {No}}$ .

Proof. By [Reference Berarducci and Mantova10, Theorem 3.8], we have

$$ \begin{align*} \omega^{\gamma} \ = \ \exp z\ = \ \{z^k,\, \exp z_L\, |\, \exp z_R\}, \end{align*} $$

where k ranges over the positive integers and $z_L$ , $z_R$ range over the predecessors of z in $\omega ^{\Delta ^{>0}}$ with $z_L<z<z_R$ . By assumption, $\exp z_L$ and $\exp z_R$ are in $\omega ^{\Delta }$ for all such $z_L$ and $z_R$ . Of course, $z^k\in \omega ^{\Delta }$ , so

$$ \begin{align*} \omega^{\gamma} \ = \ \{0,\, \omega^{\gamma_L}\, |\, \omega^{\gamma_R}\}, \end{align*} $$

where $\gamma _L$ and $\gamma _R$ range over elements of $\Delta $ with $\gamma _L<\gamma <\gamma _R$ . Since the map $x \mapsto \omega ^x:{\mathbf {No}}\to {\mathbf {No}}$ preserves simplicity by [Reference Ehrlich26, Theorem 12], it follows that $\gamma $ is the simplest element realizing a cut over $\Delta $ . Thus, $\Delta +\boldsymbol {k} \gamma $ is initial by Lemma 4.3. To see that $\Delta +\boldsymbol {k}\gamma $ is a logarithmic $\boldsymbol {k}$ -subspace of ${\mathbf {No}}$ , let $\delta \in \Delta $ and let $r \in \boldsymbol {k}$ . Since $\Delta $ is a logarithmic $\boldsymbol {k}$ -subspace of ${\mathbf {No}}$ and $\log \omega ^{\gamma } = z\in {\mathbf {No}}_{\Delta }$ , we see that $\log \omega ^{\delta +r\gamma } = \log \omega ^{\delta } + r\log \omega ^{\gamma } \in {\mathbf {No}}_{\Delta }$ .⊣