Analytic Hardy Fields

Aschenbrenner]Matthias Aschenbrenner
Kurt Gödel Research Center for Mathematical Logic

Universität Wien

1090 Wien

van den Dries]Lou van den Dries
Department of Mathematics

University of Illinois at Urbana-Champaign

Urbana, IL 61801

We show that maximal analytic Hardy fields are $\eta_1$ in the sense of Hausdorff.  We also prove  various embedding theorems about analytic Hardy fields. For example, the ordered differential field $\T$ of transseries is
shown to be isomorphic to an analytic Hardy field.

§ INTRODUCTION

This is a follow-up on [8] where the main result is that for any Hardy field $H$ and countable subsets
$A< B$ of $H$ there exists $y$ in a Hardy field extension of $H$ such that $A < y < B$. Equivalently, (the underlying ordered set of) any maximal Hardy field is $\eta_1$ in the sense of Hausdorff. In this result we do not require $H\subseteq \Cc^{\omega}$, and the glueing constructions in [8] do not give $y\in \Cc^{\omega}$, even if $H\subseteq \Cc^{\omega}$; see Notations and Conventions at the end of this introduction for the notation used here.  We call a Hardy field $H$ smooth if $H\subseteq \Cc^\infty$ and analytic if $H\subseteq \Cc^{\omega}$. By <cit.>, maximal Hardy fields, maximal smooth Hardy fields,  and maximal analytic Hardy fields are all elementarily equivalent to the ordered differential field $\T$ of transseries, and have no proper $\d$-algebraic $H$-field extension with constant field $$. We shall tacitly use these facts throughout.

Some may view non-analytic Hardy fields as somewhat artificial, since Hardy fields that occur ``in nature'' are analytic. (See
[26, 27] for Hardy fields $H⊈^∞$, and [37] for  Hardy fields $H⊆^∞$, $H⊈^ω$.) To conciliate this view and answer an obvious question we prove in Section~\ref{s4} the analytic version of [8]:

\begin{theorem}\label{aneta} If $H$ is an analytic Hardy field with countable subsets $A<B$, then there exists $y\in \Cc^{\omega}$ in a Hardy field extension of $H$ such that $A < y < B$.
\end{theorem}

\noindent
Equivalently, all maximal analytic Hardy fields are $η_1$. The theorem goes through for smooth Hardy fields  with $y∈^∞$ in the conclusion; this can be obtained by
refining the glueing constructions from~[8] (as was actually done in an early version of that paper at the cost of three extra pages).
Here we take care of the smooth and analytic versions simultaneously. Compared to [8] the new tool we use is a powerful theorem due to Whitney on approximating any $^n$-function or $^∞$-function by an analytic function, where the approximation also takes derivatives into account.  From that we obtain an analogue for germs, namely Corollary~\ref{apomega}, which in turn we use to derive Theorem~\ref{aneta} from various results in
the non-analytic setting of [8].

In the course of establishing Theorem~\ref{aneta} we also revisit results on pc-sequences and on extensions of type (b)
from [8]. In Section~\ref{s4} this also leads to:

\begin{theorem}\label{thdense} If $H$ is a maximal analytic Hardy field, then $H$ is dense in any Hardy field extension of $H$.
\end{theorem}

\noindent
If all maximal analytic Hardy fields are maximal Hardy fields, which seems to us implausible, then of course the theorems above would be trivially true. Can a maximal analytic Hardy field ever be a maximal Hardy field? For all we know answering questions of this kind might involve set-theoretic assumptions like CH. In  [8] and the present paper we ran into other set-theoretic issues of this kind, and in Section 8 we state some problems that arose this way.
% in addition to techniques from the present paper.

\medskip\noindent
Sections 5--7 prove embedding theorems about (not necessarily maximal) analytic Hardy fields. A special case of a result in Section~7:  the ordered differential field  $$ is isomorphic over $$ to an analytic Hardy field extension of $$.

\subsection*{Notations and conventions}  We take these from \cite[end of introduction]{ADHfgh}, but for the convenience of the reader we list here what is most needed.

We let $i$, $j$, $k$, $l$, $m$, $n$ range over $={0,1,2,…}$. We let $$ be the ring of germs at $+∞$ of continuous functions~$(a,+∞)→$, $a∈$, and
$^r$ for $r∈∪{∞}$ its subring of germs of $r$ times continuously differentiable functions $(a,+∞)→$, $a∈$. Thus $^<∞ :=  ⋂_n^n$ is a differential ring with the obvious derivation,
and has $^∞$ as a differential subring.  A {\em Hardy field\/}  is a differential subring of $^<∞$
whose underlying ring is a field; it is naturally also an ordered and valued field, see~[ADH, pp.~384--386]. We let $^ω$ be the differential subring of $^∞$ consisting of the germs of real analytic functions $(a,+∞)→$, $a∈$.

\endgroup

\section{Whitney's Approximation Theorem}

\noindent
In this section we let $r∈∪{∞}$ and $a,b∈$.
We shall use the one-variable case of an approximation theorem due to Whitney~\cite[Lem\-ma~6]{Whitney} to upgrade various constructions of smooth functions  to analytic functions. To formulate this theorem we introduce some notation.
Let~$U⊆$ be  open and $S⊆U$ be nonempty.
For $f$ in $(U)$ we set
%We have a     seminorm $\dabs{\,\cdot\,}_S$  on the $\R$-linear space $\Cc(U)$ given by
$$\dabs{f}_S\ :=\ \sup\big\{ \abs{f(s)}:\, s\in S\big\} \in [0,\infty], $$
so for $f,g∈(U)$ and $λ∈$ (and the convention $0·∞=∞·0=0$) we have
$$\dabs{f+g}_S\le \dabs{f}_S+\dabs{g}_S,\quad \dabs{\lambda f}_S=|\lambda|\cdot\dabs{f}_S,\quad\text{and}\quad
\dabs{fg}_S\leq\dabs{f}_S\dabs{g}_S.$$
If $∅≠S'⊆S$ then
We then put
$$\dabs{f}_{S;\,m}\ :=\ \max\big\{ \dabs{f}_S,\dots,\dabs{f^{(m)}}_S\big\}\in [0,\infty].$$
Then again for  $f,g∈(U)$ and $λ∈$  we have
$$\dabs{f+g}_{S;m}\le \dabs{f}_{S;m}+\dabs{g}_{S;m},\quad \dabs{\lambda f}_{S;m}=|\lambda|\cdot\dabs{f}_{S;m},$$
%$\dabs{\,\cdot\,}_{S;\,m}$ is a seminorm on the $\R$-linear space $\Cc^m(U)$
\begin{equation}\label{eq:prod norm}
\dabs{fg}_{S;\,m}\  \leq\ 2^m \dabs{f}_{S;\,m}\dabs{g}_{S;\,m}
\end{equation}
Let $f∈(U)$. For $U=$ we set $f_m:=f_; m$.
%Note that if $ $\supp f=f^{-1}(0)$ is bounded then
For $k≤m$ and $∅≠S'⊆S⊆U$ we have
$f_S'; k ≤f_S; m$. Moreover, $f_S; m$ does not change if
$S$ is replaced by its closure in $U$.

\begin{theorem}[Whitney]\label{thm:WAP dim 1, general} Let $(a_n)$, $(b_n)$, $(\varepsilon_n)$ be sequences in $\R$ and $(r_n)$
in $\N$ such that $a_0= b_0$, $(a_n)$ is strictly decreasing, $(b_n)$ is strictly increasing, and $\varepsilon_n>0$, $r_n\le r$ for all $n$. Set $I:=\bigcup_n K_n$, where $K_n:=[a_n, b_n]$.   Then, for any $f\in \Cc^r(I)$, there exists $g\in\Cc^\omega(I)$ such that for all $n$ we have
$\dabs{{f-g}}_{K_{n+1}\setminus K_n;\,r_n}<\varepsilon_n$.
\end{theorem}

\noindent
For a self-contained proof of Theorem~\ref{thm:WAP dim 1, general}, see the appendix to this paper.

\medskip
\noindent
For $f∈_a^m$ and nonempty $S⊆[a,+∞)$   we   put
$f_S; m:=g_S; m$ where~$g∈^m(U)$ is any extension of~$f$ to an open neighborhood $U⊆$ of $[a,+∞)$.
We shall use the following special case of Theorem~\ref{thm:WAP dim 1, general}:

\begin{cor}\label{cor:WAP dim 1} Let $f\in \Cc^r_b$, and
let   $(b_n)$ be a strictly increasing sequence
in $\R$ such that $b_0=b$ and~${b_n\to\infty}$ as $n\to\infty$, and let $(\varepsilon_n)$ be a sequence in $\R^>$  and
$(r_n)$ be a sequence in~$\N$ with~$r_n\leq r$ for all $n$.   Then there exists $g\in\Cc_b^\omega$ such that for all $n$ we have   $\dabs{{f-g}}_{[b_n,b_{n+1}];\,r_n}<\varepsilon_n$.
\end{cor}
\begin{proof}
Extend $f$ to a function in~$\Cc^r(I)$, also denoted by $f$,  where $I:=(a,+\infty)$, ${a<b}$, and take
a strictly decreasing sequence $(a_n)$ in $\R$ with $a_0=b_0$ and~${a_n\to a}$ as $n\to+\infty$.
Now apply Theorem~\ref{thm:WAP dim 1, general}.
\end{proof}

\noindent
Here is a useful  reformulation of Corollary~\ref{cor:WAP dim 1}:

\begin{cor}\label{cor:WAP dim 1, 1}
Let     $f\in \Cc^r_b$ and $\varepsilon\in\Cc_b$ be such that $\varepsilon>0$ on $[b,+\infty)$. Then there exists $g\in\Cc_b^\omega$ such that  $\abs{(f-g)^{(k)}(t)} < \varepsilon(t)$  for all $t\ge b$ and
$k\leq \min\!\big\{r,1/\varepsilon(t)\big\}$.
\end{cor}
\begin{proof}
Take a strictly increasing sequence $(b_n)$ in $\R$ with $b_0=b$ and $b_n\to \infty$ as~$n\to\infty$, and
for each $n$, set
$$\varepsilon_n:=\min\big\{\varepsilon(t): t\in[b_n,b_{n+1}]\big\}\in\R^>,\quad
r_n:= \min\left\{ r, \big\lfloor  \dabs{1/\varepsilon}_{[b_n,b_{n+1}]}\big\rfloor\right\}\in\N.$$
Corollary~\ref{cor:WAP dim 1} yields  $g\in\Cc^\omega_b$ such that $\dabs{f-g}_{[b_n,b_{n+1}];\,r_n}<\varepsilon_n$ for all~$n$. Then for~$t\in [b_n,b_{n+1}]$ and $k\leq \min\!\big\{r,1/\varepsilon(t)\big\}$ we have
$k\leq r_n$ and so
\[\abs{(f-g)^{(k)}(t)} \leq \dabs{f-g}_{[b_n,b_{n+1}];\,r_n}<\varepsilon_n\leq \varepsilon(t).\qedhere\]
\end{proof}

\noindent
This leads to an improved version of \cite[Lemma~\ref{smooth}]{ADHfgh}:

\begin{lemma}\label{smooth->analytic}
Let $f, g\in\Cc_b$ be such that $f < g$ on $[b,+\infty)$. Then there exists $y\in\Cc^\omega_b$   such that  $f < y < g$ on $[b,+\infty)$.
\end{lemma}
\begin{proof}
Let $z:=\frac{1}{2}(f+g)\in\Cc_b$ and $\varepsilon:=\frac{1}{2}(g-f)\in\Cc_b$.
Corollary~\ref{cor:WAP dim 1, 1} (with $r=0$) then yields
$y\in\Cc_{b}^\omega$ such that $\abs{y-z}<\varepsilon$ on $[b,+\infty)$, so
$f< y < g$ on $[b,+\infty)$.
\end{proof}

\noindent
Thus we can replace ``$ϕ∈𝒞^∞$''   by  ``$ϕ∈^ω$'' in the statements of Lemma~\ref{lem:smooth} and   Corollary~\ref{corsmooth} in [8].
Here is another consequence of Corollary~\ref{cor:WAP dim 1, 1}:

\begin{cor}\label{cor:WAP dim 1, 2}
Let   $f\in\Cc^r$  and $\varepsilon\in\Cc$, $\varepsilon>_{\ex} 0$. Then there exists $g\in\Cc^\omega$ such that for all $k\le r$ we have
$\abs{{(f-g)^{(k)}}} <_{\ex}  \varepsilon$.
\end{cor}
\begin{proof}
Pick $a$ and representatives of $f$ in $\Cc_a^r$ and of $\varepsilon$  in $\Cc_a$, also denoted by~$f$,~$\varepsilon$, with
$\varepsilon>0$ on $[a,+\infty)$.
Take $\varepsilon^*\in\Cc_a$ with $0<\varepsilon^*\leq \varepsilon$ on $[a,+\infty)$ and $\varepsilon^*\prec 1$.
Corollary~\ref{cor:WAP dim 1, 1} applied to $\varepsilon^*$ in place of $\varepsilon$ yields $g\in \Cc_a^\omega$ such that~$\abs{(f-g)^{(k)}(t)} < \varepsilon^*(t)$ for all $t\ge a$ and $k\leq \min\!\big\{r,1/\varepsilon^*(t)\big\}$.
Given $k\leq r$, take $b\geq a$ such that  $k\le 1/\varepsilon^*(t)$ for all~$t\geq b$; then
$\abs{(f-g)^{(k)}(t)} < \varepsilon(t)$ for such $t$.
\end{proof}

%Is there for all   $f\in\Cc^{<\infty}$ and $\varepsilon\in\Cc^\times$ a $g\in\Cc^\omega$ such that for all $k$, \marginpar{skipped for now}
%$$\abs{{(f-g)^{(k)}}}\  <_{\ex}\   \abs{\varepsilon}?$$
%Maybe first find such a $g$ in $\Cc^{\infty}$, and then apply previous corollary?

\noindent
Our next goal is to prove a version of Corollary~\ref{cor:WAP dim 1, 2} for  approximating germs in~$^<∞$ by germs in $^ω$: see Corollary~\ref{apomega} below.
First a lemma about glueing two approximations $g_-$ and $g_+$ to a function $f$ to make a single approximation $g$ to $f$ that combines properties of $g_-$ and $g_+$:

\begin{lemma}
Let $f\in \Cc_{a_0}$ and $a_0 \le a < b$. Suppose $f$ is of class $\Cc^n$ on $[a,+\infty)$ and of class $\Cc^{n+1}$ on
$[b,+\infty)$. Let also functions $\varepsilon\in \Cc_{a_0}$
and  $g_{-}, g_{+}\in \Cc^{\infty}_{a_0}$ be given such that
\begin{itemize}
\item $\varepsilon >0$ on $[a_0,+\infty)$;
\item $|(f-g_{-})^{(j)}| < \varepsilon$   on $[a,+\infty)$ for $j=0,\dots,n$; and
\item $|(f-g_{+})^{(j)}| < \varepsilon$ on $[b,+\infty)$ for $j=0,\dots,n+1$.
\end{itemize}
Then, for any $\delta\in \R^{>}$, there is a function $g\in \Cc^{\infty}_{a_0}$ and a $b'> b$ such that: \begin{enumerate}
\item[(i)] $g=g_{-}$ on $[a_0,b]$ and $g=g_{+}$ on $[b',+\infty)$;
\item[(ii)]  $|(f-g)^{(j)}| < (1+\delta)\varepsilon$ on $[a,+\infty)$ for $j=0,\dots,n$; and
\item[(iii)]  $|(f-g)^{(j)}| < \varepsilon$ on $[b',+\infty)$ for $j=0,\dots,n+1$.
\end{enumerate}
\end{lemma}
\begin{proof} Let $b'>b$, set $\beta:= \alpha_{b,b'}$ as in \cite[\eqref{eq:alphaab}]{ADHfgh}, and $g:= (1-\beta) g_{-}+\beta g_{+}$ on~$[a_0,+\infty)$, so $g\in \Cc_{a_0}^{\infty}$. Let $\delta>0$; we show that if $b'-b$ is sufficiently large, then $g$ satisfies~(i), (ii), (iii). It is clear that (i) holds, and so (iii) as well. Then the inequality in (ii) holds on~$[a,b]$ and on $[b',+\infty)$, so it suffices to consider what happens  on $[b,b']$. There we have for $j=0,\dots,n$:
$$(f-g)^{(j)}\ =
\ f^{(j)}-\big((1-\beta) g_{-}^{(j)}+\beta g_{+}^{(j)}\big)-\sum_{i=0}^{j-1}\binom{j}{i}\beta^{(j-i)}\big(g_{+}^{(i)}-g_{-}^{(i)}\big),$$
and $$f^{(j)}-\big((1-\beta) g_{-}^{(j)}+\beta g_{+}^{(j)}\big)=(1-\beta)\big(f-g_{-})^{(j)} + \beta\big(f-g_{+})^{(j)},$$ so
$$\big|f^{(j)}-\big((1-\beta) g_{-}^{(j)}+\beta g_{+}^{(j)}\big)\big|\ \le\ \max\big\{\big|(f-g_{-})^{(j)}\big|, \big|(f-g_{+})^{(j)}\big|\big\}\ <\ \varepsilon  \text{ on $[b,b']$.}$$
By \cite[\eqref{eq:alphaab(m)}]{ADHfgh} we have reals $C_m\geq 1$ (independent of $b'$) with $\abs{\beta^{(m)}}\leq C_m/(b'-b)^m$.
Hence for $j=0,\dots,n$ we have on $[b,b']$:
$$ \left|\sum_{i=0}^{j-1}\binom{j}{i}\beta^{(j-i)}\big(g_{+}^{(i)}-g_{-}^{(i)}\big)\right|\ \le\
\sum_{i=0}^{j-1}\binom{j}{i}\frac{C_{j-i}}{(b'-b)^{j-i}}\,\big|g_{+}^{(i)}-g_{-}^{(i)}\big|$$
and $\big|g_{+}^{(i)}-g_{-}^{(i)}\big|<2\varepsilon$ for $i=0,\dots, n$. So for $b'-b$  so large that
$$\sum_{i=0}^{j-1}\binom{j}{i}\frac{C_{j-i}}{(b'-b)^{j-i}}\ <\  \delta/2,$$
condition (ii) is satisfied. (See also Figure~\ref{fig:gplusgminus}.)
\end{proof}

\begin{figure}[h]
\def\betafn{10*(0.05*tanh(3+6*(x-1)) + 0.05)}
\def\gplusfn{0.28*8^(-x)+(x-0.2)^(4)+0.3}
\def\gminusfn{(\gplusfn)+0.25*x^3-0.4*x^2-0.3*x+0.3*sin(500*x)+0.3}
\begin{tikzpicture}
\begin{axis} [axis lines=center, axis y line=none, xmin=-0.5, xmax=1.3, ymin = -0.02, ymax = 1.25, width=0.8\textwidth, height = 0.5\textwidth, xlabel={$t$}, xtick={-0.35,0,1}, xticklabels={\strut $a$, \strut $b$, \strut $b'$}, ytick=1, yticklabels={$1$},   legend style={font=\small},
legend cell align=left, legend style={at={(1.1,0.75)},anchor=west}]

%  \addplot [domain=-0.35:1.1, smooth, very thick] {\betafn };
\addplot [domain=-0.45:1.1, smooth, dotted, very thick] { \gplusfn};
\addplot [domain=-0.45:1.1, smooth, dashed, very thick] {\gminusfn};
\addlegendentry{$g_+$};
\addlegendentry{$g_-$};

\addplot [domain=-0.45:1.1, smooth, very thick] { (1-(\betafn))*(\gminusfn)+(\betafn)*(\gplusfn) };

\addlegendentry{$g$};
\draw [dashed] (0,-0.02) -- (0,1.25);
\draw [dashed] (-0.35,-0.02) -- (-0.35,1.25);
\draw [dashed] (1,-0.02) -- (1,1.25);

\end{axis}
\end{tikzpicture}
\caption{}\label{fig:gplusgminus}
\end{figure}

\begin{prop}\label{apinf} Suppose $f\in \Cc^{<\infty}$ and $\varepsilon\in\Cc$,  $\varepsilon>_{\ex} 0$. Then there exists $g\in \Cc^{\infty}$ such that $|(f-g)^{(n)}| <_{\ex} \varepsilon$ for all $n$.
\end{prop}
\begin{proof}  Represent $f$ and $\varepsilon$ by continuous functions $ [a_0,+\infty)\to \R$ ($a_0\in \R$), also denoted by
$f$ and $\varepsilon$, such that $\varepsilon>0$ on $[a_0,+\infty)$. Next, take a strictly increasing sequence $(a_n)$ of real numbers starting with the already given $a_0$, such that $a_n\to \infty$ as $n\to \infty$, and $f$  is of class $\Cc^n$ on
$[a_n,+\infty)$, for each $n$.  Then Corollary~\ref{cor:WAP dim 1, 1} gives for each $n$ a function $g_n\in \Cc_{a_0}^\infty$ such that
$|(f-g_n)^{(j)}|< \varepsilon/2$ on $[a_n,+\infty)$ for~$j=0,\dots,n$.  All this remains true when increasing each  $a_n$ while keeping $a_0$ fixed and maintaining that $(a_n)$ is strictly increasing. Now use the lemma above to construct $g$ as required:  first glue
$g_0$ and $g_1$ and increase the $a_n$ for $n\ge 1$, then glue the resulting function with $g_2$ and increase the $a_n$ for $n\ge 2$, and so on, and arrange the product of the $(1+\delta)$-factors to be $<2$.
\end{proof}

\noindent
Now  Corollary~\ref{cor:WAP dim 1, 2} (for $r=∞$) and  Proposition~\ref{apinf}    yield:

\begin{cor}\label{apomega} For any germs $f\in \Cc^{<\infty}$ and  $\varepsilon\in \Cc$ with $\varepsilon>_{\ex} 0$, there exists a germ $g\in \Cc^{\omega}$ such that $|(f-g)^{(n)}| <_{\ex} \varepsilon$ for all $n$.
\end{cor}

\noindent
In the next section we apply Corollary~\ref{apomega} to bounded Hardy fields.

\section{Bounded Hardy Fields}

\noindent
As in   \cite[Section~5.4]{ADH2}  a set $H⊆𝒞$ is called
{\it bounded}\/ if for some $ϕ∈𝒞$ we have $h≤ϕ$ for all~$h∈H$, and {\it unbounded}\/ otherwise.
Every countable subset of $𝒞$ is bounded, cf.~\cite[remarks after Lemma~5.4.17]{ADH2}.
As a consequence, the union of  countably many bounded subsets of $𝒞$ is also bounded.

\medskip
\noindent
{\it In the rest of this section $H$ is a Hardy field.}\/
If~$H$ is bounded, then there is a~$ϕ∈𝒞$ with~$ϕ>_0$ and $g≺ϕ$ for all $g∈H$, so
$ε:=1/ϕ∈𝒞^×$ satisfies~$ε>_ 0$ and~$ε≺h$ for all $h∈H^×$. By \cite[Lem\-mas~5.4.18, 5.4.19]{ADH2} we have:

\begin{lemma}\label{lem:5.4.19} If $H$ is bounded, then any $\d$-algebraic Hardy field extension of $H$ is bounded, and
for any $H$-hardian $f\in \Cc^{<\infty}$, the Hardy field $H\<f\>$ is bounded.
\end{lemma}

\begin{cor}\label{lem:analytify, 1}
If $H$ is bounded  and $F$ is a Hardy field extension of $H$  and $\d$-algebraic over $H\<S\>$ for some countable $S\subseteq F$,
%such that $\operatorname{trdeg}(F|H)$ is countable.
then~$F$ is bounded.
\end{cor}

%Take $f_n\in F$ such that $F$ is $\d$-algebraic over $H\langle f_0,f_1,\dots\rangle$.
%Induction on $n$ using Lem\-ma~\ref{lem:5.4.19}
%shows that $H_n:=H\langle f_0,f_1,\dots,f_n\rangle$ is bounded, hence $H\langle f_0, f_1,\dots,\rangle$ is bounded.

%In particular, if~$\operatorname{trdeg}(H|C_H)$ is countable, then $H$ is bounded.

\begin{lemma}\label{lem:analytify, 2}
Let   $f,g\in\Cc^{<\infty}$  be such that $f$ is $H$-hardian, $\d$-transcendental over~$H$, and
$(f-g)^{(n)}\prec h$ for all $h\in H\langle f \rangle^{\times}$ and all~$n$.
Then $g$ is $H$-hardian, and there is a unique isomorphism $H\langle f\rangle\to H\langle g\rangle$
of Hardy fields fields over~$H$   sending $f$ to $g$.
\end{lemma}
\begin{proof}
Let $P\in H\{Y\}^{\neq}$,  $r:=\order P$, so  $P(f)\in H\langle f\rangle^\times$. It suffices to show that then $P(f) \sim P(g)$. By Taylor expansion [ADH, p.~210], with $\i$ ranging over $\N^{1+r}$:
$$P(g) - P(f) =  \sum_{\abs{\i}\geq 1} P_{(\i)}(f)(g-f)^{\i}\quad\text{where $P_{(\i)}=\frac{P^{(\i)}}{\i!}\in H\{Y\}$.}$$
If $\abs{\i}\geq 1$, then $(g-f)^{\i}\prec h$ for all $h\in H\langle f\rangle^{\times}$, and hence
$P_{(\i)}(f)(g-f)^{\i} \prec P(f)$. Thus~$P(g)-P(f)\prec P(f)$ as required.
\end{proof}

\noindent
With Corollary~\ref{apomega} we now obtain analytic ``copies'' of certain $H$-hardian germs:

\begin{cor}\label{cor:analytify}
Suppose $H$ is bounded  and $f$ in a Hardy field extension of $H$ is $\d$-transcendental over $H$.
Then there is an $H$-hardian $g\in\Cc^\omega$
and an isomorphism~$H\langle f\rangle\to H\langle g\rangle$
of Hardy fields over $H$   sending $f$ to $g$.
\end{cor}
\begin{proof}
By Lemma~\ref{lem:5.4.19},   the Hardy field $H\langle f\rangle$ is bounded, so we can
take~$\varepsilon\in\Cc^\times$ with $\varepsilon>_{\ex} 0$ and~$\varepsilon \prec h$   for all $h\in H\langle f\rangle^\times$.
Corollary~\ref{apomega} yields a $g\in\Cc^\omega$ such that~$\abs{(f-g)^{(n)}}\leq \varepsilon$
for all $n$, and so it remains to appeal to  Lemma~\ref{lem:analytify, 2}.
\end{proof}

\noindent
We can now also strengthen  \cite[Theorem~5.1]{ADHfgh}:

\begin{cor}\label{cor:5.1ana}
Suppose $H\supseteq\R$ is Liouville closed, and $\phi\in\Cc$, $\phi>_{\ex}H$. Then there is
an $H$-hardian $z\in\Cc^\omega$ with $z>_{\ex}\phi$.
\end{cor}
\begin{proof}
By  \cite[Theorem~\ref{sjo}]{ADHfgh} we have an $H$-hardian $y\in\Cc^\infty$ with~$y>_{\ex}\phi +1$.
Then~$y$ is $\d$-transcendental over $H$ and $H\langle y \rangle$ is bounded; cf.~\cite[Lemma~5.4.1]{ADH2} and [ADH, Lemma 16.6.10].
%As in the proof of Corollary~\ref{cor:analytify}
This yields $\varepsilon\in\Cc$ such that $\varepsilon>_{\ex} 0$ and~$\varepsilon \prec h$   for all $h\in H\langle y\rangle^\times$.
%Put $\delta:=\min\{\varepsilon,y-\phi\}\in\Cc$, so $\delta>_{\ex}0$.
Now Corollary~\ref{apomega}  gives~$z\in\Cc^\omega$ with~$\abs{y^{(n)}-z^{(n)}}<_{\ex}\varepsilon$ for all $n$.
Then $z$ is  $H$-hardian by Lemma~\ref{lem:analytify, 2}, and~$z=y+(z-y)>_{\ex}\phi$.
\end{proof}

\noindent
Thus maximal Hardy fields,  maximal $\Cc^\infty$-Hardy fields, and maximal $\Cc^{\omega}$-Hardy fields are unbounded. (Cf. also~\cite[Corollary~5.4.23 and succeeding remarks]{ADH2}.)
As~\cite[Theorem~\ref{sjo}]{ADHfgh} gave rise to \cite[Corollary~\ref{corsjo}]{ADHfgh}, so Corollary~\ref{cor:5.1ana} yields:

\begin{cor}\label{corsjoan} If $H$ is a maximal analytic Hardy field, then $\cf(H) > \omega$, and thus
$$\ci(H)\ =\ \cf(H^{<a})\ =\ \ci(H^{>a})\ >\ \omega\  \text{ for all $a\in H$.}$$
Likewise with ``smooth'' in place of ``analytic''.
\end{cor}

\noindent
Call a subset $F$ of $\Cc$ {\bf cofinal} if for each $\phi\in\mathcal C$ there exists
$f\in F$ with~$\phi\leq f$.
If~$F_1,F_2\subseteq\Cc$ and for all $f_1\in F_1$ there is an $f_2\in F_2$ with~$f_1\leq f_2$,
and $F_1$ is cofinal, then $F_2$ is cofinal.
Clearly each cofinal subset of $\Cc$ is unbounded.
The following strengthens \cite[Theorem~7]{S}:

\begin{cor}\label{CHcof}
Assume the Continuum Hypothesis {\em CH}: $2^{\aleph_0}=\aleph_1$. Then
there is a cofinal analytic Hardy field.
\end{cor}
\begin{proof}
Put $\mathfrak c:=2^{\aleph_0}$,
and let $\alpha$, $\alpha'$, $\beta$  range over ordinals~$<\mathfrak c$.
Choose an enumeration~$(\phi_\alpha)_{\alpha<\mathfrak c}$ of $\Cc$.
Suppose   $\big((H_\alpha,h_\alpha)\big)_{\alpha<\beta}$ is a family of bounded analytic Hardy fields  $H_\alpha$, each with an element $h_\alpha\in H_\alpha$,   such that
\begin{equation}\label{eq:cof}
\alpha<\alpha'<\beta\ \Rightarrow\ H_{\alpha}\subseteq H_{\alpha'}\qquad\text{and}\qquad
\alpha<\beta\ \Rightarrow\ \phi_\alpha <_{\ex} h_\alpha.
\end{equation}
Then $H:=\bigcup_{\alpha<\beta}H_\alpha$ is an analytic Hardy field, and $H$ is bounded, as the
union of countably many bounded subsets of $\mathcal C$.
By~\cite[Lemma~5.4.18]{ADH2},  $H^*:=\Li\!\big(H(\R)\big)$ is also bounded.
Take $\phi\in\Cc$ with $\phi>_{\ex} H^*$ and $\phi\geq\phi_\beta$.
Corollary~\ref{cor:5.1ana} yields an $H^*$-hardian   $h_\beta\in\Cc^\omega$ with
$h_\beta>_{\ex}\phi$. Then the analytic Hardy field~$H_\beta:=H^*\langle h_\beta\rangle$ is bounded by Lemma~\ref{lem:5.4.19}, contains $H_\alpha$ for all~$\alpha<\beta$,
and $\phi_\beta<_{\ex}h_\beta$.

Now  transfinite recursion yields a family $\big((H_\alpha,h_\alpha)\big)_{\alpha<\mathfrak c}$
where   $H_\alpha$ is a bounded analytic   Hardy field and $h_\alpha\in H_\alpha$   such that
\eqref{eq:cof} holds with $\mathfrak c$ in place of $\beta$.
Then~$\bigcup_{\alpha<\mathfrak c} H_\alpha$ is a cofinal analytic Hardy field.
\end{proof}

\noindent
See Corollary~\ref{CHcof strengthened} below for a strengthening of  Corollary~\ref{CHcof}.

\begin{remark} Vera Fischer suggested replacing
CH  in Corollary~\ref{CHcof} by $\mathfrak{b}=\mathfrak{d}$, which is strictly weaker than CH (provided of course that our base theory ZFC is consistent).
$\mathfrak{b}$ and $\mathfrak{d}$ are so-called {\em cardinal characteristics of the continuum}.  See~\cite[2.1, 2.2]{Blass} for their definitions, and~\cite[2.4]{Blass} for the inequalities $\aleph_1\leq\mathfrak{b}\leq\mathfrak{d}\leq\mathfrak{c}$.
\textit{Martin's Axiom}\/ (MA) implies $\mathfrak{b}=\mathfrak{d}=\mathfrak{c}$, see \cite[6.8, 6.9]{Blass} and~\cite[Corollary~8]{Rudin}.  If ZFC is consistent, then MA is strictly weaker than CH by~[44]. It is easy to check from their definitions that~$\mathfrak{b}$ is the least cardinality of an unbounded subset of $\C$, and $\mathfrak{d}$ is the least cardinality of a cofinal subset of~$\C$.
Replacing in the proof above~$\mathfrak{c}$ by~$\mathfrak{d}$ and taking
$(\phi_{\alpha})_{\alpha< \mathfrak{d}}$ to enumerate a cofinal subset of $\C$, the proof does indeed go through with
$\mathfrak{b}=\mathfrak{d}$ instead of CH.
\end{remark}

\subsection*{Hardy fields of countable cofinality}
Hardy fields of countable cofinality are bounded.
For later use we study here such Hardy fields in more detail. Given a valued differential field $K$, let  $C$ denote  its constant field
and $\Gamma$  its value group.

\begin{lemma}\label{lem:cf(K)}
Let $K$ be a pre-$H$-field with $\Gamma\neq\{0\}$. Then $\operatorname{cf}(K)=\operatorname{cf}(\Gamma)$.
\end{lemma}
\begin{proof}
Apply [ADH,  2.1.4]   to the increasing surjection~$f\mapsto -vf\colon K^>\to\Gamma$.
\end{proof}

% If $H\subseteq\R$, then $\operatorname{cf}(H)=\omega$. If $H\not\subseteq\R$, then $\operatorname{cf}(H)=\operatorname{cf}(\Gamma_H)$.

\begin{lemma}\label{lem:cof oags, 1}
Let $\Gamma$ be an ordered abelian group, $\Gamma\neq\{0\}$. If
$[\Gamma]$ has no largest element, then $\operatorname{cf}(\Gamma)=\operatorname{cf}([\Gamma])$;
otherwise  $\operatorname{cf}(\Gamma)=\omega$.
\end{lemma}
\begin{proof}
If $[\Gamma]$ has no largest element, then
[ADH,  2.1.4] applied to the increasing surjection $\gamma\mapsto[\gamma]\colon\Gamma^{\geq}\to[\Gamma]$
yields  $\operatorname{cf}(\Gamma)=\operatorname{cf}([\Gamma])$.
If $\gamma\in\Gamma^{>}$ is such that $[\gamma]$ is the largest element of~$[\Gamma]$,  then
$\N\gamma$ is cofinal in $\Gamma$.
\end{proof}

\noindent
Let $G$ be an abelian group, with divisible hull $\Q G=\Q\otimes_{\Z}G$. Then $\operatorname{rank}_{\Q} G:=\dim_{\Q}\Q G$ (a cardinal) is the {\it rational rank}\/ of $G$.
(NB: in [ADH, 1.7] we defined the rational rank of $G$ to be~$\infty$ if the $\Q$-linear space $\Q G$ is not finitely generated.)

\begin{lemma}\label{lem:cof oags, 2}
Let $\{0\}\neq\Delta\subseteq\Gamma$ be an extension of ordered abelian groups such that $\operatorname{rank}_{\Q}(\Gamma/\Delta)\le \aleph_0$. Then $\operatorname{cf}(\Gamma)\leq\operatorname{cf}(\Delta)$.
\end{lemma}
\begin{proof}
By Lemma~\ref{lem:cof oags, 1} we may assume that $[\Gamma]$
has no maximum.
Let $S$ be a well-ordered cofinal subset of $[\Delta]$ of order type $\operatorname{cf}([\Delta])$, so $\abs{S}=\operatorname{cf}([\Delta])$. Then~$\tilde{S}:=S\cup\big( [\Gamma]\setminus[\Delta] \big)$ is cofinal in
$[\Gamma]$, so
$\operatorname{cf}(\Gamma)=\operatorname{cf}([\Gamma])=\operatorname{cf}(\tilde S)\leq\abs{\tilde{S}}$
by Lemma~\ref{lem:cof oags, 1} and [ADH,  2.1.2].
Since~$[\Gamma]\setminus [\Delta]$ is countable  by [ADH, 2.3.9], we have~$\abs{\tilde{S}}
\leq\max\!\big\{\abs{S},\omega\big\}=\max\!\big\{\!\operatorname{cf}([\Delta]),\omega\big\}$. Now apply Lemma~\ref{lem:cof oags, 1}   to $\Delta$ in place of~$\Gamma$.
\end{proof}

\begin{lemma}\label{cofoaglem}
Let $K\subseteq L$ be an extension of pre-$H$-fields where
$\operatorname{rank}_{\Q}(\Gamma_L/\Gamma)$ is countable, and suppose
$\Gamma\neq\{0\}$ or $L$ has very small derivation and archimedean residue field.
\end{lemma}
\begin{proof}
If $\Gamma\neq\{0\}$,   then by Lemmas~\ref{lem:cf(K)} and~\ref{lem:cof oags, 2}
we have $\operatorname{cf}(L)=\operatorname{cf}(\Gamma_L)\leq \operatorname{cf}(\Gamma)=\operatorname{cf}(K)$.
Suppose $\Gamma=\{0\}$, so $L$ has very small derivation and archimedean residue field. Then  $K$ is archimedean, so
and $\Gamma_L$ is countable, hence~$\operatorname{cf}(\Gamma_L)\leq\omega$.
Therefore, if $\Gamma_L\neq\{0\}$, then~$\operatorname{cf}(L)=\operatorname{cf}(\Gamma_L)\leq\omega=\operatorname{cf}(K)$ by Lemma~\ref{lem:cf(K)} applied to~$L$ in place of $K$, and if $\Gamma_L=\{0\}$, then $\operatorname{cf}(L)=\omega=\operatorname{cf}(K)$.
\end{proof}

\noindent
By [ADH, 3.1.10] the hypothesis on $\operatorname{rank}_{\Q}(\Gamma_L/\Gamma)$ in Lemma~\ref{cofoaglem} is satisfied if
$\operatorname{trdeg}(L|K)$ is countable. Hence this lemma yields:

\begin{cor}
If   $F$ is a Hardy field extension of $H$ such that
$\operatorname{trdeg}(F|H)$ is countable, then $\operatorname{cf}(F)\leq\operatorname{cf}(H)$.
Hence if $\operatorname{trdeg}(H|C_H)$ is countable, then~$\operatorname{cf}(H)=\omega$ \textup{(}and so $H$ is bounded\textup{)}.
\end{cor}

\noindent
In \cite[Corollary~3.13]{ADHfgh} we showed that if~$H\supseteq\R$ and~$H^{>\R}$ has countable coinitiality,
and $H^{\operatorname{da}}$ is the
$\d$-closure of~$H$ in  a maximal Hardy field extension  of $H$,
then   $(H^{\operatorname{da}})^{>\R}$ also has countable coinitiality. The
property of~$H$ having countable cofinality  is equally robust:

\begin{theorem}\label{thm:cf}
Let $E$ be a differentially algebraic Hardy field extension of $H$ such that~$\exp\!\big(E(x)\big)\subseteq E(x)$. Then~$\operatorname{cf}(E) \leq \operatorname{cf}(H)$,
with equality if $\exp(H(x))\subseteq H(x)$. \end{theorem}

\noindent
Here is an immediate consequence:

\begin{cor}
If $H$ has countable cofinality, then so does $\Li\!\big(H(\R)\big)$ as well as the $\d$-closure of $H$ in any  maximal Hardy field extension  of $H$.
\end{cor}

\noindent
We precede the proof of Theorem~\ref{thm:cf} by a few lemmas.

\begin{lemma}\label{lem:cof 1}
Let $K$ be a pre-$\d$-valued field of $H$-type
and $\operatorname{dv}(K)$ its $\d$-valued   hull, and suppose   $\Gamma\neq\{0\}$.
Then  $\Gamma$ is cofinal in  $\Gamma_{\operatorname{dv}(K)}$.
\end{lemma}
\begin{proof}
This is clear if $\Gamma=\Gamma_{\operatorname{dv}(K)}$.
$\Gamma_{\operatorname{dv}(K)} = \Gamma+\Z\alpha$ where~$0<n\alpha<\Gamma^>$ for all $n\geq 1$, by [ADH, 10.3.2], so
$\Gamma$ is cofinal in $\Gamma_{\operatorname{dv}(K)}$.
\end{proof}

%\begin{lemma}\label{lem:cof 2}
%Let $K$ be a pre-$H$-field. If $\Gamma\neq\{0\}$ or $K$ has very small derivation,  then $K$ is cofinal in its $H$-field hull $H(K)$.
%If $\Gamma\neq\{0\}$, then this follows from Lemma~\ref{lem:cof 1}. If $\Gamma=\{0\}$ and $K$ has very small derivation, then $C=K$, so $K$ is an $H$-field and $H(K)=K$.

\begin{lemma}
$H$ is cofinal in $H(\R)$.
\end{lemma}
\begin{proof}
This is clear if   $H\subseteq\R$; assume $H\not\subseteq\R$.
Let $E$ be the $H$-field hull of~$H$, taken as an $H$-subfield of the Hardy field extension $H(\R)$ of~$H$.
Then $H$ is cofinal in $E$ (Lemma~\ref{lem:cof 1}), so replacing $H$ by $E$ we arrange that $H$ is an $H$-field. Now use that
$\Gamma_{H(\R)}=\Gamma_H\neq\{0\}$ by
[ADH, 10.5.15 and remark preceding 4.6.16].
% so $H$ is cofinal in~$H(\R)$.
\end{proof}

\begin{lemma}\label{lem:H(x)}
Let $K$ be a pre-$\d$-valued field of $H$-type with $\Gamma\neq\{0\}$. Let $s\in K$ and~$y'=s$, $y$ in a pre-$\d$-valued extension of $K$ of $H$-type. Then $\Gamma$ is cofinal~in~$\Gamma_{K(y)}$.
\end{lemma}
\begin{proof}
Set $L:=K(y)$ and $M:=\operatorname{dv}(L)$.
Lemma~\ref{lem:cof 1} allows us to replace $K$ by its $\d$-valued hull inside
to arrange that~$K$ is $\d$-valued.
Using [ADH, 10.5.15 and remark preceding 4.6.16] we replace $K$ by $K(C_M)$ to arrange also $L$ to be $\d$-valued with $C=C_L$.
Finally, replacing $K$ by its algebraic closure inside an algebraic closure of $L$
we arrange $K$ to be algebraically closed.
We may assume~$y\notin K$, so $y$ is transcendental over $K$. Then
$$S\ :=\ \big\{ v(s-a') : a\in K\big\}\ \subseteq\ \Gamma.$$
Assume for now that $S$ has  a maximum $\beta$. Then $\beta\notin (\Gamma^{\neq})'$ by [ADH, 10.2.5(i)], so~$\beta=\max\Psi$ or $\beta$ is a gap in $K$. If $\beta=\max\Psi$, then $\Gamma_{L}=\Gamma+\Z\alpha$ with~$\Gamma^{<} < n\alpha < 0$ for all $n\geq 1$, so~$\Gamma$ is cofinal in $\Gamma_{L}$.
Suppose $\beta$ is a gap in $K$. Take~$a\in K$ with~$\beta=v(s-a')$ and set $z:= y-a$, so $z'=s-a'$. We arrange $z\nasymp 1$
by replacing $a$ with $a+c$ for suitable $c\in C_L=C$.
%f $z\asymp 1$,  take~$c\in C_L=C$ such that~$z-c\prec 1$ and replace~$a$ by~$a+c$. Hence $z\prec 1$ or $z\succ 1$.
If $z\prec 1$, then~[ADH, 10.2.1 and its proof] gives $\Gamma_{L}=\Gamma+\Z\alpha$ with $0 < n\alpha < \Gamma^{>}$ for all $n\ge 1$, so  $\Gamma$ is cofinal in~$\Gamma_{L}$.
If $z\succ 1$, then [ADH, 10.2.3 and its proof] gives likewise that
$\Gamma$ is cofinal in~$\Gamma_{L}$.

If $S$ does not have a largest element,
then $L$ is an immediate extension of $K$: this holds by [ADH, 10.2.6]
if~$S<(\Gamma^>)'$; otherwise take~$a\in K$ with $v(s-a')\in(\Gamma^>)'$ and $y-a\nasymp 1$, and apply [ADH, 10.2.4 and 10.2.5(iii)] to $s-a'$, $y-a$ in place of~$s$,~$y$, respectively.
\end{proof}

\noindent
Lemma~\ref{lem:H(x)} and  \cite[Proposition 5.3.2(iv)]{ADH2} yield:

\begin{lemma}
If $H\subseteq\R$, then $x^\N$ is cofinal in $H(x)$, and if $H\not\subseteq\R$, then $H$ is cofinal in $H(x)$.
Hence $\operatorname{cf}(H)=\operatorname{cf}(H(x))$.
\end{lemma}

\noindent
We can now give the proof of Theorem~\ref{thm:cf}.
First, replacing $E$, $H$ by $E(x)$, $H(x)$, respectively, and using the last lemma, we arrange $x\in H$.
Let $S$ be a well-ordered cofinal subset of $H$ of order type $\operatorname{cf}(H)$.
Then $\abs{S}=\operatorname{cf}(H)$,
and $\tilde S:=\bigcup_n \exp_n(S)$ is a cofinal subset of $E$, by \cite[Lemma~5.4.1]{ADH2}.
Thus $\operatorname{cf}(E)=\operatorname{cf}(\tilde S)\leq\abs{\tilde S}=\abs{S}=\operatorname{cf}(H)$
as claimed.
If $\exp(H)\subseteq H$, then $\tilde S\subseteq H$, hence $\operatorname{cf}(E)=\operatorname{cf}(H)$.
\qed

\medskip
\noindent
For use in Section~\ref{sec:pc} we also include the following cofinality result, which is immediate from [ADH, 10.4.5(i)]:

\begin{lemma}\label{lem:10.4.5}
Let $K$ be a $\d$-valued field of $H$-type with divisible asymptotic couple~$(\Gamma,\psi)$, $\Gamma\neq\{0\}$,
and let $s\in K$ be such that
$$S:=\big\{ v(s-a^\dagger): a\in K^\times\big\} < (\Gamma^>)'$$
and $S$ has no largest element. Let $f$ be an element of an $H$-asymptotic field extension of $K$,
transcendental over $K$, with $f^\dagger = s$. Then $[\Gamma]=\big[\Gamma_{K(f)}\big]$, so $\Gamma$ is cofinal in $\Gamma_{K(f)}$.
\end{lemma}

\section{Pseudoconvergence in Analytic Hardy Fields}\label{sec:pc}

\noindent
We   complement the material on pc-sequences
from \cite[Sections~3,~4]{ADHfgh} by criteria for germs in $\Cc^{<\infty}$ to be pseudolimits of pc-sequences in Hardy fields, and then use this to show that each pc-sequence of countable length in an analytic Hardy field has an analytic pseudolimit.
{\it In this section $H$ is a Hardy field.}\/

\subsection*{Revisiting pseudoconvergence in Hardy fields}
Let $H\supseteq \R(x)$ be real closed with asymptotic integration and $(f_\rho)$ a pc-sequence in $H$ of $\d$-transcendental type over $H$ with pseudolimit $f$ in a Hardy field extension of $H$. Then~${Z(H,f)=\emptyset}$ by~[ADH, 11.4.13], so the valued field extension $H\langle f\rangle\supseteq H$ is immediate by~[ADH, 11.4.7].  In [8] we only considered pc-sequences of countable length, but here we do not assume $(f_{\rho})$ has countable length (to be exploited in the proof of Theorem~\ref{thdense}).
We begin by deriving a sufficient condition on $y\in \Cc^{<\infty}$ to be $H$-hardian with~${f_\rho\leadsto  y}$.
This will enable us to find such $y$ in $\mathcal{C}^{\omega}$. (Another possible use is to find such~$y$ with oscillating $y-f$, so that $H\<y\>$ and~$H\<f\>$ are ``incompatible'' Hardy field extensions of $H$.)
To simplify notation, set $t:= x^{-1}$.  We first observe:

\begin{lemma}\label{lem:immy}
Let $\phi$ be active in $H$, $0< \phi\prec 1$, let $\derdelta:=\phi^{-1}\der$ be the derivation of~$(\Cc^{<\infty})^\phi$, and
let  $z\in\Cc^{<\infty}$ satisfy $z^{(i)}\prec t^j$ for all $i$, $j$. Then
$\derdelta^k(z)\prec 1$ for all $k$.
\end{lemma}
\begin{proof}
This  is clear for $k=0$.  Suppose $k\geq 1$.
The identity  \eqref{eq:derdelta^k} in [8] gives~$\derdelta^k(z)=\phi^{-k}\sum_{j=1}^kR^k_j(-\phi^\dagger)z^{(j)}$
where  $R^k_j(Z)\in \Q\{Z\}$ for $j=1,\dots,k$.
This yields $\derdelta^k(z)\prec 1$  in view of $\phi^\dagger\preceq 1$ and $z^{(j)}\prec t^{2k}\prec \phi^k$
for $j=1,\dots,k$.
\end{proof}

%\begin{lemma} Suppose $y\in \Cc^{<\infty}$ is $H$-hardian and $f_{\rho}\leadsto y$. Then there is an isomorphism $H\<y\>\to H\<f\>$ of Hardy fields over $H$ sending $y$ to $f$.
%\begin{proof} From $f_{\rho}\leadsto f$ and $f_{\rho}\leadsto y$ we obtain $v(h-f)=v(h-y)$ for all $h\in H$. Then~[ADH, 11.4.7] provides a valued differential field isomorphism~${\iota\colon H\<y\>\to H\<f\>}$ over $H$ with $\iota(y)=f$.  For $Q(Y)\in H\{Y\}^{\ne}$ we take $h\in H^\times$ such that~$Q(y)\sim h$; then $Q(f)\sim h$, so \marginpar{corrected typos: $Q()$ instead of $H()$} $Q(y)$ and $Q(f)$ have the same sign as $h$. Thus $\iota$ is an isomorphism of Hardy fields.

\begin{prop} \label{propimmy1} Suppose $0\in v(f-H)$ and  $y\in \Cc^{<\infty}$ is such that for all $\fm\in H^\times$ with $v\fm\in v(f-H)$ and all $i$, $j$, $k$ we have
$$y^{(i)}-f^{(i)}\ \prec\  \fm^j t^k\ \text{ in $\Cc^{<\infty}$.}$$
Then $y$ is $H$-hardian and there is an isomorphism $H\langle y\rangle \to H\langle f\rangle$ of Hardy fields over~$H$ sending $y$ to $f$ \textup{(}and thus $f_\rho\leadsto y$\textup{)}.
\end{prop}

\begin{proof}
Let $\phi$ be active in $H$, $0<\phi\prec 1$, and $\derdelta=\phi^{-1}\der$ the derivation of $(\Cc^{<\infty})^\phi$.
Let also~$h\in H$ and $\fm\in H^\times$ with $f-h\preceq \fm$, and put $z:=\frac{y-f}{\fm}$.
By   \cite[Corollary~6.7.12]{ADH2} it suffices to show that then
$\derdelta^k\big( \frac{y-h}{\fm} \big)\preceq 1$ for all $k$; equivalently,
$\derdelta^k (z)\preceq  1$ for all $k$ (thanks to
$\frac{y-h}{\fm}-z = \frac{f-h}{\fm}\preceq 1$ and smallness of the derivation of $H\langle f\rangle^\phi$).

\claim[1]{Suppose $\fm\preceq 1$. Then $z^{(n)} \prec  \fm^jt^k$  for all $n$, $j$, $k$.}

\noindent
This holds for $n=0$ because $\fm z\prec \fm^{j+1}t^k$ for all $j$, $k$.
Let $n\ge 1$ and assume inductively that $z^{(i)}\prec \fm^jt^k$ for $i=0,\dots, n-1$ and all $j$, $k$.
Now $(\fm z)^{(n)}= y^{(n)}-f^{(n)}\prec \fm^jt^k$ for all $j$, $k$, and
$$(\fm z)^{(n)}\  =\ \ \fm^{(n)}z + \cdots + \fm z^{(n)}.$$
Since $\fm\preceq 1$, the inductive assumption gives $\fm^{(n-i)}z^{(i)}\prec \fm^jt^k$ for $i=0,\dots,n-1$ and all $j$, $k$, so
$\fm z^{(n)} \prec \fm^jt^k$  for all $j$, $k$, and thus $z^{(n)}\prec \fm^jt^k$  for all $j$, $k$.

\claim[2]{$\derdelta^k (z)\prec  1$ for all $k$.}

\noindent
If $\fm\preceq 1$, then this holds by Claim~1 and the lemma above.
In general, take $h_1\in H$ with $f-h_1\prec f-h$ and $f-h_1\preceq 1$, and then $\fm_1\in H^\times$ with $f-h_1\asymp\fm_1$.
By the special case just proved with $h_1$, $\fm_1$ in place of $h$, $\fm$ we have~$\derdelta^k \big( \frac{y-f}{\fm_1} \big)\prec  1$ for all~$k$.
$z = \big(\frac{y-f}{\fm_1}\big)\big(\frac{\fm_1}{\fm}\big)$ and $\frac{\fm_1}{\fm}  \prec 1$ (in $H$), so the claim follows using
the  Product Rule for the derivation   of $(\Cc^{<\infty})^\phi$ and smallness of the derivation of~$H^\phi$.
\end{proof}

\noindent
Here is a more useful variant for the case $0\notin v(f-H)$:

\begin{prop} \label{propimmy2} Suppose $0\notin v(f-H)$, and $y\in  \Cc^{<\infty}$ is such that
$$y^{(i)}-f^{(i)}\ \prec\   t^k \text{ for all $i$, $k$.}$$
Then $y$ is $H$-hardian and there is an isomorphism $H\langle y\rangle \to H\langle f\rangle$ of Hardy fields over~$H$ sending $y$ to $f$.
\end{prop}
\begin{proof}
Let $\phi$, $h$, $\fm$, $z$ be as in the proof of Proposition~\ref{propimmy1}; as in that proof
it suffices to show that $\derdelta^k (z)\preceq  1$ for all $k$.
Now $\fm\succ 1$,
hence with $\frak{n}:= \fm^{-1}\in H^\times$  we have~$z=\frak{n}(y-f)$ and $\frak{n}\prec 1$, so in
view of  $z^{(i)}=\frak{n}^{(i)}(y-f)+\cdots + \frak{n}(y-f)^{(i)}$ we obtain
$z^{(i)}\prec t^j$ for all $i$, $j$, and Lemma~\ref{lem:immy} then yields
$\derdelta^k (z)\prec  1$ for each $k$.
\end{proof}

\noindent
Multiplicative conjugation gives a reduction to Proposition~\ref{propimmy2}, except
when $(f_{\rho})$ is a cauchy sequence, not just a pc-sequence. We shall exploit this several times.

\subsection*{Constructing analytic pseudolimits in Hardy field extensions}
Let   $(f_{\rho})$ be a pc-sequence in $H$. Corollary~\ref{corimmom}  from [8]  says:
if $(f_\rho)$ has countable length, then~$(f_{\rho})$ pseudoconverges in some Hardy field extension of $H$. Using  Corollary~\ref{apomega} and Proposition~\ref{propimmy2}  we now deduce smooth and analytic versions of this key fact.
% from this fact itself:
%This follows from a somewhat more precise result:

\begin{prop}\label{prop:smanpc}
Suppose $H$ is an analytic Hardy field and
$(f_\rho)$   pseudoconverges in some Hardy
field extension of $H$. Suppose also that $H$ is bounded or $(f_\rho)$ does not have width~$\{\infty\}$ in the valued field $H$. Then $(f_\rho)$ pseudoconverges in an analytic Hardy field extension of $H$.
Likewise with ``smooth'' in place of ``analytic''.
\end{prop}

\begin{proof}
Assume $H$ is analytic; the smooth case goes the same way. As in [8] we can pass from $H$ to an extension of $H$ and reduce to the case that $H\supseteq \R$, $H$ is closed, and $(f_{\rho})$ has no pseudolimit in $H$. Take $f$ in a Hardy field extension of $H$ such that $f_{\rho}\leadsto f$.
Then $f\notin H$, so $f$ is $\d$-transcendental over $H$. If $H$ is bounded, then
Corollary~\ref{cor:analytify} yields an $H$-hardian $y\in \Cc^{\omega}$ with $f_{\rho}\leadsto y$.

Suppose $(f_\rho)$ does not have width $\{\infty\}$. Then take $h\in H^\times$ with ${v(hf-hf_{\rho})} < 0$ for all $\rho$. Take $\varepsilon\in \Cc$ such that $\varepsilon>_{\ex} 0$ and $\varepsilon \prec t^k$
for all $k$, for example, $\varepsilon= \ex^{-x}$. Now Corollary~\ref{apomega} gives $y\in \Cc^{\omega}$ such that $(hf)^{(i)}-y^{(i)}\prec t^k$ for all $k$. Then
$y$ is $H$-hardian and $hf_{\rho}\leadsto y$ by Proposition~\ref{propimmy2}, hence
$h^{-1}y\in \Cc^{\omega}$ is $H$-hardian and~$f_{\rho}\leadsto h^{-1}y$.
\end{proof}

\begin{cor}\label{smanpc} If $H$ is an analytic Hardy field, then
every pc-sequence in $H$ of countable length
pseudoconverges in an analytic Hardy field extension of $H$.
Likewise with ``smooth'' in place of ``analytic''.
\end{cor}
\begin{proof}
%[Proof of Corollary~\ref{smanpc}]
Suppose   $(f_\rho)$ has countable length. Then $(f_\rho)$ pseudoconverges
in a Hardy field extension of $H$, by  \cite[Corollary~\ref{corimmom}]{ADHfgh}. Moreover, if $(f_\rho)$
has width $\{\infty\}$, then~$\big(v(f-f_\rho)\big)$ is cofinal in $\Gamma_H$, so $\operatorname{cf}(H)=\operatorname{cf}(\Gamma_H)=\omega$ by Lemma~\ref{lem:cf(K)}, hence $H$ is bounded. Now use  Proposition~\ref{prop:smanpc}.
\end{proof}

\noindent
Arguing as in the proof of \cite[Corollary~\ref{cormhci}]{ADHfgh}, using Corollary~\ref{smanpc}
instead of \cite[Corol\-lary~\ref{corimmom}]{ADHfgh}, yields:

\begin{cor}\label{cormhcian}
If  $H$ is a maximal analytic or maximal smooth Hardy field, then $\ci(H^{>\R})>\omega$.
%Likewise with $\Cc^\infty$ in place of $\Cc^\omega$.
\end{cor}

\noindent
In view of \cite[Remark preceding Corollary 8.1]{ADHfgh}, Corollaries~\ref{corsjoan},~\ref{smanpc},~\ref{cormhcian} yield  a version of \cite[Corollary~\ref{hco}]{ADHfgh} for maximal analytic Hardy fields:

\begin{cor}\label{hcoan} If $H$ is a maximal analytic or maximal smooth Hardy field, then  its $H$-couple~$(\Gamma,\psi)$   is countably spherically complete and
$$\cf(\Gamma^{<})\ =\ \ci(\Gamma^{>})\ >\ \omega, \qquad \ci(\Gamma)\ =\ \cf(\Gamma)\ >\ \omega.$$
%Likewise if $H$ is a maximal smooth Hardy field.
\end{cor}

\section{Proofs of Theorems A and B} \label{s4}

\noindent
We begin with revisiting Case (\rm{b}) extensions, then prove Theorem A and use it to characterize the possible gaps in maximal analytic Hardy fields. We also determine the number of maximal analytic Hardy fields. Next we prove Theorem~B, and finish this section with two subsections on dense pairs
of closed $H$-fields.

\subsection*{Case {(\rm{b})} extensions}
Let $H\supseteq \R$ be a Liouville closed Hardy field with $H$-couple~$(\Gamma,\psi)$ over $\R$.
Suppose~$\beta$ in an $H$-couple $(\Gamma^*,\psi^*)$ over $\R$ extending $(\Gamma,\psi)$ falls under Case~(b), that is, with~$\big(\Gamma\<\beta\>, \psi_{\beta}\big)$   the $H$-couple over
$\R$ generated by~$\beta$ over $(\Gamma, \psi)$ in~$(\Gamma^*, \psi^*)$:

\begin{itemize}\item[(b)] We have a sequence $(\alpha_i)$ in
$\Gamma$ and a sequence $(\beta_i)$ in $\Gamma^*$ that is $\R$-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
$\Gamma\<\beta\>=\Gamma \oplus \bigoplus_{i=0}^\infty \R\beta_i$.
\end{itemize}

\noindent
Unlike in key parts of \cite[Section~\ref{fbg}]{ADHfgh}
we do not assume~$\beta$ is of countable type over~$\Gamma$, and this will be exploited in the proof of Theorem~\ref{thdense}.
By~\cite[Corollary~\ref{cor:lemb}]{ADHfgh}, $\beta$ also falls under Case~(b) with the same sequences $(\alpha_i)$, $(\beta_i)$ when
$(\Gamma,\psi)$ and~$(\Gamma^*,\psi^*)$ are viewed as $H$-couples over $\Q$; see \cite[Section 8]{ADHfgh} for the relevant definitions.

In the next proposition and its corollary we assume $y\in\Cc^{<\infty}$ is $H$-hardian,  $y>0$, and $vy$ realizes the same cut in~$\Gamma$ as~$\beta$. Then by~\cite[Remark~\ref{rem:prbcon}]{ADHfgh} we have a unique isomorphism over $\Gamma$ of
%$(\Gamma\<\beta\>, \psi_{\beta})$ %
the $H$-couple over $\Q$ generated by $\Gamma\cup\{\beta\}$  in~$(\Gamma^*, \psi^*)$ with
the $H$-couple of the Hardy field~$H\<y\>^{\operatorname{rc}}$ over $\Q$ sending $\beta$ to $vy$.
%Thus  we may denote the latter $H$-couple by $(\Gamma\<vy\>,\psi_y)$.
Moreover, if~$z\in\Cc^{<\infty}$ is also $H$-hardian with~$z>0$ and~$vz$ realizes the same cut in $\Gamma$ as $\beta$, then we have a unique Hardy field isomorphism $H\<y\>\to H\<z\>$ over $H$ sending $y$ to~$z$, by \cite[Proposition~\ref{prbcon}]{ADHfgh}. The problem here is to find
such~$z$ in $\Cc^{\omega}$ (in which case~$H\<z\>$ is smooth, respectively analytic, if $H$ is). This can always be done, in view of Corollary~\ref{apomega} and the following:

\begin{prop}\label{revb} There exists $\varepsilon\in \C$ such that $\varepsilon >_{\ex} 0$ and for all
$z\in \Cc^{<\infty}$, if~$(z-y)^{(i)}\prec \varepsilon$ for all $i$, then $z$ is $H$-hardian and
$vz$ realizes the same cut in~$\Gamma$~as~$\beta$.
\end{prop}
\begin{proof}
Let the sequences $(\alpha_i)$, $(\beta_i)$ be as in (b). Take $f_i\in H^{>}$ with~$vf_i=\alpha_i$, and recursively we set $y_0:= y/f_0$, $y_{i+1}:=y_i^\dagger/f_{i+1}$. Then $vy_i$ realizes the same cut in $\Gamma$ as $\beta_i$, and
$H\<y\>=H(y_0, y_1, y_2,\dots)$,
by \cite[proof of Proposition~\ref{prbcon}]{ADHfgh}.
Next set~$K:=\R\< f_0, f_1, f_2,\dots\>$ and note that
$K\<y\>=K(y_0, y_1, y_2,\dots)$, using that the right hand side contains all $y_n'$. Suppose~$z\in \Cc^{<\infty}$ is such that $(z-y)^{(i)}\prec f$ for all $i$  and all~$f\in K\langle y\rangle^{\times}$. Then~$z$ is $K$-hardian  and $\d$-transcendental over~$K$ by Lemma~\ref{lem:analytify, 2}, and the proof of that lemma also shows that $P(y)\sim P(z)$ for all
$P\in K\{Y\}^{\ne}$. Thus we have elements~$z_i\in K\<z\>$ defined recursively by  $z_0:= z/f_0$, $z_{i+1}:=z_i^\dagger/f_{i+1}$, and then~${y_i\sim z_i}$ for all $i$.
For the Hausdorff field~$H_n:=H(y_0,\dots,y_n)$ we have~$v(H_n^\times)=\Gamma\oplus\Z vy_0\oplus\cdots\oplus\Z vy_n$
by \cite[proof of Proposition~\ref{prbcon}]{ADHfgh},
and for~$h\in H^\times$ and~$i_0,\dots,i_n\in\N$ we have~$hy_0^{i_0}\cdots y_n^{i_n} \sim h  z_0^{i_0}\cdots z_n^{i_n}$~(in~$\Cc$).
Hence $z_0,\dots,z_n$ generate a Hausdorff field over $H$ with an isomorphism~$H_n\to H(z_0,\dots,z_n)$ over $H$ sending $y_i$ to $z_i$ for $i=0,\dots,n$.  These isomorphisms have therefore a common extension to an isomorphism
$H\<y\>\to H\<z\>$ of Hardy fields over $H$. In particular, $z$ is $H$-hardian, and $vz$ realizes the same cut in $\Gamma$ as $\beta$.  Now by Lemma~\ref{lem:analytify, 1} and the remarks preceding it there exists $\varepsilon\in \C$ such that $\varepsilon >_{\ex} 0$ and
$\varepsilon \prec f$ for all~$f\in K\<y\>^\times$, so any such $\varepsilon$ has the desired property.
\end{proof}

\noindent
Combining Corollary~\ref{apomega} with Proposition~\ref{revb} yields:

\begin{cor}\label{cor:revb}
There exists an $H$-hardian $z\in\Cc^\omega$ such that $z>0$ and $vz$ realizes the same cut in $\Gamma$ as $\beta$.
\end{cor}

\noindent
We can now use  \cite[Theorem~\ref{bconstruction}]{ADHfgh} to obtain an analytic strengthening of it:

\begin{cor} \label{bcon}
%Let $H\supseteq \R$ be a Liouville closed Hardy field with $H$-couple $(\Gamma,\psi)$ over $\R$. Suppose $\beta$ in an extension of $(\Gamma,\psi)$ and
Suppose $\beta$ is of countable type over $\Gamma$
% falls under Case \textup{(b)}, and
and $\beta_i^\dagger < 0$ for all $i$.
% where $(\alpha_i)$, $(\beta_i)$ are as in \textup{(b)}.
Then for some $H$-hardian $z\in \Cc^{\omega}$:  $z > 0$ and $vz$ realizes the same cut in $\Gamma$ as $\beta$.
\end{cor}
\begin{proof} \cite[Theorem 9.2]{ADHfgh} gives $H$-hardian $y>0$ such that $vy$ realizes the same cut in $\Gamma$ as $\beta$. Then Corollary~\ref{cor:revb}  gives a $z$ as required.
\end{proof}

\subsection*{Proof of Theorem A} First  an analytic/smooth version of \cite[Lemma~\ref{lemth01}]{ADHfgh}:

\begin{lemma}\label{lemth01an}
Let $H$ be a maximal analytic or maximal smooth Hardy field with $H$-couple $(\Gamma,\psi)$ over $\R$. Then no element in any
$H$-couple over $\R$ extending $(\Gamma,\psi)$ has countable type over~$\Gamma$.
%Likewise with $\Cc^\omega$ replaced by $\Cc^\infty$.
\end{lemma}
\begin{proof}
By Corollary~\ref{hcoan}, $(\Gamma,\psi)$ is countably spherically complete, and both $\Gamma$ and~$\Gamma^<$   have
uncountably cofinality. By \cite[Lemma~\ref{caseb}]{ADHfgh},
any element of any $H$-couple over $\R$ extending $(\Gamma,\psi)$ and of countable type over $\Gamma$
falls under Case~(b).
Now argue as in the proof of \cite[Lemma~\ref{lemth01}]{ADHfgh}, using
Corollary~\ref{bcon} in place of  \cite[Theorem~\ref{bconstruction}]{ADHfgh}, that there are no such elements.
%The smooth case is similar.
\end{proof}

\noindent
Theorem~\ref{aneta} from the introduction  and its smooth version follow from
Corollary~\ref{smanpc} and Lemma~\ref{lemth01an}, just as the main theorem in [8]  is derived in the beginning of \cite[Section~\ref{fbg}]{ADHfgh} from the non-smooth analogues of that corollary and lemma.
We now use this to characterize gaps in maximal analytic Hardy fields: Corollary~\ref{gapsmax} below.

\subsection*{Characters of gaps in maximal Hardy fields}
Let~$S$ be an ordered set (as in~[8] this means {\em linearly ordered set}) and $C$  a cut in $S$, that is a downward closed subset of $S$. We define the {\bf character} of~$C$ (in $S$) to be the pair $(\alpha,\beta^*)$ where~$\alpha:=\cf(C)$ and $\beta^*$ is the set $\beta:=\ci(S\setminus C)$ equipped with the reversed ordering. We then also call
$C$ an {\bf $(\alpha,\beta^*)$-cut} (in $S$); see \cite[\S{}3.2]{Harzheim}.
The characters of the cuts~$\emptyset$ and $S$ in~$S$ are~$\big(0,\ci(S)^*\big)$ and $\big(\!\cf(S),0\big)$, respectively. Note that $S$ is $\eta_1$ iff no cut in~$S$ has character $(\alpha,\beta^*)$  with~${\alpha,\beta\leq\omega}$.
A {\bf gap $A<B$ in $S$} is a pair $(A,B)$ of subsets of~$S$ such that $A<B$ and there is no $s\in S$ with $A<s<B$.
%with $A$ a cut in $S$ and $B=S\setminus $A$.
The character of such a gap~$A<B$ is defined to be the character~$(\alpha,\beta^*)$ of the cut $A^{\downarrow}=S\setminus B^\uparrow$ in $S$, and
then~$A<B$ is also called an $(\alpha,\beta^*)$-gap in $S$.

Let $G$ be an ordered abelian group.
If $v\colon G\to S_\infty$ is a surjective convex valuation on $G$ ([ADH, p.~99]) and $A<B$ is an $(\alpha,\beta^*)$-gap in $S$ where $\alpha,\beta\geq\omega$, then~$\big(v^{-1}(A)\cap G^<, v^{-1}(B)\cap G^{<}\big)$ is a  $(\alpha,\beta^*)$-gap in $G$.
If $H$ is an ordered field, then $\cf(H^<)=\ci(H^>)=\cf(H)$, and the cuts~$H^{<h}$ and $H^{\leq h}$ ($h\in H$) in~$H$ have character~$\big(\!\cf(H),1\big)$ and $\big(1,\cf(H)^*\big)$, respectively.

\begin{cor}\label{cor:gaps 1}
Let $H$ be a maximal Hardy field, or a maximal analytic Hardy field, or a maximal smooth Hardy field. Set $\kappa:=\ci(H^{>\R})$. Then $\omega<\kappa\leq\mathfrak c$, and~$H$ has gaps of character~$(\omega,\kappa^*)$, $(\kappa,\omega^*)$, and $(\kappa,\kappa^*)$.
\end{cor}
\begin{proof}  Corollary~\ref{cormhcian} and \cite[Corollary~\ref{cormhci}]{ADHfgh} give $\omega<\kappa\leq\mathfrak c$.
The gaps~${\R < H^{>\R}}$ and~${H^{<\R} < \R}$ in $H$ have character $(\omega,\kappa^*)$ and $(\kappa,\omega^*)$, respectively.
To obtain a~$(\kappa,\kappa^*)$-gap in $H$, take a coinitial sequence $(\ell_\rho)_{\rho<\kappa}$   in $H^{>\R}$ with $\ell_{\rho}\succ \ell_{\rho'}$
for all~${\rho < \rho'<\kappa}$.
Put~$\upg_\rho:=\ell_\rho^\dagger > 0$, so $(1/\ell_\rho)' = (1/\ell_\rho)^\dagger/\ell_\rho = -\upg_\rho/\ell_\rho < 0$.
%and~$\upg_\rho/\ell_\rho\prec \upg_\rho$, for each~$\rho<\kappa$, and
Set~$A:=\{\upg_\rho/\ell_\rho:\rho<\kappa\}$, $B:=\{\upg_\rho:\rho<\kappa\}$.
Then $A<B$  is a $(\kappa,\kappa^*)$-gap in $H$, using that~$H$ has asymptotic integration.
\end{proof}

\noindent
Let now $G$ be an ordered abelian group. Assume $G\ne \{0\}$ and $G^>$ has no smallest element, so~$\ci(G^>)\geq\omega$.
A gap $A<B$ in $G$ is said to be {\bf cauchy} if $A,B\neq\emptyset$, $A$ has no largest element, $B$ has no smallest element,
and for each~$\varepsilon\in G^>$ there are~$a\in A$, $b\in B$ with $b-a<\varepsilon$.
If $A<B$ is a cauchy gap in $G$, then so is $-B<-A$.

\begin{lemma}\label{cau}
Let $A<B$ be a cauchy gap in $G$ and  $(a_\rho)$ be an   increasing cofinal well-indexed sequence in $A$.
Then $(a_\rho)$ is a divergent c-sequence in $G$.
\end{lemma}
\begin{proof}
Let $\varepsilon\in G^>$ and take  $a\in A$, $b\in B$ with $b-a<\varepsilon$. Take $\rho_0$ such that $a\leq a_\rho$ for all $\rho>\rho_0$. Then $0< a_{\rho'}-a_\rho<b-a<\varepsilon$  for
$\rho_0<\rho<\rho'$. Hence $(a_\rho)$ is a c-sequence in $G$,
and  there is no $a\in G$ with $a_\rho\to a$.
\end{proof}

\begin{lemma}\label{lem:cauchy gap}
Every cauchy gap in $G$ has character $\big(\!\cf(G^<),\cf(G^<)^*\big)$. Moreover,  $G$ is complete iff $G$ has no cauchy gap.
\end{lemma}
\begin{proof}
The first claim follows from Lemma~\ref{cau} and [ADH, 2.4.11].
It also follows from this lemma that if $G$ is complete, then $G$ has no cauchy gap.
Conversely, suppose $G$ has no cauchy gap. Let $(a_\rho)$ be a c-sequence in $G$. For each $\varepsilon\in G^>$, take
$\rho_\varepsilon$ such that $\abs{a_\rho-a_{\rho'}}<\varepsilon$ for all $\rho,\rho'\geq\rho_\varepsilon$, and set
$$A\ :=\ \{ a_{\rho_\varepsilon}-\varepsilon:\  \varepsilon\in G^>\}, \qquad
B\ :=\ \{ a_{\rho_\delta}+\delta:\  \delta\in G^>\}.$$
Then $A<B$ and for all $\varepsilon\in G^{>}$ there are
$a\in A$ and $b\in B$ with $b-a<\varepsilon$. But~$A< B$ is no cauchy gap, so we have $g\in G$ with
$A\leq g\leq B$.  Then $a_{\rho} \to g$.
\end{proof}

\begin{cor}\label{cor:gaps 2}
Let $H$ be a maximal, or maximal analytic,  or maximal smooth Hardy field, and $\lambda:=\cf(H)$. Then $\omega<\lambda\leq\mathfrak c$, and $H$ has gaps of character~$(0,\lambda^*)$, $(\lambda,0)$, $(1,\lambda^*)$, $(\lambda,1)$, and if $H$ is not complete, then
$H$ has a $(\lambda,\lambda^*)$-gap.
\end{cor}
\begin{proof}
Lemma~\ref{lem:cf(K)} yields $\lambda=\cf(\Gamma)$, so $\omega<\lambda\leq\mathfrak c$ by
\cite[Corollary~\ref{hco}]{ADHfgh} and Corollary~\ref{hcoan}. For the rest use  the remarks before Corollary~\ref{cor:gaps 1} and Lemma~\ref{lem:cauchy gap}.
\end{proof}

\noindent
The main result of [8], Theorem~\ref{aneta}, and Corollaries~\ref{cor:gaps 1},~\ref{cor:gaps 2} now give:

\begin{cor}\label{gapsmax}
Assume  \textup{CH}. If $H$ is a maximal Hardy field, or a maximal analytic Hardy field, or a maximal smooth Hardy field, then the characters of gaps  in $H$ are $$\text{$(0,\omega_1^*)$, $(\omega_1,0)$,  $(1,\omega_1^*)$, $(\omega_1,1)$, $(\omega,\omega_1^*)$, $(\omega_1,\omega^*)$, and $(\omega_1,\omega_1^*)$.}$$
\end{cor}

\subsection*{The number of maximal analytic Hardy fields}
Next   analytic  and smooth versions of Theorem~\ref{th2c} from [8]:

\begin{cor}\label{cor:nomaxan}
The number of maximal analytic Hardy fields is $2^{\mathfrak c}$ where $\mathfrak c=2^{\aleph_0}$. Likewise with ``smooth'' in place of ``analytic''.
\end{cor}
\begin{proof}
We treat the number of maximal analytic Hardy fields; the smooth case is similar, using the smooth version
of Theorem~\ref{aneta}.
In the argument following the statement of \cite[Proposition~\ref{prbin}]{ADHfgh} we replace
$\mathcal H$ by the set of all analytic Hardy fields~$H\supseteq\R$ with~$\abs{*H^{\te}} < \frak{c}$.
Thus modified, this argument shows that it is enough to prove that in \cite[Proposition~\ref{prbin}]{ADHfgh} we can choose $f_0$, $f_1$ to be analytic whenever the Hardy field $H$ is analytic. For this we first note that
if $H$ in \cite[Lemma~\ref{lemte}]{ADHfgh} is analytic, then we can take $y$ there to be analytic, by appealing to Theorem~\ref{aneta} instead of \cite[Section~5 and Corollary~\ref{xo}]{ADHfgh}.
Now argue as in the remarks following~\cite[Lem\-ma~\ref{ohhb}]{ADHfgh}
using this analytic version of \cite[Lemma~\ref{lemte}]{ADHfgh}.
\end{proof}

\noindent
Corollary~7.8 of [8] has an analytic version with a similar proof:

\begin{cor}\label{anteta} If $H$ is a maximal analytic Hardy field, then  the ordered set $*H^{\te}$ is $\eta_1$, and $\abs{*H^{\te}}=\frak{c}$.
\end{cor}

\noindent
We now improve Corollary~\ref{CHcof}:  assuming CH,
there are as many cofinal maximal analytic Hardy fields as there are maximal analytic Hardy fields,
by Corollary~\ref{CHcof strengthened}.
%of the following two lemmas. \marginpar{two new lemmas and a corollary}

\begin{lemma}\label{lem:CHcof strengthened, 1}
Let $\phi\in\Cc^\omega$ be overhardian. Then there
is a set $\mathcal H_\phi$ of  analytic Hardy field extensions of~$\R\langle\phi\rangle$ with
$\abs{\mathcal H_\phi}=2^{\mathfrak c}$ such that  for each $H\in\mathcal H_\phi$,
$*\phi$ is the largest element of  $*H^{\te}$, and each Hardy field
contains at most one $H\in\mathcal H_\phi$.
\end{lemma}
\begin{proof} By \cite[Lemma~\ref{lemte}]{ADHfgh} we have $*\R\langle\phi\rangle^{\te}=\{*\phi\}$.
Let $H\supseteq\R\langle\phi\rangle$ be an analytic Hardy field
with $*\phi=\max *H^{\te}$, and
let~$P<Q$ be a countable gap in~$*H^{\te}$ with~$Q<*\phi$.
Then   \cite[Proposition~\ref{prbin}]{ADHfgh} and the argument in the proof of Corollary~\ref{cor:nomaxan} yields analytic Hardy fields~$H_0=H\langle f_0\rangle$ and~$H_1=H\langle f_1\rangle$ without a common Hardy field extension
such that for $j=0,1$, we have $f_j\in H_j^{\te}$, $P<*f_j<Q\cup\{*\phi\}$, and~$*H_j^{\te}=*H^{\te}\cup\{*f_j\}$ (thus~$*\phi=\max *H_j^{\te}$).

We now follow the argument after the statement of \cite[Proposition~\ref{prbin}]{ADHfgh}, with~$\mathcal H$ now the set of all analytic Hardy fields~$H\supseteq\R\langle\phi\rangle$ such that~$\abs{*H^{\te}} < \frak{c}$
and~$*\phi=\max*H^{\te}$. For an ordinal $\lambda$ we let $2^{\lambda}$ be the set of functions~${\lambda\to\{0,1\}}$.
With $s$ ranging over $\bigcup_{\lambda<\frak c} 2^\lambda$, we construct
a tree~$(H_s)$ in $\mathcal H$ with~$\abs{*H^{\te}}\leq\abs{{\lambda+1}}$ for~$s\in 2^\lambda$, as follows.
For $\lambda=0$ the function~$s$ has empty domain and we take~$H_s=\R\langle\phi\rangle$.
% (so $*H_s^{\te}=\{*\phi\}$).
If~$s\in 2^\lambda$ ($\lambda<\frak c$) and~$H_s\in \cH$ are given with~$\abs{*H_s^{\te}}\le \abs{\lambda+1}$, then \cite[Lemma~\ref{lmif}]{ADHfgh}  provides a countable gap~$P$,~$Q$ in~${*H_s^{\te}\setminus\{*\phi\}}$, and we let~$H_{s0}, H_{s1}\in \cH$ be obtained from $H_s$ as~$H_0$,~$H_1$ are obtained from $H$ in the remark above.
Suppose~$\lambda< \frak{c}$ is an infinite limit ordinal,
$s\in 2^\lambda$, and that for every $\alpha< \lambda$ we are given~$H_{s|\alpha}\in \cH$ with~$H_{s|\alpha}\subseteq H_{s|\beta}$ whenever
$\alpha\le \beta < \lambda$.
Then we set~$H_s:= \bigcup_{\alpha<\lambda} H_{s|\alpha}\in\cH$. Assuming also inductively that~$\abs{*H_{s|\alpha}^{\te}}\le  \abs{\alpha+1}$ for all~$\alpha<\lambda$,
we have~$\abs{*H_s^{\te}}\le \abs{\lambda}\cdot \abs{\lambda+1}=\abs{\lambda+1}$, as desired.
This finishes the construction of our tree. Then for each~$s\in 2^{\frak c}$   we have
an analytic Hardy field~$H_s:=\bigcup_{\lambda<\frak{c}} H_{s|\lambda}$ such that if~$s, s'\in 2^{\frak c}$ are different, then~$H_s$,~$H_{s'}$ have no common Hardy field extension.
Hence $\mathcal H_\phi:=\{H_s:s\in 2^{\mathfrak c}\}$ has the required properties.
\end{proof}

\begin{lemma}\label{lem:CHcof strengthened, 2}
Assume \textup{CH}.
Let $H$ be a bounded analytic Hardy field. Then
$H$ extends to a cofinal analytic Hardy field.
\end{lemma}
\begin{proof} By Lemma~\ref{lem:5.4.19} we can
replace~$H$ by $\Li\!\big(H(\R)\big)$
to arrange that $H\supseteq\R$  and~$H$ is Liouville closed. Next,
take  an enumeration    $(\phi_\alpha)_{\alpha<\mathfrak c}$   of~$\Cc$
with $\phi_0>_{\ev} H$.
Corollary~\ref{cor:5.1ana} yields an $H$-hardian~$h_0\in\Cc^\omega$ with $h_0>_{\ev} \phi_0$, and
then the analytic Hardy field $H_0:=H\langle h_0\rangle$ is bounded by Lemma~\ref{lem:5.4.19}.
Now a transfinite recursion as in the proof of  Corollary~\ref{CHcof}, beginning with $(H_0,h_0)$,  yields a
cofinal analytic Hardy field extension of~$H_0$ and thus of $H$.
%family~$\big((H_\alpha,h_\alpha)\big)_{\alpha<\frak c}$ where $H_\alpha$ is a bounded analytic Hardy field extension of $H$
%and $h_\alpha\in H_\alpha$ such that~\eqref{eq:cof} holds with $\frak c$ in place of $\beta$. Then~$\bigcup_{\alpha<\frak c} H_\alpha$ is a cofinal analytic Hardy field extension of~$H$.
\end{proof}

\noindent
Corollary~\ref{cor:5.1ana} gives an overhardian $\phi\in\Cc^\omega$. For such $\phi$ and $\mathcal H_\phi$ as in Lemma~\ref{lem:CHcof strengthened, 1}, all $H\in\mathcal H_\phi$ are bounded.
With  Lemma~\ref{lem:CHcof strengthened, 2} we can now improve Corollary~\ref{CHcof}:

\begin{cor}\label{CHcof strengthened}
Assuming \textup{CH}, there are $2^{\mathfrak c}$  cofinal maximal analytic Hardy fields.
\end{cor}

\subsection*{Maximal analytic Hardy fields approximate maximal Hardy fields}  A maximal analytic Hardy field is
an $\infty\omega$-elementary substructure of any maximal Hardy field extension, by Corollary~\ref{cor:inftyomegaelemsub} below. Maximal analytic Hardy fields are also very close to maximal Hardy fields in another way:

\begin{theorem}\label{thm:dense} Let $H$ be a maximal analytic Hardy field or a maximal smooth Hardy field. Then $H$ is dense in any Hardy field extension of $H$.
\end{theorem}

\begin{proof}  We establish two claims:

\claim[1]{If $f\in \Cc^{<\infty}$ is $H$-hardian and $H\<f\>$ is an immediate extension of $H$, then~$H$ is dense in $H\<f\>$.}

\noindent
To prove this, assume $f\in  \Cc^{<\infty}$ is $H$-hardian, $(f_{\rho})$ is a divergent pc-sequence in~$H$, and $f_{\rho}\leadsto f$. By  [ADH, 16.0.3, Section 11.4],  $(f_{\rho})$ is of $\d$-transcendental type over $H$.
If the sequence $\big(v(f-f_{\rho})\big)$ is cofinal in $\Gamma:=v(H^\times)$, then~$(f_{\rho})$
is a cauchy sequence, and so $H$ is indeed dense in $H\<f\>$, by \cite[Corollary 4.1.6]{ADH2}.
Suppose~$\big(v(f-f_{\rho})\big)$ is not cofinal in $\Gamma$. Then we have $h\in H^\times$ such that $0\notin v({hf-H})$, so
Corollary~\ref{apomega} and Proposition~\ref{propimmy2}   yield an
$H$-hardian pseudolimit of $(hf_{\rho})$ in~$\Cc^{\omega}$, contradicting the maximality of $H$. This proves Claim~1.

\claim[2]{For any Hardy field extension $K$ of $H$ we have $\Gamma_K=\Gamma$.}

\noindent
Towards a contradiction, suppose
$K$ is a Hardy field extension of $H$ and~${\beta\in \Gamma_K\setminus\Gamma}$. We arrange that $K$ is Liouville closed.
Let $(\Gamma,\psi)$ and $(\Gamma_K, \psi_K)$ be the $H$-couples of~$H$ and
$K$ over $\R$, respectively, and let~$\big(\Gamma\<\beta\>,\psi_{\beta}\big)$ be the $H$-couple over
$\R$ generated by $\beta$ over $(\Gamma,\psi)$ in $(\Gamma_K, \psi_K)$. There are several cases to consider, and we show that each is impossible.

First the case that $\big(\Gamma\<\beta\>,\psi_{\beta}\big)$ is an immediate extension of $(\Gamma,\psi)$. Then we have a
divergent pc-sequence $(\gamma_{\rho})$ in $(\Gamma,\psi)$ with $\gamma_{\rho}\leadsto \beta$. As in the beginning of~\cite[Section~\ref{vg}]{ADHfgh} we take $g_{\rho}\in H$ with $vg_{\rho}=\gamma_{\rho}$ so that $(g_{\rho}^\dagger)$ is a pc-sequence in~$H$, and arguing as in~loc.~cit.~(using $H^\dagger=H$) we see that $(g_{\rho}^\dagger)$ has no pseudolimit in $H$ (because then
$(\gamma_{\rho})$ would have one in $\Gamma$). Now take $g\in K$ with $vg=\beta$. Then~$v(g^\dagger-g_{\rho}^\dagger)=(\beta-\gamma_{\rho})^\dagger$, and the latter is eventually strictly increasing as a function of
$\rho$, and so $g_{\rho}^\dagger\leadsto g^\dagger$. Moreover, $\big(v(g^\dagger-g_{\rho}^\dagger)\big)$ is not cofinal in
$\Gamma$.  As at the end of the proof of Claim 1, with $g^\dagger$ and $(g_{\rho}^\dagger)$ in the role of $f$ and  $(f_{\rho})$, this contradicts the maximality assumption on $H$.

Now by \cite[Proposition~4.1]{ADH3} and the remark following its proof, the vector $\beta$ falls under Case (a), or Case (b), or Case
${\rm(c)}_n$ for a certain $n$. In Case (a) we have~$(\Gamma+\R\beta)^\dagger=\Gamma^\dagger$ and so $\Gamma\<\beta\>=\Gamma+\R\beta$; but $(\Gamma_K, \psi_K)$ is of Hahn type, hence~$\big(\Gamma\<\beta\>,\psi_{\beta}\big)$ is an immediate extension of $(\Gamma,\psi)$, and we have just excluded that possibility. Case ${\rm(c)}_n$ gives an element $\beta_n\in \Gamma\<\beta\>$ with
$\beta_n^\dagger\notin \Gamma$ and $\beta_n^\dagger$ falling under Case (a), and so this is also impossible.

Finally, suppose $\beta$ falls under Case (b). Take $y\in K^{>}$ with $vy=\beta$. Then  Corollary~\ref{cor:revb}  gives an $H$-hardian $z\in \Cc^{\omega}$ with $vz$ realizing the same cut in $\Gamma$ as $\beta$,  contradicting the maximality assumption on $H$. This finishes the proof of Claim 2.

\medskip\noindent
To finish the proof of the theorem, let $f\in \Cc^{<\infty}$ be $H$-hardian; it suffices to show that then $H$ is dense in $H\<f\>$.
Now by Claim 2, $H\<f\>$ is an immediate extension of $H$, and hence $H$ is indeed dense in $H\<f\>$ by Claim 1.
\end{proof}

\begin{question}
Is every maximal Hardy field dense in every Hausdorff field extension?
\end{question}

\subsection*{Dense pairs of closed $H$-fields} Let~$\mathcal L=\{0,1,{-},{+},{\,\cdot\,},{\der},{\leq},{\preceq}\}$ be the language of ordered valued differential rings; cf.~[ADH, p.~678]. We view
each ordered valued differential field as an $\mathcal L$-structure in the natural way.
We let
$\mathcal L^2$ extend $\mathcal L$ by a new unary predicate symbol~$U$.
The $\mathcal L^2$-structures
are presented as pairs~$(K,F)$ where~$K$ is an $\mathcal L$-structure and $U$ names the subset $F$ of $K$.
%is the interpretation of~$U$.
Let $T$ be the $\mathcal L$-theory of closed $H$-fields with small derivation.
%(Everything below also works if ``small'' is replaced by ``not small''.)
Recall from [ADH]  that $T$ is complete and model-complete.

\begin{theorem}\label{densepairs}
The following requirements on  $\mathcal L^2$-structures $(K,F)$
axiomatize a complete $\mathcal L^2$-theory $T^{\operatorname{d}}$:
\begin{enumerate}
\item[\textup{(1)}] $K\models T$, that is, $K$ is a closed $H$-field with small derivation;
\item[\textup{(2)}] $F$ is the underlying set of a closed $H$-subfield of $K$; and
\item[\textup{(3)}] $F\neq K$ and $F$ is dense in the ordered field $K$.
\end{enumerate}
Moreover, each $\mathcal L^2$-formula $\varphi(x)$ where $x=(x_1,\dots,x_m)$ is $T^{\operatorname{d}}$-equivalent
to a boolean combination of formulas of the form
\begin{equation}\label{eq:special fm}
\exists y_1\cdots\exists y_n\big( U(y_1)\ \&\ \cdots\ \&\ U(y_n)\ \&\ \psi(x,y)\big)
\end{equation}
where $\psi(x,y)$ with $y=(y_1,\dots,y_n)$ is an $\mathcal L$-formula.
\end{theorem}

\noindent
This follows from Fornasiero's \cite[Theorems~8.3 and 8.5]{Forn}, with  details of how it follows to appear in~[9].
We just note here that by this theorem the $\mathcal L^2$-theory~$T^{\operatorname{d}}$ is decidable. Moreover,  no pair $(K,F)\models T^{\operatorname{d}}$ induces ``new   structure'' on~$F$:

\begin{cor}\label{cor:forn}
Let $(K,F)\models T^{\operatorname{d}}$, and
let $S\subseteq K^m$ be $A$-definable in $(K,F)$, where~$A\subseteq F$.
Then $S\cap F^m$ is $A$-definable in the $\mathcal{L}$-substructure $F$ of $K$.
\end{cor}
\begin{proof}
By the theorem this reduces to the case where $S$ is defined in $(K,F)$
by a formula
as in \eqref{eq:special fm} where however $\psi(x,y)$ is now an $\mathcal L_A$-formula.
Then $S\cap F^m$ is defined in $F$ by the $\mathcal L_A$-formula $\exists y\psi(x,y)$.
\end{proof}

\noindent
Note that if $M$ is a maximal analytic or maximal smooth Hardy field and $N$ a maximal Hardy field with $M\subseteq N$, $M\neq N$, then $(N,M)\models T^{\operatorname{d}}$ by Theorem~\ref{thm:dense}. But strictly speaking, we do not know whether there exist such $M$, $N$.

To secure a model of the complete theory $T^{\operatorname{d}}$ we proceed as follows. Let $F$ be an $H$-field. Then the completion $F^{\operatorname{c}}$  of the ordered valued differential field~$F$
is an $H$-field extension of $F$, and
$F$ is dense in~$F^{\operatorname{c}}$; see [ADH, 10.5.9].
If $F$ is  closed and of countable cofinality, then  $F^{\operatorname{c}}$ is closed, by~[ADH, 14.1.6], so
if in addition~$F$  has small derivation and~$F\neq F^{\operatorname{c}}$, then~$(F^{\operatorname{c}},F)\models T^{\operatorname{d}}$.
Now $\T$ is not complete: set~$\ex_0=x$ and $\ex_{i+1}=\exp\ex_i$ for all~$i$; then $\big(\sum_{i=0}^n 1/{\ex_i}\big)_{n=0}^\infty$ is a cauchy sequence in~$\T$ but has no limit in $\T$.
Therefore $(\T^{\operatorname{c}},\T)\models T^{\operatorname{d}}$.

\section{Analytic Hardy Fields of Countable Cofinality}

\noindent
Generalizing terminology introduced in \cite[Section~8]{ADHfgh},  call a valued abelian group
{\bf countably spherically complete} if every pc-sequence in it of length~$\omega$ pseudoconverges in it.
Any $\eta_1$-ordered abelian group with a convex valuation is countably spherically complete, by [ADH, 2.4.2].
Thus maximal analytic  and maximal smooth Hardy fields are countably spherically complete.
In the  subsection below we use this fact to realize the completion of an analytic
Hardy field of countable cofinality as an analytic Hardy field.

\subsection*{Completing analytic Hardy fields of countable cofinality}
Let $K$ be an asymptotic field.  Equip  the completion $K^{\operatorname{c}}$ of the valued field~$K$ with the unique extension of the derivation of $K$
to a continuous derivation on~$K^{\operatorname{c}}$; cf.~[ADH, 4.4.11, 9.1.5].
Then~$K^{\operatorname{c}}$ is asymptotic by [ADH,  9.1.6], and if $K$ is a pre-$H$-field ($H$-field, respectively), then so is~$K^{\operatorname{c}}$ by [ADH, 10.5.9].
Let $L$ be an asymptotic field  extension of $K$  such that
$\Gamma$ is cofinal in $\Gamma_L$. By [ADH, 3.2.20], the natural inclusion~${K\to L}$ extends uniquely to an embedding $K^{\operatorname{c}}\to L^{\operatorname{c}}$
of valued fields, and it is easily checked that this  is an embedding of valued {\it differential}\/ fields.
%(Cf.~\cite[remarks before Corollary~4.2.6]{ADH2} for an analogue of these remarks for  valued differential fields with small derivation.)
If $K$ is dense in $L$, then there is a unique valued field embedding $L\to K^{\operatorname{c}}$
over $K$, by [ADH, 3.2.13], and this is also an embedding of  valued  differential  fields.

\begin{lemma}\label{lem:Kc->M}
Let $K$ be an $\upo$-free  $H$-asymptotic field whose value group  $\Gamma_K$
%is divisible and
has countable cofinality. Let $M$ be a newtonian  $H$-asymptotic field with asymptotic integration, and suppose $M$ is countably spherically complete. Then  any embedding~$K\to M$ extends to an embedding $K^{\operatorname{c}}\to M$.
\end{lemma}
\begin{proof}
Let $\iota\colon K\to M$ be an embedding; we need to extend $\iota$ to an embedding~${K^{\operatorname{c}}\to M}$.
The $\d$-valued hull  $L:=\operatorname{dv}(K)$ of $K$  is $\upo$-free by [ADH, remark after 13.6.1], and $\Gamma_L=\Gamma_K$
by [ADH, 10.3.2(i)]. By [ADH, 14.2.5], $M$ is $\d$-valued; let
$\iota_L$ be the extension of $\iota$ to an embedding $L\to M$.
Using a remark before the lemma we see that it is enough to show that
$\iota_L$ extends to an embedding $L^{\operatorname{c}}\to M$.
replacing $K$, $\iota$ by $L$, $\iota_L$,  we arrange   $K$ is $\d$-valued. Take an immediate asymptotically $\d$-algebraically maximal $\d$-algebraic extension $L$ of $K$; by a remark following the statement of
[ADH, Theorem 14.0.1] such $L$ exists and is $\upo$-free and newtonian.  Then $L$ is a newtonization of $K$ by~\cite[Theorem~3.5]{Nigel19}, so embeds into~$M$ over $K$. Passing to this newtonization
we  arrange that $K$ is
%and asymptotically $\d$-algebraically maximal.
Then $K^{\operatorname{c}}$ is
$\upo$-free by [ADH, 11.7.20] and
newtonian by [ADH, 14.1.5].

%Zorn now yields a maximal asymptotic subfield $K_1$ of~$K^{\operatorname{c}}$  which contains $K$ and an extension of $\iota$ to an embedding $\iota_1\colon K_1\to M$.
%Then $K_1$ is newtonian, and since
%there is  a unique isomorphism
%$K_1^{\operatorname{c}} \to K^{\operatorname{c}}$ over~$K_1$ by the remark before the lemma and [ADH,  3.2.16], we can   replace~$K$, $\iota$ by~$K_1$, $\iota_1$, respectively,
%to arrange that $\iota$ does not extend to an embedding of an  asymptotic subfield of
%$K^{\operatorname{c}}$,   properly containing~$K$,  into~$M$. We may then also arrange that $K$ is an asymptotic subfield of~$M$
%and $\iota$ is the natural inclusion $K\to M$.

%Towards a contradiction
Suppose $f\in K^{\operatorname{c}}\setminus K$. It suffices to show that then $\iota$ extends to an embedding~$\iota_f\colon K\<f\>\to M$. Here is why:  $K\<f\>$ is $\upo$-free by the remark before~[ADH, 11.7.20], so $K\<f\>$ has a newtonization $E$ in $K^{\operatorname{c}}$ by \cite[Theorem~B]{Nigel19}; by the same remark $E$ is $\upo$-free; moreover, $\iota_f$ extends to an embedding~$E\to M$.
Hence we can transfinitely iterate this extension process to obtain an embedding~$K^{\operatorname{c}}\to M$ extending $\iota$.

To construct $\iota_f$,
take a c-sequence~$(f_\rho)$ in~$K$ with $f_\rho\to f$ (in $K^{\operatorname{c}}$).
By [ADH, 2.2.25] the index set of $(f_\rho)$ has cofinality~$\omega$, so
%and by [ADH, 2.2.35], $(f_\rho)$ has a cofinal subsequence which is a pc-sequence in~$K$ of width~$\{\infty\}$.
by passing to a cofinal subsequence  we arrange~$(f_\rho)$ is a divergent pc-sequence in $K$ of length $\omega$ and width $\{\infty\}$ such that~$f_\rho\leadsto f$. Take $g\in M$ such that $\iota(f_\rho)\leadsto g$.
Now $K$ is asymptotically $\d$-algebraically maximal by \cite[Theorem~A]{Nigel19}, so
$(f_\rho)$ is of $\d$-transcendental type over~$K$ by~[ADH, 11.4.8, 11.4.13], hence
%Thus~$(f_\rho)$ is of transcendental type over $K$, so for each~$h\in K$ we have $v(f-h)=v(g-h)$; cf.~[ADH, 3.2].
[ADH, 11.4.7] yields  an embedding~$K\langle f\rangle\to M$ extending $\iota$ and sending
$f$ to $g$.
\end{proof}

%The hypothesis that $K$ has divisible value group in Lemma~\ref{lem:Kc->M} may be dropped, using \cite[Theorems~A, B]{Nigel19}
%instead of [ADH, 14.0.2, 14.5.4] in the proof.

\noindent
Lemma~\ref{lem:Kc->M} yields a   pre-$H$-field version  of it  without the $\upo$-free hypothesis
on $K$:

\begin{prop}\label{prop:completion Hardy field}
Let $K$ be a pre-$H$-field with $\operatorname{cf}(\Gamma_K)=\omega$ and $M$ a  countably spherically complete closed $H$-field.
Then every embedding $K\to M$ extends to  an embedding~$K^{\operatorname{c}}\to M$.
\end{prop}

\noindent
For this we arrange that $M$ extends $K$  and then have to find an
embedding~$K^{\operatorname{c}}\to M$ over $K$.
Take any expansion $\mathbf M$ of $M$ to a $\HLO$-field
and expand $K$ to  a pre-$\HLO$-field~$\mathbf K$ such that $\mathbf K\subseteq\mathbf M$, see~[ADH, 16.3.21].  Then the proposition below applied
to $\mathbf M$ in place of $\mathbf L$ yields an $\upo$-free $H$-field extension $K^*$ of~$K$ such that $K$ is cofinal in $K^*$ and an embedding $\iota^*\colon K^*\to M$ over $K$.
Lemma~\ref{lem:Kc->M} gives an extension of $\iota^*$ to an embedding $(K^*)^{\operatorname{c}}\to M$,
and by a remark before that lemma this yields an embedding $K^{\operatorname{c}}\to M$ as required.

\medskip
\noindent
It remains to establish the following ``cofinality" refinement of [ADH, 16.4.1]:

\begin{prop}\label{1641}
Let $\mathbf K$ be a pre-$\HLO$-field with $\Gamma_{\mathbf K}\neq\{0\}$. Then there exists an $\upo$-free  $\HLO$-field extension $\mathbf K^*$
of $\mathbf K$ such that;
\begin{enumerate}
\item[(i)] $\res\mathbf K^*$ is algebraic over $\res\mathbf K$;
\item[(ii)] $\mathbf K$ is cofinal in $\mathbf K^*$; and
\item[(iii)] any embedding of $\mathbf K$ into a Schwarz closed $\HLO$-field $\mathbf L$ extends to
an embedding $\mathbf K^*\to\mathbf L$.
\end{enumerate}
\end{prop}

\noindent
We revisit the proof of [ADH, 16.4.1], which consists of several lemmas and a corollary. Let
$\mathbf K=(K,I,\Lambda,\Omega)$ be a pre-$\HLO$-field with $\Gamma:=\Gamma_K\neq\{0\}$.

\begin{lemma}
Suppose $K$ is grounded, or there exists $b\asymp 1$ in $K$ such that $v(b')$ is a gap in $K$. Then $\mathbf K$
has an $\upo$-free $\HLO$-field extension $\mathbf K^*$ such that $\res\mathbf K=\res\mathbf K^*$, $\mathbf K$ is cofinal in $\mathbf K^*$, and any embedding
of $\mathbf K$ into a $\HLO$-field $\mathbf L$ closed under logarithms extends to an embedding $\mathbf K^*\to\mathbf L$.
\end{lemma}
\begin{proof}
By Lemma~\ref{lem:cof 1}, $K$ is cofinal in the $H$-field hull $F:=H(K)$ of $K$,
and hence by Lemma~\ref{lem:H(x)}, $K$ is also cofinal in the $H$-field extension $F_{\upo}$ of $F$ constructed in~[ADH, 11.7].
Thus the lemma follows from the proof of [ADH, 16.4.2].
\end{proof}

\begin{lemma}
Suppose $K$ has gap $\beta$ and $v(b')\neq\beta$ for all $b\asymp 1$ in $K$. Then there exists a grounded pre-$\HLO$-field
extension $\mathbf K_1$ of $\mathbf K$ such that $\res\mathbf K=\res\mathbf K^*$, $\mathbf K$ is cofinal in $\mathbf K_1$, and any embedding of $\mathbf K$ into a $\HLO$-field $\mathbf L$ closed under
integration extends to an embedding $\mathbf K_1\to\mathbf L$.
\end{lemma}
\begin{proof}
Take $s\in K$ with $vs=\beta$. Following the proof of [ADH, 16.4.3],
suppose~${s\notin \I(K)}$, and take $K_1$ as in Case~1 of that proof, so
$K_1=H(K)(y)$ where~$y'=s$. Now use that $\Gamma_{H(K)}=\Gamma$, and that $\Gamma_{H(K)}$ is cofinal in $\Gamma_{H_1}$ by Lemma~\ref{lem:H(x)}.
%K$ is cofinal in $H(K)$ since $H(K)$ is an immediate extension of $K$.
If~${s\in \I(K)}$ and~$K_1$ is as in Case~2, then $K_1=K(y)$ where $y'=s$, so again  $K$ is cofinal in $K_1$ by
\end{proof}

\noindent
These two lemmas yield a ``cofinality'' refinement of [ADH, 16.4.4]:

\begin{cor}\label{cor:16.4.4}
Suppose $K$ does not have asymptotic integration. Then $\mathbf K$ has
an $\upo$-free $\HLO$-field extension $\mathbf K^*$ such that
$\res\mathbf K^* = \res\mathbf K$, $\mathbf K$ is cofinal in $\mathbf K^*$, and and any embedding of $\mathbf K$
into a $\HLO$-field $\mathbf L$ closed under integration extends to an embedding $\mathbf K^*\to\mathbf L$.
\end{cor}

\noindent
The next three lemmas are ``cofinality''  refinements of [ADH, 16.4.5, 16.4.6, 16.4.7] and take care of the case where $K$ has asymptotic integration.

\begin{lemma}\label{lem:16.4.5}
Assume $K$ has asymptotic integration and is not $\upl$-free. Then~$\mathbf K$
extends to an $\upo$-free $\HLO$-field $\mathbf K^*$ such that
$\res\mathbf K^* = (\res\mathbf K)^{\operatorname{rc}}$,  $\mathbf K$ is cofinal in~$\mathbf K^*$, and any embedding
of $\mathbf K$ into a Liouville closed $\HLO$-field $\mathbf L$ extends to an embedding~$\mathbf K^*\to\mathbf L$.
\end{lemma}
\begin{proof}
As in the proof of [ADH, 16.4.5], it is enough, by Corollary~\ref{cor:16.4.4},  to show that~$\mathbf K$ has a $\HLO$-field extension
$\mathbf K_1=(K_1,\dots)$ with a gap such that  $\res\mathbf K_1 = (\res\mathbf K)^{\operatorname{rc}}$,  $K$ is cofinal in $K_1$, and any embedding
of $\mathbf K$ into a Liouville closed $\HLO$-field $\mathbf L$ extends to an embedding $\mathbf K_1\to\mathbf L$.
Take $\mathbf K_1$ as in the proof of [ADH, 16.4.5].
Put $E:=H(K)^{\operatorname{rc}}$. Then $\Gamma_{E}=\Q\Gamma$, so $K$ is cofinal in $E$.
If $E$ has a gap, then~$K_1=E$, and we are done.
Suppose $E$ has no gap. Then $K_1=E(f)$ where~$f\in K_1^\times$ and $\upl:=-f^\dagger\in K$, and $s:=-\upl$ creates a gap over $E$.
By the proof of Case~2 in~[ADH, 16.4.5], the hypothesis of
Lemma~\ref{lem:10.4.5} holds for $E$ in place of~$K$, so~$E$ is cofinal in $K_1$, and hence so is $K$.
\end{proof}

\begin{lemma}
Suppose $K$ is $\upl$-free but not $\upo$-free. Then $\mathbf K$ has an $\upo$-free $\HLO$-field
extension $\mathbf K^*$ such that $\res\mathbf K^*$ is algebraic over $\res\mathbf K$, $\mathbf K$ is cofinal in $\mathbf K^*$,
and any embedding of $\mathbf K$ into
a Schwarz closed $\HLO$-field $\mathbf L$ extends to an embedding of~$\mathbf K^*$ into $\mathbf L$.
\end{lemma}
\begin{proof}
Take $\upo\in K$ with $\upo_\rho\leadsto\upo$. Let $\mathbf K^*$ be as in the proof of [ADH, 16.4.6].
That proof shows that  $\Omega=\omega(K)^\downarrow$ or $\Omega=K\setminus\sigma\big(\Upg(K)\big){}^\uparrow$.
Suppose first that~$\Omega=\omega(K)^\downarrow$. With $\mathbf K_{\upg}=(K_{\upg},\dots)$ as in Case~1 of that proof, we have
$K_{\upg}=K\langle\upg\rangle$ where~$\upg\neq 0$, $\sigma(\upg)=\upo$, and $v\upg$ is a gap in $K_{\upg}$.
%$\Psi < v\upg < (\Gamma^>)'$.
The remarks before [ADH,  13.7.7] give $[\Gamma]=\big[\Gamma_{K_{\upg}}\big]$, so $K$ is cofinal
in $K_{\upg}$. Now follow the argument in Case~1 of loc.~cit., using Corollary~\ref{cor:16.4.4} instead of [ADH, 16.4.4].
If~$\Omega=K\setminus\sigma\big(\Upg(K)\big){}^\uparrow$, then we argue as in Case~2 of loc.~cit.,
using Lemma~\ref{lem:16.4.5} instead of [ADH, 16.4.5].
\end{proof}

\begin{lemma}
Suppose $K$ is $\upo$-free. Then $\mathbf K$ has an $\upo$-free $\HLO$-field extension $\mathbf K^*$ such that
$\res\mathbf K^* = \res\mathbf K$, $\mathbf K$ is cofinal in $\mathbf K^*$, and any embedding of $\mathbf K$ into a $\HLO$-field $\mathbf L$ extends to an embedding of $\mathbf K^*$ into $\mathbf L$.
\end{lemma}
\begin{proof}
Take $\mathbf K^*=(K^*,\dots)$ as in the proof of [ADH, 16.4.7]. Then $K^*=H(K)$, and
by Lem\-ma~\ref{lem:cof 1}, $K$ is cofinal in $H(K)$.
\end{proof}

\noindent
%Finally, suppose $\mathbf K$ is an  $\upo$-free $\HLO$-field. Then $K^{\operatorname{rc}}$ is
% $\upo$-free by [ADH, 11.7.23], and $K$ is cofinal in $K^{\operatorname{rc}}$, as
% each ordered field is cofinal in its real closure, by  [ADH, 3.5.2].
%The unique expansion $\mathbf K^{\operatorname{rc}}$ of
%$K$ to a $\HLO$-field embeds  over $K$ into every real closed $\HLO$-field extension of $\mathbf K$.
This concludes the proof of  Proposition~\ref{1641} and of Proposition~\ref{prop:completion Hardy field}. Combining the latter
with  \cite[Corollary~\ref{corimmom}]{ADHfgh}  and Corollary~\ref{smanpc} yields:

\begin{cor}\label{cor:completion Hardy field}
Let $H$ be a  Hardy field of countable cofinality and $M\supseteq H$ a maximal Hardy field.
Then there is an embedding $H^{\operatorname{c}}\to M$ over~$H$. Likewise if~$M\supseteq H$ is a maximal analytic Hardy field or a maximal smooth Hardy field.
\end{cor}

\noindent
As a consequence of Corollary~\ref{cor:completion Hardy field}, with $t:=x^{-1}$,
each maximal analytic Hardy field contains a Hardy field extending $\R(t)$ and isomorphic over~$\R(t)$ to the ordered   field~$\R(\!(t)\!)$ of Laurent series over $\R$ equipped with the continuous $\R$-linear derivation given by $t'=-t^2$.
(This may be viewed as a Hardy field version of Besicovitch's strengthening [12] of
Borel's theorem on $\Cc^\infty$-functions with prescribed Taylor series~[14].)
In Corollary~\ref{cor:embed T}   we show that even the ordered differential field~$\T$ of transseries, which vastly extends $\R(\!(t)\!)$, embeds into any given maximal analytic Hardy field. As a first step
we accomplish this below for
the $H$-subfield~$\T_{\log}$ of~$\T$ of logarithmic transseries (cf.~[ADH, p.~722]). For this it is useful to
have available some facts about short ordered sets, also needed in Section~\ref{eie}.

\subsection*{Short ordered sets}
Let $S$ be an ordered set. (As in [8], this means ``linearly ordered set''.)
Let~$S^*$ denote
$S$ equipped with the reversed ordering.
Then the following are equivalent:
\begin{enumerate}
\item[(S1)]   all well-ordered subsets of $S$
and of $S^*$ are countable;
\item[(S2)]  there are no embeddings of $\omega_1$ into $S$ or $S^*$;
\item[(S3)] $\operatorname{cf}(A),\operatorname{ci}(A) \leq\omega$ for all ordered subsets $A$ of $S$.
\end{enumerate}
Call  $S$ {\bf short} if any of the equivalent conditions (S1)--(S3) holds; cf.~\cite[1.7(i)]{DW}  and~\cite[pp.~88, 170--171]{Rosenstein}.
If  $S$ is short, then so are~$S^*$ and every ordered subset of~$S$. If~$S$ is countable, then it is short; more generally, if
$S$ is a union of countably many short ordered subsets, then $S$ is short.
If $S\to S'$ is a surjective increasing map between ordered sets and $S$ is short, then so is $S'$; similarly with ``decreasing'' instead of ``increasing''.  Shortness   enters our story via the following observation:

\begin{lemma}\label{lem:short}
Let $(G,S,v)$ be a valued abelian group where $S$ is short, and let $(a_\rho)$ be
a  pc-sequence in $(G,S,v)$. Then some final segment of $(a_\rho)$ has countable length.
\end{lemma}
\begin{proof}
Put $s_\rho:=v(a_{\rho+1}-a_\rho)$, where~$\rho+1$ is the successor of $\rho$.
After deleting an initial segment of  $(a_\rho)$  we arrange that the sequence~$(s_\rho)$ in $S$
is strictly increasing. Then the  image of the index set of $(a_\rho)$ under the embedding $\rho\mapsto s_\rho$ of ordered sets is a well-ordered subset of $S$ and hence countable.
\end{proof}

\begin{lemma}
If the order topology of $S$ is second countable, then $S$ is short.
\end{lemma}
\begin{proof}
Suppose $i\colon \omega_1\to S$ is strictly increasing.
With $\lambda$ ranging over the limit ordinals~$<\omega_1$ we then have uncountably many nonempty pairwise disjoint open intervals $\big(i(\lambda),i(\lambda+2)\big)$
in $S$, so $S$ is not second countable. An embedding $\omega_1\to S^*$ yields the same conclusion.
\end{proof}

\noindent
In particular, the real line (the ordered set of real numbers) is short.
(In fact,
by~\cite[Theorem~2]{HarringtonShelah},   each Borel ordered set is short.)
The following observation  is due to
Hausdorff~{\cite[p.~133]{Hau07}} and Urysohn~[46].

\begin{lemma}\label{lem:embed short into eta1}
Suppose $S$ is short and $T$ is an $\eta_1$-ordered set.  Then any embedding of an ordered subset of $S$ into $T$ extends to an embedding~${S\to T}$. In particular, there exists an embedding $S\to T$.
\end{lemma}
\begin{proof}
Let $A$ be an ordered subset of $S$ and $i\colon A\to T$ an embedding. Suppose~$s\in S\setminus A$.
Then $\operatorname{cf}(A^{<s}),\operatorname{ci}(A^{>s})\leq\omega$, so
we have $t\in T$ with $i(A^{<s})<t<i(A^{>s})$. Thus~$i$ extends to an
embedding $A\cup\{s\}\to T$ sending $s$ to $t$. Zorn does the rest.
\end{proof}

\begin{cor}\label{lem:eta1 size}
Every $\eta_1$-ordered set has cardinality $\geq\frak c$.
There is an~$\eta_1$-ordered set of cardinality $\frak c$.
\end{cor}
\begin{proof} For the first claim, apply Lemma~\ref{lem:embed short into eta1} to $S=\text{the real line}$.
The second claim follows from the first together with
[ADH, B.9.6]. %  or with Corollary~\ref{cor:dBRH} below (and Zorn).
\end{proof}

\noindent
Combining Lemma~\ref{lem:embed short into eta1} and Corollary~\ref{lem:eta1 size} we obtain:

\begin{cor}[Urysohn~[45, 46]]
Every short ordered set has cardinality $\leq\frak c$.
\end{cor}

\noindent
For (ordered)  Hahn products, see [ADH, 2.2, 2.4]. Shortness of $\R$ is at the root of the following result due  to Esterle
\cite[Lemme~2.2 and the remark after it]{Esterle}:

\begin{lemma}\label{lem:Esterle}
If $S$ is short, then so is the Hahn product $H[S,\R]$.
\end{lemma}

\noindent
From Lemma~\ref{lem:Esterle} and the Hahn Embedding Theorem  [ADH, 2.4.19] we obtain a characterization of short ordered abelian groups:

\begin{lemma}\label{lem:short char}
For an ordered abelian group $\Gamma$, the following are equivalent:
\begin{enumerate}
\item[\textup{(i)}] $\Gamma$ is short;
\item[\textup{(ii)}] the ordered set $[\Gamma]$ is short;
\item[\textup{(iii)}] $\Gamma$ embeds into $H[S,\R]$ for some short $S$.
\end{enumerate}
\end{lemma}

\begin{cor}\label{cor:short oag}
Let $\Delta\subseteq\Gamma$ be an extension of ordered abelian groups. Then
$$\text{$\Gamma$ is short} \quad\Longleftrightarrow\quad \text{$\Delta$ and $[\Gamma]\setminus[\Delta]$ are short.}$$
In particular,  if $\operatorname{rank}_{\Q}(\Gamma/\Delta)\le \aleph_0$, then  $\Gamma$ is short iff $\Delta$ is short,
and if $\Delta$ is convex, then $\Gamma$ is short iff $\Delta$ and $\Gamma/\Delta$ are short.
\end{cor}
\begin{proof} The direction $\Rightarrow$ is clear from Lemma~\ref{lem:short char}. For the converse, note that if
$\Delta$ and ${[\Gamma]\setminus [\Delta]}$ are short, then so is $[\Gamma]$, and hence $\Gamma$ as well by  Lemma~\ref{lem:short char}. Next, use that if~$\operatorname{rank}_{\Q}(\Gamma/\Delta)\le \aleph_0$, then $[\Gamma]\setminus [\Delta]$ is countable by [ADH, 2.3.9]. For convex~$\Delta$, see [ADH, p.~102].
\end{proof}

\begin{lemma}\label{lem:short ofield}
Let $K$ be an ordered field equipped with a convex valuation whose residue field is archimedean.
Then $K$ is short iff its value group $\Gamma$ is short.
\end{lemma}
\begin{proof}
Suppose $\Gamma$ is short. Then $\Q\Gamma$ is also short, by the previous corollary,
and the real closure of $\res K$ remains archimedean;
hence to show that $K$ is short we may replace $K$ by its real closure to arrange that $K$ is real closed.
Using [ADH, 3.3.32, 3.3.42, 3.5.1, 3.5.12] we obtain an ordered  field embedding of $K$ into the ordered Hahn field $\R(\!(t^\Gamma)\!)$.
The underlying ordered additive group of $\R(\!(t^\Gamma)\!)$ is isomorphic with the ordered
Hahn product~$H[t^{\Gamma},\R]$; see~[ADH, p.~114].
Hence   $K$ is short by Lemma~\ref{lem:short char}. Conversely, if $K$ is short then so is its ordered subset $K^>$
and then also the image $\Gamma$ of $K^>$ under the decreasing map~$f\mapsto vf\colon K^>\to\Gamma$.
\end{proof}

\noindent
Hence if an ordered field is short, then so is its real closure.
If $K$  as in Lemma~\ref{lem:short ofield} is short and $L$ is an ordered field extension of $K$ with a convex valuation
that makes it an immediate  extension of $K$, then $L$ is short.
(NB: the ordered fraction field of a short ordered integral
domain may fail to be short [18, 19].)
The following is from  \cite[\S{}2.10]{vdDMM}:

\begin{cor}\label{cor:T short}
$\T$ is short.
\end{cor}
\begin{proof}
We recall some features of the construction of $\T$ from [ADH, Appendix~A].
We have the ordered subfield
$\T_{\exp}=\bigcup_m E_m$ of $\T$ where $E_m=\R[[G_m]]$ for certain ordered subgroups $G_m$ of $\T^>$,
with~$G_0=x^{\R}$ and $G_{m+1}=G_m\exp(A_m)$ for some
subgroup $A_m$ of the additive group of~$E_m$, with $G_m$ a convex subgroup of $G_{m+1}$.
An easy induction on  $m$   shows that each $E_m$ is short, and
thus $\T_{\exp}$ is short.
Now $\T=\bigcup_n (\T_{\exp}){\downarrow^n}$ where $f\mapsto f{\downarrow^n}$ is the $n$th compositional iterate of
the automorphism $f\mapsto f{\downarrow}=f\circ\log x$ of the ordered field $\T$, hence
$\T$ is also short.
\end{proof}

\begin{question}
Is every $\d$-algebraic $H$-field extension of a short $H$-field also short?
\end{question}

\noindent
Next two algebraic variants of Lemma~\ref{lem:embed short into eta1}, attributed to Esterle in \cite[2.37]{DW}:

\begin{lemma}\label{shortetaab}
Let $\Delta$ be a short ordered abelian group and $\Gamma$ a divisible $\eta_1$-ordered abelian group.
Then any  embedding of an ordered subgroup of $\Delta$ into $\Gamma$ extends to an embedding~${\Delta\to \Gamma}$.
%In particular, there exists an embedding $\Delta\to\Gamma$.
\end{lemma}
\begin{proof} Let $\Delta_0$ be an ordered subgroup of $\Delta$ and $i\colon \Delta_0\to \Gamma$
an embedding.
The divisible hull $\Q\Delta\subseteq \Gamma$ of $\Delta$ is   short, by Corollary~\ref{cor:short oag}. Replace~$\Delta_0$, $\Delta$ by~$\Q\Delta_0$,~$\Q\Delta$ (and $i$ accordingly) to arrange   $\Delta_0$, $\Delta$ to be divisible.
Given~${\delta\in \Delta\setminus \Delta_0}$,
Lemma~\ref{lem:embed short into eta1} yields $\gamma\in\Gamma$
with~$i(\Delta_0^{<\delta})<\gamma<i(\Delta_0^{>\delta})$, and
then $i$ extends to an embedding of the ordered subgroup $\Delta_0\oplus\Q\delta$ of $\Delta$ into $\Gamma$  sending $\delta$ to $\gamma$. Zorn does the rest.
\end{proof}

\noindent
In the same way, taking real closures instead of divisible hulls in the proof:

\begin{lemma}\label{lem:embed short ordered fields} Any embedding of an ordered subfield of a short ordered field $K$ into
a real closed $\eta_1$-ordered field $L$ extends to an embedding $K\to L$.
%Let $K$ be a short ordered field and $L$ a real closed $\eta_1$-ordered field.
%Then any embedding of an ordered subfield of $K$ into $L$ extends to an embedding $K\to L$.
%In particular, there exists an embedding $K\to L$.
\end{lemma}
%\begin{proof} Like that of Lemma~\ref{shortetaab}, taking real closures in $L$ instead of divisible hulls.
%Let $i\colon F\to L$ be an embedding of an ordered subfield $F$ of $K$ into $L$.
%Similarly to the proof of the previous lemma, using the remark after the proof of Lemma~\ref{lem:short ofield}, we first arrange that    $F$, $K$ are real closed. Suppose $y\in K\setminus F$. Then~$y$ is transcendental over $F$, and we can take $z\in L$
%such that $i(F^{<y})<z<i(F^{>y})$. Then   $i$ extends uniquely to a field embedding $F(y)\to L$ with $y\mapsto z$.

\noindent
Combining Corollary~\ref{cor:T short} and the previous lemma yields:

\begin{cor}
The ordered field $\T$ embeds into each  real closed $\eta_1$-ordered field.
\end{cor}

\noindent
Lemma~\ref{lem:embed short H-fields} below is an analogue of Lemma~\ref{lem:embed short ordered fields} for $H$-fields   with small derivation.

%Pre-$\HLO$-fields with archimedean residue field can be made $\upo$-free while preserving shortness:

%Let $\mathbf K$ be a short pre-$\HLO$-field such that $\res\mathbf K$ is archimedean.  Then  there exists an $\upo$-free short $\HLO$-field extension $\mathbf K^*$ of $\mathbf K$ such that $\res\mathbf K^*$ is algebraic over $\res\mathbf K$ and  any embedding of $\mathbf K$ into a Schwarz closed $\HLO$-field $\mathbf L$ extends to an embedding~$\mathbf K^*\to\mathbf L$.
%With $\mathbf K^*$ as constructed in the proof of [ADH, 16.4.1], a routine verification shows that  $\operatorname{rank}_{\Q}(\Gamma_{K^*}/\Gamma)$ is countable, hence  $K^*$ is short by Corollary~\ref{cor:short oag} and Lemma~\ref{lem:short ofield}.

\subsection*{Realizing $\T_{\log}$ as an analytic Hardy field}
This uses the following variant of Lem\-ma~\ref{lem:Kc->M} for embedding $\upo$-free immediate extensions:

\begin{lemma}\label{lem:L->M}
Let $K$ be an $H$-asymptotic field with short value group,
$L$ an $\upo$-free   immediate extension of~$K$, and $M$ a
newtonian  $H$-asymptotic field with asymptotic integration. Suppose $M$ is countably spherically complete.
Then any embedding~$K\to M$ extends to an embedding~$L\to M$.
\end{lemma}
\begin{proof}
Let $\iota\colon K\to M$ be an embedding; we shall extend $\iota$ to an embedding ${L\to M}$. Now $L$ is pre-$\d$-valued by [ADH, 10.1.3], and as $\operatorname{dv}(L)$ is $\upo$-free by [ADH, remark after 13.6.1] and $\Gamma_{\operatorname{dv}(L)}=\Gamma$ by [ADH, 10.3.2(i)], we can replace $L$ by $\operatorname{dv}(L)$ to arrange that $L$ is $\d$-valued. Then $L$ has an immediate $\d$-algebraic newtonian
$\upo$-free extension by [ADH, remark after 14.0.1], which is then a newtonization of~$L$ by~\cite[Theorem~3.5]{Nigel19}. Replacing $L$ by this newtonization we also arrange that~$L$ is newtonian.
Using Zorn we further arrange that $\iota$ does not extend to any embedding into~$M$ of any valued differential subfield of $L$
properly containing $K$. Note that~$K$ is $\upo$-free by [ADH, remark preceding 11.7.20].
Now $M$ is $\d$-valued by [ADH, 14.2.5],
hence so is $K$ by the universal property of $\operatorname{dv}(K)$. Likewise, $K$ is newtonian, by the semiuniversal property
of the newtonization of $K$ (which exists by the same arguments as we used for $L$).
Hence  $K$ is asymptotically $\d$-algebraically maximal by~\cite[Theorem~A]{Nigel19}.
%We may also arrange $K$ to be an asymptotic subfield of $M$ and $\iota$ to be the
%natural inclusion~$K\to M$.
It remains to show that $K=L$. Suppose towards a contradiction that
$f\in L\setminus K$. Take a divergent pc-sequence
$(f_\rho)$ in~$K$ with pseudolimit~$f$. By Lemma~\ref{lem:short} we arrange that $(f_\rho)$ has
length~$\omega$, and hence we can take $g\in M$ with~$\iota(f_\rho)\leadsto g$.
As in the proof of Lemma~\ref{lem:Kc->M} we then obtain an  embedding~$K\langle f\rangle\to M$ extending $\iota$
and sending $f$ to $g$, a contradiction.
\end{proof}

%The hypothesis that $\Gamma$ is divisible in the previous lemma may be dropped; see the remark following the
%proof of Lemma~\ref{lem:Kc->M}.

\noindent
Set $\ell_0:=x\in\T$ and $\ell_{n+1}:=\log\ell_n$.
Recall that~$\T_{\log}=\bigcup_n \R[[\mathfrak L_n]]$ where $\mathfrak L_n:=\ell_0^{\R}\cdots\ell_n^{\R}$
is the subgroup of the monomial group $G^{\operatorname{LE}}$ of~$\T$ generated by the real powers of
the $\ell_i$ ($i=0,\dots,n$).
The ordered subgroup~$\mathfrak L:=\bigcup_n\mathfrak L_n$ of $G^{\operatorname{LE}}$  is divisible and short,
$\T_{\log}$ is real closed, $\upo$-free and an immediate $H$-field extension of its $H$-subfield $\R(\mathfrak L)$.
Identify   $\R(\mathfrak L)$  with an $H$-subfield of the analytic Hardy field~$\Li\!\big(\R(x)\big)$
in the obvious way.
From  \cite[Corollary~\ref{corimmom}]{ADHfgh},
Corollary~\ref{smanpc}, and Lemma~\ref{lem:L->M}, we obtain:

\begin{cor}\label{cor:Tlog}
The $H$-field $\T_{\log}$ embeds over $\R(\mathfrak{L})$ into any maximal Hardy field.
Likewise with {\em maximal analytic\/} and with {\em maximal smooth\/} in place of~{\em maximal}.
\end{cor}

\noindent
By Corollary~\ref{cor:completion Hardy field}, every embedding $i\colon \T_{\log}\to M$ as in Corollary~\ref{cor:Tlog}
extends to an embedding of the completion of $\T_{\log}$ into $M$. In the next section we
show that~$i$ even extends to an embedding of {\it every}\/  immediate $H$-field extension of $\T_{\log}$

\subsection*{Implications between ``short'', ``countable cofinality'', and ``bounded''} The first two notions are defined
for ordered sets, and ``bounded'' is defined for subsets of $\Cc$. It is clear that for ordered sets,
$$\text{short}\ \Longrightarrow\  \text{countable cofinality},$$
and that for Hausdorff fields,
$$\text{ countable cofinality}\ \Longrightarrow\  \text{bounded}.$$
These implications cannot be reversed for analytic Hardy fields: Let $H$ be a maximal analytic Hardy field. Corollary~\ref{anteta}  gives a sequence $(h_{\lambda})$ in $H$, indexed
by the ordinals $\lambda \leq \omega_1$, such that all $h_\lambda$ are transexponential and $*h_\lambda < *h_\mu$ for all~${\lambda < \mu\leq \omega_1}$.
It follows that $ \R\<h_\lambda:\,  \lambda< \omega_1\>$ is a bounded analytic Hardy field (bounded by $h_{\omega_1}$) with cofinality
$\omega_1$, and so $ \R\<h_\lambda:\, \lambda\leq \omega_1\>$  is an analytic Hardy field of cofinality $\omega$ that is not short. (We thank Philip Ehrlich and Elliot Kaplan for a useful email discussion on this topic.)

\section{Embeddings of Immediate Extensions}\label{eie}

\noindent
The goal of this section is to prove the following theorem, which partly generalizes Lemma~\ref{lem:L->M} beyond the $\upo$-free setting:

\begin{theorem}\label{thm:embedd imm}
Let $K$ be a short pre-$H$-field with archimedean residue field, and
suppose $K$ is $\upo$-free or not $\upl$-free.
$\hat{K}$ be an  immediate pre-$H$-field extension of~$K$
and let $M$ be a countably spherically complete closed $H$-field. Then every embedding~$K\to M$   extends to an embedding~$\hat{K}\to   M$.
\end{theorem}

\noindent
Using also
\cite[Corollary~\ref{corimmom}]{ADHfgh} and
Corollary~\ref{smanpc}, this yields:

\begin{cor}
Let $K$ be a short Hardy field which is $\upo$-free or not $\upl$-free, and let~$M$ be a
maximal Hardy field extending $K$. Then every immediate Hardy field extension of $K$ embeds into~$M$ over $K$. Likewise with ``maximal analytic'' as well as with ``maximal smooth'' in place of ``maximal''.
\end{cor}

\noindent
The main steps towards the proof of Theorem~\ref{thm:embedd imm} are Propositions~\ref{prop:upo-free imm} and~\ref{prop:HLOimm} below. This requires us to revisit the topic of pre-$\HLO$-fields once again.

We note also that by [4] and [ADH,10.5.8], each pre-$H$-field has an immediate strict pre-$H$-field extension that is
spherically complete.

\subsection*{Immediate pairs of pre-$\HLO$-fields}
Here we generalize [ADH, 16.4.1] to certain pairs of pre-$\HLO$-fields. A {\bf pre-$\HLO$-pair}
is a pair~$(\mathbf K,\hat{\mathbf K})$ of pre-$\HLO$-fields
with~${\mathbf K \subseteq \hat{\mathbf K}}$. Let
$(\mathbf K,\hat{\mathbf K})$ be a pre-$\HLO$-pair, with $\mathbf K=(K,\dots)$ and $\hat{\mathbf K}=(\hat{K},\dots)$. We call $(\mathbf K,\hat{\mathbf K})$ a {\bf $\HLO$-pair} if both~$\mathbf K$,~$\hat{\mathbf K}$ are $\HLO$-fields,
and we say that~$(\mathbf K,\hat{\mathbf K})$ is {\bf immediate} if the valued field extension $K\subseteq \hat K$ is immediate.
We also call~$(\mathbf K,\hat{\mathbf K})$ {\bf $\upo$-free} if both $K$, $\hat{K}$ are $\upo$-free, and similarly for
other properties of pre-$H$-fields.
A pre-$\HLO$-pair~$(\mathbf K^*,\hat{\mathbf K}^*)$
{\bf extends}~$(\mathbf K,\hat{\mathbf K})$ if~$\mathbf K\subseteq\mathbf K^*$ and
$\hat{\mathbf K}\subseteq\hat{\mathbf K}^*$.

\begin{prop}\label{prop:upo-free imm}
Suppose $(\mathbf K,\hat{\mathbf K})$ is an immediate pre-$\HLO$-pair such that if $K$ is $\upo$-free \textup{(}$\upl$-free, respectively\textup{)}, then so is~$\hat K$.
Then $(\mathbf K,\hat{\mathbf K})$ extends to an immediate $\upo$-free $\HLO$-pair $(\mathbf K^*,\hat{\mathbf K}^*)$
such that $\res{\mathbf K}^*$ is algebraic over~$\res{\mathbf K}$ and   any embedding of~$\mathbf K$ into a Schwarz closed $\HLO$-field $\mathbf L$ extends to
an embedding~$\mathbf K^*\to\mathbf L$.
\begin{CD}
\mathbf K^* @>\subseteq>{\textup{\tiny\ immediate\ }}>  \hat{\mathbf K}^* \\
@AA{\subseteq}A @A{\subseteq}AA  \\
\mathbf K   @>\subseteq>{\textup{\tiny\ immediate\ }}>  \hat{\mathbf K}
\end{CD}$$
Moreover, if~$\mathbf K$ is short and  $\res\mathbf K$ is archimedean, then we can choose such a pair~$(\mathbf K^*$, $\hat{\mathbf K}^*)$ where $\mathbf K^*$ is also short.
\end{prop}

\noindent
As with Proposition~\ref{1641} we adapt the proof of  [ADH, 16.4.1].  We assume
$(\mathbf K,\hat{\mathbf K})$ is an immediate pre-$\HLO$-pair and $\mathbf K=(K,I,\Lambda,\Omega)$, $\hat{\mathbf K}=(\hat K,\hat I,\hat \Lambda,\hat \Omega)$.
We identify~$H(K)$ in the usual way with an $H$-subfield of $H(\hat K)$, and for ungrounded $K$ we tacitly use
that the sequences $(\upl_\rho)$, $(\upo_\rho)$ in $K$ also serve for~$\hat K$.

\begin{lemma}
Suppose $K$ is grounded, or there exists $b\asymp 1$ in $K$ such that $v(b')$ is a gap in $K$. Then $(\mathbf K,\hat{\mathbf K})$ extends to an immediate $\upo$-free $\HLO$-pair $(\mathbf K^*,\hat{\mathbf K}^*)$  such that $\res\mathbf K^* = \res\mathbf K$ and any embedding
of $\mathbf K$ into a $\HLO$-field $\mathbf L$ closed under logarithms extends to an embedding $\mathbf K^*\to\mathbf L$.
\end{lemma}
\begin{proof}
Note that if $K$ is grounded, then so is $\hat K$, and any gap in $K$ remains a gap in~$\hat K$.
Put $E:=H(K)$ and $F:=H(\hat K)$, and note that the $H$-field extension $E\subseteq F$ is immediate by [ADH, 10.3.2 and remark preceding it]. Next take $e\in E$ with~$e\succ 1$ and~$v(e^\dagger)=\max\Psi_E=\max\Psi_F$.
We now construct $K^*:=E_\upo$ and $\hat K^*:=F_\upo$ as in~[ADH, 11.7] with $E, e$ and $F,e$ in the role of $F,f$ there, so $E_\upo=\bigcup_n E_n$, $F_{\upo}=\bigcup_n F_n$, $E_0=E$, $F_0=F$. We take care to do that in such a way that
by induction on $n$ using [ADH, 10.2.3 and its proof] we have for all $n$ an immediate extension $E_n\subseteq F_n$ of grounded $H$-fields with a distinguished element $e_n\in E_n^\times$ such that $e_0=e$,
${e_n\succ 1}$, $v(e_n^\dagger)=\max\Psi_{E_n}=\max\Psi_{F_n}$,  and
$$E_{n+1}\ =\ E_n(e_{n+1}),\quad  F_{n+1}\ =\ F_n(e_{n+1}), \quad e'_{n+1}\ =\ e_n^\dagger.$$
This yields an immediate extension $E_{\upo}\subseteq F_{\upo}$ of $\upo$-free $H$-fields. Expanding $K^*, \hat{K}^*$ uniquely to  $\HLO$-fields gives a pair $(\mathbf K^*,\hat{\mathbf K}^*)$ with the required properties.
\end{proof}

%Next take $f\in F$ such that $f\succ 1$ and~$vf^\dagger=\max\Psi_F$, and
%construct $\hat K^*:=E_\upo$ as in~[ADH, 11.7] with this $f$ but with~$E$ in place of $F$,
%so $E=\bigcup_n E_n$ where $E_0=E$ and $E_{n+1}=E_n(f_{n+1})$ with~${f_n\succ 1}$, $f_n^\dagger=\max\Psi_{E_n}$
%for each $n$.
%Put $F_0:=F$ and inductively   $F_{n+1}:=F_n(f_{n+1})$. An easy induction on $n$
%using [ADH, 10.2.3] and its proof shows that $F_n$ is an $H$-subfield of $E_n$
%with $E_n$ an immediate extension of~$F_n$.
%Thus~$E$ is an immediate extension of its $H$-subfield
%$K^*:=F_\upo=\bigcup_n F_n$.
%The pair $(\mathbf K^*,\hat{\mathbf K}^*)$ consisting of the unique expansions  of $K^*$, $\hat K^*$ to
%$\HLO$-fields has the required properties.

\begin{lemma}
Suppose $K$ has gap $\beta$ and $v(b')\neq\beta$ for all $b\asymp 1$ in $K$. Then $(\mathbf K,\hat{\mathbf K})$ extends to
an immediate grounded $\HLO$-pair $(\mathbf K_1,\hat{\mathbf K}_1)$
such that $\res\mathbf K_1 = \res\mathbf K$ and any embedding of $\mathbf K$ into a $\HLO$-field $\mathbf L$ closed under
integration extends to an embedding~$\mathbf K_1\to\mathbf L$.
\end{lemma}
\begin{proof}
Note that $\beta$ is a gap in $\hat K$, and  $v(b')\neq\beta$ for all $b\asymp 1$ in $\hat K$. By [ADH, 10.3.2 and remark preceding it], $H(K)$ is an immediate extension of $K$, and  $H(\hat K)$ of $\hat K$, so
$H(\hat K)$ is an immediate extension of $H(K)$.

Take~$s\in K$ with $vs=\beta$ and follow the proof of [ADH, 16.4.3].
Suppose~$s\notin I$. Then also~$s\notin\hat I$. Take $\hat K_1=H(\hat K)(y)$ as in Case~1 of that proof
applied to~$\hat{\mathbf K}$ in place of $\mathbf K$.   We have the $H$-subfield~$K_1:=H(K)(y)$ of~$\hat K_1$, and~$\hat K_1$ is an immediate extension of~$K_1$ by~[ADH, 10.2.2  and its proof]. Expanding
~$K_1$,~$\hat K_1$ uniquely to
pre-$\HLO$-fields gives a pair~$(\mathbf K_1,\hat{\mathbf K}_1)$ with the required property.
If~${s\in I}$, proceed as before, but following instead Case~2 of the proof of [ADH, 16.4.3] and with $H(K)$ and $H(\hat K)$ instead of
$K$ and $\hat K$, using~[ADH, 10.2.1 and its proof].
\end{proof}

\begin{cor}\label{cor:16.4.4 imm}
Suppose $K$ does not have asymptotic integration. Then
$(\mathbf K,\hat{\mathbf K})$ extends to an immediate $\upo$-free $\HLO$-pair
$(\mathbf K^*,\hat{\mathbf K}^*)$
such that
$\res\mathbf K^* = \res\mathbf K$, and any embedding of~$\mathbf K$
into a $\HLO$-field $\mathbf L$ closed under integration extends to an embedding~$\mathbf K^*\to\mathbf L$.
\end{cor}

\noindent
In the next three lemmas we treat the case where $K$ has asymptotic integration. For the first we adapt the proof of  [ADH, 16.4.5] and use parts of it:

\begin{lemma}\label{lem:16.4.5 imm}
Assume $K$ has asymptotic integration and is not $\upl$-free. Then
$(\mathbf K,\hat{\mathbf K})$ extends to an immediate $\upo$-free $\HLO$-pair
$(\mathbf K^*,\hat{\mathbf K}^*)$
such that
$\res\mathbf K^* = (\res\mathbf K)^{\operatorname{rc}}$, and every embedding
of $\mathbf K$ into a Liouville closed $\HLO$-field $\mathbf L$ extends to an embedding $\mathbf K^*\to\mathbf L$.
\end{lemma}
\begin{proof}
By Corollary~\ref{cor:16.4.4 imm} it is enough to show that~$(\mathbf K,\hat{\mathbf K})$ extends to an immediate $\HLO$-pair
$(\mathbf K_1,\hat{\mathbf K}_1)$ with a gap such that~$\res\mathbf K_1 = (\res\mathbf K)^{\operatorname{rc}}$ and every embedding
of~$\mathbf K$ into a Liouville closed $\HLO$-field $\mathbf L$ extends to an embedding $\mathbf K_1\to\mathbf L$.
$$E\ :=\ H(K)^{\operatorname{rc}}\ \subseteq\ F:=H(\hat K)^{\operatorname{rc}}.$$
Then $\Gamma_E=\Q\Gamma$, and $F$ is an immediate extension $H$-field extension of $E$.
We distinguish two cases:

\case[1]{$E$ has a gap.} Take $s\in E^\times$ and $n\geq 1$ such that $vs$ is a gap in~$E$ and~$s^n\in K$.
Then $E$ has exactly two $\HLO$-cuts $(I_1,\Lambda_1,\Omega_1)$, $(I_2,\Lambda_2,\Omega_2)$,
where~$I_1=\{{y\in E:y\prec s}\}$,
$I_2=\{y\in E:y\preceq s\}$, and
$F$ has
exactly two $\HLO$-cuts~$(\hat I_1,\hat \Lambda_1,\hat \Omega_1)$ and $(\hat I_2,\hat \Lambda_2,\hat \Omega_2)$,
with $\hat I_1=\{y\in F:y\prec s\}$, $\hat I_2=\{{y\in F:y\preceq s}\}$ (so $I_j=\hat I_j\cap E$ for $j=1,2$).
Take $\mathbf K_1$ as in Case~1 of the proof of~[ADH, 16.4.5]:
if~$-s^\dagger\in\Lambda$, then $\mathbf K_1:=(E,I_1,\Lambda_1,\Omega_1)$,
and if $-s^\dagger\notin\Lambda$, then $\mathbf K_1:=(E,I_2,\Lambda_2,\Omega_2)$.
Similarly, if~$-s^\dagger\in \Lambda$, then $\hat{\mathbf K}_1:=(F,\hat I_1,\hat \Lambda_1,\hat \Omega_1)$,
and if $-s^\dagger\notin \Lambda$, then  $\hat{\mathbf K}_1:=(F,\hat I_2,\hat \Lambda_2,\hat \Omega_2)$.
%Note that $-s^\dagger\in\Lambda\Leftrightarrow-s^\dagger\in\hat\Lambda$, since   $\Lambda=\hat\Lambda\cap K$.
Then $(\mathbf K_1,\hat{\mathbf K}_1)$ is an immediate $\HLO$-pair  with the desired property.

\case[2]{$E$ has no gap.} Then $E$, $F$ have asymptotic integration, and the sequence~$(\upl_\rho)$ for $K$ also serves for $E$ and for $F$.
Take $\upl\in K$ such that $\upl_\rho\leadsto\upl$. Then~$-\upl$ creates a gap over $E$ and over $F$ by [ADH, 11.5.14].
Take an element~$f\neq 0$ in some Liouville closed $H$-field extension of $F$ such that $f^\dagger=-\upl$.
Then $F(f)$ is an $H$-field  and $E(f)$ is an $H$-subfield of $F(f)$ with~$\res E(f)=\res E=\res F=\res F(f)$.
Moreover, $vf$ is a gap in $F(f)$ and in $E(f)$, and $F(f)$ is an immediate extension of~$E(f)$, by the remark after [ADH 11.5.14]
and the uniqueness part of [ADH, 10.4.5].
Now $E(f)$ has
exactly two $\HLO$-cuts $(I_1,\Lambda_1,\Omega_1)$ and $(I_2,\Lambda_2,\Omega_2)$,
$$I_1=\big\{{y\in E(f):y\prec f}\big\},\qquad I_2=\big\{y\in E(f):y\preceq f\big\},$$ and $F(f)$ has
exactly two $\HLO$-cuts~$(\hat I_1,\hat \Lambda_1,\hat \Omega_1)$ and $(\hat I_2,\hat \Lambda_2,\hat \Omega_2)$,
$$\hat I_1=\big\{y\in F(f):y\prec f\big\},\qquad \hat I_2=\big\{{y\in F(f):y\preceq f}\big\}.$$
Therefore~$I_j=\hat I_j\cap E(f)$ for $j=1,2$.
We set $\mathbf K_1:=\big(E(f),I_1,\Lambda_1,\Omega_1\big)$ and $\hat{\mathbf K}_1:=\big(F(f),\hat I_1,\hat\Lambda_1,\hat\Omega_1\big)$
if $\upl\in\Lambda$, and $\mathbf K_1:=\big(E(f),I_2,\Lambda_2,\Omega_2\big)$, $\hat{\mathbf K}_1:=\big(F(f),\hat I_2,\hat \Lambda_2,\hat \Omega_2\big)$
if $\upl\notin\Lambda$.
Then $\mathbf K_1\subseteq\hat{\mathbf K}_1$, and the immediate $\HLO$-pair $(\mathbf K_1,\hat{\mathbf K}_1)$
is as required.
\end{proof}

\begin{lemma}
Suppose $K$ is not $\upo$-free and $\hat K$ is $\upl$-free. Then $(\mathbf K,\hat{\mathbf K})$ extends to an immediate $\upo$-free $\HLO$-pair
$(\mathbf K^*,\hat{\mathbf K}^*)$ such that $\res\mathbf K^*$ is algebraic over~$\res\mathbf K$
and any embedding   $\mathbf K\to\mathbf L$ into
a Schwarz closed $\HLO$-field $\mathbf L$ extends to an embedding~$\mathbf K^*\to\mathbf L$.
\end{lemma}
\begin{proof} We adapt and use the proof of [ADH, 16.4.6].
Take $\upo\in K$ with $\upo_\rho\leadsto\upo$. Then
$\omega\big(\Upl(K)\big){}^\downarrow<\upo<\sigma\big(\Upg(K)\big){}^\uparrow$ and
either   $\Omega=\omega(K)^\downarrow$ or $\Omega=K\setminus \sigma\big(\Upg(K)\big){}^\uparrow$.
Likewise with $\hat K$ in place of $K$. Also $\upo\notin \omega(K)^\downarrow,\ \upo\notin \omega(\hat K)^\downarrow$.
There are two cases:

\case[1]{$\Omega=\omega(K)^{\downarrow}$.}
Then $\upo\notin\hat\Omega$ and so $\hat\Omega=\omega(\hat K)^{\downarrow}$.
Take a pre-$H$-field extension~$\hat K_{\upg}$ of $\hat K$
as in Case~1 of the proof of [ADH, 16.4.6] with $\hat K$ in place of~$K$.
Then $\res \hat K_{\upg}=\res \hat K=\res K$.
Put~$K_{\upg}:=K\langle\upg\rangle$, a pre-$H$-subfield of $\hat K_{\upg}$ with~$\res K_{\upg}=\res K$. Then $v\upg$ is a gap in $K_{\upg}$ and in $\hat K_{\upg}$, so
by [ADH, 13.7.6],  $\hat K_{\upg}$ is an immediate extension of $K_{\upg}$.
Expanding $K_{\upg}$ to a pre-$\HLO$-field $\mathbf K_{\upg}$ as in Case~1 of
the proof of [ADH, 16.4.6], and similarly expanding~$\hat K_{\upg}$ to a pre-$\HLO$-field~$\hat{\mathbf K}_{\upg}$, we
thus obtain the immediate  pre-$\HLO$-pair~$(\mathbf K_{\upg},\hat{\mathbf K}_{\upg})$ extending~$(\mathbf K,\hat{\mathbf K})$. Take an immediate $\upo$-free $\HLO$-pair~$(\mathbf K^*,\hat{\mathbf K}^*)$ extending~$(\mathbf K_{\upg},\hat{\mathbf K}_{\upg})$ as in Corollary~\ref{cor:16.4.4 imm} applied to $(\mathbf K_{\upg},\hat{\mathbf K}_{\upg})$ in place of~$(\mathbf K,\hat{\mathbf K})$. Then $(\mathbf K^*,\hat{\mathbf K}^*)$ has the required property.
%the proof of Case~1 of [ADH, 16.4.6]
% shows that for each , each
% embedding of $\mathbf K$ into
%a Schwarz closed $\HLO$-field $\mathbf L$ extends to an embedding of~$\mathbf K^*$ into $\mathbf L$.

\case[2]{$\Omega=K\setminus \sigma\big(\Upg(K)\big){}^\uparrow$.}
Then $\upo\in\Omega\subseteq \hat \Omega$, so $\hat\Omega=\hat K\setminus \sigma\big(\Upg(\hat K)\big){}^\uparrow$.
As in the proof of [ADH, 16.4.6] we obtain  an immediate pre-$H$-field extension~$\hat K_{\upl}:=\hat K(\upl)$ of $\hat K$ with $\upl_\rho\leadsto\upl$ and $\omega(\upl)=\upo$. Put $K_{\upl}:=K(\upl)$, an
immediate pre-$H$-field extension of $K$.
Expand $K_\upl$ to a pre-$\HLO$-field $\mathbf K_{\upl}$ as in Case~2 of the proof of~[ADH, 16.4.6],
and similarly expand $\hat K_{\upl}$  to a pre-$\HLO$-field $\hat{\mathbf K}_{\upl}$.
Then~$\mathbf K_{\upl}\supseteq\mathbf K$ and~${\hat{\mathbf K}_{\upl}\supseteq\hat{\mathbf K}}$,
and from $\upl\notin\Upl(K_\lambda)^\downarrow$ and $\upl\in \big(\hat K_{\upl}\setminus\Upd(\hat K_{\upl})^\uparrow\big)\cap K_{\upl}$ we obtain~$\hat{\mathbf K}_{\upl}\supseteq\mathbf K_{\upl}$.
Thus~$(\mathbf K_{\upl},\hat{\mathbf K}_{\upl})$ is an immediate  pre-$\HLO$-pair and extends~$(\mathbf K,\hat{\mathbf K})$. Take an immediate $\upo$-free $\HLO$-pair~$(\mathbf K^*,\hat{\mathbf K}^*)$ extending~$(\mathbf K_{\upl},\hat{\mathbf K}_{\upl})$  obtained from Lemma~\ref{lem:16.4.5 imm} applied to $(\mathbf K_{\upl},\hat{\mathbf K}_{\upl})$ in place of~$(\mathbf K,\hat{\mathbf K})$. Then $(\mathbf K^*,\hat{\mathbf K}^*)$ has the required property.
%As in the proof of Case~2 of~[ADH, 16.4.6] we see that every  embedding of $\mathbf K$ into
%a Schwarz closed $\HLO$-field~$\mathbf L$ extends to an embedding of~$\mathbf K^*$ into $\mathbf L$.
\end{proof}

\begin{lemma}
Suppose $\hat K$ is $\upo$-free.  Then
$(\mathbf K,\hat{\mathbf K})$ extends to an immediate $\upo$-free $\HLO$-pair
$(\mathbf K^*,\hat{\mathbf K}^*)$ such that  any embedding of $\mathbf K$ into
a $\HLO$-field $\mathbf L$ extends to an embedding of~$\mathbf K^*$ into $\mathbf L$.
\end{lemma}
\begin{proof}
As $\hat K$ is $\upo$-free, so is $K$.
By [ADH, 13.6.1],     $H(K)$ is $\upo$-free, and by [ADH, 10.3.2 and   remark~(a) before it], $H(K)$ is an immediate extension of $K$, and likewise with
$\hat K$ in place of~$K$. Let $\mathbf K^*$, $\hat{\mathbf K}^*$ be the unique expansions of $H(K)$, $H(\hat K)$, respectively,
to  $\HLO$-fields. Then $(\mathbf K^*,\hat{\mathbf K}^*)$ has the required properties, by the proof of [ADH, 16.4.7].
\end{proof}

\noindent
The first claim of Proposition~\ref{prop:upo-free imm} now follows.
As to the shortness  part, one checks that $\operatorname{rank}_{\Q}(\Gamma_{K^*}/\Gamma_K)\le \aleph_0$ for $(\mathbf K^*,\hat{\mathbf K}^*)$  as constructed above,    hence if $K$ is short and $\res K$ is archimedean, then $K^*$ is short by Corollary~\ref{cor:short oag} and Lemma~\ref{lem:short ofield}. \qed

\subsection*{Immediate extensions and $\HLO$-cuts}
Let  $K\subseteq\hat K$ be an extension of  pre-$H$-fields. Given a $\HLO$-cut
$(\hat I,\hat\Lambda,\hat\Omega)$ in $\hat K$, we obtain the $\HLO$-cut
$$(\hat I,\hat\Lambda,\hat\Omega)\cap K:=(\hat I\cap K,  \hat\Lambda\cap K, \hat\Omega\cap K)$$ in $K$.
Recall from [ADH, remark before 16.3.19] that a pre-$H$-field has at least one and at most two $\HLO$-cuts.
In the rest of this subsection we assume that  $K\subseteq\hat K$ is immediate  and~$(I,\Lambda,\Omega)$ is a
$\HLO$-cut in $K$, and we ask when there is a
$\HLO$-cut~$(\hat I,\hat\Lambda,\hat\Omega)$ in $\hat K$   such that~$(I,\Lambda,\Omega)=(\hat I,\hat\Lambda,\hat\Omega)\cap K$.

\begin{prop}\label{prop:HLOimm}
The following are equivalent:
\begin{enumerate}
\item[(i)]
There is a $\HLO$-cut $(\hat I,\hat\Lambda,\hat\Omega)$ in $\hat K$  with
$(I,\Lambda,\Omega)=(\hat I,\hat\Lambda,\hat\Omega)\cap K$;
\item[(ii)] $K$ is not $\upl$-free, or $K$ is $\upo$-free, or $\hat K$ is $\upl$-free, or
\end{enumerate}
\end{prop}

\noindent
This is a consequence of Lemmas~\ref{lem:HLOimm, 1}--\ref{lem:HLOimm, 6} below, which also address the uniqueness of the
$\HLO$-cut  in $\hat K$  in part~(i) of the proposition. For the next two labeled displays, let $K$ be ungrounded.
Then by [ADH, 11.8.14] we have
\begin{equation}\label{eq:HLOim} \Upl(K)^\downarrow\ =\  \Upl(\hat K)^\downarrow\cap K, \qquad
\Upd(K)^\uparrow\ =\  \Upd(\hat K)^\uparrow \cap K, \quad
\Upg(K)^\uparrow\ =\  \Upg(\hat K)^\uparrow \cap K,\end{equation}
and by [ADH, 11.8.14, remark before 11.8.21, and 11.8.29]:
\begin{equation}\label{eq:HLOimm}
\omega\big(\Upl(\hat K)\big){}^\downarrow\cap K\ =\ \omega\big(\Upl(K)\big){}^\downarrow, \qquad
\big(\hat K\setminus\sigma\big(\Upg(\hat K)\big){}^\uparrow\big)\cap K\ =\ K\setminus\sigma\big(\Upg(K)\big){}^\uparrow.\end{equation}
By [ADH, 11.8.2] we also have $\I(K)=\I(\hat K)\cap K$ if $K$ has asymptotic integration.

\begin{lemma}\label{lem:HLOimm, 1}
Suppose $K$ does not have asymptotic integration or $\hat K$ is $\upo$-free.  Then there is a unique  $\HLO$-cut $(\hat I,\hat\Lambda,\hat\Omega)$ in $\hat K$ with~$(I,\Lambda,\Omega)=(\hat I,\hat\Lambda,\hat\Omega)\cap K$.
\end{lemma}
\begin{proof}
Note that   $K$ has asymptotic integration iff $\hat K$ has,
and if $K$ has a gap $\beta$ and~$v(a')\neq\beta$ for all $a\asymp 1$ in $K$, then $\beta$ remains a gap in $\hat K$ and
$v(b')\neq\beta$ for all~$b\asymp 1$ in $\hat K$. If $\hat K$ is $\upo$-free, then so is $K$.
Now use~[ADH, 16.3.11--16.3.14].
\end{proof}

\begin{lemma}\label{lem:HLOimm, 4}
Suppose $K$ is $\upo$-free, but $\hat K$ is not.
Then there are exactly two  $\HLO$-cuts $(\hat I,\hat\Lambda,\hat\Omega)$ in $\hat K$ with~$(I,\Lambda,\Omega)=(\hat I,\hat\Lambda,\hat\Omega)\cap K$.
\end{lemma}
\begin{proof}
By [ADH, 16.3.14], $(I,\Lambda,\Omega)$ is the unique $\HLO$-cut in $K$, so
$(I,\Lambda,\Omega)=(\hat I,\hat\Lambda,\hat\Omega)\cap K$ for every $\HLO$-cut $(\hat I,\hat\Lambda,\hat\Omega)$ in $\hat K$.
Moreover, $\hat K$  has exactly two $\HLO$-cuts, by  [ADH, 16.3.16]
if $\hat K$ is $\upl$-free, and by [ADH, 16.3.17, 16.3.18]  if not.
\end{proof}

\noindent
In particular, if $K$ has no asymptotic integration or $K$ is $\upo$-free then we have
a $\HLO$-cut $(\hat I,\hat\Lambda,\hat\Omega)$ in $\hat K$  with
$(I,\Lambda,\Omega)=(\hat I,\hat\Lambda,\hat\Omega)\cap K$. The next lemmas deal
with the case where $K$ has asymptotic integration and $K$ is not $\upo$-free.

\begin{lemma}\label{lem:HLOimm, 2}
Suppose $K$ has asymptotic integration and  is not $\upl$-free.
%$2\Psi$ has a supremum in~$\Gamma$.
Then there is
exactly one  $\HLO$-cut $(\hat I,\hat\Lambda,\hat\Omega)$ in $\hat K$ with~$(I,\Lambda,\Omega)=(\hat I,\hat\Lambda,\hat\Omega)\cap K$.
\end{lemma}
\begin{proof}
Suppose first that $2\Psi$ has no supremum in $\Gamma$.
Then by [ADH, 16.3.17] there are exactly two $\HLO$-cuts
$(I_1,\Lambda_1,\Omega_1)$, $(I_2,\Lambda_2,\Omega_2)$
in~$K$, with $\Lambda_1=\Upl(K)^\downarrow$, $\Lambda_2=K\setminus\Upd(K)^\uparrow$, and
Similarly there are exactly two $\HLO$-cuts
$(\hat I_1,\hat \Lambda_1,\hat \Omega_1)$, $(\hat I_2,\hat \Lambda_2,\hat \Omega_2)$
in $\hat K$, with $\hat \Lambda_1=\Upl(\hat K)^\downarrow$, $\hat \Lambda_2=\hat K\setminus\Upd(\hat K)^\uparrow$.
Now use that by \eqref{eq:HLOim} we have  $\Upl(K)^\downarrow = \Upl(\hat K)^\downarrow\cap K$
and $K\setminus\Upd(K)^\uparrow = \big(\hat K\setminus\Upd(\hat K)^\uparrow\big)\cap K$.
The case where $2\Psi$ has a supremum in $\Gamma$ is similar, using  [ADH, 16.3.18]
instead of  [ADH, 16.3.17].
\end{proof}

\begin{lemma}\label{lem:HLOimm, 5}
Suppose $K$ is not $\upo$-free and $\hat K$ is $\upl$-free. Then
there is
exactly one  $\HLO$-cut $(\hat I,\hat\Lambda,\hat\Omega)$ in $\hat K$ such that~$(I,\Lambda,\Omega)=(\hat I,\hat\Lambda,\hat\Omega)\cap K$.
\end{lemma}
\begin{proof}
By [ADH, 16.3.16],  $K$ being $\upl$-free, but not $\upo$-free, it has exactly two $\HLO$-cuts, namely
$ \big(\!\I(K), \Upl(K)^{\downarrow}, \omega\big(\Upl(K)\big){}^\downarrow\big)$ and $\big(\!\I(K), \Upl(K)^{\downarrow}, K\setminus \sigma\big(\Upg(K)\big){}^\uparrow\big)$,
and similarly with $\hat K$ in place of $K$.
Now use \eqref{eq:HLOim} and \eqref{eq:HLOimm}.
\end{proof}

\begin{lemma}\label{lem:HLOimm, 6}
Suppose $K$ is $\upl$-free, but not $\upo$-free, and $\hat K$ is not $\upl$-free.
Then there is a $\HLO$-cut $(\hat I,\hat\Lambda,\hat\Omega)$ in $\hat K$ such that~$(I,\Lambda,\Omega)=(\hat I,\hat\Lambda,\hat\Omega)\cap K$ iff $\Omega\neq\omega(K)^\downarrow$, and in this case
there are exactly two such $\HLO$-cuts in $\hat K$.
\end{lemma}
\begin{proof}
By [ADH, 16.3.16]
$K$ has exactly two $\HLO$-cuts $(I_1,\Lambda_1,\Omega_1)$, $(I_2,\Lambda_2,\Omega_2)$~
$$I_1\ =\ I_2\ =\ \I(K),\quad \Lambda_1\ =\ \Lambda_2\ =\Upl(K)^{\downarrow},\quad
\Omega_1\ =\ K\setminus \sigma\big(\Upg(K)\big){}^\uparrow\ \ne\  \Omega_2\ =\  \omega(K)^{\downarrow}.$$
Now $K$ is $\upl$-free, so $2\Psi$ has no supremum in $\Gamma$ by [ADH, 9.2.17, 11.6.8], hence
by~[ADH, 16.3.17], $\hat K$ has exactly two $\HLO$-cuts $(\hat I_1,\hat\Lambda_1,\hat\Omega_1)$,
$(\hat I_2,\hat\Lambda_2,\hat\Omega_2)$, where
$$\hat I_1= \hat I_2 = \I(\hat K),\quad \hat\Lambda_1 = \Upl(\hat K)^{\downarrow}, \quad
\hat\Lambda_2 = \hat K\setminus \Upd(\hat K)^{\uparrow}, \quad
\hat\Omega_1 = \hat\Omega_2 = \hat K\setminus \sigma\big(\Upg(\hat K)\big){}^{\uparrow}.$$
Thus $(\hat I_j,\hat\Lambda_j,\hat\Omega_j)\cap K=(I_1,\Lambda_1,\Omega_1)$ for $j=1,2$ by \eqref{eq:HLOimm}. This yields the lemma.
\end{proof}

\subsection*{Proof of Theorem~\ref{thm:embedd imm}}
Let $K$, $\hat K$, $M$ be as in the statement of the theorem,
and let $i\colon K\to M$ be an embedding.
If $\hat K$ is $\upo$-free, then Lemma~\ref{lem:L->M} and [ADH, 10.5.8] give an extension of $i$ to an embedding~${\hat{K}\to  M}$ as required.

In the rest of the proof we therefore assume that $\hat K$ is not $\upo$-free.
If $\hat K$ is $\upl$-free, then, taking $\upo\in\hat K$ with~$\upo_\rho\leadsto\upo$, [ADH, 11.7.13] yields an immediate
pre-$H$-field extension~$\hat K_{\upl}:=\hat K(\upl)$ of $\hat K$ with $\upl_\rho\leadsto\upl$ and $\omega(\upl)=\upo$, so that replacing $\hat K$ by~$\hat K_{\upl}$ we arrange  that $\hat K$ is  not even $\upl$-free.

Suppose $K$ is not $\upl$-free.
Let $\mathbf M$ be the unique expansion of $M$ to a $\HLO$-field, and expand $K$ to a pre-$\HLO$-field $\mathbf K$ such that $i$ is an embedding~$\mathbf K\to\mathbf M$ of pre-$\HLO$-fields. Proposition~\ref{prop:HLOimm} yields an expansion of $\hat K$ to a pre-$\HLO$-field $\hat{\mathbf K}$ such
that~${\mathbf K\subseteq\hat{\mathbf K}}$, and
then Proposition~\ref{prop:upo-free imm} gives an immediate $\upo$-free short $\HLO$-pair~$(\mathbf K^*,\hat{\mathbf K}^*)$ extending $(\mathbf K,\hat{\mathbf K})$ with $\res\mathbf K^*$ algebraic over $\res\mathbf K$ and an extension of $i$ to an embedding $i^*:\mathbf K^*\to\mathbf M$.  The case of $\upo$-free $\hat K$ treated earlier applied instead to
$\hat{K}^*$ now yields an extension  of $i^*$ to an embedding $\hat{K}^*\to M$.

Next, suppose $K$ is $\upo$-free.
Then the pc-sequence $(\upl_\rho)$ in $K$ is of $\d$-tran\-scen\-den\-tal type over $K$, by [ADH, 13.6.3].
Take $\upl\in\hat K$ such that~$\upl_\rho\leadsto\upl$.
Now $K$ is short and $M$ is countably spherically complete, so by Lemma~\ref{lem:short}  we have
$\upl^*\in M$ with~${i(\upl_\rho)\leadsto\upl^*}$.
By [ADH, 11.4.7, 11.4.13, 10.5.8] we obtain a unique extension of~$i$
to an embedding $j\colon K\langle\upl\rangle\to M$ such that $j(\upl)=\upl^*$.
The case of non-$\upl$-free $K$ applied instead to $K\langle\upl\rangle$ yields an extension of $j$ to
an embedding~$\hat K\to M$. \qed

\section{Embeddings into Analytic Hardy Fields}\label{sec:embedd}

\noindent
Below we use Theorem~\ref{aneta} to derive results about
back-and-forth equivalence and isomorphism for maximal analytic Hardy fields, as was done in   \cite[Section~\ref{sec:iso}]{ADHfgh} for maximal Hardy fields.
Let~$\No$ be the ordered field of surreal numbers     equipped with
the derivation $\der_{\BM}$ of Berarducci and Mantova~[11]. Then $\No$ is a closed $H$-field, by [5]. Moreover, given an  uncountable cardinal $\kappa$,  the surreal
numbers of   length~$<\kappa$ form an ordered differential subfield~$\No(\kappa)$ of $\No$
with $\No(\kappa)\preceq\No$, by~\cite[Corollary~4.6]{ADH1+}.
As in the argument leading up to \cite[Corollary~\ref{iso4}]{ADHfgh}, combining Theorem~\ref{aneta}
%(instead of the main theorem of loc.~cit.)
\cite[Corollary~\ref{iso3}]{ADHfgh} yields:

\begin{cor}
Let $M$ be a maximal analytic or maximal smooth Hardy field. Then the ordered differential fields $M$ and  $\No(\omega_1)$ are back-and-forth equivalent. Hence~$M \equiv_{\infty\omega} \No(\omega_1)$, and
assuming~$\operatorname{CH}$,   $M\cong \No(\omega_1)$.
%Likewise if~$M$ is a maximal smooth Hardy field.
\end{cor}

\noindent
The ordered field $\No(\omega_1)$
is not complete:  Set $a_{\nu}:=\sum_{\mu< \nu} \omega^{-\mu}$ with $\mu,\nu$ ranging over countable ordinals. Then $(a_{\nu})$ is a cauchy sequence in $\No(\omega_1)$ without a limit in
$\No(\omega_1)$. Thus, assuming CH, no real closed $\eta_1$-ordered  field extension of $\R$ of cardinality $\mathfrak{c}$ is complete,
in particular, no maximal Hardy field is complete.  %\marginpar{added ``real closed'', and paranthetical remark}
(This also follows from \cite[Theorem~3.12(ii)]{DW}: if $G$ is a complete $\eta_1$-ordered abelian group, then $\abs{G}>\aleph_1$.)
%with $\abs{G}\geq 2^{\aleph_1}$ if $\ci(G^>)=\omega_1$

\medskip\noindent
Let $K$ be an    $H$-field  with small derivation and constant field~$\R$.
Then \cite[Theorem~3]{ADH1+} yields an embedding $K\to\No$
of ordered differential fields.
The argument in the proof of~\cite[Theorem~3]{ADH1+} shows that if $\kappa>\abs{K}$ is a regular  cardinal, then we can choose~$\iota$ so that~$\iota(K)\subseteq\No(\kappa)$.
If  $\operatorname{trdeg}(K|\R)$ is countable, then~$K$
actually embeds into~$\No(\omega_1)$.