% LaTex2e Feb 2012
% Absoluteness of convexly orderable
% A. Miller
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\begin{document}
\begin{center}
Absoluteness of convexly orderable
\end{center}
\begin{flushright}
A. Miller \\
Feb 2012
\end{flushright}
\bigskip
During a talk by Vincent Guingona \cite{vincent}, Uri
Andrews raised the question of whether the property
of being convexly orderable
is set theoretically absolute. We show that it
is absolute when the language is countable but it isn't for
uncountable languages.
\begin{define}
An $L$-structure $M$ is convexly orderable iff
there is a linear order $\leq$ on $M$ and
a function $\phi\mapsto k_\phi$ from $L$-formulas
to $\om$ such that for any $\phi(x,\vec{y})$ every
set of the form $\phi(M,\vec{a})$ for some $\vec{a}\in M$
can be written as the union of $\leq k_\phi$ convex
sets.
\end{define}
\begin{theorem}
Suppose $V$ is a transitive model of set theory, $M\in V$,
and $V$ models that $M$ is a structure in a countable
language $L$
which is not convexly orderable. Then any transitive model
of set theory $W\supseteq V$ also models that $M$ is
not convexly orderable.
\end{theorem}
\proof
In $V$ let $\phi_n$ for $n<\om$ list all $L$-formulas
and suppose that in $W$ there is a function
$f:\om\to\om$ and linear order on $M$ such that
each set of the form $\phi_n(M,\vec{a})$ is the union
of $\leq f(n)$ convex sets. For each $N<\om$ note
that $V$ models there is a linear order on $M$ such that
for every $n0$ such that
$\bigcup_{i>n$ such that
$$\bigcup_{i