provides an intermediary for $A$, $B$, which in view of Proposition 6.5 gives
the desired result. ∎

### Proof of Theorem 6.2

We assume $\operatorname{cf}(A)=\operatorname{ci}(B)=\omega$. Our job is to
obtain a $y\in\mathcal{C}^{<\infty}$ such that
$A<_{\operatorname{e}}y<_{\operatorname{e}}B$ and $y$ generates a Hardy field
over $H$. Take any $g\in A$. Then $g>{\mathbb{R}}$, so $g^{\prime}$ is active
in $H$, and we pass to the compositional conjugate $H\circ g^{\text{inv}}$,
which is again a Liouville closed Hardy field containing ${\mathbb{R}}$ as a
subfield, and having $A\circ g^{\operatorname{inv}},B\circ
g^{\operatorname{inv}}$ as a wide gap with $x=g\circ g^{\operatorname{inv}}\in
A\circ g^{\operatorname{inv}}$. Now Corollary 6.7 yields a
$y\in\mathcal{C}^{\infty}$ such that $A\circ
g^{\operatorname{inv}}<_{\operatorname{e}}y<_{\operatorname{e}}B\circ
g^{\operatorname{inv}}$ and $y$ generates a Hardy field over $H\circ
g^{\operatorname{inv}}$. It follows that $y\circ g\in\mathcal{C}^{<\infty}$,
$A<_{\operatorname{e}}y\circ g<_{\operatorname{e}}B$, and $y\circ g$ generates
a Hardy field over $H$. This concludes the proof. ∎

If $A$ in Theorem 6.2 contains an element of $\mathcal{C}^{\infty}$, then we
can take $y$ in the conclusion of that theorem to be in $\mathcal{C}^{\infty}$
as well: in the proof, take $g\in\mathcal{C}^{\infty}$.

##  7\. The Number of Maximal Hardy Fields

Since $\mathcal{C}$ has cardinality $\mathfrak{c}=2^{\aleph_{0}}$, the number
of Hardy fields (and thus of maximal Hardy fields) is at most
$2^{\mathfrak{c}}$. By Proposition 3.7 in [13] there are
$\geqslant\mathfrak{c}$ many maximal Hardy fields. In this short section we
show:

######  Theorem 7.1.

The number of maximal Hardy fields is equal to $2^{\mathfrak{c}}$.

This is mainly an application of the previous two sections. Let $S$ be an
ordered set. Define a countable gap in $S$ to be a pair $P$, $Q$ of countable
subsets of $S$ such that $P<Q$ and there is no $s\in S$ with $P<s<Q$; for
example, if $P$ is a countable cofinal subset of $S$, then $P$, $\emptyset$ is
a countable gap in $S$. Also, $S$ is $\eta_{1}$ iff it has no countable gap.
We thank Ilijas Farah for pointing out that the following well-known lemma
might be useful in proving statements like Theorem 7.1 via a suitable binary
tree construction:

######  Lemma 7.2.

If $S$ has cardinality $<\mathfrak{c}$, then $S$ has a countable gap.

###### Proof.

Suppose $S$ has no countable gap. Then $S$ is in particular dense: for any
${p<q}$ in $S$ there is an $s\in S$ with $p<s<q$. Thus we can embed the
ordered set $({\mathbb{Q}};{<})$ of rational numbers into $S$. Identifying
${\mathbb{Q}}$ with its image under such an embedding, there is for every
$r\in{\mathbb{R}}\setminus{\mathbb{Q}}$ an $s\in S$ such that for all
$t\in{\mathbb{Q}}$: $s>t$ in $S$ iff $r>t$ in ${\mathbb{R}}$. Thus the
cardinality of $S$ is at least that of ${\mathbb{R}}$, which is
$\mathfrak{c}$. ∎

Below $H\supseteq{\mathbb{R}}$ is a Hardy field. We set

$H^{\operatorname{te}}\ :=\ \\{f\in H:\,\text{$f$ is overhardian}\\},$

the transexponential (or overhardian) part of $H$. By Corollary 5.11 we have

$H^{\operatorname{te}}\ =\ \big{\\{}f\in H:\,\text{$f>\exp_{n}(x)$ for all
$n$}\big{\\}},$

so $H^{\operatorname{te}}$ is closed upward in $H^{>{\mathbb{R}}}$. On
$H^{\operatorname{te}}$ we define the equivalence relation $\sim_{\exp}$ of
exponential equivalence by

$\displaystyle f\sim_{\exp}g$
$\displaystyle\quad:\Longleftrightarrow\quad\text{$f\leqslant\exp_{n}(g)$ and
$g\leqslant\exp_{n}(f)$ for some $n$}$ $\displaystyle\quad\
\Longleftrightarrow\quad\text{$f\leqslant\exp_{m}(g)$ and
$g\leqslant\exp_{n}(f)$ for some $m$, $n$.}$

Let $*f$ be the exponential equivalence class of $f\in H^{\operatorname{te}}$,
a convex subset of $H^{\operatorname{te}}$. We linearly order the set
$*H^{\operatorname{te}}$ of exponential equivalence classes by:

$*f<*g\quad:\Longleftrightarrow\quad\text{$\exp_{n}(f)<g$ for all
$n$}\qquad(f,g\in H^{\operatorname{te}}).$

For a Hardy field extension $H_{1}$ of $H$ we have
$H_{1}^{\operatorname{te}}\cap H=H^{\operatorname{te}}$, and we identify
$*H^{\operatorname{te}}$ with a subset of $*H_{1}^{\operatorname{te}}$ via the
order-preserving embedding

$*f\,(\text{in $*H^{\operatorname{te}}$})\ \mapsto\ *f\,(\text{in
$*H_{1}^{\operatorname{te}}$})\quad\text{ for }f\in H^{\operatorname{te}}.$

Note that ${\mathbb{R}}(x)^{\operatorname{te}}\ =\ \emptyset$. If $H$ is
Liouville closed, then $\exp(*f)=\log(*f)=*f$ for $f\in
H^{\operatorname{te}}$. We record a few other properties of $\sim_{\exp}$ used
later:

######  Lemma 7.3.

Let $f\in H^{\operatorname{te}}$. Then

  1. (i)

$(*f)\cdot(*f)=*f$;

  2. (ii)

if $g\in H^{>{\mathbb{R}}}$ and $[vf]=[vg]$, then $g\in H^{\operatorname{te}}$
and $*f=*g$;

  3. (iii)

$(*f)^{\dagger}=*f$ and $\der(*f)=*f$.

###### Proof.

Parts (i) and (ii) follow easily from the definitions. For (iii), note first
that $f^{\dagger}$ is overhardian by a remark before Corollary 5.7 and $\log
f<f^{\dagger}$ by Lemma 5.9, so $f<\exp(f^{\dagger})$, and
$f^{\dagger}<f<\exp(f)$, and thus $f^{\dagger}\sim_{\exp}f$. This yields
$(*f)^{\dagger}=f*$, hence $\der(*f)=*f$ by (i). ∎

Using results of the previous section we shall prove:

######  Proposition 7.4.

Suppose $P$, $Q$ is a countable gap in $*H^{\operatorname{te}}$. Then $H$ has
Hardy field extensions $H_{0}=H\langle f_{0}\rangle$, $H_{1}=H\langle
f_{1}\rangle$ with $f_{0}\in H_{0}^{\operatorname{te}}$, $f_{1}\in
H_{1}^{\operatorname{te}}$, such that

$P<*f_{0}<Q,\qquad P<*f_{1}<Q,$

$H_{0}$ and $H_{1}$ have no common Hardy field extension, and

$*H_{0}^{\operatorname{te}}=*H^{\operatorname{te}}\cup\\{*f_{0}\\},\qquad*H_{1}^{\operatorname{te}}=*H^{\operatorname{te}}\cup\\{*f_{1}\\}.$

We accept this for the moment, and indicate how it enables a binary tree
construction leading to Theorem 7.1. Let $\lvert X\rvert$ denote the
cardinality of the set $X$, and identify as usual a cardinal with the least
ordinal of that cardinality, where an ordinal $\lambda$ is considered as the
set of ordinals $<\lambda$. Let $\mathcal{H}$ be the set of all Hardy fields
$H\supseteq{\mathbb{R}}$ such that
$\lvert*H^{\operatorname{te}}\rvert<\mathfrak{c}$. We build by transfinite
recursion a binary tree in $\mathcal{H}$ by assigning to each ordinal
$\lambda<\mathfrak{c}$ and function $s\colon\lambda\to\\{0,1\\}$ a Hardy field
$H_{s}\in\mathcal{H}$ with
$\lvert*H_{s}^{\operatorname{te}}\rvert\leqslant\lvert\lambda\rvert$. For
$\lambda=0$ the function $s$ has empty domain and we take
$H_{s}={\mathbb{R}}$. Suppose $s\colon\lambda\to\\{0,1\\}$ as above and
$H_{s}\in\mathcal{H}$ are given with
$\lvert*H_{s}^{\operatorname{te}}\rvert\leqslant\lvert\lambda\rvert$. Then
Lemma 7.2 provides a countable gap $P$, $Q$ in $*H_{s}^{\operatorname{te}}$.
Let $s0,s1\colon\lambda+1\to\\{0,1\\}$ be the obvious extensions of $s$, and
let $H_{s0},H_{s1}\in\mathcal{H}$ be obtained from $H_{s}$ as $H_{0}$, $H_{1}$
are obtained from $H$ in Proposition 7.4. Let $\lambda<\mathfrak{c}$ be an
infinite limit ordinal and $s\colon\lambda\to\\{0,1\\}$; assume that for every
$\alpha<\lambda$ there is given $H_{s|\alpha}\in\mathcal{H}$ with
$H_{s|\alpha}\subseteq H_{s|\beta}$ whenever $\alpha\leqslant\beta<\lambda$.
Then we set $H_{s}:=\bigcup_{\alpha<\lambda}H_{s|\alpha}$. Assuming also
inductively that
$\lvert*H_{s|\alpha}^{\operatorname{te}}\rvert\leqslant|\alpha|$ for all
$\alpha<\lambda$, we obtain
$\lvert*H_{s}^{\operatorname{te}}\rvert\leqslant\lvert\lambda\rvert\cdot\lvert\lambda\rvert=\lvert\lambda\rvert$,
as desired. This finishes the construction of our tree. It yields for any
function $s\colon\mathfrak{c}\to\\{0,1\\}$ a Hardy field
$H_{s}:=\bigcup_{\lambda<\mathfrak{c}}H_{s|\lambda}$, and the way we
constructed the tree guarantees that if
$s,s^{\prime}\colon\mathfrak{c}\to\\{0,1\\}$ are different, then $H_{s}$ and
$H_{s^{\prime}}$ have no common Hardy field extension. Thus there are
$2^{\mathfrak{c}}$ many maximal Hardy fields.

It remains to prove Proposition 7.4. This goes via some lemmas.

######  Lemma 7.5.

Let $K$ be an asymptotic field with value group $\Gamma$ and
$\Psi:=\psi(\Gamma^{\neq})$ its $\Psi$-set. Let $L$ be an asymptotic field
extension of $K$ of finite transcendence degree over $K$. Then
$\Psi_{L}\setminus\Psi$ is finite, where $\Psi_{L}$ is the $\Psi$-set of $L$.

###### Proof.

By [ADH, 3.1.11] (Zariski-Abhyankar), $\Gamma_{L}/\Gamma$ has finite rational
rank. Now use that if $\gamma_{1},\dots,\gamma_{n}\in\Gamma_{L}^{\neq}$ and
$\gamma_{1}^{\dagger},\dots,\gamma_{n}^{\dagger}$ are pairwise distinct and
not in $\Psi$, then $\gamma_{1},\dots,\gamma_{n}$ are ${\mathbb{Z}}$-linearly
independent modulo $\Gamma$. ∎

######  Lemma 7.6.

Let $L\supseteq H\supseteq{\mathbb{R}}$ be Hardy fields where $L$ is
$\operatorname{d}$-algebraic over $H$. Then
$*H^{\operatorname{te}}=*L^{\operatorname{te}}$. (In particular,
$*H^{\operatorname{te}}=*\operatorname{Li}(H)^{\operatorname{te}}$.)

###### Proof.

Let $y\in L^{\operatorname{te}}$; we claim that $y\sim_{\exp}h$ for some $h\in
H^{\operatorname{te}}$. To prove this, set $\gamma:=vy$. Then
$\gamma<\psi(\gamma)<\dots<\psi^{n}(\gamma)<\psi^{n+1}(\gamma)<\dots<0$, so
Lemma 7.5 gives $n\geqslant 1$ with
$\psi^{n}(\gamma)\in\Gamma:=v(H^{\times})$, say $\psi^{n}(\gamma)=vh$ with
$h\in H^{>}$. Then $h$ is overhardian and $y\sim_{\exp}h$ by Lemma 7.3. ∎

######  Lemma 7.7.

Let $H\supseteq{\mathbb{R}}$ be a Hardy field and let $P$, $Q$ be a countable
gap in $*H^{\operatorname{te}}$. Then $H$ has a Hardy field extension
$H\langle y\rangle$ with overhardian $y\in\mathcal{C}^{<\infty}$ such that
${P<*y<Q}$. For any such $y$ we have $*H\langle
y\rangle^{\operatorname{te}}=*H^{\operatorname{te}}\cup\\{*y\\}$.

###### Proof.

Using Lemma 7.6 we arrange that $H$ is Liouville closed, in particular,
$\exp_{n}(x)\in H$ for all $n$. Assume for now that $Q\neq\emptyset$. Then
$P$, $Q$ gives rise to a wide gap $A$, $B$ in $H$ by

$\displaystyle A\ $ $\displaystyle:=\
\big{\\{}\\!\exp_{n}(x):\,n=0,1,2,\dots\big{\\}}\cup\big{\\{}\\!\exp_{n}(a):n=0,1,2,\dots,\
a\in H^{\operatorname{te}},\ *a\in P\big{\\}},$ $\displaystyle B\ $
$\displaystyle:=\ \big{\\{}\\!\log_{n}(b):n=0,1,2,\dots,\ b\in
H^{\operatorname{te}},\ *b\in Q\big{\\}},$

with $\text{cf}(A)=\omega$ and $\text{ci}(B)=\omega$. Then Corollary 6.7
yields an overhardian $y\in\mathcal{C}^{<\infty}$ with
$A<_{\operatorname{e}}y<_{e}B$. Given any such $y$ it generates a Hardy field
$H\langle y\rangle$ over $H$ by Lemma 6.3, with $P<*y<Q$. Moreover, by the
proof of that lemma,

$v\big{(}H\langle y\rangle^{\times}\big{)}\ =\
v(H^{\times})\oplus\bigoplus_{n}{\mathbb{Z}}v\big{(}y^{\langle
n\rangle}\big{)},$

with convex subgroups $\Delta$ of $v(H^{\times})$ (as defined in that proof)
and $\Delta+D$ of $v\big{(}H\langle y\rangle^{\times}\big{)}$ with
$D:=\bigoplus_{n}{\mathbb{Z}}v\big{(}y^{\langle n\rangle}\big{)}$. Let $f\in
H\langle y\rangle^{\operatorname{te}}$. There are three possibilities:

  1. (1)

$vf\in\Delta$. Then $*f\in*H^{\operatorname{te}}$.

  2. (2)

$vf\in\Delta+D$, $vf<\Delta$. Then $vf=mvy^{\langle i\rangle}+o(vy^{\langle
i\rangle})$ for some $i$ and some $m\geqslant 1$, hence $*f=*y$ by Lemma 7.3.

  3. (3)

$vf<\Delta+D$. Then an easy argument gives $b>A$ in $H$ with $vf=vb+o(vb)$,
hence $*f=*b\in*H^{\operatorname{te}}$ by Lemma 7.3.

If $Q=\emptyset$, then we set $A:=H^{>{\mathbb{R}}}$, $B:=\emptyset$, and
proceed as before, using results from Section 5 instead of Corollary 6.7 to
obtain the existence of an overhardian $y\in\mathcal{C}^{<\infty}$ with
$A<_{\operatorname{e}}y$, and using instead of Lemma 6.3 the remark following
the proof of that lemma. ∎

The following consequence of Lemmas 7.7 and 7.2 is worth recording. (It also
uses the fact that any Hardy field, as a subset of $\mathcal{C}$, has
cardinality $\leqslant\mathfrak{c}$.)

######  Corollary 7.8.

If $H$ is a maximal Hardy field, then the ordered set $*H^{\operatorname{te}}$
is $\eta_{1}$, and $\lvert*H^{\operatorname{te}}\rvert=\mathfrak{c}$.

As to the $H$-field ${\mathbb{T}}$ of transseries that was studied extensively
in [ADH], we usually think of ${\mathbb{T}}$ as rather large, but
$H^{\operatorname{te}}=\emptyset$ for any Hardy field $H\supseteq{\mathbb{R}}$
which embeds into ${\mathbb{T}}$ (as $H$-fields); those $H$ are dwarfed by any
maximal Hardy field.

Next, given $\phi\in\mathcal{C}^{<\infty}$, call $\phi$ hardy-small if
$\phi^{(n)}\prec 1$ for all $n$, and hardy-bounded if $\phi^{(n)}\preccurlyeq
1$ for all $n$. For example, $\sin x$ is hardy-bounded. Here are some simple
observations about these notions: If $\phi,\theta\in\mathcal{C}^{<\infty}$ are
hardy-small, then so is $\phi+\theta$. If $\phi\in\mathcal{C}^{<\infty}$ is
hardian and $\phi\prec 1$, then $\phi$ is hardy-small. If
$\phi\in\mathcal{C}^{<\infty}$ is hardian and $\phi\preccurlyeq 1$, then
$\phi$ is hardy-bounded. If $\phi\in\mathcal{C}^{<\infty}$ is hardy-bounded
and $\theta\in\mathcal{C}^{<\infty}$ is hardy-small, then $\phi\theta$ is
hardy-small. If $\phi,\theta\in\mathcal{C}^{<\infty}$ are hardy-bounded, then
so are $\phi+\theta$ and $\phi\theta$. A routine computation gives:

######  Lemma 7.9.

If $\phi\in\mathcal{C}^{<\infty}$ is hardy-small, then
$(1+\phi)^{-1}=1+\theta$ for some hardy-small
$\theta\in\mathcal{C}^{<\infty}$, and and so $(1+\phi)^{-1}$ is hardy-bounded.

For the proof of Proposition 7.4 we shall use (see also [7, Corollary
5.4.14]):

######  Lemma 7.10 (Boshernitzan [11, Theorem 13.6]).

Suppose $\phi\in\mathcal{C}^{<\infty}$ is overhardian and
$\theta\in\mathcal{C}^{<\infty}$ is hardy-bounded. Then $\phi+\theta$ is
overhardian.

###### Proof.

Note that $\phi+\theta\in(\mathcal{C}^{<\infty})^{\times}$ by Corollary 5.7.
Moreover,

$(\phi+\theta)^{\dagger}\ =\
\left[\phi\cdot\left(1+\frac{\theta}{\phi}\right)\right]^{\dagger}\ =\
\phi^{\dagger}+\left(1+\frac{\theta}{\phi}\right)^{\dagger}\ =\
\phi^{\dagger}+\frac{(\theta/\phi)^{\prime}}{1+(\theta/\phi)}.$

Now $\phi^{-1}$ is hardy-small, so $\theta/\phi$ and $(\theta/\phi)^{\prime}$
are hardy-small. Hence by Lemma 7.9, $\big{(}1+(\theta/\phi\big{)}^{-1}$ is
hardy-bounded, and so $\frac{(\theta/\phi)^{\prime}}{1+(\theta/\phi)}$ is
hardy-small. Therefore, as $\phi^{\dagger}$ is still overhardian, we can
iterate the above to obtain

$(\phi+\theta)^{\langle n\rangle}\ =\ \phi^{\langle
n\rangle}+\theta_{n},\quad\text{ with hardy-small $\theta_{n}$ for $n\geqslant
1$.}$

Thus $\phi+\theta$ is overhardian in view of Corollary 5.7. ∎

In particular, if $\phi\in\mathcal{C}^{<\infty}$ is overhardian, then so is
$\phi+\sin x$. Thus for $H$, $P$, $Q$ as in the hypothesis of Proposition 7.4
and taking $y$ as in Lemma 7.7, the conclusion of that proposition holds for
$f_{0}:=y$ and $f_{1}:=y+\sin x$ by the proof of Lemma 7.7.

##  8\. The $H$-couple of a Maximal Hardy Field

Taking into account Lemma 1.1, proving Theorem A has now been reduced to
showing that the value group of every maximal Hardy field is $\eta_{1}$. Let
$H$ be a maximal Hardy field. Then $H$ is an asymptotic field in the sense of
[ADH, 9.1], so it has an $H$-asymptotic couple $(\Gamma,\psi)$ where $\Gamma$
is the value group of $H$.

Let us consider more generally any asymptotic field $K$ with its asymptotic
couple $(\Gamma,\psi)$. Recall that $\psi\colon\Gamma^{\neq}\to\Gamma$ is
given by $\psi(\gamma)=v(g^{\dagger})$, with $g\in K^{\times}$ such that
$v(g)=\gamma$, and that $\psi(\gamma)$ is also written as $\gamma^{\dagger}$.
Recall also that $\psi$ is a valuation on $\Gamma$. Let $(\gamma_{\rho})$ be a
pc-sequence in $\Gamma$ with respect to the valuation $\psi$ on $\Gamma$. Take
$g_{\rho}\in K^{\times}$ with $v(g_{\rho})=\gamma_{\rho}$. Then
$(g_{\rho}^{\dagger})$ is a pc-sequence in $K$, since
$v(g_{\sigma}^{\dagger}-g_{\rho}^{\dagger})=(\gamma_{\sigma}-\gamma_{\rho})^{\dagger}$
for $\sigma>\rho$, provided $\rho$ is sufficiently large. Suppose
$g_{\rho}^{\dagger}\leadsto g^{\dagger}$ with nonzero $g$ in some asymptotic
field extension $L$ of $K$ (possibly $L=K$). We claim that then for
$\gamma=vg$ we have $\gamma_{\rho}\leadsto\gamma$. This is because eventually
$(\gamma-\gamma_{\rho})^{\dagger}=v(g^{\dagger}-g_{\rho}^{\dagger})$, and the
latter is eventually strictly increasing, using also that eventually
$\gamma_{\rho}\neq\gamma$. In particular, if $K=K^{\dagger}$ and every pc-
sequence in $K$ of length $\omega$ has a pseudolimit in $K$, then every pc-
sequence in $\Gamma$ of length $\omega$ has a pseudolimit in $\Gamma$. Thus by
Corollaries 3.2, 4.8, and 5.2:

######  Corollary 8.1.

If $H$ is a maximal Hardy field with asymptotic couple $(\Gamma,\psi)$, then
every pc-sequence in $(\Gamma,\psi)$ of length $\omega$ has a pseudolimit in
$(\Gamma,\psi)$, and

$\operatorname{cf}(\Gamma^{<})\ =\ \operatorname{ci}(\Gamma^{>})\ >\
\omega,\qquad\operatorname{ci}(\Gamma)\ =\ \operatorname{cf}(\Gamma)\ >\
\omega.$

Corollary 8.1 includes [7, Proposition 5.6.6]: every maximal Hardy field
contains a germ $\ell$ which is translogarithmic, that is,
${\mathbb{R}}<\ell\leqslant\ell_{n}$ for all $n$, where $\ell_{n}$ is
inductively defined by $\ell_{0}:=x$ and $\ell_{n+1}:=\log\ell_{n}$.

### Ordered vector spaces and $H$-couples over an ordered field

In the rest of this section we fix an ordered field ${{\boldsymbol{k}}}$ (only
the case ${{\boldsymbol{k}}}={\mathbb{R}}$ is really needed) and use notation,
terminology, and results from [6]. Let $\Gamma$ be an ordered vector space
over ${{\boldsymbol{k}}}$ (as defined there). For $\alpha\in\Gamma$ we defined
its ${{\boldsymbol{k}}}$-archimedean class

$[\alpha]_{{{\boldsymbol{k}}}}\ :=\
\big{\\{}\gamma\in\Gamma:\,\text{$|\gamma|\leqslant c|\alpha|$ and
$|\alpha|\leqslant c|\gamma|$ for some
$c\in{{\boldsymbol{k}}}^{>}$}\big{\\}},$

and we linearly ordered the set $[\Gamma]_{{{\boldsymbol{k}}}}$ of
${{\boldsymbol{k}}}$-archimedean classes. We defined $\Gamma$ to be a Hahn
space if for all $\alpha,\gamma\in\Gamma^{\neq}$ with
$[\alpha]_{{{\boldsymbol{k}}}}=[\gamma]_{{{\boldsymbol{k}}}}$ there is a
scalar $c\in{{\boldsymbol{k}}}^{\times}$ such that
$[\alpha-c\gamma]_{{{\boldsymbol{k}}}}<[\alpha]_{{{\boldsymbol{k}}}}$. (If
${{\boldsymbol{k}}}={\mathbb{R}}$, then the ${{\boldsymbol{k}}}$-archimedean
class $[\alpha]_{{{\boldsymbol{k}}}}$ of an element $\alpha$ in an ordered
vector space over ${{\boldsymbol{k}}}$ equals its archimedean class
$[\alpha]$, and every ordered vector space over ${{\boldsymbol{k}}}$ is a Hahn
space.) For an ordered vector space $\Delta$ over ${{\boldsymbol{k}}}$
extending $\Gamma$ we identify $[\Gamma]_{{{\boldsymbol{k}}}}$ with a subset
of $[\Delta]_{{{\boldsymbol{k}}}}$ via the order-preserving embedding
$[\gamma]_{{{\boldsymbol{k}}}}\mapsto[\gamma]_{{{\boldsymbol{k}}}}\colon[\Gamma]_{{{\boldsymbol{k}}}}\to[\Delta]_{{{\boldsymbol{k}}}}$.

Let now $(\Gamma,\psi)$ be an $H$-couple over ${{\boldsymbol{k}}}$, as defined
in [6], so for all $\alpha,\beta\in\Gamma^{\neq}$,

$[\alpha]_{{{\boldsymbol{k}}}}\leqslant[\beta]_{{{\boldsymbol{k}}}}\
\Longrightarrow\ \psi(\alpha)\geqslant\psi(\beta).$

We defined $(\Gamma,\psi)$ to be of Hahn type if for all
$\alpha,\beta\in\Gamma^{\neq}$ with $\psi(\alpha)=\psi(\beta)$ there exists a
scalar $c\in{{\boldsymbol{k}}}^{\times}$ such that
$\psi(\alpha-c\beta)>\psi(\alpha)$; a consequence of “Hahn type” is that for
all $\alpha,\beta\in\Gamma^{\neq}$,

$[\alpha]_{{\boldsymbol{k}}}\leqslant[\beta]_{{\boldsymbol{k}}}\
\Longleftrightarrow\ \psi(\alpha)\geqslant\psi(\beta),$

and so the underlying ordered vector space $\Gamma$ over ${{\boldsymbol{k}}}$
is a Hahn space. We defined $(\Gamma,\psi)$ to be closed if
$\Psi:=\psi(\Gamma^{\neq})$ is downward closed in the ordered set $\Gamma$,
and $(\Gamma,\psi)$ has asymptotic integration.

Let now $K$ be a Liouville closed $H$-field. Recall from [6] that its value
group $\Gamma$ is then an ordered vector space over its (ordered) constant
field $C$, with scalar multiplication given by $c\,vf=vg$ whenever $f,g\in
K^{\times}$ and $cf^{\dagger}=g^{\dagger}$. Its asymptotic couple
$(\Gamma,\psi)$ with this scalar multiplication is a closed $H$-couple over
$C$ of Hahn type. For a Liouville closed Hardy field $H\supseteq{\mathbb{R}}$
its constant field is ${\mathbb{R}}$, and we construe its asymptotic couple as
an $H$-couple over ${\mathbb{R}}$ as indicated.

### Elements of countable type

Let $\Gamma$ be an ordered vector space over ${{\boldsymbol{k}}}$. Let $\beta$
be an element in an ordered vector space over ${{\boldsymbol{k}}}$ that
extends $\Gamma$. Then we say that $\beta$ has countable type over $\Gamma$ if
$\beta\notin\Gamma$ and
$\operatorname{cf}(\Gamma^{<\beta}),\operatorname{ci}(\Gamma^{>\beta})\leqslant\omega$;
in that case every element in
$(\Gamma+{{\boldsymbol{k}}}\beta)\setminus\Gamma$ has countable type over
$\Gamma$. See [ADH, 2.2] for immediate extensions of valued abelian groups and
[ADH, 2.4] for the ${{\boldsymbol{k}}}$-valuation of an ordered vector space
over ${{\boldsymbol{k}}}$.

######  Lemma 8.2.

Suppose $\beta$ has countable type over $\Gamma$ and the ordered vector space
$\Gamma+{{\boldsymbol{k}}}\beta$ over ${{\boldsymbol{k}}}$ is an immediate
extension of $\Gamma$ with respect to the ${{\boldsymbol{k}}}$-valuation. Then
$\beta$ is a pseudolimit of a divergent pc-sequence in $\Gamma$ of length
$\omega$.

###### Proof.

The assumptions yield a countable (necessarily infinite) set
$A\subseteq\Gamma$ such that for every $\gamma\in\Gamma$ there exists an
$\alpha\in A$ with
$[\alpha-\beta]_{{{\boldsymbol{k}}}}<[\gamma-\beta]_{{{\boldsymbol{k}}}}$.
This easily yields a divergent pc-sequence $(\alpha_{n})$ in $\Gamma$ with all
$\alpha_{n}\in A$ such that $\alpha_{n}\leadsto\beta$. ∎

######  Lemma 8.3.

Suppose $\operatorname{cf}(\Gamma),\operatorname{cf}(\Gamma^{<})>\omega$, and
$\beta$ has countable type over $\Gamma$. Then

$\operatorname{cf}(\Gamma^{<\beta})\ =\ \operatorname{ci}(\Gamma^{>\beta})\ =\
\omega.$

###### Proof.

If $\beta<\Gamma$, then
$\operatorname{ci}(\Gamma^{>\beta})=\operatorname{ci}(\Gamma)=\operatorname{cf}(\Gamma)>\omega$,
contradicting $\operatorname{ci}(\Gamma^{>\beta})\leqslant\omega$. Thus
$\Gamma^{<\beta}\neq\emptyset$. If
$\operatorname{cf}(\Gamma^{<\beta})\neq\omega$, then $\Gamma^{<\beta}$ has a
largest element $\gamma$, so $\Gamma^{>\beta}=\Gamma^{>\gamma}$, contradicting
$\operatorname{ci}(\Gamma^{>\gamma})=\operatorname{cf}(\Gamma^{<})>\omega$.
Thus $\operatorname{cf}(\Gamma^{<\beta})=\omega$; likewise,
$\operatorname{ci}(\Gamma^{<\beta})=\omega$. ∎

For us the relevant fact relating “countable type” to the $\eta_{1}$-property
is as follows: given an $H$-couple $(\Gamma,\psi)$ over ${{\boldsymbol{k}}}$,

$\text{$\Gamma$ is
$\eta_{1}$}\quad\Longleftrightarrow\quad\begin{cases}&\parbox{200.0003pt}{there
is no $H$-couple over ${{\boldsymbol{k}}}$ extending~{}$(\Gamma,\psi)$ with an
element of countable type over $\Gamma$.}\end{cases}$

(For “$\Leftarrow$” use model-theoretic compactness.)

######  Lemma 8.4.

Let $(\Gamma,\psi)$ be a closed $H$-couple over ${{\boldsymbol{k}}}$, and
suppose $\beta$ in an $H$-couple over ${{\boldsymbol{k}}}$ extending
$(\Gamma,\psi)$ has countable type over $\Gamma$ and
$\beta^{\dagger}\notin\Gamma$. Then $\beta^{\dagger}$ has countable type over
$\Gamma$.

###### Proof.

Without loss of generality we assume $\beta>0$. Consider first the case where
we have a strictly increasing sequence $(\alpha_{m})$ in $\Gamma^{>}$ and a
strictly decreasing sequence $(\gamma_{n})$ in $\Gamma^{>}$, such that
$\alpha_{m}<\beta<\gamma_{n}$ for all $m$, $n$, and $(\alpha_{m})$ is cofinal
in $\Gamma^{<\beta}$, and $(\gamma_{n})$ is coinitial in $\Gamma^{>\beta}$.
Then $(\alpha_{m}^{\dagger})$ is decreasing, $(\gamma_{n}^{\dagger})$ is
increasing, $\alpha_{m}^{\dagger}>\beta^{\dagger}>\gamma_{n}^{\dagger}$ for
all $m$, $n$. Using that the $H$-couple $(\Gamma,\psi)$ is closed we also
obtain that $(\alpha_{m}^{\dagger})$ is coinitial in
$\Gamma^{>\beta^{\dagger}}$ and that $(\gamma_{n}^{\dagger})$ is cofinal in
$\Gamma^{<\beta^{\dagger}}$. Thus $\beta^{\dagger}$ has countable type over
$\Gamma$. Next consider the case $\beta>\Gamma$. Then the cofinality of
$\Gamma$ is $\omega$, $\beta^{\dagger}<\Gamma$, and so $\beta^{\dagger}$ has
countable type over $\Gamma$, since the coinitiality of $\Gamma$ is also
$\omega$.

The case that there are $\alpha,\gamma\in\Gamma^{>}$ with
$\alpha<\beta<\gamma$ and there is a largest $\alpha\in\Gamma^{>}$ with
$\alpha<\beta$ or a least $\gamma\in\Gamma^{>}$ with $\beta<\gamma$ cannot
occur, since for such a largest $\alpha$ we would have $\alpha<\beta<2\alpha$,
so $\alpha^{\dagger}=\beta^{\dagger}$, contradicting
$\beta^{\dagger}\notin\Gamma$ (and a least such $\gamma$ yields the same
contradiction).

It remains to consider the case $0<\beta<\Gamma^{>}$. Then $\beta$ being of
countable type over $\Gamma$ yields a strictly decreasing sequence
$(\gamma_{n})$ in $\Gamma^{>}$ that is coinitial in $\Gamma^{>}$. Then
$(\gamma_{m}^{\dagger})$ is increasing, $(\gamma^{\prime}_{n})$ is decreasing,
$\gamma_{m}^{\dagger}<\beta^{\dagger}<\gamma^{\prime}_{n}$ for all $m$, $n$,
and $(\gamma_{m}^{\dagger})$ is cofinal in $\Gamma^{<\beta^{\dagger}}$ and
$(\gamma^{\prime}_{n})$ is coinitial in $\Gamma^{>\beta^{\dagger}}$. So here
$\beta^{\dagger}$ is also of countable type over $\Gamma$. ∎

### Good approximations

Let $\Gamma$ be an ordered vector space over ${{\boldsymbol{k}}}$, and let
$\alpha$, $\gamma$ range over $\Gamma$. An extension of $\Gamma$ is an ordered
vector space over ${{\boldsymbol{k}}}$ extending $\Gamma$.

######  Lemma 8.5.

Let $\beta\notin\Gamma$ be an element in an extension of $\Gamma$. Then for
any $\alpha$,

$[\beta-\alpha]_{{{\boldsymbol{k}}}}\notin[\Gamma]_{{{\boldsymbol{k}}}}\
\Longrightarrow\ [\beta-\alpha]_{{{\boldsymbol{k}}}}=\min_{\gamma}\
[\beta-\gamma]_{{{\boldsymbol{k}}}}.$

If $\Gamma+{{\boldsymbol{k}}}\beta$ is a Hahn space, then this implication
turns into an equivalence.

###### Proof.

If $[\beta-\gamma]_{{{\boldsymbol{k}}}}<[\beta-\alpha]_{{{\boldsymbol{k}}}}$,
then $(\beta-\alpha)-(\beta-\gamma)=\gamma-\alpha$ yields
$[\beta-\alpha]_{{{\boldsymbol{k}}}}=[\gamma-\alpha]_{{{\boldsymbol{k}}}}\in[\Gamma]_{{{\boldsymbol{k}}}}$;
this gives (the contrapositive of) “$\Rightarrow$”. Suppose
$\Gamma+{{\boldsymbol{k}}}\beta$ is a Hahn space. If
$[\beta-\alpha]\in[\Gamma]_{{{\boldsymbol{k}}}}$, say
$[\beta-\alpha]_{{{\boldsymbol{k}}}}=[\gamma]_{{{\boldsymbol{k}}}}$, then
$\big{[}\beta-(\alpha+c\gamma)\big{]}_{{{\boldsymbol{k}}}}<[\beta-\alpha]_{{{\boldsymbol{k}}}}$
for some $c\in{{\boldsymbol{k}}}^{\times}$; this proves (the contrapositive
of) “$\Leftarrow$”. ∎

Suppose $\beta\notin\Gamma$ lies in an extension of $\Gamma$. Then a good
approximation of $\beta$ in $\Gamma$ is by definition an $\alpha$ such that
$[\beta-\alpha]_{{{\boldsymbol{k}}}}\notin[\Gamma]_{{{\boldsymbol{k}}}}$. Note
that a good approximation of $\beta$ in $\Gamma$ exists iff
$[\Gamma+{{\boldsymbol{k}}}\beta]_{{{\boldsymbol{k}}}}\neq[\Gamma]_{{{\boldsymbol{k}}}}$.
Together with Lemma 8.2 this yields:

######  Corollary 8.6.

Suppose $\beta$ lies in an extension $\Gamma^{*}$ of $\Gamma$ and $\Gamma^{*}$
is a Hahn space. Assume also that there is no divergent pc-sequence of length
$\omega$ in $\Gamma$ and that $\beta$ has countable type over $\Gamma$. Then
$\beta$ has a good approximation in $\Gamma$.

######  Lemma 8.7.

Suppose $\beta\notin\Gamma$ in an extension of $\Gamma$ has a good
approximation $\alpha$ in $\Gamma$. Then the following holds:

  1. (i)

if $[\beta]_{{{\boldsymbol{k}}}}\in[\Gamma]_{{{\boldsymbol{k}}}}$, then
$\alpha\neq 0$,
$[\beta-\alpha]_{{{\boldsymbol{k}}}}<[\beta]_{{{\boldsymbol{k}}}}=[\alpha]_{{{\boldsymbol{k}}}}$;
and

  2. (ii)

for all $\gamma$, if
$\operatorname{sign}(\beta-\gamma)\neq\operatorname{sign}(\beta-\alpha)$, then
$[\alpha-\gamma]_{{{\boldsymbol{k}}}}=[\beta-\gamma]_{{{\boldsymbol{k}}}}>[\beta-\alpha]_{{{\boldsymbol{k}}}}$.

###### Proof.

Part (i) is clear. For (ii), assume $\alpha<\beta<\gamma$; the case
$\gamma<\beta<\alpha$ reduces to this case by taking negatives. Then
$\gamma-\alpha>\beta-\alpha>0$, so
$[\gamma-\alpha]_{{{\boldsymbol{k}}}}>[\beta-\alpha]_{{{\boldsymbol{k}}}}$,
since
$[\beta-\alpha]_{{{\boldsymbol{k}}}}\notin[\Gamma]_{{{\boldsymbol{k}}}}$. Thus
$[\beta-\gamma]_{{{\boldsymbol{k}}}}=\big{[}(\beta-\alpha)+(\alpha-\gamma)\big{]}_{{{\boldsymbol{k}}}}=[\alpha-\gamma]_{{{\boldsymbol{k}}}}$.
∎

In the rest of this section $(\Gamma,\psi)$ is an $H$-couple over
${{\boldsymbol{k}}}$, and $\alpha$, $\gamma$ range over $\Gamma$. By an
_extension_ of $(\Gamma,\psi)$ we mean an $H$-couple over ${{\boldsymbol{k}}}$
that extends $(\Gamma,\psi)$. We consider $(\Gamma,\psi)$ as a valued ordered
vector space over ${{\boldsymbol{k}}}$ with the valuation on $\Gamma$ given by
$\psi$, so $\alpha\sim\gamma$ means
$(\alpha-\gamma)^{\dagger}>\alpha^{\dagger}$. For $\alpha\neq 0$ we set
$\alpha^{\sim}:=\\{\gamma:\,\alpha\sim\gamma\\}$.

######  Lemma 8.8.

Suppose $(\Gamma,\psi)$ is closed and $\alpha\neq 0$. Then

$\operatorname{cf}(\alpha^{\sim})\ =\ \operatorname{ci}(\alpha^{\sim})\ =\
\operatorname{cf}(\Gamma^{<})\ =\ \operatorname{ci}(\Gamma^{>}).$

###### Proof.

We have
$\alpha^{\sim}=\big{\\{}\alpha+\gamma:\,\gamma^{\dagger}>\alpha^{\dagger}\big{\\}}$.
The map $\alpha+\gamma\mapsto\alpha-\gamma$ is a decreasing permutation of
$\alpha^{\sim}$, so
$\operatorname{cf}(\alpha^{\sim})=\operatorname{ci}(\alpha^{\sim})$. We also
have the decreasing map

$\alpha+\gamma\mapsto\gamma^{\dagger}\colon\alpha^{\sim}\cap\Gamma^{>\alpha}\to\Gamma^{>\alpha^{\dagger}}$

whose image is coinitial in $\Gamma^{>\alpha^{\dagger}}$, since
$(\Gamma,\psi)$ is closed. Hence
$\operatorname{cf}(\alpha^{\sim})=\operatorname{ci}(\Gamma^{>\alpha^{\dagger}})=\operatorname{ci}(\Gamma^{>})$
by [ADH, 2.1.4]. ∎

######  Lemma 8.9.

Suppose $(\Gamma,\psi)$ is of Hahn type, closed, and
$\operatorname{cf}(\Gamma),\operatorname{cf}(\Gamma^{<})>\omega$. Let $\beta$
in an extension of $(\Gamma,\psi)$ have countable type over $\Gamma$ with
$[\beta]_{{{\boldsymbol{k}}}}\notin[\Gamma]_{{{\boldsymbol{k}}}}$. Then
$\beta^{\dagger}\notin\Gamma$, and so $\beta^{\dagger}$ has countable type
over $\Gamma$ by Lemma 8.4.

###### Proof.

We may replace $\beta$ by $-\beta$, and so we arrange $\beta>0$. For
$0<\alpha<\beta<\gamma$ we have
$[\alpha]_{{{\boldsymbol{k}}}}<[\beta]_{{{\boldsymbol{k}}}}<[\gamma]_{{{\boldsymbol{k}}}}$,
and so $\alpha^{\dagger}\geqslant\beta^{\dagger}\geqslant\gamma^{\dagger}$,
but $\alpha^{\dagger}>\gamma^{\dagger}$ by the Hahn type assumption. Suppose
towards a contradiction that $\beta^{\dagger}\in\Gamma$. We distinguish two
cases. First case: $\alpha^{\dagger}=\beta^{\dagger}$ for some $\alpha$ with
$0<\alpha<\beta$. Then $\beta^{\dagger}>\gamma^{\dagger}$ for all
$\gamma>\beta$, but then $\operatorname{ci}(\Gamma^{>\beta})=\omega$ (by Lemma
8.3) and $(\Gamma,\psi)$ being closed gives for such $\alpha$ that
$\operatorname{cf}(\Gamma^{<\alpha^{\dagger}})=\operatorname{cf}(\Gamma^{<\beta^{\dagger}})\leqslant\omega$,
contradicting $\operatorname{cf}(\Gamma^{<\alpha^{\dagger}})>\omega$. Second
case: $\beta^{\dagger}=\gamma^{\dagger}$ for some $\gamma>\beta$. This leads
to a contradiction in a similar way. ∎

We say that $(\Gamma,\psi)$ is countably spherically complete if every pc-
sequence in it of length $\omega$ pseudoconverges in it. In particular, if
$(\Gamma,\psi)$ is the $H$-couple of a maximal Hardy field (with
${{\boldsymbol{k}}}={\mathbb{R}}$), then $(\Gamma,\psi)$ is of Hahn type,
closed, countably spherically complete, and
$\operatorname{cf}(\Gamma),\operatorname{cf}(\Gamma^{<})>\omega$. (See
Corollary 8.1.)

If $(\Gamma,\psi)$ is of Hahn type, then the valuation $\psi$ on $\Gamma$ is
equivalent to the ${{\boldsymbol{k}}}$-valuation of $\Gamma$ [ADH, p. 82]. If
in addition $(\Gamma,\psi)$ is countably spherically complete, then by
Corollary 8.6, any $\beta$ in an extension of $(\Gamma,\psi)$ and of countable
type over $\Gamma$ and such that $\Gamma+{{\boldsymbol{k}}}\beta$ is a Hahn
space has a good approximation in $\Gamma$.

In the next lemma only part (i) of the conclusion is needed later. The other
parts are included for their independent interest.

######  Lemma 8.10.

Suppose $(\Gamma,\psi)$ is of Hahn type, closed, and
$\operatorname{cf}(\Gamma),\operatorname{cf}(\Gamma^{<})>\omega$. Let $\beta$
in an extension of $(\Gamma,\psi)$ have countable type over $\Gamma$, with
$[\beta]_{{{\boldsymbol{k}}}}\in[\Gamma]_{{{\boldsymbol{k}}}}$, and let
$\alpha_{0}$ be a good approximation of $\beta$ in $\Gamma$. Then

  1. (i)

$\beta_{*}:=(\beta-\alpha_{0})^{\dagger}\notin\Gamma$, and $\beta_{*}$ has
countable type over $\Gamma$;

  2. (ii)

if $\alpha_{0}<\beta$, then there is a sequence $(\gamma_{n})$ in
$\Gamma^{>\beta}$ such that

$[\beta-\alpha_{0}]_{{{\boldsymbol{k}}}}<[\gamma_{n}-\beta]_{{{\boldsymbol{k}}}}<[\beta]_{{{\boldsymbol{k}}}},\
\text{ for all $n$,}$

and $\big{(}[\gamma_{n}-\beta]_{{{\boldsymbol{k}}}}\big{)}$ is strictly
decreasing and coinitial in
$[\Gamma]_{{{\boldsymbol{k}}}}^{>[\beta-\alpha_{0}]_{{{\boldsymbol{k}}}}}$;

  3. (iii)

if $\beta<\alpha_{0}$, then there is a sequence $(\gamma_{n})$ in
$\Gamma^{<\beta}$ such that

$[\beta-\alpha_{0}]_{{{\boldsymbol{k}}}}<[\beta-\gamma_{n}]_{{{\boldsymbol{k}}}}<[\beta]_{{{\boldsymbol{k}}}},\
\text{ for all $n$,}$

and $\big{(}[\beta-\gamma_{n}]_{{{\boldsymbol{k}}}}\big{)}$ is strictly
decreasing and coinitial in
$[\Gamma]_{{{\boldsymbol{k}}}}^{>[\beta-\alpha_{0}]_{{{\boldsymbol{k}}}}}$;
and

  4. (iv)

$\alpha_{0}\sim\beta$, that is,
$\beta_{*}>\alpha_{0}^{\dagger}=\beta^{\dagger}$.

###### Proof.

Applying Lemma 8.9 to $\beta-\alpha_{0}$ in the role of $\beta$ gives (i). As
to (ii), let $\alpha_{0}<\beta$ and suppose
$[\alpha]_{{{\boldsymbol{k}}}}>[\beta-\alpha_{0}]_{{{\boldsymbol{k}}}}$; then
$[\alpha]_{{{\boldsymbol{k}}}}=[\gamma-\beta]_{{{\boldsymbol{k}}}}$ for some
$\gamma>\beta$: taking $\alpha>0$, this holds with
$\gamma:=\alpha_{0}+\alpha$. Hence

$\big{\\{}[\gamma-\beta]_{{{\boldsymbol{k}}}}:\,\gamma>\beta\big{\\}}\ =\
[\Gamma]_{{{\boldsymbol{k}}}}^{>[\beta-\alpha_{0}]_{{{\boldsymbol{k}}}}}$

by Lemma 8.7(ii). Using also (i) we have a decreasing bijection

$[\gamma-\beta]_{{{\boldsymbol{k}}}}\mapsto(\gamma-\beta)^{\dagger}\
:\,[\Gamma]_{{{\boldsymbol{k}}}}^{>[\beta-\alpha_{0}]_{{{\boldsymbol{k}}}}}\to\Gamma^{<\beta_{*}}\qquad(\gamma>\beta).$

Thus
$\operatorname{ci}\\!\big{(}[\Gamma]_{{{\boldsymbol{k}}}}^{>[\beta-\alpha_{0}]_{{{\boldsymbol{k}}}}}\big{)}=\operatorname{cf}(\Gamma^{<\beta_{*}})=\omega$
by (i) and Lemma 8.3 applied to $\beta_{*}$ in the role of $\beta$, and
$[\beta]_{{{\boldsymbol{k}}}}>[\beta-\alpha_{0}]_{{{\boldsymbol{k}}}}$ by
Lemma 8.7(i). This proves (ii), and taking negatives we obtain (iii). For (iv)
first note that
$\operatorname{cf}(\alpha_{0}^{\sim})=\operatorname{cf}(\Gamma^{<})>\omega$ by
Lemma 8.8. If $\alpha_{0}<\gamma<\beta$, then $\alpha_{0}\sim\gamma$:
otherwise $\alpha_{0}<\gamma<\beta$ and
$[\gamma-\alpha_{0}]_{{{\boldsymbol{k}}}}\geqslant[\alpha_{0}]_{{{\boldsymbol{k}}}}>[\beta-\alpha_{0}]_{{{\boldsymbol{k}}}}$,
which is impossible. The set $\alpha_{0}^{\sim}$ must contain elements
$>\beta$, since otherwise $\alpha_{0}^{\sim}$ would be a cofinal subset of
$\Gamma^{<\beta}$, contradicting $\operatorname{cf}(\Gamma^{<\beta})=\omega$.
Thus $\alpha_{0}\sim\beta$. ∎

### Case (b) extensions

In this subsection $(\Gamma,\psi)$ is an $H$-couple over ${{\boldsymbol{k}}}$
with asymptotic integration, and $\beta\notin\Gamma$ is in an $H$-couple
$(\Gamma^{*},\psi^{*})$ over ${{\boldsymbol{k}}}$ that extends
$(\Gamma,\psi)$. Let $(\Gamma\langle\beta\rangle,\psi_{\beta})$ be the
$H$-couple over ${{\boldsymbol{k}}}$ generated by $\beta$ over $(\Gamma,\psi)$
in $(\Gamma^{*},\psi^{*})$. The structure of the extension
$\big{(}\Gamma\langle\beta\rangle,\psi_{\beta}\big{)}$ of $(\Gamma,\psi)$ is
described in detail in [6, Section 4]: the possibilities are listed in [6,
Proposition 4.1] as (a), (b), ${\rm(c)}_{n}$, and ${\rm(d)}_{n}$. Case (b) is
as follows:

  * (b)

We have a sequence $(\alpha_{i})$ in $\Gamma$ and a sequence $(\beta_{i})$ in
$\Gamma^{*}$ that is ${{\boldsymbol{k}}}$-linearly independent over $\Gamma$,
such that $\beta_{0}=\beta-\alpha_{0}$ and
$\beta_{i+1}=\beta_{i}^{\dagger}-\alpha_{i+1}$ for all $i$, and such that
$\Gamma\langle\beta\rangle=\Gamma\oplus\bigoplus_{i=0}^{\infty}{{\boldsymbol{k}}}\beta_{i}$.

######  Lemma 8.11.

Suppose $(\Gamma,\psi)$ is of Hahn type, closed, countably spherically
complete, and
$\operatorname{cf}(\Gamma),\operatorname{cf}(\Gamma^{<})>\omega$. Assume also
that $\Gamma^{*}$ is a Hahn space, and $\beta$ has countable type over
$\Gamma$. Then $\beta$ falls under Case (b).

###### Proof.

Suppose $\beta$ falls under Case (a). This means
$(\Gamma+{{\boldsymbol{k}}}\beta)^{\dagger}=\Gamma^{\dagger}$. In particular,
$\beta^{\dagger}\in\Gamma$, hence
$[\beta]_{{{\boldsymbol{k}}}}\in[\Gamma]_{{{\boldsymbol{k}}}}$ by Lemma 8.9,
so $(\beta-\alpha)^{\dagger}\notin\Gamma$ for some $\alpha$, by Lemma 8.10(i)
and the remark preceding that lemma, contradicting
$(\Gamma+{{\boldsymbol{k}}}\beta)^{\dagger}=\Gamma^{\dagger}$.

Next, assume $\beta$ falls under Case ${\rm(c)}_{n}$. Then we have
$\alpha_{0},\dots,\alpha_{n}\in\Gamma$, and nonzero
$\beta_{0},\dots,\beta_{n}\in\Gamma^{*}$ such that
$\beta_{0}=\beta-\alpha_{0}$, $\beta_{i+1}=\beta_{i}^{\dagger}-\alpha_{i+1}$
for $0\leqslant i<n$, the vectors
$\beta_{0},\dots,\beta_{n},\beta_{n}^{\dagger}$ are
${{\boldsymbol{k}}}$-linearly independent over $\Gamma$, and
$(\Gamma+{{\boldsymbol{k}}}\beta_{n}^{\dagger})^{\dagger}=\Gamma^{\dagger}$.
As $\beta$ has countable type over $\Gamma$, an induction using Lemma 8.4
gives that $\beta_{0},\dots,\beta_{n},\beta_{n}^{\dagger}$ have countable type
over $\Gamma$. But then Case (a) would apply to $\beta_{n}^{\dagger}$ in the
role of $\beta$, and we already excluded that possibility.

The cases ${\rm(d)}_{n}$ are excluded because $(\Gamma,\psi)$ is closed, as
noted after the proof of Proposition 4.1 in [6]. ∎

Here is more information about Case (b):

######  Lemma 8.12.

Let $(\alpha_{i})$ and $(\beta_{i})$ be as in (b). Then:

  1. (i)

$\beta_{i}^{\dagger}\notin\Gamma$ for all $i$, and thus
$[\beta_{i}]_{{{\boldsymbol{k}}}}\notin[\Gamma]_{{{\boldsymbol{k}}}}$ for all
$i$;

  2. (ii)

$\alpha_{0}$ is a good approximation of $\beta$ in $\Gamma$;

  3. (iii)

$\alpha_{i+1}$ is a good approximation of $\beta_{i}^{\dagger}$ in $\Gamma$,
for all $i$;

  4. (iv)

$\beta_{i}^{\dagger\dagger}\leqslant\beta_{i+1}^{\dagger}$ for all $i$;

  5. (v)

$(\beta_{i}^{\dagger})$ is strictly increasing, and thus
$\big{(}[\beta_{i}]_{{{\boldsymbol{k}}}}\big{)}$ is strictly decreasing;

  6. (vi)

$\big{[}\Gamma\langle\beta\rangle\big{]}_{{{\boldsymbol{k}}}}=[\Gamma]_{{{\boldsymbol{k}}}}\cup\big{\\{}[\beta_{i}]_{{{\boldsymbol{k}}}}:\,i\in{\mathbb{N}}\big{\\}}$,
and thus $\Psi_{\beta}=\Psi\cup\big{\\{}\beta_{i}^{\dagger}:\
i\in{\mathbb{N}}\big{\\}}$;

  7. (vii)

there is no $\delta\in\Gamma\langle\beta\rangle$ with
$\Psi<\delta<(\Gamma^{>})^{\prime}$;

  8. (viii)

$\Gamma^{<}$ is cofinal in $\Gamma\langle\beta\rangle^{<}$.

If $(\Gamma,\psi)$ is closed and $\eta$ in an extension of $(\Gamma,\psi)$
realizes the same cut in $\Gamma$ as $\beta$, then there is an isomorphism
$\big{(}\Gamma\langle\beta\rangle,\psi_{\beta}\big{)}\to\big{(}\Gamma\langle\eta\rangle,\psi_{\eta}\big{)}$
of $H$-couples over ${{\boldsymbol{k}}}$ that is the identity on $\Gamma$ and
sends $\beta$ to $\eta$. If $(\Gamma,\psi)$ is of Hahn type, then so is
$(\Gamma\langle\beta\rangle,\psi_{\beta})$.

###### Proof.

Except for (ii), (iii), (iv), and the isomorphism claim this is in [6, Lemma
4.2]. Now (ii) holds by
$[\beta-\alpha_{0}]_{{{\boldsymbol{k}}}}=[\beta_{0}]_{{{\boldsymbol{k}}}}\notin[\Gamma]_{{{\boldsymbol{k}}}}$,
and (iii) by
$[\beta_{i}^{\dagger}-\alpha_{i+1}]_{{{\boldsymbol{k}}}}=[\beta_{i+1}]_{{{\boldsymbol{k}}}}\notin[\Gamma]_{{{\boldsymbol{k}}}}$.
As to (iv), this is because
$[\beta_{i}^{\dagger}]_{{{\boldsymbol{k}}}}\geqslant[\beta_{i}^{\dagger}-\alpha_{i+1}]_{{{\boldsymbol{k}}}}=[\beta_{i+1}]_{{{\boldsymbol{k}}}}$
by (iii) and Lemma 8.5.

Now assume $(\Gamma,\psi)$ is closed and $\eta$ in an extension
$(\Gamma_{1},\psi_{1})$ of $(\Gamma,\psi)$ realizes the same cut in $\Gamma$
as $\beta$, in particular, $\eta\notin\Gamma$. The case
$(\Gamma_{1},\psi_{1})=(\Gamma^{*},\psi^{*})$ is actually part of [6, Lemma
4.2], and one can reduce to that case: the theory of closed $H$-couples over
${{\boldsymbol{k}}}$ has QE in the language specified in [6, Section 3], and
so there is an $H$-couple $(\Gamma_{1}^{*},\psi_{1}^{*})$ extending
$(\Gamma,\psi)$ with embeddings
$(\Gamma^{*},\psi^{*})\to(\Gamma_{1}^{*},\psi_{1}^{*})$ and
$(\Gamma_{1},\psi_{1})\to(\Gamma_{1}^{*},\psi_{1}^{*})$ over $\Gamma$. ∎

We add the following observations:

######  Corollary 8.13.

Suppose $(\alpha_{i})=(\alpha_{0},\alpha_{1},\alpha_{2},\dots)$ and
$(\beta_{i})=(\beta_{0},\beta_{1},\beta_{2},\dots)$ are as in (b). Then
$-\beta$ falls under Case (b) with associated sequences
$(-\alpha_{0},\alpha_{1},\alpha_{2},\dots)$ and
$(-\beta_{0},\beta_{1},\beta_{2},\dots)$. Also, for any $i$,
$\beta_{i}^{\dagger}$ falls under Case (b) with associated sequences
$(\alpha_{i+1},\alpha_{i+2},\alpha_{i+3},\dots)$ and
$(\beta_{i+1},\beta_{i+2},\beta_{i+3},\dots)$.

######  Corollary 8.14.

Suppose $(\Gamma,\psi)$ is closed, $\beta$ has countable type over $\Gamma$,
$\alpha<\beta<\gamma$ for some $\alpha$, $\gamma$, and $(\alpha_{i})$,
$(\beta_{i})$ are as in (b). Then
$\operatorname{cf}(\Gamma^{<\beta_{i}})=\operatorname{ci}(\Gamma^{>\beta_{i}})=\omega$
for all $i$.

###### Proof.

Induction using Lemma 8.4 shows that every $\beta_{i}$ has countable type over
$\Gamma$ and for every $i$ there are $\alpha$, $\gamma$ with
$\alpha<\beta_{i}<\gamma$. It follows from Lemma 8.12(viii) that for any
$\eta\in\Gamma\langle\beta\rangle\setminus\Gamma$ the ordered set
$\Gamma^{<\eta}$ has no largest element and the ordered set $\Gamma^{>\eta}$
has no least element. Applying this to the $\beta_{i}$ gives the desired
result. ∎

In the next corollary we let ${{\boldsymbol{k}}}_{0}$ be an ordered subfield
of ${{\boldsymbol{k}}}$. Then $(\Gamma,\psi)$, $(\Gamma^{*},\psi^{*})$ are
also $H$-couples over ${{\boldsymbol{k}}}_{0}$.

######  Corollary 8.15.

Let $(\alpha_{i})$ be a sequence in $\Gamma$ and $(\beta_{i})$ be a sequence
in $\Gamma^{*}$. Then $\beta$ falls under Case (b) with respect to
$(\alpha_{i})$, $(\beta_{i})$ iff $\beta$ falls under Case (b) with respect to
$(\alpha_{i})$, $(\beta_{i})$ when $(\Gamma,\psi)$ and $(\Gamma^{*},\psi^{*})$
are viewed as $H$-couples over ${{\boldsymbol{k}}}_{0}$.

###### Proof.

Use Lemma 8.12(i),(v) and $\beta_{i}^{\dagger}=\beta_{i+1}+\alpha_{i+1}$. ∎

Although the element $\beta$ of $(\Gamma^{*},\psi^{*})$ does not determine
uniquely the sequence $(\beta_{i})$ in Case (b), it follows from Lemma
8.12(i),(v),(vi) that $\beta$ does determine uniquely the sequences
$(\beta_{i}^{\dagger})$ and $\big{(}[\beta_{i}]_{{{\boldsymbol{k}}}}\big{)}$.
Without changing $\beta$ we still have considerable flexibility in choosing
the $\alpha_{i}$ and $\beta_{i}$:

######  Lemma 8.16.

Let $(\alpha_{i})$, $(\beta_{i})$ be as in (b). Let $\alpha_{0}^{*}$ be a good
approximation of $\beta$ in $\Gamma$, and $\alpha_{i+1}^{*}$ a good
approximation of $\beta_{i}^{\dagger}$ in $\Gamma$, for all $i$. Set
$\beta_{0}^{*}:=\beta-\alpha_{0}^{*}$ and
$\beta_{i+1}^{*}:=\beta_{i}^{\dagger}-\alpha_{i+1}^{*}$. Then
$(\alpha_{i}^{*})$ and $(\beta_{i}^{*})$ are also as in (b), with
$[\beta_{i}^{*}]_{{{\boldsymbol{k}}}}=[\beta_{i}]_{{{\boldsymbol{k}}}}$ and
$\beta_{i}^{*}-\beta_{i}\in\Gamma$ for all $i$.

###### Proof.

We have $\beta_{i}^{*}-\beta_{i}=\alpha_{i}-\alpha_{i}^{*}\in\Gamma$ for each
$i$ and so
$\Gamma\langle\beta\rangle=\Gamma\oplus\bigoplus_{i=0}^{\infty}{{\boldsymbol{k}}}\beta_{i}^{*}$.
From Lemma 8.5 and Lemma 8.12(ii),(iii) we get
$[\beta_{i}^{*}]_{{{\boldsymbol{k}}}}=[\beta_{i}]_{{{\boldsymbol{k}}}}$ for
all $i$, and so
$\beta_{i+1}^{*}=\beta_{i}^{\dagger}-\alpha_{i+1}^{*}=(\beta_{i}^{*})^{\dagger}-\alpha_{i+1}^{*}$
as required. ∎

Next we consider a shift $(\Gamma,\psi-\gamma)$ of $(\Gamma,\psi)$ and replace
$\beta$ by $\beta-\gamma$, viewed as an element of the extension
$(\Gamma^{*},\psi^{*}-\gamma)$ of $(\Gamma,\psi-\gamma)$:

######  Lemma 8.17.

Let $(\alpha_{i})$, $(\beta_{i})$ be as in (b). Then $\beta-\gamma$ falls
under (b) with respect to the indicated shifts, as witnessed by the sequences
$(\alpha_{i}-\gamma)$, $(\beta_{i})$.

At the end of the introduction we defined $\alpha^{\langle n\rangle}$. This
comes into play now.

######  Lemma 8.18.

Let $(\alpha_{i})$ and $(\beta_{i})$ be as in (b), and suppose that
$\beta_{i}^{\dagger}<0$ for all $i$. Then $\beta_{i}^{\langle
n+1\rangle}\leqslant\beta_{i+n}^{\dagger}<0$ for all $i$ and all $n$.

###### Proof.

This is trivial for $n=0$. Suppose $\beta_{i}^{\langle
n+1\rangle}\leqslant\beta_{i+n}^{\dagger}$. Then by Lemma 8.12(iv),

$\beta_{i}^{\langle n+2\rangle}\ \leqslant\ \beta_{i+n}^{\dagger\dagger}\
\leqslant\ \beta_{i+n+1}^{\dagger}.\qed$

We next discuss a situation where we can arrange that $\beta_{i}^{\dagger}<0$
for all $i$.

######  Remark 8.19.

Suppose $\operatorname{cf}(\Gamma^{<})>\omega$ and $(\alpha_{i})$,
$(\beta_{i})$ are as in (b). Then
$\operatorname{cf}(\Psi)=\operatorname{cf}(\Gamma^{<})>\omega$, so we have
$\gamma\in\Psi$ with $\beta_{i}^{\dagger}<\gamma$ for all $i$, hence
$\beta_{i}^{\dagger}-\gamma<0$ for all $i$. Thus $\beta-\gamma$ falls under
Case (b) with respect to the shifts $(\Gamma,\psi-\gamma)$ and
$(\Gamma^{*},\psi^{*}-\gamma)$ and for the associated sequences
$(\alpha_{i}-\gamma)$, $(\beta_{i})$ we have $(\psi^{*}-\gamma)(\beta_{i})<0$
for all $i$, so that the hypothesis of Lemma 8.18 is satisfied for this
shifted situation.

### Constructing a case (b)-extension

Let $K$ be a Liouville closed $H$-field; below we view its asymptotic couple
$(\Gamma,\psi)$ as an $H$-couple over ${{\boldsymbol{k}}}:={\mathbb{Q}}$.
Assume $\beta\notin\Gamma$ in an extension $(\Gamma^{*},\psi^{*})$ of
$(\Gamma,\psi)$ falls under Case (b). We show:

######  Proposition 8.20.

There exists an $H$-field extension $K\langle y\rangle$ of $K$ such that:

  1. (i)

$y>0$ and $vy\notin\Gamma$ realizes the same cut in $\Gamma$ as $\beta$;

  2. (ii)

for any $H$-field extension $M$ of $K$ and any $z\in M^{>}$ such that
$vz\notin\Gamma$ and $vz$ realizes the same cut in $\Gamma$ as $\beta$, there
is an $H$-field embedding $K\langle y\rangle\to M$ over $K$ sending $y$ to
$z$.

###### Proof.

Model-theoretic compactness gives a Liouville closed $H$-field extension $L$
of $K$ with $y\in L^{>}$ such that $vy\notin\Gamma$ realizes the same cut in
$\Gamma$ as $\beta$. Lemma 8.12 then yields an isomorphism
$\big{(}\Gamma\langle\beta\rangle,\psi_{\beta}\big{)}\to\big{(}\Gamma\langle
vy\rangle,\psi_{y}\big{)}$ of $H$-couples over ${\mathbb{Q}}$ that is the
identity on $\Gamma$ and sends $\beta$ to $vy$. (Here $\big{(}\Gamma\langle
vy\rangle,\psi_{y}\big{)}$ is the $H$-couple over ${\mathbb{Q}}$ generated by
$\Gamma\cup\\{vy\\}$ in the $H$-couple of $L$ over ${\mathbb{Q}}$.) It follows
that $\Gamma\langle vy\rangle/\Gamma$ has infinite dimension as a vector space
over ${\mathbb{Q}}$, so $y$ is differentially transcendental over $K$ in view
of $\Gamma\langle vy\rangle\subseteq v(K_{y}^{\times})$ where $K_{y}$ is the
real closure of $K\langle y\rangle$ in $L$. We claim that $K\langle y\rangle$
has the properties stated in the proposition; in particular, we show that
$K\langle y\rangle$ is an $H$-subfield of $L$, not just an asymptotic
(ordered) subfield of $L$.

Let $(\alpha_{i})$ and $(\beta_{i})$ be as in (b); for each $i$, take
$f_{i}\in K^{>}$ such that $vf_{i}=\alpha_{i}$. We define $y_{i}\in K\langle
y\rangle$ by recursion: $y_{0}:=y/f_{0}$, and
$y_{i+1}=y_{i}^{\dagger}/f_{i+1}$; to make this recursion possible we
simultaneously show by induction on $i$ that $y_{i}\neq 0$ and
$vy_{i}\notin\Gamma$ realizes the same cut in $\Gamma$ as $\beta_{i}$, and
$v(y_{i}^{\dagger})\notin\Gamma$ realizes the same cut in $\Gamma$ as
$\beta_{i}^{\dagger}$. This is all straightforward using the above isomorphism

(8.1)
$\big{(}\Gamma\langle\beta\rangle,\psi_{\beta}\big{)}\to\big{(}\Gamma\langle
vy\rangle,\psi_{y}\big{)},$

which sends $\beta_{i}$ to $vy_{i}$ for all $i$. Likewise we obtain that for
all $n$,

$\displaystyle K_{n}\ $ $\displaystyle:=\
K\big{(}y,y^{\prime},\dots,y^{(n)}\big{)}\ =\ K(y_{0},\dots,y_{n})\ =\
K\big{(}y,\dots,y^{\langle n\rangle}\big{)},\text{ and}$ $\displaystyle
v(K_{n}^{\times})\ $ $\displaystyle=\
\Gamma\oplus{\mathbb{Z}}vy_{0}\oplus\cdots\oplus{\mathbb{Z}}vy_{n}\ \subseteq\
\Gamma\langle vy\rangle,$

with the above isomorphism (8.1) restricting to an isomorphism

$\Gamma\oplus{\mathbb{Z}}\beta_{0}\oplus\cdots\oplus{\mathbb{Z}}\beta_{n}\
\to\ \Gamma\oplus{\mathbb{Z}}vy_{0}\oplus\cdots\oplus{\mathbb{Z}}vy_{n},\quad\
\beta_{i}\mapsto vy_{i}\quad(i=0,\dots,n)$

of ordered abelian groups. Hence the residue field $\operatorname{res}(K_{n})$
of the valued subfield $K_{n}$ of $L$ is algebraic over
$\operatorname{res}(K)$ by [ADH, 3.1.11] (Zariski-Abhyankar), and so
$\operatorname{res}(K)$ being real closed gives
$\operatorname{res}(K_{n})=\operatorname{res}(K)$. Then from $K\langle
y\rangle=\bigcup_{n}K_{n}$ we obtain $\operatorname{res}\\!\big{(}K\langle
y\rangle\big{)}=\operatorname{res}(K)$, so $K\langle y\rangle$ is an
$H$-subfield of $L$ with the same constant field as $K$, by [ADH, 9.1.2]. So
far we only used $y\neq 0$ rather than $y>0$.

Next, let $M$ be any $H$-field extension of $K$ and $z\in M^{\times}$ such
that $vz\notin\Gamma$ realizes the same cut as $\beta$ in $\Gamma$. By
increasing $M$ we can assume $M$ is Liouville closed, and then all the above
goes through with $z$ instead of $y$. In particular, setting $z_{0}:=z/f_{0}$
and $z_{i+1}:=z_{i}^{\dagger}/f_{i+1}$, we obtain for each $n$ an isomorphism
of the valued subfield $K_{n}$ of $L$ onto the valued subfield
$K(z_{0},\dots,z_{n})$ of $M$ over $K$, sending $y_{i}$ to $z_{i}$ for
${i=0,\dots,n}$. These have a common extension to a valued differential field
isomorphism $K\langle y\rangle\to K\langle z\rangle$ over $K$ sending $y$ to
$z$. For this isomorphism to preserve the ordering, we now assume besides
$y>0$ that also $z>0$. Induction on $i$ then shows that $y_{i}$ and $z_{i}$
are both positive, or both negative, for each $i$: use that all $f_{i}>0$ and
that for any $g$ in any $H$-field we have:

$g\succ 1\ \Rightarrow\ g^{\dagger}>0,\qquad g\prec 1\ \Rightarrow\
g^{\dagger}<0.$

The valuation determines for every polynomial $P(Y_{0},\dots,Y_{n})\in
K[Y_{0},\dots,Y_{n}]^{\neq}$ the unique dominant term in
$P(y_{0},\dots,y_{n})$ and in $P(z_{0},\dots,z_{n})$ in the same way, so this
isomorphism $K\langle y\rangle\to K\langle z\rangle$ is also order-preserving.
∎

######  Remark 8.21.

Let $K\langle y\rangle$ be an $H$-field extension of $K$ with $y>0$ such that
$vy$ realizes the same cut in $\Gamma$ as $\beta$, with real closure
$F:=K\langle y\rangle^{\operatorname{rc}}$. By the proof above $F$ has the
same constant field as $K$, and the $H$-couple of $F$ over ${\mathbb{Q}}$ is
generated over $(\Gamma,\psi)$ by $vy$, as witnessed by an isomorphism
$\big{(}\Gamma\langle\beta\rangle,\psi_{\beta})\to(\Gamma_{F},\psi_{F})$ over
$\Gamma$ sending $\beta$ to $vy$.

With $f_{i}$, $y_{i}$ as in the proof above (so
$y_{i}^{\dagger}=f_{i+1}y_{i+1}$ for all $i$), we think informally of the
element $y$ in Proposition 8.20 as given in terms of the $f_{i}$ by

$y\ =\ f_{0}y_{0}\ =\
f_{0}\operatorname{e}^{{}^{\textstyle\int\\!f_{1}y_{1}}}\ =\
f_{0}\operatorname{e}^{{}^{\textstyle\int\\!f_{1}\operatorname{e}^{{}^{\textstyle\int\\!f_{2}y_{2}}}}}\
=\cdots\ =\ \
f_{0}\operatorname{e}^{{}^{\textstyle\int\\!f_{1}\operatorname{e}^{{}^{\textstyle\int\\!f_{2}\operatorname{e}^{{}^{\textstyle\int{}^{\textstyle\iddots}}}}}}}$

In the next section we show how to construct such a $y$ analytically when $K$
is a Liouville closed Hardy field containing ${\mathbb{R}}$, under additional
hypotheses on $\beta$.

##  9\. Filling Gaps of Type (b)

In Section 8—see in particular the remark at the beginning of that section and
the remark preceding Lemma 8.4—we showed that Theorem A reduces to:

######  Lemma 9.1.

Let $H$ be a maximal Hardy field with $H$-couple $(\Gamma,\psi)$ over
${\mathbb{R}}$. Then no element in any extension of $(\Gamma,\psi)$ has
countable type over $\Gamma$.

###### Proof.

Suppose towards a contradiction that $\beta$ in some extension of
$(\Gamma,\psi)$ has countable type over $\Gamma$. Then $\beta$ falls under
Case (b) by the remarks that precede Lemma 8.10 and by Lemma 8.11. Let
$(\alpha_{i})$, $(\beta_{i})$ be as in (b). Then $(\beta_{i}^{\dagger})$ is
strictly increasing by Lemma 8.12(v). Since
$\operatorname{cf}(\Gamma^{<})=\operatorname{cf}(\Psi)>\omega$, we can take
$\gamma\in\Psi$ such that $\beta_{i}^{\dagger}<\gamma$ for all $i$. Take $g\in
H^{>}$ with $vg=\gamma$, and $\ell\in H$ with $\ell^{\prime}=g$, so
$\ell>{\mathbb{R}}$. Composing with $\ell^{\operatorname{inv}}$ yields a
maximal Hardy field $H\circ\ell^{\operatorname{inv}}$ whose $H$-couple over
${\mathbb{R}}$ we identify with the shift $(\Gamma,\psi-\gamma)$ of
$(\Gamma,\psi)$. As indicated in Remark 8.19 this allows us to replace $H$ by
$H\circ\ell^{\operatorname{inv}}$ and $\beta$ by $\beta-\gamma$. By renaming
we thus arrange that $\beta_{i}^{\dagger}<0$ for all $i$. This situation is
impossible by Theorem 9.2 below. ∎

Theorem 9.2 is of interest independent of Theorem A and Lemma 9.1, since it
involves a new way of constructing certain Hardy field extensions.

######  Theorem 9.2.

Let $H\supseteq{\mathbb{R}}$ be a Liouville closed Hardy field with $H$-couple
$(\Gamma,\psi)$ over ${\mathbb{R}}$. Suppose $\beta$ in an extension of
$(\Gamma,\psi)$ and of countable type over $\Gamma$ falls under Case (b), and
$\beta_{i}^{\dagger}<0$ for all $i$, where $(\alpha_{i})$, $(\beta_{i})$ are
as in (b). Then there exists ${y\neq 0}$ in a Hardy field extension of $H$
such that $vy$ realizes the same cut in $\Gamma$ as $\beta$.

The special cases $\beta<\Gamma$ and $\beta>\Gamma$ of Theorem 9.2 are taken
care of by Section 5: say $\beta<\Gamma$; then
$\operatorname{ci}(\Gamma)=\operatorname{cf}(H)=\omega$, and so there are
overhardian $y>_{\operatorname{e}}H$, and any such $y$ has the desired
property by Corollary 5.10.

The rest of this section proves Theorem 9.2 in the case where
$\alpha<\beta<\gamma$ for some $\alpha,\gamma\in\Gamma$. As we saw, this is
also the final step in proving Theorem A.

### Some useful inclusions

Let $H\supseteq{\mathbb{R}}$ be a Liouville closed Hardy field with $H$-couple
$(\Gamma,\psi)$ over ${\mathbb{R}}$, and let $\alpha$, $\gamma$ range over
$\Gamma$. Let $\beta$ in an extension of $(\Gamma,\psi)$ of Hahn type be such
that $\alpha<\beta<\gamma$ for some $\alpha$, $\gamma$,
$[\beta]\notin[\Gamma]$ (so $\beta^{\dagger}\notin\Gamma$), and
$\beta^{\langle n\rangle}<0$ for all $n\geqslant 1$. (We do allow $\beta>0$,
but use $-|\beta|$ below to arrange a value $<0$, with $(-|\beta|)^{\langle
n\rangle}=\beta^{\langle n\rangle}$ for $n\geqslant 1$.) Set

$A\ :=\ \big{\\{}h\in H^{>{\mathbb{R}}}:\,-|\beta|<vh\big{\\}},\qquad B:=\
\big{\\{}h\in H^{>{\mathbb{R}}}:vh<-|\beta|\big{\\}}.$

Then $A\cup B=H^{>{\mathbb{R}}}$, $A<B$, and so there is no $h\in H$ with
$A<h<B$. Also

$vA\cup vB\ =\ \Gamma^{<},\quad vB\ <\ -|\beta|\ <\ vA\ <\ 0,$

and so there is no $\alpha$ with $vB<\alpha<vA$.

######  Lemma 9.3.

The sets $A$ and $B$ have the following properties:

  1. (i)

$\operatorname{e}_{n}:=\exp_{n}(x)\in A$ for all $n$, and $B\neq\emptyset$;

  2. (ii)

$A=\operatorname{sq}(A)$ and $B=\sqrt{B}$.

###### Proof.

As to (i), an easy induction shows that
$\operatorname{e}_{n}^{\dagger}=\operatorname{e}_{1}\cdots\operatorname{e}_{n-1}$
for $n\geqslant 1$ and $\operatorname{e}_{n}^{\langle
m\rangle}\sim\operatorname{e}_{n-m+1}^{\dagger}$ for $n\geqslant m\geqslant
1$. In particular, $\operatorname{e}_{n}^{\langle n\rangle}\sim 1$ for
$n\geqslant 1$. Since $\beta^{\langle n\rangle}<0$ for all $n\geqslant 1$,
this gives $v(\operatorname{e}_{n})>\beta$ for all $n$. Item (ii) follows from
$[\beta]\notin[\Gamma]$. ∎

We now set

$\displaystyle A^{\dagger}\ $ $\displaystyle:=\ \big{\\{}a^{\dagger}:\,a\in
A,\ a^{\dagger}\succ 1\big{\\}},\quad B^{\dagger}\ :=\
\big{\\{}b^{\dagger}:\,b\in B\big{\\}},\text{ so in view of
$\beta^{\dagger}\notin\Gamma$:}$ $\displaystyle A^{\dagger}\ $
$\displaystyle=\ \big{\\{}h\in
H^{>{\mathbb{R}}}:\,\beta^{\dagger}<vh\big{\\}},\quad B^{\dagger}\ =\
\big{\\{}h\in H^{>{\mathbb{R}}}:\,vh<\beta^{\dagger}\big{\\}}.$

Thus $A^{\dagger}\cup B^{\dagger}=H^{>{\mathbb{R}}}$,
$A^{\dagger}<B^{\dagger}$, and there is no $h\in H$ with
$A^{\dagger}<h<B^{\dagger}$. Also

$v(A^{\dagger})\cup v(B^{\dagger})\ =\ \Gamma^{<},\qquad v(B^{\dagger})\ <\
\beta^{\dagger}\ <\ v(A^{\dagger})\ <\ 0,$

and there is no $\alpha$ with $v(B^{\dagger})<\alpha<v(A^{\dagger})$. Note
also that $\operatorname{e}_{n}^{\dagger}\in A^{\dagger}$ for all $n\geqslant
2$.

######  Corollary 9.4.

$\log A\subseteq A^{\dagger}\subseteq A$ and $\log B\supseteq
B^{\dagger}\supseteq B$.

###### Proof.

If $h\in H$, $h\geqslant\operatorname{e}_{2}$, then $\log h\preccurlyeq(\log
h)^{\prime}=h^{\dagger}$. Then by Lemma 9.3(i) we have $\log A\subseteq
A^{\dagger}$. Now use $\log A<\log B$ and $\log A\cup\log B=A^{\dagger}\cup
B^{\dagger}=H^{>{\mathbb{R}}}$. As to $A^{\dagger}\subseteq A$: if $h\in
H^{>{\mathbb{R}}}$ and $h^{\dagger}\succ 1$, then $vh^{\dagger}=o(vh)$ by
[ADH, 9.2.10(iv)]. ∎

To indicate the dependence of $A$, $B$, $A^{\dagger}$, $B^{\dagger}$ on
$\beta$ we may denote these sets by

$A(\beta),\quad B(\beta),\quad A^{\dagger}(\beta),\quad B^{\dagger}(\beta).$

In fact, these four sets depend only on $[\beta]$ rather than $\beta$, in view
of $[\beta]\notin[\Gamma]$.

Recall that $\beta^{\dagger}\notin\Gamma$ and $\beta^{\dagger}<0$, so if
$\beta^{\dagger}$ has a good approximation in $\Gamma$, it has a good
approximation $\leqslant 0$ in $\Gamma$. Note: if
$[\beta^{\dagger}]\notin[\Gamma]$, then $0$ is a good approximation of $\beta$
in $\Gamma$ and any good approximation $\alpha\leqslant 0$ to
$\beta^{\dagger}$ in $\Gamma$ satisfies $\beta^{\dagger}<\alpha$.

Suppose now that $\alpha\leqslant 0$ is a good approximation of
$\beta^{\dagger}$ in $\Gamma$, so $[\beta^{\dagger}-\alpha]\notin[\Gamma]$.
Set $\beta_{\operatorname{next}}:=\beta^{\dagger}-\alpha$, and assume also
that $\beta_{\operatorname{next}}^{\langle n\rangle}<0$ for all $n\geqslant
1$. This means that the conditions we imposed earlier on $\beta$ are now also
satisfied by $\beta_{\operatorname{next}}$. Since
$[\beta_{\operatorname{next}}]$ does not depend on the particular good
approximation $\alpha\leqslant 0$ of $\beta^{\dagger}$ in $\Gamma$,

$A(\beta_{\operatorname{next}})\ =\ \big{\\{}h\in
H^{>{\mathbb{R}}}:vh>-|\beta_{\operatorname{next}}|\big{\\}}$

doesn’t either, and the assumption that $\beta_{\operatorname{next}}^{\langle
n\rangle}<0$ for all $n\geqslant 1$ will still be satisfied for any such
$\alpha$.

######  Lemma 9.5.

$A(\beta_{\operatorname{next}})\subseteq\log A(\beta)$.

###### Proof.

Let $h\in A(\beta_{\operatorname{next}})$; it suffices to show that then
$\operatorname{e}^{h}\in A(\beta)$. Suppose towards a contradiction that
$\operatorname{e}^{h}\in B$. Then $v\operatorname{e}^{h}<-|\beta|$, so
$v(\operatorname{e}^{h})^{\dagger}=vh^{\prime}<\beta^{\dagger}<0$. If
$[\beta^{\dagger}]\in\Gamma$, then
$[\beta^{\dagger}]>[\beta_{\operatorname{next}}]$, so $h^{\prime}\in
B(\beta_{\operatorname{next}})$, and thus $h\in
B(\beta_{\operatorname{next}})$ by Lemma 1.13(ii) applied to
$B(\beta_{\operatorname{next}})$ in the role of $B$. If
$[\beta^{\dagger}]\notin\Gamma$, then
$[\beta^{\dagger}]=[\beta_{\operatorname{next}}]$, and again $h^{\prime}\in
B(\beta_{\operatorname{next}})$, so $h\in B(\beta_{\operatorname{next}})$. In
both cases we contradict $h\in A(\beta_{\operatorname{next}})$. ∎

The diagram in Figure 10 depicts the gaps

$(A,B)=\big{(}A(\beta),B(\beta)\big{)},\quad(A^{\dagger},B^{\dagger}),\quad(\log
A,\log
B),\quad\big{(}A(\beta_{\operatorname{next}}),B(\beta_{\operatorname{next}})\big{)}$

in $H$ and hypothetical $H$-hardian germs $y$, $y_{\operatorname{next}}$ with
$A<y<B$ and
$A(\beta_{\operatorname{next}})<y_{\operatorname{next}}<B(\beta_{\operatorname{next}})$,
as well as $y^{\dagger}$ and $\log y$.

$A$$B$$y$$A^{\dagger}$$B^{\dagger}$$y^{\dagger}$$\log A$$\log B$$\log
y$$A(\beta_{\operatorname{next}})$$B(\beta_{\operatorname{next}})$${\mathbb{R}}$$y_{\operatorname{next}}$
Figure 10. Various gaps in $H$ associated to $(A,B)$

######  Lemma 9.6.

We have $[\beta]>[\beta_{\operatorname{next}}]$. If there is no $\gamma$ such
that $[\beta]>[\gamma]>[\beta_{\operatorname{next}}]$, then
$A(\beta)=A(\beta_{\operatorname{next}})$,
$B(\beta)=B(\beta_{\operatorname{next}})$, and $A$, $B$ is a wide gap.

###### Proof.

From [ADH, 9.2.10(iv)] and $\beta^{\dagger}<0$ we get
$\beta^{\dagger}=o(\beta)$, so
$[\beta]>[\beta^{\dagger}]\geqslant[\beta_{\operatorname{next}}]$. Suppose
there is no $\gamma$ with $[\beta]>[\gamma]>[\beta_{\operatorname{next}}]$.
Then clearly $A(\beta)=A(\beta_{\operatorname{next}})$ and
$B(\beta)=B(\beta_{\operatorname{next}})$, so $A\subseteq\log A$ by Lemma 9.5.
Thus $A$, $B$ is a wide gap. ∎

By Lemma 9.3 we have $\operatorname{e}_{n}\in A(\beta_{\operatorname{next}})$
for all $n$. In combination with the next result this gives further
information about the behavior of $A$ and $B$ and of the gap between them. For
$p,\phi\in\mathcal{C}$ with $\phi>_{\operatorname{e}}0$ we have the germ
$\phi^{p}\in\mathcal{C}$. Let $p\in H$; then $h\in H^{>}$ gives
$h^{p}=\exp(p\log h)\in H^{>}$, and for $S\subseteq H^{>}$ we set
$S^{p}:=\\{h^{p}:\,h\in S\\}\subseteq H^{>}$.

######  Proposition 9.7.

The sets $A$, $A^{\dagger}$, $A(\beta_{\operatorname{next}})$, $B$ have the
following properties:

  1. (i)

$A(\beta_{\operatorname{next}})\cdot A^{\dagger}\subseteq A^{\dagger}$;

  2. (ii)

if $p\in A(\beta_{\operatorname{next}})$, then $A^{p}\subseteq A$ and
$B^{1/p}\subseteq B$;

  3. (iii)

if $p\in A(\beta_{\operatorname{next}})$, $\phi\in\mathcal{C}$, and
$A<_{\operatorname{e}}\phi<_{\operatorname{e}}B$, then
$A<_{\operatorname{e}}\phi^{1/p}<_{\operatorname{e}}\phi<_{\operatorname{e}}\phi^{p}<_{\operatorname{e}}B$.

###### Proof.

For (i) we distinguish two cases. Suppose first that $\beta^{\dagger}<\alpha$.
Let $p\in A(\beta_{\operatorname{next}})$, $h\in A^{\dagger}$; we need to show
$ph\in A^{\dagger}$, that is, $v(ph)>\beta^{\dagger}$, equivalently,
$vp>\beta^{\dagger}-vh$. Since
$vp>\beta_{\operatorname{next}}=\beta^{\dagger}-\alpha$, we do have
$vp>\beta^{\dagger}-vh$ if $vh\geqslant\alpha$. If $vh<\alpha$, then
$\beta^{\dagger}<vh<\alpha$, so $vh$ is also a good approximation of
$\beta^{\dagger}$ in $\Gamma$, and then replacing $\alpha$ by $vh$ yields
$vp>\beta^{\dagger}-vh$ in view of remarks made earlier about
$A(\beta_{\operatorname{next}})$.

Next, suppose $\alpha<\beta^{\dagger}$, so $[\beta^{\dagger}]\in[\Gamma]$ by
an earlier remark. Let $p\in A(\beta_{\operatorname{next}})$, $h\in
A^{\dagger}$; as before we need to show $vp>\beta^{\dagger}-vh$. Now
$\alpha<\beta^{\dagger}<vh$ gives
$[\alpha-\beta^{\dagger}]<[\beta^{\dagger}-vh]$ by Lemma 8.7(ii). Since
$\alpha-\beta^{\dagger}$ and $\beta^{\dagger}-vh$ are both negative, this
yields $\alpha-\beta^{\dagger}>\beta^{\dagger}-vh$, which together with
$vp>\alpha-\beta^{\dagger}$ gives $vp>\beta^{\dagger}-vh$.

As to (ii), let $p\in A(\beta_{\operatorname{next}})$ and $h\in A$. We have
$h^{p}>{\mathbb{R}}$ and

$(h^{p})^{\dagger}\ =\ (p\log h)^{\prime}\ =\ p^{\prime}\log h+ph^{\dagger},$

and $ph^{\dagger}\preccurlyeq 1$ or $ph^{\dagger}\in A^{\dagger}$ by (i). Also
$p^{\prime}\in A(\beta_{\operatorname{next}})$ or $0<p^{\prime}\preccurlyeq
1$, by Lemma 1.12, and $\log p\in A^{\dagger}$ by Corollary 9.3, so
$p^{\prime}\log h\preccurlyeq 1$ or $p^{\prime}\log h\in A^{\dagger}$ by (i).
Hence $(h^{p})^{\dagger}\in A^{\dagger}$, and thus $h^{p}\in A$. Next, let
$p\in A(\beta_{\operatorname{next}})$ and $h\in B$. Then $h^{1/p}\notin B$
would mean $h^{1/p}\in A$ or $0<h^{1/p}\preccurlyeq 1$, and in either case
$h=(h^{1/p})^{p}$ would give $h\in A$ or $h\preccurlyeq 1$, contradicting
$h\in B$. This concludes the proof of (ii).

Property (iii) is a routine consequence of (ii). ∎

Part (iii) of Proposition 9.7 is only relevant if there is any
$\phi\in\mathcal{C}$ with $A<_{\operatorname{e}}\phi<_{\operatorname{e}}B$.
There are indeed such $\phi$ if
$\operatorname{cf}(A)=\operatorname{ci}(B)=\omega$, by Corollary 2.8.

To describe $A(\beta_{\operatorname{next}})$ directly in terms of
$A^{\dagger}$, take $f\in H^{>}$ with $vf=\alpha$. Then:

######  Lemma 9.8.

If $\beta^{\dagger}<\alpha$, then $f\in A^{\dagger}$ or $f\asymp 1$, and

$A(\beta_{\operatorname{next}})\ =\ H^{>{\mathbb{R}}}\cap
f^{-1}A^{\dagger},\qquad B(\beta_{\operatorname{next}})\ =\
f^{-1}B^{\dagger}.$

If $\alpha<\beta^{\dagger}$, then $f\in B^{\dagger}$,
$[\beta^{\dagger}-\alpha]<[\beta^{\dagger}]=[\alpha]\in[\Gamma]$, and

$A(\beta_{\operatorname{next}})\ =\ H^{>{\mathbb{R}}}\cap
f(B^{\dagger})^{-1},\quad f(A^{\dagger})^{-1}\text{ is a coinitial subset of
}B(\beta_{\operatorname{next}}).$

###### Proof.

If $\beta^{\dagger}<\alpha$, then the inclusion
$A(\beta_{\operatorname{next}})\supseteq H^{>{\mathbb{R}}}\cap
f^{-1}A^{\dagger}$ and the equality
$B(\beta_{\operatorname{next}})=f^{-1}B^{\dagger}$ are almost obvious, and one
can use $\alpha\leqslant 0$ to prove the inclusion
$A(\beta_{\operatorname{next}})\subseteq H^{>{\mathbb{R}}}\cap
f^{-1}A^{\dagger}$.

Next, suppose $\alpha<\beta^{\dagger}$. Then for the inclusion
$A(\beta_{\operatorname{next}})\supseteq H^{>{\mathbb{R}}}\cap
f(B^{\dagger})^{-1}$, use $\beta^{\dagger}<\alpha-\beta^{\dagger}$, and for
the statement about $B(\beta_{\operatorname{next}})$, note that

$\displaystyle B(\beta_{\operatorname{next}})\ $ $\displaystyle=\
\big{\\{}h\in H^{>{\mathbb{R}}}:\,vh<\alpha-\beta^{\dagger}\big{\\}},\text{
and }$ $\displaystyle f(A^{\dagger})^{-1}\ $ $\displaystyle=\ \big{\\{}h\in
H^{>{\mathbb{R}}}:\,\alpha<vh<\alpha-\beta^{\dagger}\big{\\}}.\qed$

Using the first part of Lemma 9.8 we obtain:

######  Corollary 9.9.

Suppose $\beta_{\operatorname{next}}<0$ and
$z_{0},z_{1}\in\mathcal{C}^{<\infty}$ are such that

$z_{0}>_{\operatorname{e}}0,\qquad z_{0}^{\dagger}\ =\ fz_{1},\qquad
A(\beta_{\operatorname{next}})<_{\operatorname{e}}z_{1}<_{\operatorname{e}}B(\beta_{\operatorname{next}}).$

Then $A<_{\operatorname{e}}z_{0}<_{\operatorname{e}}B$.

###### Proof.

Let $h\in A$. Then $f^{-1}h^{\dagger}\in A(\beta_{\operatorname{next}})$ or
$f^{-1}h^{\dagger}\preccurlyeq 1$, so
$f^{-1}h^{\dagger}<_{\operatorname{e}}z_{1}=f^{-1}z_{0}^{\dagger}$, and thus
$h^{\dagger}<_{\operatorname{e}}z_{0}^{\dagger}$. Then Lemma 2.16 gives
$c\in{\mathbb{R}}^{>}$ with $ch<_{\operatorname{e}}z_{0}$. Applying this
argument to $h^{2}$ instead of $h$ gives $d\in{\mathbb{R}}^{>}$ with
$dh^{2}<_{\operatorname{e}}z_{0}$, which in view of
$h<_{\operatorname{e}}dh^{2}$ gives $h<_{\operatorname{e}}z_{0}$. In the same
way one shows that if $h\in B$, then $z_{0}<_{\operatorname{e}}h$. ∎

Likewise, using the second part of Lemma 9.8:

######  Corollary 9.10.

If $\beta_{\operatorname{next}}>0$ and $z_{0},z_{1}\in\mathcal{C}^{<\infty}$
are such that

$z_{0},z_{1}>_{\operatorname{e}}0,\qquad z_{0}^{\dagger}\ =\ f/z_{1},\qquad
A(\beta_{\operatorname{next}})<_{\operatorname{e}}z_{1}<_{\operatorname{e}}B(\beta_{\operatorname{next}}),$

then $A<_{\operatorname{e}}z_{0}<_{\operatorname{e}}B$.

### What remains to be done

Let $H\supseteq{\mathbb{R}}$ be a Liouville closed Hardy field with $H$-couple
$(\Gamma,\psi)$ over ${\mathbb{R}}$, and let $\alpha$, $\gamma$ range over
$\Gamma$. Suppose $\beta$ in an extension of $(\Gamma,\psi)$ is of countable
type over $\Gamma$ and falls under Case (b), with $(\alpha_{i})$,
$(\beta_{i})$ as in (b), and $\beta_{i}^{\dagger}<0$, $\alpha_{i+1}\leqslant
0$ for all $i$. Assume also that $\alpha<\beta<\gamma$ for some $\alpha$,
$\gamma$. Then $\big{(}\Gamma\langle\beta\rangle,\psi_{\beta}\big{)}$ is of
Hahn type, by the last claim in Lemma 8.12, and so all $\beta_{i}$ lie in this
extension of $(\Gamma,\psi)$ of Hahn type; this is significant because of the
initial assumption on $\beta$ in the previous subsection. Note also that for
all $i$ there are $\alpha$, $\gamma$ with $\alpha<\beta_{i}<\gamma$, and that
by Lemma 8.18 we have $\beta_{i}^{\langle n\rangle}<0$ for all $i$ and all
$n\geqslant 1$. Thus we can apply the previous subsection to each $\beta_{i}$
in the role of $\beta$ there. Set

$A_{i}\ :=\ \big{\\{}h\in H^{>{\mathbb{R}}}:\,vh>-|\beta_{i}|\big{\\}},\qquad
B_{i}\ :=\ \big{\\{}h\in H:vh<-|\beta_{i}|\big{\\}},$

so $A_{i}=A(\beta_{i})$, $B_{i}=B(\beta_{i})$, and
$\beta_{i+1}=(\beta_{i})_{\operatorname{next}}$ in the notation of the
previous subsection. Thus by lemmas in that subsection:

  1. (i)

$\operatorname{e}_{n}\in A_{i}$ for all $i$, $n$;

  2. (ii)

$A_{i}$ and $\operatorname{sq}(A_{i})$ are cofinal, and $B_{i}$ and
$\sqrt{B_{i}}$ are coinitial;

  3. (iii)

$h\in A_{i}\Rightarrow h^{\operatorname{e}_{n}}\in A_{i}$, and $h\in
B_{i}\Rightarrow h^{1/\\!\operatorname{e}_{n}}\in B_{i}$;

By Corollary 8.14 we have
$\operatorname{cf}(\Gamma^{<\beta_{i}})=\operatorname{ci}(\Gamma^{>\beta_{i}})=\omega$
for all $i$. Hence $\operatorname{cf}(A_{i})=\operatorname{ci}(B_{i})=\omega$
for all $i$. What remains to be done is to show the existence of a $y>0$ in a
Hardy field extension of $H$ such that $vy$ realizes the same cut in $\Gamma$
as $\beta$.

For each $i$, take $f_{i}\in H^{>}$ with $vf_{i}=\alpha_{i}$, and
$f_{i}\geqslant 1$ for $i\geqslant 1$. To get the right idea for our reverse
engineering, suppose $y>0$ is $H$-hardian and $vy$ realizes the same cut in
$\Gamma$ as $\beta$. As in the proof of Proposition 8.20, let $y_{i}\in
H\langle y\rangle$ be given by $y_{0}:=y/f_{0}$, and
$y_{i+1}=y_{i}^{\dagger}/f_{i+1}$. Then $vy_{i}$ realizes the same cut in
$\Gamma$ as $\beta_{i}$, so $y_{i}\succ 1$ if $\beta_{i}<0$ and $y_{i}\prec 1$
if $\beta_{i}>0$. To have only positive infinite germs, set

$z_{i}\ :=\ |y_{i}|\ \text{ if }\ \beta_{i}<0,\qquad z_{i}\ :=\ |y_{i}|^{-1}\
\text{ if }\ \beta_{i}>0.$

One verifies easily that then
$A_{i}<_{\operatorname{e}}z_{i}<_{\operatorname{e}}B_{i}$, and

$\beta_{i+1}<0\ \Longrightarrow\
z_{i}^{\dagger}\,=\,f_{i+1}z_{i+1},\qquad\beta_{i+1}>0\ \Longrightarrow\
z_{i}^{\dagger}\,=\,f_{i+1}/z_{i+1}.$

We first deal with a “wide gap” case:

######  Lemma 9.11.

Suppose for some $n$ there is no $\gamma$ with
$[\beta_{n}]>[\gamma]>[\beta_{n+1}]$. Then there exists $H$-hardian $y>0$ such
that $vy$ realizes the same cut in $\Gamma$ as $\beta$.

###### Proof.

Let $n$ be as in the hypothesis. Then $A_{n}$, $B_{n}$ is a wide gap by Lemma
9.6, hence Section 6 gives an $H$-hardian $z_{n}$ such that
${A_{n}<_{\operatorname{e}}z_{n}<_{\operatorname{e}}B_{n}}$. Let
$L:=\operatorname{Li}\\!\big{(}H\langle z_{n}\rangle\big{)}$. Then we have
$z_{n-1},\dots,z_{0}\in L^{>}$ such that for all $i<n$:

$\beta_{i+1}<0\ \Longrightarrow\
z_{i}^{\dagger}\,=\,f_{i+1}z_{i+1},\qquad\beta_{i+1}>0\ \Longrightarrow\
z_{i}^{\dagger}\,=\,f_{i+1}/z_{i+1}.$

Downward induction on $i$ using Corollaries 9.9 and 9.10 then gives
$A_{i}<_{\operatorname{e}}z_{i}<_{\operatorname{e}}B_{i}$ for all $i\leqslant
n$. Thus if $\beta_{0}<0$, then $v(z_{0})$ realizes the same gap in $\Gamma$
as $\beta_{0}$, and so $y:=f_{0}z_{0}$ has the desired property. If
$\beta_{0}>0$, then $v(z_{0})$ realizes the same gap in $\Gamma$ as
$-\beta_{0}$, and so $y:=f_{0}/z_{0}$ has the desired property. ∎

It remains to consider the case that for all $i$ there exists $\gamma$ with
$[\beta_{i}]>[\gamma]>[\beta_{i+1}]$. We assume this for the rest of this
section. The goal of our reverse engineering will be to construct germs
$z_{i}$ as in the next lemma:

######  Lemma 9.12.

Let the germs $z_{i}\in\mathcal{C}^{<\infty}$ be such that for all $i$,
$A_{i}<_{\operatorname{e}}z_{i}<_{\operatorname{e}}B_{i}$ and

$\beta_{i+1}<0\ \Longrightarrow\ z_{i}^{\dagger}\ =\
f_{i+1}z_{i+1},\qquad\beta_{i+1}>0\ \Longrightarrow\ z_{i}^{\dagger}\ =\
f_{i+1}/z_{i+1}.$

Then there exists $H$-hardian $y>0$ such that $vy$ realizes the same cut in
$\Gamma$ as $\beta$.

###### Proof.

Note that for each $n$ we have the ordered subgroup
$\Gamma\oplus{\mathbb{Z}}\beta_{0}\oplus\cdots\oplus{\mathbb{Z}}\beta_{n}$ of
$\Gamma\langle\beta\rangle$, and likewise with ${\mathbb{Q}}$ instead of
${\mathbb{Z}}$. We prove by induction on $n$ that $z_{0},\dots,z_{n}$ generate
a Hausdorff field $H_{n}:=H(z_{0},\dots,z_{n})$ over $H$, with

$v(H_{n}^{\times})\ =\
\Gamma\oplus{\mathbb{Z}}vz_{0}\oplus\cdots\oplus{\mathbb{Z}}vz_{n},$

and with an ordered abelian group isomorphism that is the identity on
$\Gamma$:

$\Gamma\oplus{\mathbb{Z}}\beta_{0}\oplus\cdots\oplus{\mathbb{Z}}\beta_{n}\
\to\ \Gamma\oplus{\mathbb{Z}}vz_{0}\oplus\cdots\oplus{\mathbb{Z}}vz_{n},\quad\
-|\beta_{i}|\mapsto vz_{i}\quad(i=0,\dots,n).$

For $n=0$ this follows from Lemma 1.11. Assume that the above holds for a
certain $n$. Then for the real closure $H_{n}^{\operatorname{rc}}$ of $H_{n}$
as a Hausdorff field extension of $H_{n}$,

$v\big{(}H_{n}^{\operatorname{rc},\times}\big{)}\ =\
\Gamma\oplus{\mathbb{Q}}vz_{0}\oplus\cdots\oplus{\mathbb{Q}}vz_{n},$

with an ordered abelian group isomorphism that is the identity on $\Gamma$:

$\Gamma\oplus{\mathbb{Q}}\beta_{0}\oplus\cdots\oplus{\mathbb{Q}}\beta_{n}\
\to\ \Gamma\oplus{\mathbb{Q}}vz_{0}\oplus\cdots\oplus{\mathbb{Q}}vz_{n},\quad\
-|\beta_{i}|\mapsto vz_{i}\quad(i=0,\dots,n).$

Thus
$\big{[}v\big{(}H_{n}^{\operatorname{rc},\times}\big{)}\big{]}=[\Gamma]\cup\big{\\{}[vz_{0}],\dots,[vz_{n}]\big{\\}}$
by Lemma 8.12.

Claim : For each $f\in H_{n}^{\operatorname{rc},>}$, either $f\preccurlyeq
h$ for some $h\in A_{n+1}$, or $f\succcurlyeq h$ for some ${h\in B_{n+1}}$.

Otherwise we have $f\in H_{n}^{\operatorname{rc}}$ with $A_{n+1}<f<B_{n+1}$,
so $[vf]\notin[\Gamma]$ and $vf$ realizes the same cut in $\Gamma$ as
$-|\beta_{n+1}|$. Taking $\gamma$ with $[\beta_{n}]>[\gamma]>[\beta_{n+1}]$ we
obtain ${[vz_{0}]>\cdots>[vz_{n}]>[\gamma]>[vf]}$, contradicting $vf\in
v\big{(}H_{n}^{\operatorname{rc},\times}\big{)}$.

The claim and Lemma 1.11 give a Hausdorff field extension
$H_{n}^{\operatorname{rc}}(z_{n+1})$ of $H_{n}^{\operatorname{rc}}$, and the
resulting Hausdorff field extension $H_{n+1}=H_{n}(z_{n+1})$ of $H$ has the
properties that the inductive step requires. This concludes the proof by
induction.

An easy induction on $n$ now shows that for $z:=z_{0}$ the elements
$z,z^{\prime},\dots,z^{(n)}$ of $\mathcal{C}^{<\infty}$ generate the Hausdorff
field $H(z,z^{\prime},\dots,z^{(n)})=H_{n}$ over $H$, and so we have a Hardy
field $H\langle z\rangle$ over $H$. If $\beta_{0}<0$, then $y:=f_{0}z_{0}$ has
the desired property, and if $\beta_{0}>0$, then $y:=f_{0}/z_{0}$ has the
desired property. ∎

### First step in reverse engineering

To construct germs $z_{i}$ as in Lemma 9.12 we first take for each $i$ a
continuous function $[0,+\infty)\to{\mathbb{R}}^{>}$ that represents the germ
$f_{i}\in H$ and to be denoted also by $f_{i}$, and with $f_{i}\geqslant 1$ on
$[0,+\infty)$ for $i\geqslant 1$. Next, let $(a_{i})$ be a strictly increasing
sequence of real numbers $\geqslant 0$ tending to $+\infty$ such that
$f_{0},\dots,f_{m}$ are of class $\mathcal{C}^{m}$ on $[a_{m},+\infty)$. Let
there also be given for each $m\geqslant 1$ a continuous function
$z_{m-1,m}\colon[a_{m-1},a_{m}]\to{\mathbb{R}}^{>}$. Then we define the
continuous function $z_{k,m}\colon[a_{k},a_{m}]\to{\mathbb{R}}^{>}$ for
$0\leqslant k<m$ by downward recursion: $z_{m-1,m}$ for $m\geqslant 1$ is
already given to us, and for $1\leqslant k<m$,

$z_{k-1,m}(t)\ :=\ \begin{cases}\ z_{k-1,k}(t)&\text{for $a_{k-1}\leqslant
t\leqslant a_{k}$,}\\\ \
z_{k-1,k}(a_{k})\cdot\exp\displaystyle\int_{a_{k}}^{t}f_{k}(s)z_{k,m}(s)\,ds&\text{for
$a_{k}\leqslant t\leqslant a_{m}$, if $\beta_{k}<0$,}\\\\[10.00002pt] \
z_{k-1,k}(a_{k})\cdot\exp\displaystyle\int_{a_{k}}^{t}\frac{f_{k}(s)}{z_{k,m}(s)}\,ds&\text{for
$a_{k}\leqslant t\leqslant a_{m}$, if $\beta_{k}>0$.}\end{cases}$

Downward induction on $k$ gives $z_{k,m}=z_{k,m+1}$ on $[a_{k},a_{m}]$ for
$k<m$. This fact gives for each $k\in{\mathbb{N}}$ a continuous function
$z_{k}\colon[a_{k},+\infty)\to{\mathbb{R}}^{>}$ such that $z_{k}=z_{k,m}$ on
$[a_{k},a_{m}]$, for all $m>k$. Thus for $k\geqslant 1$ and $t\geqslant a_{k}$
we have

$\displaystyle\beta_{k}<0\ $ $\displaystyle\Longrightarrow\ z_{k-1}(t)\ =\
z_{k-1}(a_{k})\cdot\exp\int_{a_{k}}^{t}f_{k}(s)z_{k}(s)\,ds,$
$\displaystyle\beta_{k}>0\ $ $\displaystyle\Longrightarrow\ z_{k-1}(t)\ =\
z_{k-1}(a_{k})\cdot\exp\int_{a_{k}}^{t}\frac{f_{k}(s)}{z_{k}(s)}\,ds,$

so $z_{k-1}$ is of class $\mathcal{C}^{1}$ on $[a_{k},+\infty)$, and:

$\displaystyle\beta_{k}<0\ $ $\displaystyle\Longrightarrow\
z_{k-1}^{\dagger}=f_{k}\,z_{k}\text{ on $[a_{k},+\infty)$,}$
$\displaystyle\beta_{k}>0\ $ $\displaystyle\Longrightarrow\
z_{k-1}^{\dagger}=f_{k}/z_{k}\text{ on $[a_{k},+\infty)$.}$

Hence induction on $m$ gives that $z_{k}$ is of class $\mathcal{C}^{m}$ on
$[a_{k+m},+\infty)$ (for all $k$, $m$), and thus (the germ of) each $z_{k}$
lies in $\mathcal{C}^{<\infty}$.

The above is a general construction of functions whose germs satisfy the
equalities in Lemma 9.12. More work is needed to satisfy also the inequalities
$A_{i}<_{\operatorname{e}}z_{i}<_{\operatorname{e}}B_{i}$ in that lemma. We
now turn to this task.

### Second step in reverse engineering

Assume in this subsection that $z_{k-1,k}\geqslant 1$ on $[a_{k-1},a_{k}]$,
for all $k\geqslant 1$. Then $z_{k}\geqslant 1$ on $[a_{k},+\infty)$ for all
$k$. For $k\geqslant 1$ we have $z_{k-1}(t),z_{k-1}^{\dagger}(t)>0$ for all
$t\geqslant a_{k}$, so $z_{k-1}$ is strictly increasing on $[a_{k},+\infty)$.
For each $k$, let $p_{k},q_{k}\colon[a_{k},+\infty)\to{\mathbb{R}}^{>}$ be
continuous functions such that

$p_{m-1}\ \leqslant\ z_{m-1,m}\ \leqslant\ q_{m-1}\ \text{ on
$[a_{m-1},a_{m}$], for all $m\geqslant 1$.}$

We try to find conditions on the families $(p_{k})$ and $(q_{k})$ so that
these inequalities extend to $p_{k}\leqslant z_{k,m}\leqslant q_{k}$ on
$[a_{k},a_{m}]$ for all $k$, $m$ with $k<m$ (and thus $p_{k}\leqslant
z_{k}\leqslant q_{k}$ on $[a_{k},+\infty)$ for all $k$). Let $1\leqslant k<m$
and assume inductively that $p_{k}\leqslant z_{k,m}\leqslant q_{k}$ on
$[a_{k},a_{m}]$. On $[a_{k-1},a_{k}]$ we have $p_{k-1}\leqslant
z_{k-1,k}\leqslant q_{k-1}$, so $p_{k-1}\leqslant z_{k-1,m}\leqslant q_{k-1}$,
as desired.

First suppose $\beta_{k}<0$. Then for $a_{k}\leqslant t\leqslant a_{m}$ we
have

$z_{k-1,m}(t)\ =\ z_{k-1,k}(a_{k})\exp\int_{a_{k}}^{t}f_{k}(s)z_{k,m}(s)ds,$

hence

$p_{k-1}(a_{k})\exp\int_{a_{k}}^{t}f_{k}(s)p_{k}(s)ds\ \leqslant\
z_{k-1,m}(t)\ \leqslant\
q_{k-1}(a_{k})\exp\int_{a_{k}}^{t}f_{k}(s)q_{k}(s)ds,$

and so the desired $p_{k-1}\leqslant z_{k-1,m}\leqslant q_{k-1}$ on
$[a_{k-1},a_{m}]$ would follow if

(${\rm{I}}_{k}$) $\displaystyle p_{k-1}(t)\ $ $\displaystyle\leqslant\
p_{k-1}(a_{k})\exp\int_{a_{k}}^{t}f_{k}(s)p_{k}(s)\,ds$ for all $t\geqslant
a_{k}$, (${\rm{II}}_{k}$) $\displaystyle q_{k-1}(t)\ $
$\displaystyle\geqslant\
q_{k-1}(a_{k})\exp\int_{a_{k}}^{t}f_{k}(s)q_{k}(s)\,ds$ for all $t\geqslant
a_{k}$.

Now assume $\beta_{k}>0$. Then for $a_{k}\leqslant t\leqslant a_{m}$ we have

$z_{k-1,m}(t)\ =\
z_{k-1,k}(a_{k})\exp\int_{a_{k}}^{t}\frac{f_{k}(s)}{z_{k,m}(s)}\,ds,$

and so

$p_{k-1}(a_{k})\exp\int_{a_{k}}^{t}\frac{f_{k}(s)}{q_{k}(s)}\,ds\ \leqslant\
z_{k-1,m}(t)\ \leqslant\
q_{k-1}(a_{k})\exp\int_{a_{k}}^{t}\frac{f_{k}(s)}{p_{k}(s)}\,ds,$

and so the desired $p_{k-1}\leqslant z_{k-1,m}\leqslant q_{k-1}$ on
$[a_{k-1},a_{m}]$ would follow if

(${\rm{III}}_{k}$) $\displaystyle p_{k-1}(t)\ $ $\displaystyle\leqslant\
p_{k-1}(a_{k})\exp\int_{a_{k}}^{t}\frac{f_{k}(s)}{q_{k}(s)}\,ds$ for all
$t\geqslant a_{k}$, (${\rm{IV}}_{k}$) $\displaystyle q_{k-1}(t)\ $
$\displaystyle\geqslant\
q_{k-1}(a_{k})\exp\int_{a_{k}}^{t}\frac{f_{k}(s)}{p_{k}(s)}\,ds$ for all
$t\geqslant a_{k}$.

The above leads to the following:

######  Lemma 9.13.

Let $p_{i}\in A_{i}$ and $q_{i}\in B_{i}$ for $i=0,1,2,\dots$ be given such
that

$\displaystyle\beta_{i+1}\ <\ 0\ $ $\displaystyle\Longrightarrow\
p_{i}^{\dagger}\ \leqslant\ f_{i+1}p_{i+1},$
$\displaystyle\hskip-40.00006ptq_{i}^{\dagger}\ \geqslant\ f_{i+1}q_{i+1}\ $
(in $H$), $\displaystyle\beta_{i+1}\ >\ 0\ $ $\displaystyle\Longrightarrow\
p_{i}^{\dagger}\ \leqslant\ f_{i+1}/q_{i+1},$
$\displaystyle\hskip-40.00006ptq_{i}^{\dagger}\ \geqslant\ f_{i+1}/p_{i+1}\ $
(in $H$).

Then there are germs $z_{i}\in\mathcal{C}^{<\infty}\ (i=0,1,2,\dots)$ such
that for all $i$,

$p_{i}\leqslant z_{i}\leqslant q_{i}\ \text{ in
}\mathcal{C},\quad\beta_{i+1}<0\ \Longrightarrow\
z_{i}^{\dagger}=f_{i+1}z_{i+1},\quad\beta_{i+1}>0\ \Longrightarrow\
z_{i}^{\dagger}=f_{i+1}/z_{i+1}.$

###### Proof.

Take a strictly increasing sequence $(a_{i})$ of real numbers $\geqslant 0$
tending to $+\infty$ and represent $p_{i}$, $q_{i}$ for each $i$ by
$\mathcal{C}^{1}$-functions $[a_{i},+\infty)\to{\mathbb{R}}^{>}$, also to be
denoted by $p_{i}$, $q_{i}$, such that for all $m$,

  * •

$f_{0},\dots,f_{m}$ are of class $\mathcal{C}^{m}$ on $[a_{m},+\infty)$;

  * •

$1\leqslant p_{m}\leqslant q_{m}$ on $[a_{m},+\infty)$;

  * •

$\beta_{m+1}<0\ \Longrightarrow\ p_{m}^{\dagger}\leqslant f_{m+1}p_{m+1},\quad
q_{m}^{\dagger}\geqslant f_{m+1}q_{m+1}$ on $[a_{m+1},+\infty)$;

  * •

$\beta_{m+1}>0\ \Longrightarrow\ p_{m}^{\dagger}\leqslant f_{m+1}/q_{m+1},\
q_{m}^{\dagger}\geqslant f_{m+1}/p_{m+1}$ on $[a_{m+1},+\infty)$.

Upon replacing $(a_{i})$ by a strictly increasing sequence $(b_{i})$ of reals
with $a_{i}\leqslant b_{i}$ for all $i$ and the $p_{i}$, $q_{i}$ by their
restrictions to $[b_{i},+\infty)$, for each $i$, the conditions above are
obviously still satisfied. For $t\geqslant a_{m}$ we have

$p_{m}(t)\ =\
p_{m}(a_{m+1})\exp\int_{a_{m+1}}^{t}p_{m}^{\dagger}(s)\,ds,\qquad q_{m}(t)\ =\
q_{m}(a_{m+1})\exp\int_{a_{m+1}}^{t}q_{m}^{\dagger}(s)\,ds,$

and so for all $k\geqslant 1$ conditions (${\rm{I}}_{k}$) and
(${\rm{II}}_{k}$) are satisfied if $\beta_{k}<0$, and conditions
(${\rm{III}}_{k}$) and (${\rm{IV}}_{k}$) are satisfied if $\beta_{k}>0$. Thus
by the above we can take any continuous function
$z_{m-1,m}\colon[a_{m-1},a_{m}]\to{\mathbb{R}}$ with $p_{m-1}\leqslant
z_{m-1,m}\leqslant q_{m-1}$ on $[a_{m-1},a_{m}]$ for $m=1,2,\dots$ to give
germs $z_{i}$ for $i=0,1,\dots$ as required. ∎

### Final step in reverse engineering

This step involves a diagonalization. We take $p_{i,n}\in A_{i}$ and
$q_{i,n}\in B_{i}$ (for $i=0,1,2,\dots$, $n=0,1,2,\dots$) such that for all
$i$, $n$:

  * •

$p_{i,n}\prec p_{i,n+1}$, and $\\{p_{i,0},p_{i,1},p_{i,2},\dots\\}$ is cofinal
in $A_{i}$;

  * •

$q_{i,n}\succ q_{i,n+1}$, and $\\{q_{i,0},q_{i,1},q_{i,2},\dots\\}$ is
coinitial in $B_{i}$;

  * •

if $\beta_{i+1}<0$, then $p_{i+1,n}=p_{i,N}^{\dagger}/f_{i+1}$ and
$q_{i+1,n}=q_{i,N}^{\dagger}/f_{i+1}$ for some $N=N(i,n)>n$;

  * •

if $\beta_{i+1}>0$, then $p_{i+1,n}=f_{i+1}/q_{i,N}^{\dagger}$ and
$q_{i+1,n}=f_{i+1}/p_{i,N}^{\dagger}$ for some $N=N(i,n)>n$.

It follows from Lemma 9.8 that there is such a family
$\big{(}(p_{i,n},q_{i,n})\big{)}$. Setting $p_{i}:=p_{i,i}$ and
$q_{i}:=q_{i,i}$ we note that the hypotheses of Lemma 9.13 are satisfied, and
this gives us germs $z_{i}\in\mathcal{C}^{<\infty}$ for $i=0,1,2,\dots$ such
that for all $i$,

  1. (1)

$p_{i}\leqslant z_{i}\leqslant q_{i}$ in $\mathcal{C}$;

  2. (2)

$\beta_{i+1}<0\Longrightarrow
z_{i}^{\dagger}=f_{i+1}z_{i+1},\qquad\beta_{i+1}>0\Longrightarrow
z_{i}^{\dagger}=f_{i+1}/z_{i+1}$.

We claim that then $A_{i}<_{\operatorname{e}}z_{i}<_{\operatorname{e}}B_{i}$
for all $i$. (Establishing this claim achieves our goal by Lemma 9.12.) To
prove this claim, suppose for a certain pair $i$, $n$ with $i<n$ we have
$p_{i+1,n}\leqslant z_{i+1}\leqslant q_{i+1,n}$. (See Figure 11.) As a
subclaim we show that then $p_{i,n}\leqslant z_{i}\leqslant q_{i,n}$. Consider
first the case $\beta_{i+1}<0$. Then for $N:=N(i,n)>n$,

$f_{i+1}^{-1}p_{i,N}^{\dagger}\ \leqslant\ z_{i+1}\ =\
f_{i+1}^{-1}z_{i}^{\dagger}\ \leqslant\ f_{i+1}^{-1}q_{i,N}^{\dagger},\ \text{
so }\ p_{i,N}^{\dagger}\ \leqslant z_{i}^{\dagger}\ \leqslant\
q_{i,N}^{\dagger},$

which by Lemma 2.16 gives constants $c_{1},c_{2}>0$ with
$c_{1}p_{i,N}\leqslant z_{i}\leqslant c_{2}q_{i,N}$, and so
$p_{i,N-1}\leqslant z_{i}\leqslant q_{i,N-1}$. Now $N-1\geqslant n$, and thus
$p_{i,n}\leqslant z_{i}\leqslant q_{i,n}$ as promised. The case
$\beta_{i+1}>0$ is handled in the same way. Given $i<n$ we have
$p_{n,n}\leqslant z_{n}\leqslant q_{n,n}$, and so $p_{i,n}\leqslant
z_{i}\leqslant q_{i,n}$ by iterated application of the subclaim. For any fixed
$i$ this yields $A_{i}<_{\operatorname{e}}z_{i}<_{\operatorname{e}}B_{i}$ by
the cofinality and coinitiality requirements we imposed on the $p_{i,n}$ and
$q_{i,n}$. This proves the claim, and concludes the proof of Theorem 9.2, and
thus of Theorem A. ∎

${\begin{array}[]{ ccccccccccccccccccc
}\framebox{$p_{0,0}$}&\prec&p_{0,1}&\prec&\cdots&\prec&p_{0,n}&\prec&\cdots&z_{0}&\cdots&\prec&q_{0,n}&\prec&\cdots&\prec&q_{0,1}&\prec&\framebox{$q_{0,0}$}\\\
p_{1,0}&\prec&\framebox{$p_{1,1}$}&\prec&\cdots&\prec&p_{1,n}&\prec&\cdots&z_{1}&\cdots&\prec&q_{1,n}&\prec&\cdots&\prec&\framebox{$q_{1,1}$}&\prec&q_{1,0}\\\
\vdots&&\vdots&\ddots&&&\vdots&&&\vdots&&&\vdots&&&\iddots&\vdots&&\vdots\\\
\vdots&&\vdots&&&\ddots&\vdots&&&\vdots&&&\vdots&\iddots&&&\vdots&&\vdots\\\
p_{n,0}&\prec&p_{n,1}&\prec&\cdots&\prec&\framebox{$p_{n,n}$}&\prec&\cdots&z_{n}&\cdots&\prec&\framebox{$q_{n,n}$}&\prec&\cdots&\prec&q_{n,1}&\prec&q_{n,0}\\\
\vdots&&\vdots&&&&\vdots&\ddots&&\vdots&&\iddots&\vdots&&&&\vdots&&\vdots\end{array}}$
Figure 11.

##  10\. Isomorphism of Maximal Hardy Fields

The cardinality of any Hardy field extending ${\mathbb{R}}$ is
$2^{\aleph_{0}}$. By Theorem A, all maximal Hardy fields are $\eta_{1}$ and
thus $\aleph_{1}$-saturated as real closed ordered fields; in particular,
under $\operatorname{CH}$ they are all isomorphic as ordered fields. However,
they are not $\aleph_{1}$-saturated as ordered differential fields, since
their constant field ${\mathbb{R}}$ isn’t. Thus to show they are isomorphic
(under $\operatorname{CH}$), we need to argue in a different way, and this is
what we do in this section.

######  Lemma 10.1.

Let $K$ be a countable closed $H$-field with archimedean constant field $C$,
let $L$ be a closed $H$-field with constant field ${\mathbb{R}}$ and small
derivation, and assume $L$ is $\eta_{1}$. Let $E$ be an $\upomega$-free
$H$-subfield of $K$, and let $i\colon E\to L$ be an $H$-field embedding. Then
$i$ extends to an $H$-field embedding $K\to L$.

---
$\textstyle{L}$$\textstyle{K\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{E\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\subseteq}$$\scriptstyle{i}$

###### Proof.

We identify $C$ in the usual (and only possible) way with a subfield of
${\mathbb{R}}$, and note that then $i$ is the identity on
$C_{E}\subseteq{\mathbb{R}}$. Then [ADH, 10.5.15, 10.5.16] yield an extension
of $i$ to an $H$-field embedding $E(C)\to L$ that is the identity on
$C\subseteq{\mathbb{R}}$. The $H$-subfield $E(C)$ of $K$ is
$\operatorname{d}$-algebraic over $E$, so is $\upomega$-free. Replacing $E$ by
$E(C)$ we reduce to the case that $C_{E}=C$. Recall that $\eta_{1}$-ordered
sets are $\aleph_{1}$-saturated. Hence [ADH, 16.2.3] applies and gives the
desired conclusion. ∎

######  Lemma 10.2.

Let $L_{1}$, $L_{2}$ be closed $H$-fields with small derivation and common
constant field ${\mathbb{R}}$, and assume that $L_{1}$ and $L_{2}$ are
$\eta_{1}$. Then the collection of $H$-field isomorphisms $K_{1}\to K_{2}$
between countable closed $H$-subfields $K_{1}$ of $L_{1}$ and $K_{2}$ of
$L_{2}$ is nonempty and is a back-and-forth system between $L_{1}$ and
$L_{2}$. In particular, $L_{1}$ and $L_{2}$ are back-and-forth equivalent.

###### Proof.

The theory of closed $H$-fields with small derivation has a (countable) prime
model, by [ADH, p. 705], and so there is an $H$-field isomorphism between
copies of that prime model in $L_{1}$ and in $L_{2}$. Also, any countable
subset of a closed $H$-field $L$ is contained in a countable closed
$H$-subfield of $L$, by downward Löwenheim-Skolem [ADH, B.5.10]. It remains to
use Lemma 10.1. ∎

A standard argument (cf. proof of [ADH, B.5.3]) using Lemma 10.2 now yields:

######  Corollary 10.3.

Let $L_{1}$, $L_{2}$ as in Lemma 10.2 have cardinality $2^{\aleph_{0}}$.
Assume $\operatorname{CH}$. Then $L_{1}$ and $L_{2}$ are isomorphic as
$H$-fields.

Next we recall that Berarducci and Mantova [9] defined a derivation
${}_{\operatorname{BM}}$ on the real closed field No of surreal numbers and
proved that No with ${}_{\operatorname{BM}}$ is a Liouville closed $H$-field
with ${\mathbb{R}}\subseteq\text{{\bf No}}$ as its field of constants. Below
we consider No as an $H$-field in this way, and recall also that its
derivation ${}_{\operatorname{BM}}$ is small. We proved in [5, Theorems 1 and
2] that No is even a closed $H$-field, that its real closed subfield
$\text{{\bf No}}(\omega_{1})$ is closed under ${}_{\operatorname{BM}}$, and
that $\text{{\bf No}}(\omega_{1})$ as a differential subfield of No is a
closed $H$-field as well. Moreover, $\text{{\bf No}}(\omega_{1})$ has
cardinality $2^{\aleph_{0}}$, and is $\eta_{1}$ as an ordered set. In
combination with Theorem A and Corollary 10.3, with $L_{1}$ any maximal Hardy
field and $L_{2}=\text{{\bf No}}(\omega_{1})$, this yields Corollary B in the
introduction; more precisely, also using [8, Theorem 3], we have:

######  Corollary 10.4.

Let $M$ be a maximal Hardy field. Then the ordered differential fields $M$ and
$\text{{\bf No}}(\omega_{1})$ are back-and-forth equivalent. Hence $M$ and
$\text{{\bf No}}(\omega_{1})$ are $\infty\omega$-equivalent, and assuming
$\operatorname{CH}$, $M$ and $\text{{\bf No}}(\omega_{1})$ are isomorphic.

We finish with a lemma on $\infty\omega$-elementary embeddings:

######  Lemma 10.5.

Let $L_{1}$ and $L_{2}$ be as in Lemma 10.2, with $L_{1}$ an $H$-subfield of
$L_{2}$. Then $L_{1}\preccurlyeq_{\infty\omega}L_{2}$.

###### Proof.

Let $\Phi$ be the back-and-forth system from Lemma 10.2. By downward
Löwenheim-Skolem there is for all $a_{1},\dots,a_{n}\in L_{1}$ a countable
closed $H$-subfield $K_{1}$ of $L_{1}$ containing $a_{1},\dots,a_{n}$, and
then the identity map $K_{1}\to K_{1}$ belongs to $\Phi$. This yields
${L_{1}\preccurlyeq_{\infty\omega}L_{2}}$ by [8, Theorem 4]. ∎

This will be used in the follow-up paper on maximal analytic Hardy fields.

## References

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  * [2] N. Alling, On the existence of real-closed fields that are $\eta_{\alpha}$-sets of power $\aleph_{\alpha}$, Trans. Amer. Math. Soc. 103 (1962), 341–352.
  * [3] M. Aschenbrenner, L. van den Dries, Liouville closed $H$-fields, J. Pure Appl. Algebra 197 (2005), 83–139.
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