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\put(143,66){\makebox(0,0)[tl]{\footnotesize Proceedings of the Ninth Prague Topological Symposium}}
\put(143,50){\makebox(0,0)[tl]{\footnotesize Contributed papers from the symposium held in}}
\put(143,34){\makebox(0,0)[tl]{\footnotesize Prague, Czech Republic, August 19--25, 2001}}
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\title[Nets in Ordered Fields]{Using nets in Dedekind, monotone, or Scott
incomplete ordered fields and definability issues}
\author{Mojtaba Moniri}
\author{Jafar S. Eivazloo}
\address{Department of Mathematics, Tarbiat Modarres University, Tehran,
Iran}
\email{moniri\_m@modares.ac.ir}
\email{eivazl\_j@modares.ac.ir}
\thanks{Mojtaba Moniri and Jafar S. Eivazloo,
{\em Using nets in Dedekind, monotone, or Scott incomplete ordered fields
and definability issues},
Proceedings of the Ninth Prague Topological Symposium, (Prague, 2001),
pp.~195--203, Topology Atlas, Toronto, 2002; {\tt math.arXiv:GN/0204135}}
\begin{abstract}
Given a Dedekind incomplete ordered field, a pair of convergent nets of
gaps which are respectively increasing or decreasing to the same point is
used to obtain a further equivalent criterion for Dedekind completeness of
ordered fields: Every continuous one-to-one function defined on a closed
bounded interval maps interior of that interval to the interior of the
image.
Next, it is shown that over all closed bounded intervals in any monotone
incomplete ordered field, there are continuous not uniformly continuous
unbounded functions whose ranges are not closed, and continuous 1-1
functions which map every interior point to an interior point (of the
image) but are not open.
These are achieved using appropriate nets cofinal in gaps or coinitial in
their complements.
In our third main theorem, an ordered field is constructed which has
parametrically definable regular gaps but no $\emptyset$-definable
divergent Cauchy functions (while we show that, in either of the two cases
where parameters are or are not allowed, any definable divergent Cauchy
function gives rise to a definable regular gap).
Our proof for the mentioned independence result uses existence of infinite
primes in the subring of the ordered field of generalized power series
with rational exponents and real coefficients consisting of series with no
infinitesimal terms, as recently established by D.~Pitteloud.
\end{abstract}
\subjclass[2000]{03C64, 12J15, 54F65}
\keywords{Ordered Fields, Gaps, Completeness Notions, Definable Regular
Gaps, Definable Cauchy Functions, Generalized Power Series}
\maketitle
\section{A Dedekind Incompleteness Feature via Convergent Nets of Gaps}
A cut of an ordered field $F$ is a subset which is downward closed in
$F$. By a nontrivial cut, we mean a nonempty proper cut.
A nontrivial cut is a gap if it does not have a least upper bound in the
field.
An ordered field is Archimedean (has no infinitesimals) just in case it
can be embedded in ${\mathbb R}$.
The following fact presents some of the well known characterizations of
the ordered field of real numbers.
A more delicate equivalent condition is presented in Theorem 1.2.
\begin{fac}\label{easy2nd}
The real ordered field ${\mathbb R}$ is, up to isomorphism, the
unique ordered field which satisfies either of the following
equivalent conditions:
\begin{itemize}
\item[(i)] Dedekind Completeness, i.e. not having any gaps,
\item[(ii)] connectedness,
\item[(iii)] not being totally disconnected,
\item[(iv)] every (nonempty) convex subset being an interval (of one of
the usual kinds),
\item[(v)] all convex subsets being connected,
\item[(vi)] all intervals being connected,
\item[(vii)] all continuous functions on the field mapping any convex
subset onto a convex subset,
\item[(viii)] all continuous functions on the field satisfying the
intermediate value property.
\end{itemize}
\end{fac}
These are fairly well known, let us give the argument for (iii).
Pick a Dedekind incomplete ordered field $F$. In the Archimedean
case, $F$ is a proper subfield of ${\mathbb R}$ which therefore
misses some points in any real interval. In the non-Archimedean
case, and given any two points $a**0}$,
$C+\epsilon\not\subseteq C$,
\item[(iii)]
All functions $f:F\rightarrow F$ which are Cauchy at
$\infty_{F}$, are convergent there.
\end{itemize}
We abbreviate the (equivalent non first order) axiomatizations obtained
from the theory $OF$ of ordered fields by adding condition (ii) or
(iii) as $S_{rc}COF$ and $S_{cf}COF$ respectively.
Observe that an ordered field of cofinality $\lambda$ is Scott complete if
and only if it has no divergent Cauchy nets of length $\lambda$.
If an ordered field $F$ is not Scott complete, then for all $a****0$, there
exists $\alpha_{0} <\lambda$ such that $a_{\alpha_{0}
+1}-a_{\alpha_{0}}<\delta$.
To see that such a reduction indeed causes no loss of generality, take a
net $(c_{\alpha})_{\alpha<\lambda}$ converging to $0$ such that
$00$, by the assumption we made above, there exists
$\alpha_{0} <\lambda$ such that $a_{\alpha_{0} +1}-a_{\alpha_{0}}<\delta$.
Now by $u_{\alpha +1}-u_{\alpha} \geq 1$ and the way $f$ is constructed,
we have
$$f(a_{\alpha_{0} +1})-f(a_{\alpha_{0}})\geq 1 >\epsilon.$$
(ii).
Consider the interval $[a,b]$ with a point $c\in (a,b)$.
We now follow a construction similar to the one in Theorem
~\ref{involved2nd}.
Using the same notation as there, the change we make is the following.
We linearly inject $T_{\alpha}$, for $1\leq \alpha <\lambda$, into
$T_{(\alpha\cdot 2 )+1}$ but $T_{0}$ onto $[a,c)$.
The latter can be done as in part (i) in a piece-wise manner.
Here we may assume that $T_{0}$ is the upward closure of a strictly
decreasing divergent $\lambda$-net in $(b_{1},b_{0}]$.
Then $c$ will be a boundary point of the image of the open interval
$(a_{1},b_{1})$ under $f$, so that image is not open.
On the other hand, by construction, $f$ sends interior to interior.
\end{proof}
\section{P-Definable Regular Gaps Not Traversed by $\emptyset$-Definable
Cauchy Functions}
We now consider two notions of being definably (with or without
parameters) Scott complete for ordered fields, those corresponding to
$S_{rc}COF$ and $S_{cf}COF$. They are ordered fields with no definable
regular gaps, respectively no definable functions which are Cauchy at
infinity but divergent there. We denote the corresponding theories by
$D_{p}S_{rc}COF$, $D_{p}S_{cf}COF$, $D_{\emptyset}S_{rc}COF$, and
$D_{\emptyset}S_{cf}COF$.
It is easy to see that, as long as there are no definability concerns, all
regular gaps can be traversed by suitable (divergent Cauchy) functions and
vice versa, every divergent Cauchy function induces a regular gap. The
latter converse is shown in Theorem ~\ref{conditional}(i) to be true in
either of the p-definable or $\emptyset$-definable cases. For the former
direction, however, we are only able in Theorem ~\ref{conditional}(ii) to
prove an independence result in a mixed $\emptyset$-definable /
p-definable case.
\begin{lem}\label{shepherdson}
If the ordered field $F$ is a proper dense sub-field of its real closure,
then $F\nvDash D_{p}S_{rc}COF$.
\end{lem}
\begin{proof}
Pick out an element $r\in RC(F)\setminus F$ and consider the set
$$C=\{x\in F: \ RC(F)\vDash x0}$.
By the Cauchy condition for $\mathcal{F}$, there exists $\theta_{0}<\lambda$
such that $\forall \alpha,\beta\in[[F^{G}]]$ with
$\alpha,\beta\geq\chi_{\{a_{\theta_{0}}\}}$, we have
$|\mathcal{F}(\alpha)-\mathcal{F}(\beta)|<\epsilon\chi_{\{g\}}$.
This shows that
$|\mathcal{F}(\alpha)(g)-\mathcal{F}(\beta)(g)|<\epsilon$.
Therefore $\forall\theta_{1},\theta_{2}\geq\theta_{0}$,
$$|f_{g}(\theta_{1})-f_{g}(\theta_{2})| =
|(\mathcal{F}(\chi_{\{a_{\theta_{1}}\}}))(g) -
(\mathcal{F}(\chi_{\{a_{\theta_{2}}\}}))(g)|<\epsilon,$$
since
$\chi_{\{a_{\theta_{1}}\}},\chi_{\{a_{\theta_{2}}\}}
\geq \chi_{\{a_{\theta_{0}}\}}$.
Hence the net $(f_{g}(\theta))_{\theta <\lambda}$ is Cauchy in $F$ and so
convergent there, since $F$ is Scott complete.
\end{proof}
Let $\gamma:G\rightarrow F$ be defined by
$\gamma (g)=\lim(f_{g}(\theta))_{\theta <\lambda}$.
\begin{claim}\label{c2}
$(\forall \eta<\lambda)
(\exists \theta_{0}<\lambda)
(\forall \theta\geq\theta_{0})
(\forall g\in G^{<-a_{\eta}})
\mathcal{F}(\chi_{\{a_{\theta}\}})(g) = \gamma(g)$.
\end{claim}
\begin{proof}[Proof of Claim \ref{c2}]
For any $\epsilon=\chi_{\{-a_{\eta}\}}$, with $\eta <\lambda$, there
exists $\theta_{0}<\lambda$ such that
$\forall\theta_{1},\theta_{2}<\lambda$ with
$\theta_{1},\theta_{2}\geq\theta_{0}$, we have
$$|\mathcal{F}(\chi_{\{a_{\theta_{1}}\}}) -
\mathcal{F}(\chi_{\{a_{\theta_{2}}\}})|<\chi_{\{-a_{\eta}\}}.$$
This shows that for all $\theta_{1},\theta_{2}\geq\theta_{0}$ and
$g<-a_{\eta}$, we have
$$(\mathcal{F}(\chi_{\{a_{\theta_{1}}\}}))(g) =
(\mathcal{F}(\chi_{\{a_{\theta_{2}}\}}))(g)=\gamma (g).$$
Therefore, for all $\theta\geq\theta_{0}$ and $g\in G^{<-a_{\eta}}$, we
have $\mathcal{F}(\chi_{\{a_{\theta}\}})(g)=\gamma(g)$.
\end{proof}
\begin{claim}\label{c3}
$\gamma\in [[F^{G}]]$.
\end{claim}
\begin{proof}[Proof of Claim \ref{c3}]
It suffices to show that the support of $\gamma$ is well ordered.
For all $g\in G$, there exists $\eta <\lambda$ such that
$g<-a_{\eta}$. Let $\theta_{0}$ be as in Claim \ref{c2}.
As $g$ can not be the initial term of any infinite strictly decreasing
sequence in the support of $\mathcal{F}(\chi_{\{a_{\theta_{0}}\}})$,
Claim \ref{c2} shows that the same holds for the support of $\gamma$.
\end{proof}
\begin{claim}\label{c4}
The function $\mathcal{F}$ on $[[F^{G}]]$ tends to $\gamma$ at infinity.
\end{claim}
\begin{proof}[Proof of Claim \ref{c4}]
It is enough, by the Cauchy criterion for $\mathcal{F}$, to apply $\mathcal{F}$
on those $f$'s in $[[F^{G}]]$ that are of the form
$\chi_{\{a_{\theta}\}}$, for $\theta <\lambda$ and let $\theta$ tend to
$\lambda$.
The result is then immediate from Claim \ref{c2}.
\end{proof}
The above claims give the result.
\end{proof}
\begin{lem}\label{ifthencf}
Suppose that $F$ is a Scott complete ordered field, $G$ is a 2-divisible
ordered abelian group, and $K$ is a dense ordered sub-field of $[[F^{G}]]$
which contains $F$. Assume furthermore that $K$ is closed under the
automorphism on $[[F^{G}]]$ sending $\chi_{\{g\}}$ to $\chi_{\{g+g\}}$ and
its inverse.
Then $K$ satisfies $D_{\emptyset}S_{cf}COF$.
\end{lem}
\begin{proof}
If $K=[[F^{G}]]$, then the conclusion will trivially
hold since $[[F^{G}]]$ is Scott complete. Suppose $K\subsetneq
[[F^{G}]]$. Assume for the purpose of a contradiction that
$\mathcal{F}$ is a $\emptyset$-definable divergent Cauchy function on $K$.
Fix a net $(k_{\alpha})_{\alpha<\lambda}$ of elements of $K$,
where $\lambda$ is cofinality of $[[F^{G}]]$, which is cofinal in
the latter. For any $f\in [[F^{G}]]$, let $\lambda (f)$ be the
least ordinal less than $\lambda$ such that $f0}$. From the Cauchy
criterion for $f$, there exists $d\in F$ such that $\forall x,y\geq d$,
$|f(x)-f(y)|<\frac{\epsilon}{2}$. Let $z=f(d)-\frac{\epsilon}{2}$. Then
$F\vDash\psi (z)$ and $F\vDash\neg\psi (z+\epsilon)$.
Now let $\sup(C)=\alpha$.
To show $\lim f(x)=\alpha$, as $x$
becomes arbitrarily large in $F$, let $\epsilon\in F^{>0}$ and $d$
be as before.
By the definition of $\alpha$, there exists $z\in C$
with $z>\alpha-\frac{\epsilon}{2}$.
From this, we get $\exists
x_{0}\geq d$ with $\alpha-\frac{\epsilon}{2}**