Completions of countable non-standard models of $\rationals$
Peter Laubenheimer
Thomas Schick
Uni G{\"o}ttingen
Germany
www \protect\href{http://www.uni-math.gwdg.de/schick}{http://www.uni-math.gwdg.de/schick}
\protect\href{http://www.uni-math.gwdg.de/stuhler}{http://www.uni-math.gwdg.de/stuhler} \protect
Fax: ++49 -551/39 2985 }
In this note, we study non-standard models of the rational numbers with countably many elements. These are ordered fields, and so it makes sense to complete them, using non-standard Cauchy sequences. The main part of this note shows that these completions are real closed, i.e.~each positive number is a square, and each polynomial of odd degree has a root. This way, we give a direct proof of a consequence of a theorem of Hauschild. In a previous version of this note, not being aware of these results, we missed to mention this reference. We thank Matthias Aschenbrenner for pointing out this and related work. We also give some information about the set of real parts of the finite elements of such completions ---about the more interesting results along this we have been informed by Matthias Aschenbrenner and Martin Goldstern. The main idea of our proof relies on a way to describe real zeros of a polynomial in terms of first order logic. This is achieved by carefully using the sign changes of such a polynomial.
math0604466.tex
\documentclass[a4paper,twoside]{article}

%%%%%%%%%%  DO NOT FORGET TO CHANGE DATE %%%%%%%%%

\newcommand{\heuteIst}{April 20, 2006 }

%\usepackage[active]{srcltx}
%%%%%%%%%%  DO NOT FORGET TO CHANGE DATE %%%%%%%%%

\usepackage{amsmath,amsthm,amsfonts,latexsym,amscd,amssymb,enumerate}
%\usepackage[notref,notcite]{showkeys}
%\usepackage[backref,pdftex]{hyperref}
\usepackage[hypertex,backref]{hyperref}

\setcounter{secnumdepth}{2}
\setcounter{tocdepth}{3}

\swapnumbers
\theoremstyle{plain}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{conjecture}[theorem]{Conjecture}

\theoremstyle{definition}
\newtheorem{statement}[theorem]{Statement}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{convention}[theorem]{Convention}
\newtheorem{situation}[theorem]{Situation}

\theoremstyle{remark}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{question}[theorem]{Question}

%\input{theorems}   %Reihenfolge wichtig!
%\input{Math_declarations.tex}


\newcommand{\profinQ}{\mathfrak{Q}}
\newcommand{\reals}{\mathbb{R}}
\newcommand{\complexs}{\mathbb{C}}
\newcommand{\naturals}{\mathbb{N}}
\newcommand{\bbbn}{{\mathbb N}}
\newcommand{\integers}{\mathbb{Z}}
\newcommand{\rationals}{\mathbb{Q}}
\newcommand{\bbbq}{{\mathbb Q}}
\newcommand{\quaternions}{\mathbb{H}}
\newcommand{\hyperbolic}{\mathbb{H}}
\newcommand{\K}{\mathbb{K}}
\newcommand{\realsorcomplexs}{\mathbb{K}}
\newcommand{\finitefield}{\mathbb{F}}
\newcommand{\Proj}{\mathbb{P}}
%\newcommand{\1}{\mathbf{1}}  % if this does not work: \boldsymbol{1}
\DeclareMathOperator{\id}{id}
\newcommand{\boundary}[1]{\partial#1}
\newcommand{\boundedops}{\mathcal{B}}
\newcommand{\bundleover}{\!\!\downarrow\!\!}
\newcommand{\abs}[1]{\left\lvert#1\right\rvert} %absolute value
%\newcommand{\bigabs}[1]{\left\lvert#1\right\rvert} %absolute value
\newcommand{\norm}[1]{\left\lVert#1\right\rVert}
\newcommand{\generate}[1]{\langle#1\rangle}
\newcommand{\tensor}{\otimes}
\newcommand{\into}{\hookrightarrow}
\newcommand{\onto}{\twoheadrightarrow}
\newcommand{\iso}{\cong}
\newcommand{\disjointunion}{\amalg}
\newcommand{\subgroup}{\leq}
\newcommand{\supergroup}{\geq}
\newcommand{\superset}{\supset}
\newcommand{\normalsubgroup}{\lhd}
\newcommand{\semiProd}{\rtimes}
\newcommand{\semiprod}{\semiProd}
\newcommand{\dividesnot}{\nmid}
\newcommand{\divides}{\mid}
\DeclareMathOperator{\Aut}{Aut}
\DeclareMathOperator{\supp}{supp}   %support
\DeclareMathOperator{\clos}{clos}  %closure
\DeclareMathOperator{\im}{im}      %image
\DeclareMathOperator{\vol}{vol}    %Volume
\DeclareMathOperator{\diam}{diam}  %Diameter
\DeclareMathOperator{\dist}{dist}  %Distance
\DeclareMathOperator{\ord}{ord}    %order
\DeclareMathOperator{\End}{End}    %Endomorphisms
\DeclareMathOperator{\Hom}{Hom}    %Homomorphisms
\DeclareMathOperator{\Span}{span}
\DeclareMathOperator{\spin}{spin}
\DeclareMathOperator{\diag}{diag}
\DeclareMathOperator{\spec}{Spec}
\DeclareMathOperator{\trace}{trace}
\DeclareMathOperator{\tr}{tr}
\DeclareMathOperator{\pr}{pr}
%\DeclareMathOperator{\res}{res}  %Restriction
\DeclareMathOperator{\grad}{grad}  %Restriction
% \DeclareMathOperator{\Sp}{Sp}    %Spur % schon definiert???
\DeclareMathOperator{\fl}{f\/l}  %flip
\DeclareMathOperator{\coker}{coker}
\DeclareMathOperator{\Char}{Char}
\DeclareMathOperator{\ind}{ind}
\DeclareMathOperator{\sgn}{sgn}
\DeclareMathOperator{\Ric}{Ric}  % Ricci curvature
\DeclareMathOperator{\scal}{scal}  % Scalar curvature
\DeclareMathOperator*{\dirlim}{dirlim}
\DeclareMathOperator*{\invlim}{invlim}
\DeclareMathOperator{\rank}{rank}
\DeclareMathOperator{\coind}{coind}
\DeclareMathOperator{\res}{res}
\newcommand{\Kommentar}[1]{\textbf{(Kommentar: #1)}}
%\newcommand{\Kommentar}[1]{}
\newcommand{\forget}[1]{}
\newenvironment{Forget}{}{}
\newcommand{\innerprod}[1]{\langle #1 \rangle}

{\catcode`@=11\global\let\c@equation=\c@theorem}
\renewcommand{\theequation}{\thetheorem}
% Hier werden Gleichungen und Theoreme zusammen gezaehlt. Soll ein anderer Zaehler statt theorem verwendet werden (entspr. dem \newtheorem-Befehl), muss 2-mal theorem durch diesen Zaehler ersetzt werden. (Die Zeilen entsprechen der Zaehlung von \newtheorem{equation}[theorem]).

%\renewcommand{\labelenumi}{(\arabic{enumi})}

\renewcommand{\theenumi}{(\arabic{enumi})}
\renewcommand{\labelenumi}{\theenumi}
% Diese neue Variante schreibt jetzt auch arabische Ziffern in
% Klammern beim zitieren!

\allowdisplaybreaks[2]


\begin{document}
\pagestyle{myheadings}
\markboth{P.~Laubenheimer, U.~Stuhler and T.~Schick}{Completions of countable
  non-standard models of the rational numbers}

\date{Last compiled \today; last edited  \heuteIst or later}


\title{Completions of countable non-standard models of $\rationals$}

\author{Peter Laubenheimer 
\and Thomas Schick\\
Uni G{\"o}ttingen\\
Germany
\and Ulrich Stuhler\thanks{
email: \protect\href{mailto:schick@uni-math.gwdg.de}{
  schick@uni-math.gwdg.de}, \protect\href{mailto:stuhler@uni-math.gwdg.de}{stuhler@uni-math.gwdg.de}
\protect\\
www \protect\href{http://www.uni-math.gwdg.de/schick}{http://www.uni-math.gwdg.de/schick},
\protect\href{http://www.uni-math.gwdg.de/stuhler}{http://www.uni-math.gwdg.de/stuhler}
\protect\\
Fax: ++49 -551/39 2985
}}
\maketitle

\begin{abstract}
  In this note, we study non-standard models of the rational numbers
  with countably many elements.

  These are ordered fields, and so it makes sense to complete them,
  using non-standard Cauchy sequences. The main part of this note
  shows that these completions are
  real closed, i.e.~each positive number is a square, and each
  polynomial of odd degree has a root. This way, we give a direct proof of a
  consequence of a theorem of Hauschild. In a previous version of this note,
  not being aware of these results, we missed to mention this reference. We
  thank Matthias Aschenbrenner for pointing out this and related work.

  We also give some information about the set of real parts of the finite
  elements of such completions ---about the more interesting results along
  this we have been informed by Matthias Aschenbrenner and Martin Goldstern.

  The main idea of our proof relies on a way to describe
  real zeros of a polynomial in terms of first order logic. This is
  achieved by carefully using the sign changes of such a polynomial.
\end{abstract}


\section{About the "'size"' of ${}^*\rationals$}
\label{sec:about-size-rationals}

Let $^*\rationals$ be a countable field which is elementary equivalent, but not
isomorphic to $\rationals$. Recall that ``elementary equivalent''
means that an (arithmetic) expression of first order is true in
$^*\rationals$ if and only if it is true in $\rationals$.

We have a canonical embedding of $\rationals$ into
$^*\rationals$. We continue to denote the image with $\rationals$. Recall
that any non-standard model of $\rationals$ contains an element $e$
such that $e>q$ for each $q\in\rationals$.

The shortest way to construct a model for ${}^*\rationals$ uses model
theory. We simply take as axioms all axioms of $\rationals$ and
additionally the following countable number of axioms: the existence
of an element $e$ with $e>1$, $e>2$, \ldots.

Each finite subset of this axioms is satisfied by the standard
$\rationals$. By the compactness theorem in first order model theory,
there exists a model which also satisfies the given infinite set of
axioms. By the theorem of L{\"o}wenheim-Skolem, we can choose
such models of countable cardinality.


Each non-standard model ${}^*\rationals$ contains the (externally
  defined) subset ${}^*\rationals_{fin}:=\{ x\in {}^*\rationals\mid
  \exists n\in\rationals: -n\le x\le n\}$.

  Every element $x\in {}^*\rationals_{fin}$ defines a Dedekind cut:
  $\rationals=\{q\in\rationals\mid q\le x\} \cup \{q\in\rationals\mid
  q>x\}$. We therefore get a order preserving map
  \begin{equation}\label{eq:finite_part}    fp\colon {}^*\rationals_{fin} \to \reals
  \end{equation}
  which restricts to the standard inclusion of the standard rationals
  and which respects addition and multiplication. An element of
  ${}^*\rationals_{fin}$ is called \emph{infinitesimal}, if it is
  mapped to $0$ under the map $fp$.

  We first address the question what is the possible range of $fp$.

  \begin{proposition}
    Choose an arbitrary subset $M\subset\reals$. Then there is a model
    ${}^*\rationals^M$ such that $fp({}^*\rationals_{fin}^M) \superset
    M$. Moreover, the cardinality of ${}^*\rationals^M$ can be chosen
    to coincide with $M$, if $M$ is infinite.
  \end{proposition}
  \begin{proof}
    Choose $M\subset \reals$. For each $m\in M$ choose
    $q_1^m<q_2^m<\cdots <p_2^m<p_1^m$ with $\lim q_k^m=\lim p_k^m =m$.

    We add to the axioms of $\rationals$ the following axioms:
    $\forall m\in M \exists e_m$ such that $q_k^m<e_m<p_k^m$ for all
    $k\in\naturals$.

    Again, the standard $\rationals$ is a model for each finite subset
    of these axioms, so that the compactness theorem implies the
    existence of ${}^*\rationals^M$ as required, where the cardinality
    of ${}^*\rationals^M$ can be chosen to be the cardinality of the
    set of axioms, i.e.~of $M$, if $M$ is infinite. Note that by
    construction $fp(e_m)=m$ for all $m\in M$.
  \end{proof}

\begin{remark}
  It follows in particular that for each countable subset of $\reals$
  we can find a countable model of ${}^*\rationals$ such that the
  image of $fp$ contains this subset. Note, on the other hand, that
  the image will only be countable, so that the different models will
  have very different ranges.

  We prove in Section \ref{sec:cauchy-completions} that
  $fp({}^*\rationals)$ contains for any countable model
  ${}^*\rationals$ every real algebraic number. It remains an open
  problem to determine precisely the possible image sets.

  Matthias Aschenbrenner kindly explained us, that the set of computable real
  numbers always is contained in the image. Here, a real number $r\in\reals$
  is \emph{computable} if the set of pairs $\{(m,n)\in \naturals\mid n\ne 0,
  m/n<\abs{r}\}$ is computable. Moreover, he also explained that the image is
  itself a real closed subfield of $\reals$. As explained by Martin Goldstern,
  the first result can be strengthened to the fact that every \emph{definable}
  real number is contained in the image. A real number $r$ is
  definable if the 
  set of pairs $\{(m,n)\in \naturals^2\mid n\ne 0, m/n<\abs{r}\}$ is
  definable by a formula $A(x,y)$ of first order logic over the rationals. 
\end{remark}


\section{Cauchy completions}
\label{sec:cauchy-completions}



\begin{definition}
  A Cauchy-Sequence in $^*\rationals$ is a sequence
  $(a_k)_{k\in\naturals}$ such that for every $\epsilon\in
  {}^*\rationals$, $\epsilon>0$ there is an $n_\epsilon\in\naturals$
  such that $\abs{a_m-a_n}<\epsilon$ for each $m,n>n_\epsilon$.

  We define the completion $\overline{{}^*\rationals}$ in the usual
  way as equivalence classes of Cauchy sequences.
\end{definition}

\begin{remark}\label{rem:inf_close}
  This is a standard construction and works for all ordered
  fields. The result is again a field, extending the original
  field. Note that, in our case, each point in
  $\overline{{}^*\rationals}$ is infinitesimally close to a point in
  ${}^*\rationals$, since~there is an $0<\epsilon\in{}^*\rationals$ which is smaller
  than every positive rational number.
\end{remark}

  In many non-standard models of $\rationals$, there are no sequences
  tending to zero which are not eventually zero. This is not the case
  if ${}^*\rationals$ is countable, so that
  $\overline{{}^*\rationals}$ is potentially different from
  ${}^*\rationals$ (and we will actually see that it is different
  from ${}^*\rationals$ as a consequence of  Theorem \ref{theo:real_closed}.

\begin{lemma}\label{rem:def_of_zero_sequence}
 Assume that  ${}^*\rationals$ is countable. Let $\{q_1,q_2,\dots\}={}^*\rationals_{>0}$ be a
  enumeration of the positive elements of ${}^*\rationals$.

  Define the sequence $c_k$ inductively as follows: $c_1:=q_1$, $c_k$
  is the next element in the list after $c_{k-1}$ which is smaller
  than $c_{k-1}$. We end up with a decreasing sequence
  which eventually is smaller than each positive element of
  ${}^*\rationals$, i.e.~tends to zero.
\end{lemma}

\begin{remark}
  Every  element of ${}^*\rationals$ which is algebraic over
  $\rationals$ already belongs to $\rationals$.

  This follows since for a given (irreducible) polynomial
  $p\in\rationals[x]$ the statement: ``there is no $a\in\rationals$
  with $p(a)=0$'' is of first order and therefore remains true in
  ${}^*\rationals$. 
\end{remark}

\begin{lemma}\label{lem:extend_fp}
  For every model ${}^*\rationals$, $fp\colon {}^*\rationals_{fin}\to\reals$ extends
  to $fp\colon \overline{{}^*\rationals}_{fin}\to\reals$, but the
  image is unchanged.
\end{lemma}
\begin{proof}
  Every $x\in\overline{{}^*\rationals}$ is the limit of a
  sequence of elements in ${}^*\rationals$, i.e.~is inienitesimally
  close to elements in ${}^*\rationals$. Consequently, the finite part
  is the limit of elements of ${}^*\rationals$, and the map $fp$
  extends by continuity. Since the image of an infinitesimal element
  under $fp$ is $0\in\reals$, for each $x\in
  \overline{{}^*\rationals}_{fin}$ there is an
  $x'\in{}^*\rationals_{fin}$ with $fp(x)=fp(x')=0$ (since $x-x'$ is infinitesimal).
\end{proof}

Nevertheless, the passage from ${}^*\rationals$ to
$\overline{{}^*\rationals}$ adds many roots of polynomials:

\begin{theorem}\label{theo:real_closed}
Assume that ${}^*\rationals$ is a non-standard model of the rationals
which has countably many elements. Then the field $\overline{{}^*\rationals}$ is real closed, i.e.
 \begin{enumerate}
 \item If $c\in \overline{{}^*\rationals}$ satisfies $c>0$ then there is
   $\sqrt{c}\in\overline{{}^*\rationals}$ with $\sqrt{c}>0$ and $\sqrt{c}^2=c$.
 \item If $p\in \overline{{}^*{\rationals}}[x]$ is a polynomial of odd degree, then
   there is $a\in \overline{{}^*\rationals}$ with $p(a)=0$.
 \end{enumerate}
\end{theorem}

This is a direct consequence of the following theorem \ref{theo:sign_changes},
because every
polynomial of odd degree, as well as the polynomial $x^2-a$ for $a$
positive has a sign change. However, it can also be regarded as a special case
of the following theorem of Hauschild (compare \cite{Priess}).
\begin{theorem}
  \label{theo:hauschild}
  The completion (defined using Cauchy sequences indexed by ordinals) of an
  ordered field $K$ is real closed if and only if for every polynomial $f\in
  K[x]$ and all $a,b,\epsilon\in K$ with $a<b$ and every $\epsilon>0$ there is
  $c\in K$ with $a<c<b$ and $\abs{f(c)}<\epsilon$.
\end{theorem}

Since the condition is valid in $\rationals$ and is ``a'' statement of logic
of first order, the condition is valid for \emph{any} non-standard model
${}^*\rationals$, as well, and it follows that the completion is real closed.

This result was kindly pointed out to us by Matthias Aschenbrenner, and
therefore makes the following arguments superfluous. Nonetheless, it might be
a nice illustration of this kind of argument to follow the argument, which is
neither particularly new nor particularly general.

\begin{theorem}\label{theo:sign_changes} Suppose, $f(x)\in\ ^*\bar{\bbbq}[x]$, a polynomial of degree $m$, $a,b \in\ ^*\bar{\bbbq}$ such that $f(a)f(b) < 0$, that is, $f$ is changing sign between $a$ and $b$, then there exists $x\in [a,b]$, $f(x)=0$.
\end{theorem}


\begin{remark}
  Let $fp\colon {}^*\rationals\to\reals$ be the finite part map of
  \eqref{eq:finite_part}. We denote the elements of $f^{-1}(r)$ for $r\in\reals$
  \emph{infinitesimally close} to $r$.

  Note that by Lemma \ref{lem:extend_fp} the map $fp$ extends to
  $\overline{{}^*\rationals}$, but its image is unchanged.
  Theorem \ref{theo:real_closed} therefore implies in particular that
  for each polynomial of odd degree over ${}^*\rationals$ we can find
  points which are ``infinitesimally close'' to a root.
\end{remark}

\begin{proof}[Proof of Theorem \ref{theo:sign_changes}]

\noindent
{\bf I)} In a first step we show the theorem under the additional assumption, that $f(x)\in\ ^*\bbbq[x]$.\\
Proof of this: By choosing $a',b'$ sufficiently close to $a,b$, such that $[a',b']\subset (a,b)$, we can assume additionally, that $a,b\in\ ^* \bbbq$ as well. We consider the following statement: Given $^*n\in\ ^*\bbbn$, $a,b\in\ ^*\bbbq$, such that $f(a)f(b)< 0$, there exist $a_1,b_1\in\ ^*\bbbq$, $[a_1,b_1]\subset[a,b]$, $|a_1-b_1|\le\frac{1}{^*n}$, such that $f(a_1)f(b_1) < 0$.

This is a first order statement, which obviously is true in $\bbbq$ itself. Therefore it is true in $^*\bbbq$. Now we choose a sequence of nonstandard numbers $(^*n_i\mid i=1,2,\ldots)$ in $^*\bbbn$, such that $\lim\limits_{i\to\infty}\ ^*n_i=\infty$ resp. $\lim\limits_{i\to\infty}\frac{1}{^*n_i}=0$. We can choose a sequence of nested intervalls $[a,b]\supset[a_1,b_1]\supset\ldots ,$ such that $|a_i-b_i|\le\frac{1}{^*n_i}$ and $f(a_i)f(b_i) < 0$ for all $i=1,2,\ldots$. Then $(a_i\mid i=1,2,\ldots)$, and $(b_i\mid i=1,2,\ldots)$ are Cauchy convergent sequences in $^*\bbbq$, which have a common limit $\lim\limits_{i\to\infty}a_i=\lim\limits_{i\to\infty}b_i=x\in[a,b]$. By renaming the $a_i,b_i$ we can additionally assume, that $f(a_i) < 0$ for all $i \ge 1$, $f(b_i) > 0$ for all $i\ge 1$. (It is not necessary, that $a_i < b_i$!) Then
$$
f(x)=f(\lim\limits_{i\to\infty}a_i)=\lim\limits_{i\to\infty} f(a_i) \le 0\quad,
$$
but also 
$$
f(x)=f(\lim\limits_{i\to\infty}b_i)=\lim\limits_{i\to\infty}f(b_i) \ge 0\quad .
$$ 
Therefore $f(x)=0$ and I) is proved.

\medskip\noindent
{\bf II)}\, Now we show the theorem in full generality. We do this by induction with respect to $m=\deg f(x)$.

\medskip
i)\, \,  $m=1$ is trivial.

\medskip
ii) \, \,  \emph{Induction step.} We assume the theorem to hold for all polynomials of degree $\le (m-1)$. We will show, that it holds for $f(x)$, $\deg f(x)=m$. We consider the derivative $f'(x)$ on the intervall $[a,b]$. Suppose $\{y_i\mid 0=1,\ldots,r\}$ is the set of zeros of $f'$ on $[a,b]$ in $^*\bar{\bbbq}$, so $r\le (m-1)$. We obtain a partition of the intervall $[a,b]$ 
$$
a:=y_0\le y_1 < y_2 < \ldots < y_r \le b=: y_{r+1}\quad.
$$
By the induction hypothesis, $f'(x)$ does not change sign on any of the open intervalls $(y_i,y_{i+1})$. However, there must be a change of sign of $f$ in at least one of the intervalls $[y_i,y_{i+1}]$.


Therefore, by replacing, if necessary, $[a,b]$ by one of the intervalls
$[y_i,y_{i+1}]$,  we can assume from the beginning, that $f$ on $[a,b]$
satisfies the following conditions:
\begin{enumerate}
\item \quad $f(a) < 0$, $f(b)> 0$
\item \quad Either $f'(x) > 0$ for all $x\in (a,b)$ or $f'(x) < 0$ for all $x\in (a,b)$.
\end{enumerate}
Replacing, if necessary, $f$ by $(-f)$, we can assume also $a < b$. Additionally, we can assume: $a,b\in\, ^*\bbbq$.

Of course, additionally we can exclude the second possibility in 2) as follows:\\ choose $\varepsilon\in\ ^*\bbbq$, such that $\varepsilon(b-a) < \frac{f(b)-f(a)}{2}$ and approximate  $f(x)$ by $g(x)\in\ ^*\bbbq[x]$, $\deg(g)=m$, coefficientwise so sharp, such that $|f'(x)-g'(x)| < \varepsilon$ holds on $[a,b]$. This implies in particular, that $g'(x) < \varepsilon$ for $x\in [a,b]$. Then as this is again a first order statement, which does hold for $\bbbq$, we have
$$
g(b)-g(a) < \varepsilon(b-a)\quad,
$$
so 
$$
g(b)-g(a) < \frac{f(b)-f(a)}{2}=: c\quad.
$$
If $g$ is chosen, such that additionally $|g(b)-f(b)| < \frac c4$, $|g(a)-f(a)| < \frac c4$ hold, we obtain a contradition. Therefore we can even assume
$$
f(a) < 0\, \, f(b) > 0\quad ,\leqno 1)
$$
$$
f'(x) > 0\, \, \mbox{\rm for all}\, \, x\in (a,b)\, \, .\leqno 2)
$$
Using the same kind of argument, we obtain additionally
$$
f(x)\, \, \mbox{\rm is monoton increasing on}\, \, [a,b]\, \, .\leqno 3)
$$
We can still improve the situation by finding $m,M\in\ ^*\bbbq$, $0 < m < M$ such that $m < f'(x) < M$ holds for all $x\in [a,b]$, (again it may be necessary to replace $[a,b]$ by a subintervall).


This can be seen as follows: Consider the second derivative $f''(x)$ on $[a,b   ]$. By induction hypothesis there is a finite set $\{z_1,\ldots,z_s\}$ of zeros of $f''(x)$ on $[a,b]$, $s\le m-2$. Using the considerations above, it follows, that $f'(x)$ is monoton on each of the intervalls $[z_i,z_{i+1}],\, i=0,\ldots,s$, where $z_0:=a$, $z_{s+1}:=b$. Therefore we can choose 
$$
m:= \inf\{f'(z_i)\mid i=0,\ldots,s+1\}
$$
and
$$
M:= \sup\{f'(z_i)\mid i=0,\ldots,s+1\}
$$
and the claim above is shown.


Now, finally the situation is sufficiently under control to proof the theorem: As earlier, we choose a sequence of nonstandard numbers $(^*n_i\mid i=1,2,\ldots)$, $^*n_i\in\ ^*\bbbn$ satisfying $\lim\limits_{i\to\infty}\frac{1}{^*n_i}=0$.


We choose polynomials $g_i(x)\in\ ^*\bbbq[x]$ $(i \ge 1)$ satisfying
$$
|f(x)-g_i(x)|\le \frac{1}{^*n_i}\leqno 1)
$$
$$
|f'(x)-g'_i(x)|\le \frac{1}{^*n_i}\leqno 2)
$$
for $x\in[a,b]$.


Additionally, we can assume
\begin{enumerate}
\item[3)] \, $g_i(a) < 0,\, g_i(b) > 0$ for $i\in\bbbn$
\item[4)] \, $g_i(x)$ is monoton increasing on $[a,b]$  and one has
$$
m < g'_i(x) < M \, \, \mbox{\rm for}\, \, x\in [a,b]
$$
$i\in\bbbn$ arbitrary.
\end{enumerate}
(4) is satisfied by replacing $m,M$ for example by $\frac m2,\, 2M$) and throwing away finitely many $g_i$ if necessary.)


By I) above, we can conclude, that there exist $x_i\in\ ^*\bar{\bbbq}$, $x_i\in[a,b]$, uniquely determined, such that $g_i(x_i)=0$, $i=1,2,\ldots$\ , .


We show next that $(x_i\mid i\in \bbbn)$ is a Cauchy sequence in $^*\bar{\bbbq}$. \\
We conclude as follows: Consider $x_i,x_j\in [a,b]$. Assume first, that $g_j(x_i) < 0$ which implies $x_i < x_j$. We have
$$
0=g_j(x_j) \ge g_j(x_i)+(x_j-x_i)m
$$
This implies
$$
0 < x_j-x_i < \frac{-g_j(x_i)}{m}
$$
However,
\begin{eqnarray*}
|g_j(x_i)|&=&|g_j(x_i)-g_i(x_i)|\\
 &\le&|g_j(x_i)-f(x_i)|+|f(x_i)-g_i(x_i)|\\
 &\le&\frac{2}{\min\{^*n_i,\ ^*n_j\}}
\end{eqnarray*}
Interchanging $i,j$ we obtain the same estimate, if $g_j(x_i) > 0$. (Then $x_i>x_j$, so $g_i(x_j) < g_i(x_i)=0$).


In any case we obtain $|x_j-x_i| \le \frac{2}{m\min\{^*n_i,\ ^*n_j\}}$, 
which implies that $(x_i\mid i=1,2,\ldots)$ is a Cauchy sequence in
$[a,b]$. So $\lim\limits_{i\to\infty}x_i=: x\in [a,b]$ already
exists. We obtain $f(x)=f(\lim\limits_{i\to\infty}
x_i)=\lim\limits_{i\to\infty}f(x_i)$, but $|f(x_i)|\le
g_i(x_i)+\frac{1}{^*n_i}\le \frac{1}{^*n_i}$, so $f(x)=0$
follows. \hfill$\Box$
 
\end{proof}
% We now prove Theorem \ref{theo:real_closed}. Fix $n\in\naturals$ odd. 


% Consider the following statement:
% \begin{statement}

% As a shortcut to be used later, we define for a polynomial $p(x)$ of
% degree $n$ and $a<b$ the statement
% OS(p,a,b) (``there is exactly an odd number of sign changes of $p$ in $[a,b]$), defined as follows
% \begin{equation}
%   \begin{split}
%     \exists  1\le & k\le n \text{ odd }: \forall \epsilon>0 \; \exists a<\alpha_1<\beta_1<\cdots <
%     \alpha_k<\beta_k<b:\\
%     &\qquad\abs{\beta_i-\alpha_i}<\epsilon \text{ for } i=1,\dots,k\\
%     & p(\alpha_i)p(\beta_i)<0,\; p(t_1)p(\beta_i)\le
%     p(t_2)p(\beta_i)\,\forall \alpha_i\le t_1\le t_2\le \beta_i, i=1,\dots,k;\\
%     & p(t)p(\alpha_1)>0 \forall a\le t\le \alpha_1; p(t)\cdot
%     p(\beta_k)>0\,\forall \beta_k\le t\le b, \\
%     & p(\beta_i)p(t)>0\,\forall
%     \beta_i\le t\le \alpha_{i+1}, i=1,\dots,k-1
% \end{split}
% \end{equation}

% We now formulate the main statement about zeros of polynomials of 
% \emph{odd}
% degree $n$ with
% rational coefficients:
% \begin{equation}\label{eq:main_statement}
%   \begin{split}
%     \forall C &>0: \exists r>0 \text{ and } \forall \epsilon>0 \exists \delta>0:\hfill \\
%     & \text{if } a_0,\dots, a_n\in\rationals \text{ with }
%     p(x):=\sum_{i=0}^n
%     a_kx^k,  a_n=1, \abs{a_k}\le C\hfill \\
%     & \quad\text{then }p(-r)<-1, p(r)>1 \text{ and }\exists l_o\in
%     \{1,\dots,n\}, \exists l_e\in
%     \{0,\dots,n-1\}\\
%     &\quad\exists \text{ intervals
%     }[b_1,c_1],\dots,[b_{l_o+l_e},c_{l_o+l_e}] \subset [-r,r],
%     \frac{3\epsilon}{8n}\le c_i-b_i\le \frac{\epsilon}{n}
%     \forall i=1,\ldots,l\\
%   &  \qquad\qquad\text{ and pairwise of distance }>\frac{\epsilon}{2n}\\
%     &\text{ with the property:}\\
%   &\quad  \forall \tilde a_0,\dots,\tilde a_n: \abs{a_k-\tilde a_k}<\delta,
%     \tilde p(x):=\sum_{i=0}^n \tilde a_i x^i\\ 
%    &
%       \implies \;OS(\tilde
%       p,b_i+\frac{\epsilon}{16n},c_i-\frac{\epsilon}{16n}) \text{ for  $i=1,\dots,l_o$};\\
%    & \qquad\text{and if $-r\le \alpha<\beta<\alpha+\frac{\epsilon}{16n^2}\le
%       r$
%       satisfies $\tilde p(\alpha)\cdot \tilde p(\beta)\le 0$}\\
%    &\qquad \text{ then }\exists i\in\{1,\dots,l_e+l_o\}\text{ with }\tilde
%     [\alpha,\beta]\subset
%     [b_i+\frac{\epsilon}{16n},c_i-\frac{\epsilon}{16n}]
% \end{split}
% \end{equation}

% In other words,  exactly for $i=1,\dots,l_o$, the polynomial $\tilde p(x)$ has an
% odd number of sign changes in the interval $[b_i,c_i]$, i.e.~an odd number of real zeros (counted
% with multiplicity), and every sign change (i.e.~every real zero of odd
% order)
% happens in one of the intervals $[b_i,c_i]$ for $i=1,\dots,l_o+l_e$.
% \end{statement}
% This is a (sloppily formulated) statement of first order logic.

% \begin{lemma}\label{lem:statement_true}
%   If $n$ is odd, the statement \eqref{eq:main_statement} is true over $\rationals$,
%   consequently it is true over each non-standard model ${}^*\rationals$.
% \end{lemma}
% \begin{proof}
%   For simplicity, we use real and complex analysis to prove this
%   lemma. First recall that the map $\reals^n\to \complexs^n/\Sigma_n$
%   which assigns to $(a_0,\dots,a_{n-1})$ the unordered tuple of the
%   complex zeros (with multiplicities) of $p(x)= x^n+\sum_{k=0}^{n-1} a_kx^k$ is continuous, and therefore
%   uniformly continuous on $[-C,C]^n$. For a given $C>0$, $\epsilon>0$
%   choose therefore $\delta>0$ such that whenever $a_i,\tilde a_i\in
%   [-2C,2C]$ with $\abs{a_i-\tilde a_i}<\delta$, $i=0,\dots,n-1$, for each zero of $p(x):=\sum_{i=1}^n
%   a_ix^i$ ($a_n=1$) there is a zero of $\tilde p(x):=\sum_{i=1}^n
%   \tilde a_ix^i$ ($\tilde a_n=1$) of distance $<\frac{\epsilon}{16 n^2}$ and vice
%   versa.

%   Moreover, the map 
%   \begin{equation*}
% [-2C,2C]^n\times \reals\to \reals;
%   (a_0,\dots,a_{n-1},x)\to \sum_{i=0}^{n-1} a_ix^i+x^n
% \end{equation*}
%  is proper, and
%   tends to $\pm\infty$ for $x\to \pm\infty$. Therefore, we find $r>2$
%   depending on $C>0$, such that $p(-s)<-1$ and $p(s)>1$ for all
%   polynomials $p(x)=x^n+\sum_{i+0}^{n-1} a_ix^i$ with $a_i\in [-2C,2C]$
%   as above and for all $s\ge r/2-1$. 

%   We now construct the intervals as required. Fix $a_0,\dots,
%   a_{n-1}\in [-C,C]$. Let $t_1<\dots<t_v\in
%   [-r,r]$ be the real parts (this time without multiplicities) of all zeros of $p(x)=\sum_{i=0}^n a_ix^i$
%   ($a_n=1$) which lie in the interval $[-r,r]$. Note that $v\ge 1$ by
%   the mean value theorem (more precisely, there is an odd number of
%   real zeros of $p(x)$ (counted with multiplicity) contained in $[-r,r]$) and $v\le n$.

%   Consider on $\{t_1,\dots,t_v\}$ the equivalence relation generated
%   by $t_i\sim t_j$ if $\abs{t_i-t_j}<\epsilon/n$. The set
%   $\{t_1,\dots,t_v\}$ splits into $l\ge 1$ equivalence classes
%   (ordered by the size of its members), for $1\le i\le l$, the $i$-th
%   equivalence class
%   consists of a consecutive sequence $t_{j_i}<\cdots< t_{k_i}$ where
%   the distance of two neighbors is $<\epsilon/n$, with $1=j_1\le k_1 <
%   j_2\le k_2<\cdots <j_{l}\le k_{l}=v$, and where
%   $t_{j_i}-t_{k_{i-1}}>\epsilon/n$ for $i=2,\dots, l$.

%   Set 
%   \begin{equation*}
% b_i:=t_{j_i}-\frac{3\epsilon}{16n},\qquad
%   c_i:=t_{k_i}+\frac{3\epsilon}{16n}.
% \end{equation*}
% Since there are at most $n$
%   elements in each equivalence class, $c_i-b_i\le \epsilon$ for each
%   $i\in\{1,\dots,l\}$. Moreover, 
%   \begin{equation*}
% b_{i+1}-c_i \ge
%   \frac{\epsilon}{n}-2\frac{3\epsilon}{16n} \ge \frac{\epsilon}{2n}\quad
%   \forall i\in\{2,\dots,l\}.
% \end{equation*}

%   We define now $l_o>0$ such that exactly $l_0$ of the intervals $[b_i,c_i]$
%    contain an odd number of real zeros (counted with multiplicity) of
%   $p(x)$. Since the total number of zeros of $p(x)$ is odd, and the
%   non-real ones come in complex conjugate pairs, $l_o$ indeed has to
%   be different from zero. If we consider $\tilde p(x)= x^n+\sum_{i=0}^{n-1} \tilde a_i
%   x^i$ with $\abs{a_i-\tilde a_i}<\delta$, then each of the zeros of
%   $\tilde p(x)$ has distance at most $\frac{\epsilon}{16n^2}$ from one
%   of the zeros of $p(x)$, and $\tilde p(x)$ has no real zero outside
%   $[-r,r]$. 

%   We claim that exactly the same $l_0$ of the intervals $[b_i,c_i]$ as
%   before contain an odd number of zeros of $\tilde p(x)$. This is the
%   case because:
%   \begin{enumerate}
%   \item by the choice of $\delta$, a zero of $p(x)$ corresponding to a
%     real  zero of $\tilde p(x)$ can not be a real zero which lies in a
%     different interval, and vice versa.
%   \item Non-real zeros of $p(x)$ or $\tilde p(x)$ only occur in
%     complex conjugate pairs,
%   \item all real zeros of $p(x)$ and of $\tilde p(x)$ are contained in
%     $[-r/2-1,r/2+2]\subset [-r,r]$
%   \end{enumerate}
%   such that the number of real zeros of
%   $p(x)$ and of $\tilde p(x)$  in any of the intervals $[b_i,c_i]$ can
%   only differ by an even integer.

%   Finally, note that by our construction of the $t_i$ each real zero
%   of $\tilde p(x)$ has real value which differs at most
%   $\frac{\epsilon}{4n^2}$ from one of the $t_i$ and therefore lies in
%   one of the intervals $[b_i,c_i]$.

%   To finish the statement, we now renumber the intervals, so that the
%   intervals with $i=1,\dots,l_o$ correspond to the intervals with an
%   odd number of real zeros. 

%   It remains now to translate from real
%   zeros of $\tilde p(x)$ of odd order to sign changes. Note that the
%   absence of real zeros outside the intervals
%   $[b_i+\frac{2\epsilon}{16n},c_i-\frac{2\epsilon}{16n}]$ implies the
%   absence of sign changes for points of distance less than
%   $\frac{\epsilon}{16n^2}$ outside the intervals $[b_i+\frac{\epsilon}{16n},c_i-\frac{\epsilon}{16n}]$.

%  A real zero of odd order of a real polynomial corresponds to a sign
%   change, whereas a zero of even order definitely does not give a sign
%   change. Consequently,  the
%   odd number of real zeros in each of the intervals
%   $[b_i+\frac{2\epsilon}{16n},c_i-\frac{2\epsilon}{16n}]$,
%   $i=1,\dots,l_o$ is (using also the
%   intermediate value theorem) equivalent to the existence of an odd
%   number of sign changes in the intervals
%   $[b_i+\frac{\epsilon}{16n},c_i-\frac{\epsilon}{16n}]$, $i=1,\dots,l_o$.

% \end{proof}

% We continue to work with a fixed odd natural number $n$.
% \begin{lemma}\label{lem:interval_containment}
%   Given $C>0, a_0,\dots, a_{n-1}\in [-C,C], \epsilon>0$ choose $r$,
%   $\delta>0$, $l_o,l_e$ and intervals $[b_i,c_i]$ as in Lemma
%   \ref{lem:statement_true}.

%   Choose then $\tilde a_0,\dots,\tilde a_{n-1}\in [-C,C]$ with
%   $\abs{a_i-\tilde a_i}< \delta$ and $\tilde
%   \epsilon<\frac{\epsilon}{16n^2}$, and choose $\tilde \delta$, $\tilde
%   l_0,\tilde l_e$ and intervals $[\tilde b_i,\tilde c_i]$ according to
%   Lemma \ref{lem:statement_true} and this data.

%   Then for each $i=1,\dots, l_o$ at least one of the intervals
%   $[\tilde b_j,\tilde c_j]$ with $j\in \{1,\dots,\tilde l_o\}$ is contained
%   in $[b_i,c_i]$.

%   This statement is true in $\rationals$ and every non-standard model
%   of the rationals.
% \end{lemma}
% \begin{proof}
%   This just follows from the logic  of Statement
%   \eqref{eq:main_statement}: for $i=1,\dots,l_o$ each of the intervals $[b_i+\frac{\epsilon}{16n},c_i-\frac{\epsilon}{16n}]$
%   contains an odd number of sign changes of $\tilde
%   p$. Any one of these sign changes is contained in one of the
%   intervals $[\tilde b_j,\tilde c_j]$. Because of the size of the
%   intervals $[\tilde b_j,\tilde c_j]$ ($<\epsilon/16n^3$), any of those intervals containing a sign
%   change which happens inside
%   $[b_i+\frac{\epsilon}{16n},c_i-\frac{\epsilon}{16n}]$ is entirely
%   contained in $[b_i,c_i]$. Since we start with an odd number of sign
%   changes in $[b_i,c_i]$, at least one of the $[\tilde b_j,\tilde c_j]$ which
%   contains sign changes from $[b_i,c_i]$ for $i=1,\dots,l_o$ will
%   itself contain an odd number of sign changes, as claimed.
% \end{proof}


% \begin{lemma}\label{lem:odd_degree}
%   Let ${}^*\rationals$ be countable, with completion
%   $\overline{{}^*\rationals}$. Then every polynomial of odd degree
%   over $\overline{{}^*\rationals}$ has a zero in $\overline{{}^*\rationals}$.
% \end{lemma}
% For the proof, let
% $q(x) = x^n+\sum_{k=0}^{n-1} a_k x^k\in\overline{{}^*\rationals}[x]$ be a
% polynomial of odd degree $n\in\naturals$.

% Choose $C>0$, $C\in{}^*\rationals$ such that $\abs{a_k}\le C/2$ for
% $k=0,\dots,n-1$. 

% Choose a sequence $\{c_j\}_{j\ge0}\subset
% {}^*\rationals_{>0}$ such that $c_j\xrightarrow{j\to \infty}0$
% and such that $c_j$ is monotonously decreasing, according to
% lemma \ref{rem:def_of_zero_sequence}. Here we use that
% ${}^*\rationals$ is countable.

% Replace $c_j$ inductively by $0<\epsilon_j\le c_j$ in such a way that
% $\epsilon_{j+1} \le \frac{\epsilon_j}{16
%   n^2}$, but still is a sequence tending to zero.

% Using Lemma \ref{lem:statement_true}, to the given $C>0$ choose now
% $r\in{}^*\rationals_{>0}$ and to each $j\in\naturals$ choose
% $\delta_j\in{}^*\rationals$, $\delta_j>0$ so that the Statement
% \eqref{eq:main_statement} is satisfied for the given $C$ and
% $\epsilon_j$. By making $\delta_j$ smaller, if necessary, arrange that
% $\delta_{j+1}\le \delta_j/2$ for each $j\in\naturals$ and that
% $\delta_j\xrightarrow{j\to\infty}0$. 

% Moreover, for each $j\in\naturals$, using the fact that each
% $a_k\in\overline{{}^*\rationals}$ lies in the completion of
% ${}^*\rationals$, choose $a_k^j\in{}^*\rationals$ such that
% $\abs{a_k-a_k^j}<\delta_j/2$. By our choice of $\delta_j$, then
% $a_k^j\xrightarrow{j\to\infty} a_k$ for $k=0,\dots,n-1$.

% Set $q_j(x):=x^n+\sum_{k=0}^{n-1} a_k^j x^k$.

% Choose now inductively, using Lemma \ref{lem:statement_true} and Lemma
% \ref{lem:interval_containment}, $-r<b_j<c_j<r$ such that
% \begin{enumerate}
% \item\label{item:small} $c_j-b_j<\frac{\epsilon_j}{n}$
% \item $OS(q_j,b_j+\frac{\epsilon_j}{16n},c_j-\frac{\epsilon_j}{16n})$,
%   and $q_j$ has no sign changes on $[b_j,b_j+\frac{\epsilon_j}{16n}]$
%   and on $[c_j-\frac{\epsilon_j}{16n},c_j]$
%   i.e.~the interval $[b_j,c_j]$ contains precisely an odd number, in
%   particular at
%   least one, sign change of $q_j$.
% \item\label{item:contained} $[b_{j+1},c_{j+1}]\subset [b_j,c_j]$ for
%   all $j\in\naturals$.
% \end{enumerate}

% It follows that $q_j(b_j)\cdot q_j(c_j)<0$ for each $j\in\naturals$. Pick
% $d_j\in \{b_j,c_j\}$ with $q_j(d_j)<0$ and $e_j\in\{b_j,c_j\}$ with $q_j(e_j)>0$.

% From \ref{item:small} and \ref{item:contained} it follows that the
% sequences $(b_j)_{j\in\naturals}$ and $(c_j)_{j\in\naturals}$ are
% Cauchy sequences which have a common limit $z\in
% \overline{{}^*\rationals}$. Then also $\lim_{j\to\infty} d_j=z$ and
% $\lim_{j\to\infty} e_j=z$.

% By continuity of the map 
% \begin{equation*}
% \overline{{}^*\rationals}^n\to
% \overline{{}^*\rationals}; (x,b_0,\dots,b_{n-1})\mapsto
% x^n+\sum_{k=0}^{n-1} b_k x^k
% \end{equation*}
% it follows that
% \begin{equation*}
%   \begin{split}
%     q(z)& = \lim_{j\to\infty} q_j(d_j)\le 0\\
%     q(z) &= \lim_{j\to\infty} q_j(e_j)\ge 0,
% \end{split}
% \end{equation*}

% \begin{lemma}\label{lem:roots}
%   If ${}^*\rationals$ is countable, then for each
%   $c\in\overline{{}^*\rationals}$ with $c>0$ there is a
%   $z\in\overline{{}^*\rationals}$ with $z^2=c$.
% \end{lemma}
% \begin{proof}
%   We leave the details of the proof to the reader, the method is
%   exactly the same as in the proof of Lemma \ref{lem:odd_degree}. If
%   $a\in\rationals$, $a>0$, we know that we have exactly two sign
%   changes of the polynomial $p(x)=x^2-a$ in the interval $[-a,a]$, and
%   their location depends continuously on $a$. This  can be exploited
%   exactly as in the case of polynomials of odd degree.
% \end{proof}

% This finishes the proof of our main Theorem \ref{theo:real_closed}. 

\medskip

We now want to address the question, which is the image of the map
$fp\colon \overline{{}^*\rationals}_{fin}\to\reals$. Let
$K\subset\overline{{}^*\rationals}$ be the set of all elements which
are algebraic over $\rationals$. Recall that this implies that every
element of $\overline{{}^*\rationals}$ which is algebraic over $K$
already belongs to $K$. Since $\overline{{}^*\rationals}$ is a real
closed ordered field, the subfield $K$ is real closed ordered, as well.

Given 
$p(x)=x^n+\sum_{k=0}^{n-1}a_kx^n\in\rationals[x]$, the statement 
\begin{equation*}
  \abs{p(t)} \ge 1\quad\forall t\in\rationals\text{ with }\abs{t}\ge n+
  \sum_{k=0}^n \abs{a_k}
\end{equation*}
is true and therefore remains true for
$t\in{}^*\rationals$. Since each element in $K$ is the zero of a
polynomial over $\rationals$, $K\subset\overline{{}^*\rationals}_{fin}$.

Consequently, the restricted map $fp\colon K\to \reals$, being
multiplicative and additive and now defined on all of $K$, is a field
extension into the subfield $K_\reals$ of $\reals$ consisting of algebraic
numbers over $\rationals$.

Since both $K$ and $K_{\reals}$ are real closed, adjoining $\sqrt{-1}$
produces the algebraic closure of $\rationals$ and therefore the
induced map $K[\sqrt{-1}]\to K_{\reals}[\sqrt{-1}]$ is an isomorphism,
so that $fp\colon K\to K_{\reals}$ also  is an isomorphism.


\begin{corollary}
  If ${}^*\rationals$ is countable, then the image of 
  \begin{equation*}
    fp\colon {}^*\rationals_{fin}\to \reals
  \end{equation*}
  contains every real number which is algebraic over $\rationals$.
\end{corollary}
\begin{proof}
  This follows from the preceeding discussion because
  $fp(\overline{{}^*\rationals}_{fin})=fp({}^*\rationals_{fin})$. 
\end{proof}

\begin{question}
  We have seen that for every countable subset $Z$ of $\reals$, we can
  find a countable model ${}^*\rationals$ such that the countable set
  $fp({}^*\rationals_{fin})$ contains $Z$.

  On the other hand, $fp({}^*\rationals_{fin})$ always contains the subfield
  $K_{\reals}\subset\reals$ of algebraic numbers over $\rationals$. Even
  stronger, using the results explained to us by Matthias Aschenbrenner and
  Martin Goldstern, $fp({}^*\rationals_{fin})$ always contains all definable
  real numbers, and is real closed.

  It would be interesting to find more precise information about the
  possible image sets.

  In particular: can every subfield of $\reals$ with the above properties be
  obtained as image of $fp$? Does this perhaps depend on the model of set
  theory one uses?
\end{question}

\begin{thebibliography}{1}

\bibitem{MR1098499}
Richard Kaye.
\newblock {\em Models of {P}eano arithmetic}, volume~15 of {\em Oxford Logic
  Guides}.
\newblock The Clarendon Press Oxford University Press, New York, 1991.
\newblock Oxford Science Publications.

\bibitem{Priess}
Sibylla Prie{\ss}-Crampe.
\newblock {\em Angeordnete {S}trukturen}, volume~98 of {\em Ergebnisse der
  Mathematik und ihrer Grenzgebiete}.
\newblock Springer-Verlag, Berlin, 1983.
\end{thebibliography}

\end{document}

%%% Local Variables: 
%%% mode: latex
%%% TeX-master: t
%%% End: 


math0604466.extracted_bib
\begin{thebibliography}{1}

\bibitem{MR1098499}
Richard Kaye.
\newblock {\em Models of {P}eano arithmetic}, volume~15 of {\em Oxford Logic
  Guides}.
\newblock The Clarendon Press Oxford University Press, New York, 1991.
\newblock Oxford Science Publications.

\bibitem{Priess}
Sibylla Prie{\ss}-Crampe.
\newblock {\em Angeordnete {S}trukturen}, volume~98 of {\em Ergebnisse der
  Mathematik und ihrer Grenzgebiete}.
\newblock Springer-Verlag, Berlin, 1983.
\end{thebibliography}