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\title{An oscillation theorem for groups of isometries\footnote{Key words 
and phrases: Distortion; oscillation; Ramsey theory; groups of isometries; 
Polish groups. 2000 Mathematics Subject Classification: Primary 
03E15, 03C15,   05D10, 22F50; Secondary: 37F20, 46B03, 54C25, 
54D80, 54E40.}}  
       % Enter your title between curly braces



\author{Greg Hjorth\footnote{Partially supported by NSF grant DMS
0140503}}        % Enter your name between curly braces
\date{\today}          % Enter your date or \today between curly braces 

\maketitle





\abstract{This paper answers a question raised by 
Vladimir Pestov. 

Let $G$ be a non-trivial group of isometries 
of a complete, separable metric space. Let $\hat{G}$ be the 
semi-group arising from its completion with respect to the left 
uniformity. Then there is a uniformly continuous function from 
$\hat{G}$ to the unit interval which assumes both extreme points 
on any non-empty right ideal.} 


\bigskip 
\bigskip 


This paper has one main theorem. 
If $G$ is a group of isometries of a space $X$, 
$\hat{G}$ the 
semi-group that arises as its completion in a certain natural sense, 
then we can find uniformly continuous functions on $X^2$ that 
display high oscillation on the image of any $\rho\in \hat{G}$. 
There are some details to be added to make this statement precise, 
but they are unsurprising, and can be postponed. 
There are some further issues about when $X^2$ can be 
replaced by $X$, and what one can say for specific classes of 
$X$ and $G$, and these need also to be eventually addressed. 

Before any of these 
housekeeping chores, I want to give an 
extended aside on  
two 
seemingly 
unrelated results in combinatorics, 
which can actually be 
viewed as lying at the heart of this theorem. 
Although these results are 
well known, the proofs are sufficiently short to  
be included, and the proof of the second of these 
will be needed at a later point. 

\begin{theorem} (Ramsey's theorem) \label{ramsey}
Let 
\[f: \N^2\rightarrow \{0, 1\}.\] 
Then there is an infinite 
\[A\subset \N\] 
and $i\in \{0, 1\}$  
such that for any $a<b$ we have 
\[f(a, b) =i.\] 
\end{theorem} 

\begin{proof} 
For each $a\in \N$ and infinite $B\subset \N$ 
we may certainly find 
an $i\in \{0, 1\}$ and 
an infinite $B'\subset B$ 
such that for all 
$b\in B'$ 
\[f(a, b)=i_a.\] 
Applying this we  successively choose $a_0\in \N, B_{a_0}\subset \N, 
a_1\in B_{a_0}, B_{a_1}\subset B_{a_0}, 
a_2\in B_{a_1},...$ and so on, to obtain an increasing infinite 
sequence of $a_i$'s and a sequence of infinite sets 
\[B_{a_0}\supset B_{a_1}\supset B_{a_2}\supset...\] 
such that at each $n$ we have some $i_{a_n}$ such that 
every $b\in B_{a_n}$ has $f(a_n , b)=i_{a_n}$. 

At this point we can find some infinite subsequence 
$(a_{n(k)})_{k\in \N}$ such that the corresponding 
$(i_{a_{n(k)}})_{k\in \N}$ is constant.\end{proof} 


The next theorem is sometimes associated with the name of 
Devlin, though his real contribution is not this folklore result 
but the far more non-trivial task of computing certain 
precise bounds for partitions of $\Q^n$ at $n>2$. 
There are various proofs for this theorem. 
The specific proof given below will be important. 

\begin{theorem} (Folklore) \label{devlin} 
There is a function 
\[f: \Q\rightarrow \{0, 1\}\] 
such that for any order preserving injection 
\[\rho: \Q\rightarrow \Q\] 
there exist 
\[q_1< q_2, r_1<r_2,\] 
all in the image of $\rho$, with 
\[f(q_1, q_2)\neq f(r_1, r_2).\] 
\end{theorem} 

\begin{proof} 
Let $(x_n)_{n\in \N}$ be some enumeration of 
$\Q$. For any $x_n<x_m$ in the rationals set 
\[f(x_n, x_m)=0\] 
if $n<m$ and equal to 1 otherwise. 

Now let $\rho: \Q\rightarrow \Q$ be an order preserving 
injection. Since $\rho$ is order preserving we can find 
two real numbers $\alpha_1<\alpha_2$ which are limits of 
increasing sequences from the image of $\rho$. 
Then choose 
sequences $(k(n))_{n\in \N}, (\ell(n))_{n\in \N}$ with 
\[x_{k(n)}\rightarrow \alpha_1, \] 
\[x_{\ell(n)}\rightarrow \alpha_2,\] 
each $x_{k(n)}, x_{\ell(n)}$ in the image of 
$\rho$, each $x_{k(n)}<\alpha_1, x_{\ell(n)}<\alpha_2$. 

We then let $n$ be arbitrary and choose $n'$ so that 
$\ell(n')>k(n)$ and then $n''$ with $k(n'')>\ell(n')$ 
to obtain 
\[f(x_{k(n)}, x_{\ell(n')})=0,\] 
\[f(x_{k(n'')}, x_{\ell(n')})=1.\] 
\end{proof} 

The conclusions of Ramsey's and Devlin's theorem 
contrast one another. We obtain one kind of behavior for 
the ordering of $(\N, <)$ and the total opposite for 
$(\Q, <)$. 

A distinctive property of $(\Q, <)$ is that for any two 
finite increasing sequences of the same size 
\[q_1<q_2<...<q_\ell,\]
\[r_1<r_2<...<r_\ell,\] 
we can find an automorphism of $(\Q, <)$ which sends the 
first sequence to the second. In this sense $(\Q, <)$ is  
{\it ultrahomogeneous}. 
Clearly one could phrase a 
related notion for graphs -- a graph $(V, E)$ is 
{\it ultrahomogeneous} if any 
isomorphism between finite 
subgraphs extends to an automorphism of the whole 
graph -- and after that arbitrary countable structures 
in some relational language. 
Here there should be the analog of considering 
the values of $\pi$ on increasing pairs, and for 
general ultrahomogenous structures the notion is that of 
{\it having the same type}; for the moment 
this can be roughly defined as having the 
same orbit 
under the automorphism group of the structure. 
Then 
instead of letting $\rho$ range over order preserving 
maps we need to consider $\rho$ which preserve the 
entire flank of relations associated to the 
structure. 

\begin{theorem} \label{ultraintro}
Let $\m$ be an ultrahomogeneous relational structure whose 
automorphism group has size at least two. 
Then there is a subset $S\subset\m^2$ on which the 
automorphism group acts transitively and a function 
\[f: S\rightarrow \{0, 1\}\] 
such that for any monomorphism 
\[\rho: \m\rightarrow \m\] 
we have $a_1, a_2, b_1, b_2$ in the image of 
$\rho$ with 
\[(a_1, a_2), (b_1, b_2)\in S,\]
\[f(a_1, a_2)\neq f(b_1, b_2).\] 
\end{theorem} 



\ref{ultraintro} is not the ultimate goal, and 
while a weaker form can be derived 
from the final theorem, its proof is easier. 
The main result of 
this paper is to prove a kind of 
generalization for 
separable complete metric spaces. 

Let $X$ be a complete separable metric 
space; at this stage we will assume just 
for convenience that the 
metric is bounded by one. 
Let $G$ be a group of isometries, and for $\{x_i: i\in \N\}$ a 
dense subset of $X$ we let 
\[\hat{D}(\pi_1, \pi_2)=
\sum_{n\in \N}2^{-n}d(\pi_1(x_n), \pi_2(x_n))\] 
provide a metric on $G$ and we let $\hat{G}$ be the completion of 
$G$ with respect to this metric; $\hat{G}$ can be identified with 
a semigroup of isometries from $X$ to $X$. 


\begin{theorem} For $G\leq {\rm Isom}(X)$ as above, 
$G$ of size at least two, 
there are $x_0, x_1\in X$ and uniformly continuous 
\[f: \overline{\{(\pi\cdot x_0, \pi \cdot x_1): 
\pi \in G\}}\rightarrow [0, 1]\] 
such that for any $\rho\in \hat{G}$ there exist 
\[(y_0, y_1), (z_0, z_1)\in \{(\rho(\pi(x_0)), \rho(\pi(x_1))): 
\pi \in G\}\] 
with 
\[f(y_0, y_1)=0,\] 
\[f(z_0, z_1)=1.\] 
\end{theorem} 


A corollary answers a question raised in 
\cite{kepeto}. 

\begin{corollary} \label{corollaryintro}
Let $G$ be a Polish group of size bigger 
than one and let $\hat{G}$ be its completion with respect to 
the left uniformity. Then there is a bounded, uniformly left 
continuous 
\[f:\hat{G}\rightarrow[0, 1]\] 
such for any non-empty right ideal 
\[{\mathcal I}\subset \hat{G}\] we can find 
\[\rho_0\neq \rho_1\in {\mathcal I}\] 
with $f(\rho_0)=0, f(\rho_1)=1.$ 
\end{corollary} 

The connection between \ref{corollaryintro} and 
\ref{ultraintro} is that the latter can be used to 
prove the former in the case that $G$ is a closed 
subgroup of $S_\infty$, the group of all permutations 
of $\N$ in the topology of pointwise convergence. 


We prove \ref{ultraintro}  
in  
$\S$\ref{ultrahomogeneous}. It consists in 
analyzing the relation of 
which tuples inside the structure can, up to some finite 
error, be defined from which other 
tuples. There are several cases, most of which turn out to be 
trivial, and in the one non-trivial case we construct the 
{\it irrepressibly} oscillating function 
$f$ as in the proof of \ref{devlin}. 
$\S$\ref{firstorder} 
explores 
how 
variations might succeed or fail for other classes of 
countable structures. $\S$\ref{firstorder}, unlike the 
rest of the paper, requires some familiarity with first 
order logic. 

$\S$\ref{lemmas} gives some background information 
about groups of isometries and their ideals. 
We present the theorem in $\S$\ref{biglentil}. 
The central idea of the proof is the notion of 
$\epsilon$-capturing. This is something like an 
analog for metric spaces of the concept of definability 
over countable structures, and indeed 
there are parallels between 
the proof of this theorem and the proof of \ref{ultraintro}. 
There are also certain 
technical problems arising from the use of  
approximation arguments which have no 
precursor in the case of countable structures.  
It may be easier to understand the proof 
in $\S$\ref{biglentil}  
having first seen the proof of \ref{ultra} 
in $\S$\ref{ultrahomogeneous}. 

The \ref{corollaryintro},  
along with associated notational background, is presented in  
$\S$\ref{polishgroups}. 
$\S$\ref{derive} describes 
the connection between the 
main theorem and \ref{ultra}. 
$\S$\ref{tomandjerry} discusses some examples and specific 
situations in 
which a more optimal result may be possible. 
$\S$\ref{ragged} explores 
connections with distortion theorems such as the 
Odell-Schlumprecht theorem for Hilbert space. 


\bigskip

{\bf Acknowledgments:} I am extremely grateful to 
Vladimir Pestov for a number of conversations around the 
subject matter of this paper. My interest began 
after attending a talk of his at the 2003 AMS Annual 
Meeting in Phoenix. 

I am also indebted to Alexander Kechris and Vladimir Pestov 
for reading through earlier handwritten versions of this 
argument, and providing various corrections and helpful comments. 

\section{Devlin type theorems} 
\label{structure}



\subsection{Ultrahomogeneous structures} 
\label{ultrahomogeneous} 

\begin{definition} Let $\m=(M; R_0, R_1, R_2, ...)$ be a relational 
structure, with underlying set $M$, and each $R_i$ a relation of 
arity $n(i)$, which is to say a subset of $M^{n(i)}$. 
(We allow either finitely many relations or countably 
infinitely many.) 
We will make the technically convenient assumption that one of the relations 
is equality. 

$\pi: M\rightarrow M$ is said to be an {\it automorphism} of 
$\m$ if it is a bijection and for any 
$\vec a=(a_0,a_1,...,a_{\ell -1})\in M^{n(i)}$ we 
have 
\[\vec a\in R_i\Leftrightarrow (\pi(a_0), \pi(a_1),...,
\pi(a_{n(i)-1})\in R_i.\]
$\pi:\m\rightarrow \m$ is a 
{\it monomorphism} if it simply respects the relations in this way 
without necessarily being onto. (Since equality is hardwired into our
language, 
such a $\pi$ will necessarily be one to one.) We use Aut$(\m)$ to denote the 
automorphism group and for $\vec a$ a finite sequence from $M$ we use
Aut$(\m, \vec a)$ 
to denote the subgroup that fixes $\vec a$ pointwise. 

For $\vec a=(a_0,a_1,...,a_{\ell -1})\in M^\ell$ 
we say that the {\it quantifier free type of $\vec a$ in $M$}, written 
q.f.type$_\m(\vec a)$, is the set of pairs $(\vec s, i)$, where $\vec
s\in \{0, 1, ... \ell\}^{n(i)}$, 
and $(a_{s(0)}, a_{s(1)}, ..., a_{s(n(i)-1)})\in R_i$. In other words, the 
quantifier free type of a tuple $\vec a$ is the complete description of
which finite 
sequences chosen from $\vec a$, with repetitions allowed, satisfy which
relations from 
the language of $\m$. 

We then say that a structure $\m$ is {\it ultrahomogeneous} if whenever 
$\vec a, \vec b$ are sequences of the same length in $\m$ with the same
quantifier free type, 
then there is an automorphism 
\[\pi:\m\rightarrow \m\] 
with $\pi(a_i)=b_i$ at each applicable $i$. 
\end{definition} 

In the following we will generally observe the usual lazy, convenient,
but arguably inaccurate, 
custom, and identify $\m$ with its underlying set $M$. 

\begin{example} The {\it random graph} can be defined in the following
way. Consider the complete 
loopless undirected graph on the vertex set $\N$, with exactly one 
edge between any two distinct vertices. 
Assign to each a $\frac{1}{2}$ probability of 
being removed, and then go through the graph removing various edges with
independent 
probability. 

As can be found in \cite{hodges}, 
the isomorphism type of the resulting structure is 
constant on a conull set, and we use the term {\it random graph} to
describe any representative 
of this isomorphism class. This structure is ultrahomogeneous, as can be
shown by a 
``back-and-forth" argument, presented for instance in \cite{hodges}. 
\end{example} 

\begin{example} $(\Q, <)$, the rationals in their usual ordering are
ultrahomogeneous. 
Again, see \cite{hodges} for details. 
\end{example} 

\begin{example} Two examples which will be especially relevant are the
{\it rational 
Urysohn metric space} and the {\it bounded rational Urysohn metric space}. 
The first of these can be characterized up to isometry as follows. It is
the countable metric 
space such that the distance between any two points is rational and it
satisfies the 
following extension axiom: given any 
finite sequence $x_0, x_1, ..., x_{\ell -1}$ from the space and rational
distances 
$q_0, q_1, ..., q_{\ell -1}$ with 
\[d(x_i, x_j)\leq q_i + q_j,\]
\[q_i \leq d(x_i, x_j) + q_j,\]  
at all $i, j$, there exists 
$y$ in the space such that each $d(x_i, y)=q_i$. 
Its cousin, the bounded Urysohn space, is similar, but allowing only
distances 
$\leq 1$ and only minding this extension axiom when each $q_i\leq 1$. 
It is again a routine back-and-forth argument to show isomorphisms between 
finite pieces extend to automorphisms of the whole space. 
\end{example} 

\begin{example} A degenerate example is provided by the structure where every 
point has a different name. So let $\m=(\N; R_0, R_1, R_2,...)$ be the
relational 
structure which has each $R_i$, $i>0$, of arity $1$ with 
\[R_i=\{i+1\}.\] 
In this way distinct points have distinct quantifier free types, 
and thus the only way for q.f.type$_\m(\vec a)$ to equal 
q.f.type$_\m(\vec b)$ is for the sequences to be equal and then the 
identity map is the required automorphism. 
\end{example} 

\begin{example} A non-example is provided by $(\N; <)$. Clearly the
quantifier free 
type of the sequence $(2, 3)$ is the same as $(5, 10)$, since all we can
say is that the 
first point 
is smaller than another and these two sequences have that much in common. 
However there is no automorphism of $(\N; <)$ other than the trivial one 
which fixes every point. 
\end{example} 


\begin{definition} For $\m$ a countable relational structure and 
$\vec a$ a finite sequence from $\m$, we say that $b$ is 
{\it in the algebraic closure of $\vec a$,} written 
\[b\in{\rm acl}_{\m}(\vec a),\] 
if the image of $b$ under Aut$(\m, \vec a)$, the group of automorphisms of 
$\m$ fixing $\vec a$ pointwise, is finite. 
\end{definition} 

Note that in the case that $\m$ is ultrahomogeneous, q.f.type$_\m(\vec
a^\smallfrown b)$ 
determines orbit in the automorphism group, and thus $b\in {\rm
acl}_\m(\vec a)$ 
if and only there are just finitely many $c$ with 
\[{\rm q.f.type}_\m(\vec a^\smallfrown b) = 
{\rm q.f.type}_\m(\vec a^\smallfrown c).\] 


\begin{theorem} \label{ultra} 
Let $\m$ be an ultrahomogeneous structure with $|{\rm Aut}(\m)|>1$. Then
there is a 
pair $\vec a =(a_0, a_1)$ from $\m$ and 
\[f: \m^2\rightarrow \{0, 1\}\] 
such that for any monomorphism 
\[\rho : \m\rightarrow \m\] 
we can find $\vec b, \vec c$ in the image of $\rho$, both having the same 
quantifier free type 
as $\vec a$, but with 
\[f(\vec b)=0, \]
\[f(\vec c)=1.\] 
\end{theorem} 

\begin{proof} 
We split the proof into cases, the first being the simplest. 

\medskip 

\noindent{\bf Case(I):} Every element of $\m$ is in acl$(\emptyset)$, the 
algebraic closure 
of the empty sequence. 

\medskip 

Fix some $\pi\in {\rm Aut}(\m)$ not equal to the identity. 
Fix $a\in \m$ with $\pi(a)\neq a$ and let $b=\pi(a)$. 

Our case assumptions imply that every point has only finitely many other
points with 
the same quantifier free type. Since every monomorphism is injective and
preserves 
quantifier free type, it follows that all monomorphisms are onto, and so 
we have $a$ and $b$ in the image of any such $\rho$. Thus if we simply define 
\[f(a, x) =0\] 
\[f(b, x)=1\] 
for any $x\in\m$, and $f$ arbitrary in all other cases, then $f$ along
with the 
quantifier free type of $(a, a)$ is as required. 

\medskip 


\noindent{\bf Case(II):} Some $a\in\m$ is not in the algebraic closure 
of the empty sequence. 

\medskip 

Let $A$ be the collection of all points in $\m$ with the same 
quantifier free type as $a$. There is a further split into subcases. 

\medskip 

\noindent{\bf Subcase(IIa):} ${\rm acl}_\m(a)\cap A$ is infinite. 

\medskip 

This implies that for any $b\in A$ we again have ${\rm acl}_\m(b)\cap A$
infinite, 
since the automorphism group of $\m$ acts transitively on $A$. 
Let $(c_i)_{i\in\N}$ enumerate $A$ and choose successively points 
$d_i\neq e_i\in A$ such that 
\[d_i, e_i\in {\rm acl}_\m(c_i)\cap A\] 
and 
\[d_i, e_i\notin\{d_j, e_j: j<i\}.\] 
We can do this since each ${\rm acl}_\m(c_i)\cap A$ is infinite. 

Then, parallel to before, we define at each $i$ 
\[f(d_i, x)=0,\]
\[f(e_i, x)=1\] 
all $x$, arbitrary else wise. 
Any monomorphism must pick up some $c_i$ in its image, and after that 
some corresponding $d_i, e_i$. 


\medskip 

There is a sense in which these two cases are trivial, since we 
could have found the function $f:\m\rightarrow\{0, 1\}$, which implies 
the existence of a function on pairs as described by the theorem. The one 
case left is non-trivial in that it actually requires pairs, and that  
case tracks the argument from 
\ref{devlin}. 

\medskip 

\noindent{\bf Subcase(IIb):} ${\rm acl}_\m(a)\cap A$ is finite. 

\medskip 

Then since the automorphism group acts transitively on $A$ there will be some 
single bound $n_0$ such that for every $b\in A$ 
\[|{\rm acl}_\m(a)\cap A|\leq n_0.\] 
Since $A$ is infinite we can find a sequence of length $n_0+1$ 
\[b_0, b_1, ..., b_{n_0},\] 
and another point 
\[c\in A\setminus( {\rm acl}_\m(b_0)\cup {\rm acl}_\m(b_1)\cup...\cup 
{\rm acl}_\m(b_{n_0})).\] 
One of these $b_i$'s will fail to be in the algebraic closure of $c$, and 
in this 
way we have a pair $(c, b_i)$, which it will be convenient to rename,  
\[(a_0, a_1),\] 
with 
\[a_0\notin{\rm acl}_\m(a_1),\] 
\[a_1\notin{\rm acl}_\m(a_0).\] 
We let $(x_i)_{i\in\N}$ enumerate $A$. For $(x_i, x_j)$ having the same
type as 
\[(a_0, a_1)\] 
we let 
\[f(x_i, x_j)=0\] 
if $i<j$ and 
\[f(x_i, x_j)=1\] 
if $i>j$. $f$ is again arbitrary off of 
the type of $(a_0, a_1)$. 

The claim is that $f$ and the type of $(a_0, a_1)$ is as required, and 
so suppose $\rho:\m\rightarrow\m$ is a monomorphism. We let 
\[x_i =\rho(a_0),\]
\[x_j =\rho(a_1).\] 
There are two possibilities, the first being $i<j$ and $f(\rho(a_0),
\rho(a_1))=0$. 
Then note that since there are infinitely many $d$'s with $(d, a_1)$
having the 
same quantifier free type as $(a_0, a_1)$ we can find some such $d$ with 
\[\rho(d)\notin \{x_0, x_1, ..., x_j\}\] 
and then $\rho(d)=x_k$ some $k>j$ and 
\[f(\rho(d), \rho(a_1))=1\] 
as required. 

The other possibility is $i>j$, but then, similarly, 
since $a_1\notin {\rm acl}_\m(a_0)$ we 
can replace $a_1$ with some suitable $d$ to make 
$f(\rho(a_0), \rho(d))=0$. 
\end{proof} 

\subsection{Examples and counterexamples 
for first order logic and beyond} 
\label{firstorder}

This section differs from the other parts of the paper in 
having considerable prerequisites from logic which are 
not always made explicit.  
This section can be bypassed. None of the results  
resurface later in the paper. 

We consider various ways in which \ref{ultra} 
might be modified. 
From the point of view of a logician it is natural to consider the situation 
of elementary embeddings between first order structure. While there are two 
theorems which follow directly from \ref{ultra}, there is a counterexample to 
the most naive formulations. 


\begin{example} We build a structure $\m$ admitting many pairs with the 
same first order type and 
such that for any 
set of tuples with the same type 
\[S\subset \m^2\] 
and 
\[f: S\rightarrow \{0, 1\}\] 
we can find an elementary embedding of 
$\m$ into itself with the image homogeneous for 
$f$. 

We first let $\{Q_n:n\in \N\}$ be mutually dense co-dense subsets   
of the half open interval $(0, 1]$ such that any two 
distinct rationals are separated by some $Q_n$; we do this so 
that for any distinct non-overlapping finite sets $I_0, I_1\subset \N$ the 
intersection 
\[\bigcup_{n\in I_0} Q_n \setminus (\bigcup_{n\in I_1} Q_n)\] 
is dense in the interval $(0, 1)$. 
(The required $Q_n$'s can be produced by a routine 
diagonalization argument, which we omit.)  
We then extend these subsets to $P_n\subset \R$ by setting 
$\ell + r\in P_n$ if and only if $r\in Q_n$, where $\ell\in \Z$, 
$r\in (0, 1]$. 
We let $\m$ be the 
structure whose underlying set is $\Q^+$, the positive 
rationals, equipped with the usual ordering $<$ of the 
rationals, but also the countably many unary predicates, 
$\{P_n: n\in \N\}$, 
with the interpretation given above. 
Note then that for each $q\in \m$ and semi-open interval 
$(\ell, \ell +1]$, $\ell \in \N$, 
there is exactly one $q'\in(\ell, \ell +1]$ 
with the same type as $q$. 


Now given $S\subset \m^2$ consisting of 
pairs of the same type, we can find some $q, q'\in \Q$ such that 
$S(a,b)$ implies $a\equiv q$ mod $\Z$ and $b\equiv q'$ mod $\Z$. 
After possibly 
replacing $S$ by its reflection we can assume 
$S(x_0, x_1)\Rightarrow x_0 < x_1$. Then let 
\[f: S\rightarrow \{0, 1\}.\] 
Thus the domain of $f$ consists of all pairs 
\[(\ell + q, k + q')\] 
which have $\ell + q < k + q'$. 

From $f$ we define a new function 
\[f_\N: [\N]^2\rightarrow \{0, 1\}\] 
with 
\[f_\N(\ell, k)=f(\ell + q, k + q').\] 
Applying Ramsey's theorem we find a 
\[A\subset \N\] 
and a single $i\in \{0, 1\}$ 
such that whenever $k<\ell $ are both in $A$ 
\[f_\N(k, \ell)=i.\] 
%We can assume $0\notin A$. 
We let 
\[k_0<k_1<k_2<...\] 
enumerate $A$ in increasing order. 
Then let $\hat{\m}$ be the set of rationals 
which are in a semi-open interval of the 
form 
\[[k_{2n}, k_{2n} +q)\] 
($(k_0, k_0 + q)$ 
in the case $n=0$) 
or 
\[[k_{2n+1} + q, k_{2n+1}+1).\] 
It follows from the assumptions on
$A$ that $f$ is constant on $S\cap \hat{\m}^2$.
Standard model theoretic arguments show that 
\[\hat{\m}{\Large \prec } \m,\] 
and the inclusion map from $\hat{\m}$ into ${\m}$ 
is elementary.   
(Using a back and forth argument one 
can show that if $\l_n$ is the 
finite language consisting of $(<, P_0, P_1,...,P_n)$ 
then for any $\vec a\in \hat{\m}$ 
\[(\hat{\m}, \vec a)|_{\l_n}\cong (\m, \vec a)|_{\l_n},\] 
which clearly suffices.) 
\end{example} 

To have a sensible analog of \ref{ultra} we 
must either pass to infinitary logic or content ourselves 
with atomic models. 

Recall that $\l_{\omega_1, \omega}$ is the infinitary 
logic obtained by allowing countable conjunctions and 
disjunctions. 

\begin{proposition} 
\label{infinitary} 
Let $\m$ be a countable first order structure with non-trivial 
automorphism group. 
Then there is an automorphism invariant set $S\subset \m^2$ 
and 
\[f: S\rightarrow \{0, 1\}\] 
such that for any $\l_{\omega_1, \omega}$ elementary substructure 
\[\n{\Large \prec}_{\l_{\omega_1, \omega}} \m\] 
$f$ picks up both possible values on 
\[\n^2 \cap S.\] 
\end{proposition} 

\begin{proof} Following Scott's theorem, as found in 
say \cite{keisler}, we can find a countable 
fragment $F\subset \l_{\omega_1, \omega}$ such that 
for any $\vec a\in \m$ there is some $\varphi(\vec x)\in F$ 
such that for any other $\vec b$
\[(\m, \vec a)\cong (\m, \vec b)\] 
if and only if $\m\models \varphi(\vec b)$. 

Then let $\{R_\varphi: \varphi\in F\}$ be a collection of 
fresh predicates, and let $\hat{\m}$ be the expansion of 
$\m$ obtained by adding these predicates to $\m$ with the 
obvious interpretation. It follows from the assumptions on 
$F$ that this new expanded model $\hat{\m}$ is ultrahomogeneous, 
and in fact this is true in the strong sense that 
any tuple $\vec a$ has its orbit under Aut$(\m)$ determined 
by a single quantifier free formula $R_\varphi(\vec a)$. 
It also follows from the usual kinds of back and forth arguments 
that for any $\l_{\omega_1, \omega}$ elementary substructure 
\[\n{\Large \prec}_{\l_{\omega_1, \omega}} \m\] 
we can find an $\l_{\omega_1, \omega}$-elementary 
\[\rho: \m\cong \n.\] 

This all said, the proposition 
follows from 
\ref{ultra}. 
\end{proof}   

\begin{proposition} 
\label{atomic} 
Let $\m$ be a countable atomic model with non-trivial 
automorphism group.  
Then there is a type  $\tau(x_0, x_1)$ and 
\[f: \{(a, b)\in \m^2: \m \models \tau(a, b)\}\rightarrow \{0, 1\}\] 
such that for an elementary 
\[\n{\Large \prec} \m\]
we have $f$ picking up both values on $\n^2$. 
\end{proposition} 

\begin{proof} 
This follows from the proof of the last proposition. 
$\m$ being atomic means that the fragment 
$F$ from that argument 
can be taken to be the collection of 
first order formulas. 
\end{proof} 

Another detail which arises from the proof of 
\ref{ultra} is that frequently we can take the 
irrepressible function $f$ to be one which is 
defined not on pairs but on single points. In both case(I) 
and case(IIa) from \ref{ultra} we actually obtain 
\[f: \m\rightarrow \{0, 1\}\] 
such that for any 
elementary $\rho:\m\rightarrow \m$ one has both 
values assumed by $f$ on $\rho[\m]$. There are some 
atomic structures for 
which the oscillating $f$ is unavoidably defined 
on pairs -- for instance, $(\N; =)$, a countably infinite 
set equipped only with the relation of equality, or even 
$(\Q; <)$. 
On the other hand the partially ordered space obtained from 
$(\Q; <)$ by bifurcating every point into an antichain of size 
two admits an irrepressibly oscillating 
$f$ defined just on single points. 

It is not clear whether one can characterize the atomic models 
for which the $f$ of \ref{atomic} can be defined on $\m$ as against 
$\m^2$.  



\section{The oscillation theorem} 
\label{oscillation} 


\subsection{Groups and semigroups of isometries: Definitions, lemmas, and 
so on} 

\label{lemmas}

\begin{definition} 
Given a  
metric space $(X, d)$
we let  Isom($X$) be the group of isometries of $X$ onto 
itself and In($X$) the semi-group of 
isometries from $X$ into itself.  We define the 
{\it canonical topologies} on In($X$) as 
follows. Given a $z_0, z_1,..., z_m, y_0, ..., y_m 
\in X, \epsilon>0$ we let 
\[{\cal O}(z_0,..., z_m, y_0,..., y_m, \epsilon)\] 
be the set of 
$\rho\in{\rm In}(X)$ with 
\[d(\rho(z_i), y_i)<\epsilon\] 
at each $i\leq m$, and we give In($X$) the topology 
with the basis all open sets of this form. 
Isom($X$) is then given the induced subspace topology. 
\end{definition} 

\begin{lemma} 
If $X$ is separable and completely metrizable, then 
both {\rm In}($X$) and {\rm Isom}($X$) are Polish in the indicated topology. 
\end{lemma} 

\begin{proof} Let 
$\{x_i: i\in \N\}$ be some arbitrary dense subset of 
$X$. We define $D$ on Isom($X$) with 
\[D(\pi_1, \pi_2) = 
\sum_{n\in \N} 2^{-n} {\rm max}(1, d (\pi_1(x_n), \pi_2 (x_n)) + 
d(\pi_1^{-1}(x_n), \pi_2^{-1} (x_n)))\] 
and $\hat{D}$ on In($X$) by 
\[\hat{D}(\rho_1, \rho_2)= 
\sum_{n\in \N} 2^{-n}{\rm max}( 1,d(\rho_1(x_n), \rho_2(x_n))).\] 
It is routine to verify that these metrics have the required 
properties of completeness and separability. 
\end{proof} 

\begin{definition} 
For $G\leq {\rm Isom}(X)$ a subgroup, we let 
$\hat{G}$ be its closure in In($X$). 
\end{definition} 

Even when $G$ is a group, there is no necessity for 
its closure in the semi-group In($X$) to form a group. 
For instance, if $X=\N$ and $d$ is the complete metric, 
then for $G=S_\infty$ the group of all permutations of 
$\N$ we have $\hat{G}$ equal to all injections from 
$\N$ to $\N$. In general we also have 
$\hat{G}$ much smaller than 
In($X$) even when $G$ equals Isom($X$). For instance 
if $X$ equals $\N$ with $d(m, n)=|m-n|$, then the 
isometry group is trivial, whilst In($X$) consists 
of infinitely many possible shifts. 

For the rest of this section fix some complete 
separable metric space $(X, d)$ and let $G$ be some 
group of isometries. 

\begin{definition} For $\rho_0\in \hat{G}$ we let 
$\j(\rho_0)$ be the $\hat{D}$ closure of the set 
\[\{\rho_0\cdot \pi: \pi\in G\}.\] 
\end{definition} 

\begin{lemma} \label{I} 
$\rho$ is in $\j(\rho_0)$ if and only if for every 
$\epsilon >0$ and $x_1, ..., x_m\in X$ there exists 
$\pi\in G$ such that 
\[d(\rho_0(\pi(x_i)), \rho(x_i))<\epsilon\] 
at every $i\leq m$. 
\end{lemma} 

\begin{proof} 
At once from the definition of the topology. 
\end{proof} 

\begin{lemma} \label{II} 
Let $\rho_0\in \hat{G}$ and use 
$X_0$ to denote $\rho_0[X]$. 
Let $x_1, ..., x_m\in X, y_1, ..., y_m\in X_0$. 
Then the following three things are equivalent: 
 
\leftskip 0.4in 

\noindent(i) There exists $\rho\in \{\rho_0\cdot \pi: 
\pi\in G\}$  with
\[d(\rho(x_i), y_i)<\epsilon\]
all $i\leq m$. 

\noindent (ii) There there exists $\rho\in \j(\rho_0)$ with 
\[d(\rho(x_i), y_i)<\epsilon\] 
all $i\leq m$. 

\noindent (iii) There exists 
$\pi\in G$ with 
\[d(\pi(x_i), y_i)<\epsilon\]
all $i\leq m$. 

\leftskip 0in 

\end{lemma} 

\begin{proof} 
(i) and (ii) are clearly equivalent in virtue of the 
definition of $\j(\rho_0)$, thus it suffices to show 
(ii) if and only if (iii). 
The only if direction is immediate from 
$\rho_0$ being in the $\hat{D}$-closure of 
$G$, and so let us assume instead that $\pi\in G$ and 
$d(\pi(x_i), y_i)<\epsilon$  
all $i\leq m$. 
Let $z_i$ be such that $\rho_0(z_i)=y_i$ all $i\leq m$. 
Choose $\sigma_n\in G$ with $\sigma_n\rightarrow \rho_0$ in 
$\hat{G}$; this gives in particular that at each $i$ 
\[\sigma_n(z_i)\rightarrow y_i\] 
at each $i$, and then since $G$ acts by isometries 
\[\sigma_n^{-1}(y_i)\rightarrow z_i,\] 
and then at large enough $n$ 
\[d(\sigma_n^{-1}(\pi(x_i)), z_i)\leq 
d(\sigma_n^{-1}(\pi(x_i)), \sigma_n^{-1}(y_i)) + 
d(\sigma_n^{-1}(y_i), z_i)\]
\[= d(\pi(x_i), y_i) +
d(\sigma_n^{-1}(y_i), z_i)<\epsilon.\] 
Then since $\rho_0$ is an isometry, at large 
enough $n$ 
\[ d(\rho_0(\sigma_n^{-1}(\pi(x_i))), \rho_0(z_i)) 
=d(\rho_0(\sigma_n^{-1}(\pi(x_i))),y_i)<\epsilon,\] 
as required for 
$\rho=\rho_0\circ \sigma_n^{-1}$ to 
verify the lemma. 
\end{proof} 

\begin{corollary} 
\label{corollaryI} 
$\j(\rho_0)$ is closed under multiplication. 
\end{corollary} 

\begin{proof} 
Let $\gamma_0, \gamma_1\in \j(\rho_0)$. 
Since $\rho_0$ is a limit of elements in 
$G$ we can find sequences $(\sigma_n^0)_n, (\sigma_n^1)_n$ 
from $G$ with  
\[\sigma_n^0\rightarrow \gamma_0,\] 
\[\sigma_n^1\rightarrow \gamma_1.\] 
Appealing to \ref{I} 
we need only show that for all $x_1, ..., x_m\in X$, 
$\epsilon>0$, 
if we set $y_i=\gamma_0(\gamma_1(x_i)))$, then 
we can produce $\pi\in G$ with 
\[d(\rho_0(\pi(x_i)), y_i)<\epsilon\] 
at each $i\leq m$. By appealing to 
\ref{II}, it suffices to show that there is 
some $\hat{\pi}\in G$ with 
\[d(\hat{\pi}(x_i), y_i)<\epsilon\] 
all $i\leq m$. 

But if $z_i=\gamma_1(x_i)$, then we obtain 
\[\sigma^1_n(x_i)\rightarrow \gamma_1(x_i)\] 
and then 
\[\sigma^2_n(z_i)\rightarrow \gamma_2(z_i)=y_i\] 
by the definition of $\hat{D}$ convergence. From this 
and the assumption that each $\sigma_n^1$ is an isometry we 
obtain 
\[\sigma^1_n\circ \sigma^2_n(x_i)\rightarrow y_i\] 
as $n\rightarrow \infty$, and so we take $\hat{\pi}$ 
to be $\sigma^1_n\circ \sigma^2_n$ for $n$ sufficiently 
large. 
\end{proof} 

\begin{definition} $\j\subset \hat{G}$ is a 
{\it right ideal} if it is closed under multiplication on 
the right by $G$ and it is closed under composition. 
\end{definition} 

Thus \ref{corollaryI} shows that any $\j(\rho_0)$ is a 
right ideal. 



\subsection{The theorem} 
\label{biglentil}



%Recall from 
%$\S$\ref{lemmas} that we view the semi-group of isometries 
%of a space as coming equipped with a canonical topology. 
%This also induces a topology on Isom($X$), the group of 
%isometries of $X$. 

%The new idea in the proof of this theorem is given by the 
%following definition. 

\begin{definition} Let $G$ be a group of isometries of 
$(X, d)$ and $\epsilon, \delta>0$.  
For $x,y\in X$ we say that $x$ {\it $\epsilon,\delta$-captures} 
$y$ relative to $G$ if there is some $N\in \N$ such that we 
cannot find $N$ disjoint balls of the form 
\[B_{\epsilon} (\pi(y))\] 
for $\pi\in G$ with 
\[d(\pi(x), x)<\delta.\] 
\end{definition} 

Here $B_\epsilon(\pi(y))$ denotes the ball of all points 
$z$ with $d(\pi(y), z)<\epsilon$. Thus in loose terms 
we could say that $x\:$ $\epsilon,\delta$-captures $y$ if it 
not possible to keep $x$ very still while simultaneously moving 
$y$ around to many different places. 



\begin{theorem} 
\label{maintheorem} 
Let $(X, d)$ be a complete 
metric space. Let $G\leq {\rm Isom}(X, d)$ be a group 
of isometries and let $\hat{G}\subset {\rm In}(X, d)$ be 
its closure. 
Assume $G$ has size bigger than one. 

Then there exists $x_0, x_1\in X$ and uniformly continuous 
\[f: \overline{\{(\pi\cdot x_0, \pi \cdot x_1):
\pi \in G\}}\rightarrow [0, 1]\]
such that for any $\rho_0\in \hat{G}$ there exist
\[(y_0, y_1), (z_0, z_1)\in \{(\rho_0(\pi(x_0)), \rho_0(\pi(x_1))):
\pi \in G\}\]                                   
with
\[f(y_0, y_1)=0,\]
\[f(z_0, z_1)=1.\]
\end{theorem} 

\begin{proof} 
It suffices to show 
that there is a uniformly continuous 
$f: X\times X\rightarrow [0, 3]$ and $x_0, x_1\in X$ 
such that for any  $\rho_0\in \hat{G}$ there are
$\pi_1, \pi_2 \in G$ 
with
\[f(\rho_0(\pi_1(x_0)), \rho_0(\pi_1(x_1)))<1,\] 
\[f(\rho_0(\pi_2(x_0)), \rho_0(\pi_2(x_1)))>2.\] 
Given an $f$ with 
this property, we can obtained the $f$ as above by truncating to 
the interval $[1, 2]$ and subtracting 1. 
Thus we head towards constructing 
$f: X\times X\rightarrow [0, 3]$. 
Once the problem has been put in this form we can always 
assume that $G$ is a closed subgroup of Isom($X$), since 
this present form of the statement of the theorem will 
pass downwards from the closure of $G$ to $G$ itself. 






\medskip 

{\bf Case(I):} Every $x\in X$ has precompact orbit under 
$G$. 

\medskip 

That means we can write 
\[X=\bigcup_{x\in X} C_x,\] 
each $C_x\subset X$ with $C_x\supset [x]_G$, compact and 
\[\pi[C_x]\subset C_x\] 
all $\pi\in G$, and hence 
\[\rho[C_x]\subset C_x\] 
all $\rho\in \hat{G}$. But since any isometry of a 
compact metric space into itself will be onto, it follows that 
any $\rho\in \hat{G}$ will actually be an isometry, and 
we are done once we choose some $(x_0, x_1)\in X\times X$ with 
non-trivial orbit under $G$ and an {\it onto} 
uniformly continuous $f: \overline{(\pi(x_0), \pi(x_1)): \pi\in G\}}\rightarrow [0,3].$ 


\medskip 

{\bf Case(II):} Some $x\in X$ does not have precompact orbit 
in $X$. 

\medskip 

Fix such an $x_0$ without precompact orbit. Fix $\epsilon$ 
such that there are infinitely many disjoint balls of the form 
\[B_\epsilon(\pi(x_0))\] 
for $\pi\in G$. The rest of the argument involves rather messy 
approximations, but in general terms we want to replace the notion 
of algebraic closure by the notion of being 
$\frac{\epsilon}{20},\frac{\epsilon}{100}$-captured. To avoid 
constantly needing to mention these unpleasant constants, from now 
on we use the following completely ad hoc definition. 

\begin{definition} We say that $x$ {\it $\epsilon$-captures} 
$y$ if it $\frac{\epsilon}{20},\frac{\epsilon}{100}$-captures 
$y$ in the original sense given prior to the statement of the 
theorem. 
\end{definition} 

{\bf Subcase(IIa):} There exists $x_1\in G\cdot x_0$ such that 
$x_1$ does not $\epsilon$-capture $x_0$ and 
$x_0$ does not $\epsilon$-capture $x_1$. 

\medskip 

Note that since $G$ acts by isometries, for any $x\in G\cdot x_0$ 
there exists $x'\in G\cdot x=G\cdot x_0$ with neither of 
$x, x'$ $\epsilon$-capturing the other. 



Fix such an $x_1$ as in subcase hypothesis. 
Let $\{z_n: n\in \N\}$ be dense in $G_\cdot x_0$. 
At each $n\in \N$ we define continuous 
\[g_n: X\rightarrow [0, 3]\] 
by 
\[g_n(y)={\rm max}\{0, 3-\frac{150}{\epsilon} 
(d(y, B_{\frac{\epsilon}{50}}(z_n)))\}.\] 
(Here $d(y, B_{\frac{\epsilon}{50}}(z_n))$ denotes 
the infimum over $d(y, z)$ for $z$ chosen with 
$d(z, z_n)<\frac{\epsilon}{50}$.) 
Thus $g_n$ is uniformly continuous uniformly in 
$n$, and constantly assumes the value 3 on 
$ B_{\frac{\epsilon}{50}}(z_n)$ and the value 
zero outside 
$ B_{\frac{\epsilon}{25}}(z_n)$. 

We then define $f: X\times X\rightarrow [0,3]$ by 
\[f(y, y')=
{\rm max}(0, {\rm sup}\{g_n(y)-{\rm max}_{m\leq n}g_m(y'): n\in \N\}.\] 
Given the uniformity of the continuity of the $g_n's$ we obtain 
that $f$ is uniformly continuous. Unwrapping the definitions 
we have that $f$ constantly assume 3 on 
\[ B_{\frac{\epsilon}{50}}(z_n)\times (X\setminus\bigcup_{m\leq n} 
 B_{\frac{\epsilon}{25}}(z_m))\] 
and zero on 
\[ (X\setminus\bigcup_{m <n}
 B_{\frac{\epsilon}{25}}(z_m)) 
\times  B_{\frac{\epsilon}{50}}(z_n).\] 

I claim that $f$ is as required. 

For this purpose fix $\rho_0\in \hat{G}$ and 
\[\sigma_n\rightarrow \rho_0\] 
in $G$. We need to show that $f$ attains 
$[0, 1)$ and $(2, 3]$ inside 
\[\{(\rho_0\circ \pi(x_0), \rho_0\circ \pi(x_1)): 
\pi\in G\}.\]   

Choose $m(x_0), m(x_1)$ so that 
\[\rho_0(x_0)\in B_{\frac{\epsilon}{100}}(z_{m(x_0)}),\] 
\[\rho_0(x_1)\in B_{\frac{\epsilon}{100}}(z_{m(x_1)}).\]
Let $k=$max$\{m(x_0), m(x_1)\}+1$. Let 
$\pi_1, ...,\pi_k, \pi_1',..., \pi_k'\in G$ be such that 
at each $i\leq k$ we have 
\[d(\pi_i(x_0), x_0)<\frac{\epsilon}{100},\] 
\[d(\pi_i'(x_1), x_1)<\frac{\epsilon}{100},\]
but for $i\neq j$ 
\[B_{\frac{\epsilon}{20}}(\pi_i(x_1))\cap 
B_{\frac{\epsilon}{20}}(\pi_j(x_1))=0,\] 
\[B_{\frac{\epsilon}{20}}(\pi_i'(x_0))\cap
B_{\frac{\epsilon}{20}}(\pi_j'(x_0))=0.\]
Since $\rho_0$ is an isometry arising as a limit 
of invertible isometries in $G$ these imply 
$B_{\frac{\epsilon}{20}}(\rho(\pi_i(x_1)))\cap
B_{\frac{\epsilon}{20}}(\rho(\pi_j(x_1))=0$ and 
$B_{\frac{\epsilon}{20}}(\rho(\pi_i'(x_0)))\cap
B_{\frac{\epsilon}{20}}(\rho(\pi_j'(x_0))=0$. 

\def\hp{\hat{\pi}}

Choose $\delta>0$ with 
\[\rho_0(x_0)\in 
B_{(\frac{\epsilon}{100}-2\delta)} (z_{m(x_0)}),\]   
\[\rho_0(x_1)\in 
B_{(\frac{\epsilon}{100}-2\delta)} (z_{m(x_1)}),\]  
and $\delta<\frac{\epsilon}{200}$. 
Choose $n$ large enough that at each $i\leq k$ 
\[d(\sigma_n(\pi_i(x_0)), \rho_0(\pi_i(x_0))),\:\: 
d(\sigma_n(\pi_i(x_1)), \rho_0(\pi_i(x_1)))<\frac{\epsilon}{200},\] 
\[d(\sigma_n(x_0), \rho_0(x_0)), \: \: d(\sigma_n(x_1), \rho_0(x_1)) 
<\delta.\] 
Let $\hp_i=\sigma_n \pi_i \sigma_n^{-1}$, 
$\hp_i'=\sigma_n \pi_i' \sigma_n^{-1}$. 

Then at each $i\leq k$ we have 
\[d(\hp_i(\rho_0(x_0)), \rho_0(x_0))\leq 
d(\hp_i(\rho_0(x_0)), \hp_i(\sigma_n(x_0))) 
+ d(\hp_i(\sigma_n(x_0)), \sigma_n(x_0)) 
+ d(\sigma_n(x_0), \rho_0(x_0)),\] 
which in virtue of $\hp_i$ being an isometry equals 
\[d(\rho_0(x_0), \sigma_n(x_0))
+ d(\hp_i(\sigma_n(x_0)), \sigma_n(x_0))
+ d(\sigma_n(x_0), \rho_0(x_0)),\]
which in turn by the definition of $\hp_i$ equals 
\[d(\rho_0(x_0), \sigma_n(x_0))              
+ d(\sigma_n(\pi_i(x_0)), \sigma_n(x_0))
+ d(\sigma_n(x_0), \rho_0(x_0)).\]
Then since $\sigma_n$ is an isometry this equals 
\[d(\rho_0(x_0), \sigma_n(x_0))
+ d(\pi_i(x_0), x_0) 
+ d(\sigma_n(x_0), \rho_0(x_0))\]
\[<\delta + \frac{\epsilon}{100} + \delta
=2\delta + \frac{\epsilon}{100}.\] 

A similar calculation gives 
\[d(\hp_i(\rho_0(x_1)), \rho_0(\pi_i(x_1)))\leq 
d(\hp_i(\rho_0(x_1)), \hp_i(\sigma_n(x_1))) + 
d(\hp_i(\sigma_n(x_1)), \rho_0(\pi_i(x_1)))\]
\[=d(\rho_0(x_1), \sigma_n(x_1)) +
d(\hp_i(\sigma_n(x_1)), \rho_0(\pi_i(x_1)))<
\frac{\epsilon}{200} + \frac{\epsilon}{200} 
=\frac{\epsilon}{100}.\] 
And of course we have the same calculations on the 
other side, replacing $\hp_i$ by $\hp_i'$, $x_0$ by 
$x_1$, and $x_1$ by $x_0$. 
Thus at some high enough $m$ we have 
\[d(\hp_i(\sigma_m(x_0)), \rho_0(x_0))<2\delta + \frac{\epsilon}{100},\] 
\[d(\hp_i(\sigma_m(x_1)), \rho_0(\pi_i(x_1)))< \frac{\epsilon}{100},\] 
\[d(\hp_i'(\sigma_m(x_1)), \rho_0(x_1))<2\delta + \frac{\epsilon}{100},\]
\[d(\hp_i'(\sigma_m(x_0)), \rho_0(\pi_i(x_0)))< \frac{\epsilon}{100}.\]



At this point we can appeal to lemma \ref{II} and find 
$\rho_1, ..., \rho_k, \rho_1', ...,\rho_k'\in 
\{\rho_0\circ \pi: \pi\in G\}$ such that at each $i$ 
\[d(\rho_i(x_0), \rho_0(x_0))<2\delta + \frac{\epsilon}{100},\]
\[d(\rho_i(x_1), \rho_0(\pi_i(x_1)))< \frac{\epsilon}{100},\]
\[d(\rho_i'(x_1), \rho_0(x_1))<2\delta + \frac{\epsilon}{100},\]
\[d(\rho_i'(x_0)), \rho_0(\pi_i(x_0)))< \frac{\epsilon}{100}.\] 
Then 
\[B_{\frac{\epsilon}{25}}(\rho_i(x_1))\subset 
B_{(\frac{\epsilon}{25}+\frac{\epsilon}{100})}(\rho_0(\pi_i(x_1)))
=B_{\frac{\epsilon}{20}}(\rho_0(\pi_i(x_1))),\] 
and hence 
\[B_{\frac{\epsilon}{25}}(\rho_i(x_1))\cap 
B_{\frac{\epsilon}{25}}(\rho_j(x_1))=0\] 
for $i\neq j$. On the other hand 
\[d(\rho_i(x_0), z_{m(x_0)}) \leq d(\rho_i(x_0), \rho_0(x_0))
+ d(\rho_0(x_0), z_{m(x_0)}) \]
\[< 
(2\delta + \frac{\epsilon}{100}) 
+ (\frac{\epsilon}{100}-2\delta) 
=\frac{\epsilon}{50}\] 
\[\therefore \rho_i(x_0)\in B_{\frac{\epsilon}{50}}(z_{m(x_0)}).\] 
Similarly at each $i\neq j$ 
\[B_{\frac{\epsilon}{25}}(\rho_i'(x_0))\cap
B_{\frac{\epsilon}{25}}(\rho_j'(x_0))=0\]
whilst 
\[\rho_i'(x_1)\in B_{\frac{\epsilon}{50}}(z_{m(x_1)}).\] 


Since each $z_m$ can be in at most one of the 
balls $B_{\frac{\epsilon}{25}}(\rho_i(x_1))$, 
$i\leq k$, and since $k>m(x_0)$, 
we may find specific $i_0$ with 
\[\rho_{i_0}(x_1)\notin B_{\frac{\epsilon}{25}}(z_m)\] 
all $m\leq m(x_0)$. Similarly we can fix a $j_0$ with 
\[\rho_{j_0}'(x_0)\notin B_{\frac{\epsilon}{25}}(z_m)\]
all $m\leq m(x_1)$. We then let 
\[(y_0, y_1)=(\rho_{i_0}(x_0), \rho_{i_0}(x_1)), \]
\[(z_0, z_1)= (\rho_{j_0}'(x_1), \rho_{j_0}'(x_1)).\] 

Since $y_1\notin\bigcup_{m\leq m(x_0)} B_{\frac{\epsilon}{25}} (z_m)$ 
whilst $y_0\in B_{\frac{\epsilon}{50}}(z_{m(x_0)})$ we get 
\[f(y_0, y_1)=3,\] 
whilst since  
$z_0\notin\bigcup_{m\leq m(x_1)} B_{\frac{\epsilon}{25}} (z_m)$ 
whilst $z_1\in B_{\frac{\epsilon}{50}}(z_{m(x_1)})$ we 
alternatively obtain 
\[f(z_0, z_1)=0.\] 

\medskip 

{\bf Subcase(IIb):} There does 
{\it not} 
exist $x_1\in G\cdot x_0$ with 
neither $x_0, x_1$ $\epsilon$-capturing the other. 

\medskip 

{\bf Subsubcase(IIbi):} The set of $y\in G\cdot x_0$ such that 
$x_0$ $\epsilon$-captures $y$ can be covered by finitely many balls of 
radius $\frac{\epsilon}{4}$. 

\medskip 

Fix $N$ such that the $y$'s in $G\cdot x_0$ that are $\epsilon$-captured 
by $x_0$ can be covered by $N$ balls of radius $\frac{\epsilon}{4}$. Since 
$G$ is acting by isometries we have that for every 
$x\in G\cdot x_0$ the set of $y$'s $\epsilon$-captured by 
$x$ can be covered by $N$ many balls of radius 
$\frac{\epsilon}{4}$. 

\medskip 

{\bf Claim:} For any $\rho\in \hat{G}$ and $x'\in G\cdot x_0$ 
there exists $y\in \rho[X]$ with $d(y, x')<\frac{\epsilon}{10}$. 

\medskip 

{\bf Proof of Claim:} Fix $\sigma_n\rightarrow \rho$ in $G$. 
Fix $y_1, ..., y_N, y_{N+1}$ in $G\cdot x_0$ with 
\[B_\epsilon(y_i)\cap B_\epsilon(y_j)=0\] 
for $i\neq j$. 
We then pass to a large enough $M$ that 
\[d(\sigma_n(y_i), \sigma_m(y_i))<\frac{\epsilon}{100}\] 
all $m, n\geq M$, $i\leq N+1$ and let 
\[z=\sigma^{-1}_M(x').\] 

By the subsubcase assumptions at (IIbi) 
there will be some $i_0\leq N+1$ 
with $x'$ not $\epsilon$-capturing $\sigma_M(y_{i_0})$. But then 
the subcase assumptions at (IIb) we have $\sigma_M(y_{i_0})$ 
$\epsilon$-capturing $x'=\sigma_M(z)$. 

Then for all $n\geq M$ 
\[
d((\sigma_n\circ \sigma_M^{-1})(\sigma_M(y_{i_0})), 
\sigma_M(y_{i_0}))
=d(\sigma_n(y_{i_0}), \sigma_M(y_{i_0})) 
<\frac{\epsilon}{100},\] 
and thus by $\sigma_M(y_{i_0})$ $\epsilon$-capturing 
$x'=\sigma_M(z)$ we have 
\[d(\sigma_n(z), x')=d(\sigma_n(\sigma_M^{-1}(x')),
x')<\frac{\epsilon}{20}.\] 
Since this happens at all $n$ sufficiently large, 
\[d(\rho_{i_0}(z), x')\leq \frac{\epsilon}{20},\] 
and we may complete the claim by taking $y=\rho(z)$. 
\hfill (Claim$\square$) 

\medskip 

But now we simply choose $q_0, q_1\in G\cdot x_0$ with 
\[B_\epsilon(q_0)\cap B_\epsilon(q_1)=0\] 
and define  
\[f_0: X\rightarrow [0,3]\] 
\[w\mapsto {\rm min}(0, {\rm max}(3, \frac{6}{\epsilon} 
d(w, B_{\frac{\epsilon}{2}}(q_1)).\] 
We let $x_1$ be arbitrary and define  
\[f: \overline{(\pi(x_0), \pi(x_1)): \pi \in G\}} 
\rightarrow [0, 3]\] 
with $f(w_0, w_1)=f_0(w_0)$.  


For any $\rho_0\in \hat{G}$ we can appeal to the claim and 
lemma \ref{II} to find $\pi_0, \pi_1\in G$ with 
\[\rho_0(\pi_0(x_0))\in B_{\frac{\epsilon}{2}}(q_0), \] 
\[\rho_0(\pi_1(x_0))\in B_{\frac{\epsilon}{2}}(q_1),\] 
and take 
$(y_0, y_1)= (\rho_0(\pi_0(x_0)), \rho_0(\pi_0(x_1))), 
(z_0, z_1)= (\rho_0(\pi_1(x_0)), \rho_0(\pi_1(x_1)))$. 



\medskip 

{\bf Subsubcase(IIbii):} The set of $y\in G\cdot x_0$ 
that are $\epsilon$-captured by $x_0$ {\it cannot} be 
covered by finitely many balls of radius $\frac{\epsilon}{4}$. 

\medskip 


Let $\{x_n: n\in \N\}$ enumerate a dense subset of $G\cdot x_0$. 

\medskip 

{\bf Claim:} There exists sequences $(k'_n)_{n\in \N}$, $(k_n'')_{n\in \N}, 
(\ell'_n)_{n\in \N}, (\ell_n'')_{n\in \N}$ and  arrays 
\[(y'_{j, i, m})_{m\in \N, i<k_m', j<\ell_m'},\] 
\[(y''_{j, i, m})_{m\in \N, i<k_m'', j<\ell_m''}\] 
of elements in $G\cdot x_0$ such that at each $n\in \N$: 

\leftskip 0.4in 

\noindent (a) if $\pi\in G$ has $d(\pi(x_m), x_m)<\frac{\epsilon}{100}$ and 
$j'\leq\ell_m', j''\leq \ell_m''$ then at some $p'\leq k_m', p''\leq k_m''$ 
\[d(\pi(y_{j', 0, m}'), y_{j', p', m}')<\frac{\epsilon}{10},\]
\[d(\pi(y_{j'', 0, m}''), y_{j'', p'', m}'')<\frac{\epsilon}{10};\] 

\noindent (b) for $j\neq j'\leq \ell_m'$ 
\[d(y_{j, 0, m}', y_{j', 0, m}')\geq \frac{\epsilon}{4};\] 

\noindent (c) for $j\neq j'\leq \ell_m''$ 
\[d(y_{j, 0, m}'', y_{j', 0, m}'')\geq \frac{\epsilon}{4};\] 

\noindent (d) 
\[\ell_n'> \sum_{m<n}\ell_m'\cdot k_m'+\sum_{m<n} \ell_m''\cdot k_m'';\] 

\noindent (e) 
\[\ell_n''> \sum_{m\leq n}\ell_m'\cdot k_m'+\sum_{m<n} \ell_m''\cdot k_m''.\] 

\leftskip 0in 

{\bf Proof of Claim:} This is 
a matter of grinding down the assumptions of 
subsubcase(IIbii). 

We build the sequences inductively. 
Given that we have decided on 
\[(y'_{j, i, m})_{m<n, i<k_m', j<\ell_m'}, 
(y''_{j, i, m})_{m<n, i<k_m'', j<\ell_m''}\] 
we choose some very large 
$\ell_n'$ as indicated at (d) and then a sequence of $(y_{j, 0, n}')_{j<\ell_n'}$ 
which are all $\epsilon$-captured by $x_0$ and are 
distance at least $\frac{\epsilon}{4}$ apart. 
At each $j<\ell_n'$ we can choose finitely many $y_{j, i, m}'$'s in (a) by the definition of 
$\epsilon$-capturing. We take $k_n'$ to be the maximum of the number of $y_{j, i, m}'$'s we need 
at the various $i<\ell_n'$. Then construction of the $(y_{j, i, n})_{i<k_n'', j<\ell_n''}$ is 
exactly similar. 
\hfill (Claim$\square$) 

\medskip 


Our interest in this claim will be in terms of an application of the pigeonhole principle. 
If $\pi$ is an isometry of $X$ then at some $q'<\ell_n'$ we have 
\[\pi(y'_{q', 0, n})\notin \bigcup_{m<n, i<k_m', j<\ell_m'} B_{\frac{\epsilon}{9}}(y_{j, i, m}') \cup 
 \bigcup_{m<n, i<k_m'', j<\ell_m''} B_{\frac{\epsilon}{9}}(y_{j, i, m}'') \] 
and at some $q''<\ell_n'$ we have 
\[\pi(y'_{q'', 0, n})\notin \bigcup_{m\leq n, i<k_m', j<\ell_m'} B_{\frac{\epsilon}{9}}(y_{j, i, m}') \cup 
 \bigcup_{m<n, i<k_m'', j<\ell_m''} B_{\frac{\epsilon}{9}}(y_{j, i, m}'').\] 

Now at each $m$ we let 
\[A_m =\bigcup_{i<k_m', j<\ell_m'} B_{\frac{\epsilon}{9}}(y_{j, i, m}'),\] 
\[B_m =\bigcup_{i<k_m'', j<\ell_m''} B_{\frac{\epsilon}{9}}(y_{j, i, m}''),\] 
and then 
\[C_n = \bigcup_{i<k_n', j<\ell_n'} B_{\frac{\epsilon}{10}}(y_{j, i, n}')\setminus 
(\bigcup_{m<n} A_m \cup \bigcup_{m<n} B_m),\] 
\[D_n = \bigcup_{i<k_n'', j<\ell_n''} B_{\frac{\epsilon}{10}}(y_{j, i, n}'')\setminus 
(\bigcup_{m\leq n} A_m \cup \bigcup_{m<n} B_m).\]

It follows from clauses (b) and (d) above that each $C_n$ is non-empty, and from (c) and (e) that 
each $D_n$ is non-empty. The way in which these sets have been defined guarantees that 
\[d(C_n, D_m)\geq \frac{\epsilon}{9}-\frac{\epsilon}{10}=\frac{\epsilon}{90}\] 
at {\it every} $m, n$, and thus $d(C, D)\geq \frac{\epsilon}{90}$, and thus we can find a 
uniformly continuous 
\[f_0: X\rightarrow [0, 3]\] 
with $f_0|_C\equiv 3$, $f_0|_D\equiv 0$. 

\medskip 

{\bf Claim:} For any $\rho_0\in \hat{G}$ there exists $z'\in \overline{C}\cap \rho_0[G\cdot x_0]$. 

\medskip 

{\bf Proof of Claim:} Choose $\sigma_n\rightarrow \rho_0$ in $G$, and go to $N_0$ such that at all 
later $n\geq N_0$ we have 
\[d(\sigma_n(x_0), \sigma_{N_0}(x_0))<\frac{\epsilon}{300}.\] 
Choose $N_1$ with $d(\sigma_{N_0}(x_0), x_{N_1})<\frac{\epsilon}{300}$ and then observe that at 
all $n\geq N_0$
\[d(\sigma_n\sigma_{N_0}^{-1}(x_{N_1}), x_{N_1})\leq d(\sigma_n\sigma_{N_0}^{-1}(x_{N_1}), 
\sigma_n (x_0)) + 
d(\sigma_n(x_0), \sigma_{N_0}(x_0)) + 
d(\sigma_{N_0}(x_0), x_{N_1})\]
\[=d(\sigma_n\sigma_{N_0}^{-1}(x_{N_1}), 
\sigma_n\sigma_{N_0}^{-1}(\sigma_{N_0}(x_0)) + 
d(\sigma_n(x_0), \sigma_{N_0}(x_0)) + 
d(\sigma_{N_0}(x_0), x_{N_1})\]
\[= d(x_{N_1}, 
\sigma_{N_0}(x_0)) + 
d(\sigma_n(x_0), \sigma_{N_0}(x_0)) + 
d(\sigma_{N_0}(x_0), x_{N_1})\]
\[<\frac{\epsilon}{3} + \frac{\epsilon}{3} + \frac{\epsilon}{3}=
\frac{\epsilon}{100}.\] 
Thus by assumption (a) on our construction we can find at each $n\geq N_0$ 
and $j<\ell_{N_1}'$ we can find 
some corresponding $p_j<k_{N_1}'$ with 
\[d(\sigma_n \sigma_{N_0}^{-1}(y_{j, 0, N_1}'), y_{j, p_j, N_1}')
<\frac{\epsilon}{10}.\] 
(b) gives for  $j<j'\leq \ell_{N_1}'$ we have 
\[d(\sigma_n \sigma_{N_0}^{-1}(y_{j, 0, N_1}'), 
\sigma_n \sigma_{N_0}^{-1}(y_{j', 0, N_1}'))
\geq \frac{\epsilon}{4},\] 
and so by applying the pigeonhole principle to clause (d) from our 
construction we can find some specific $j(n)<\ell_{N_1}'$ with 
\[\sigma_n \sigma_{N_0}^{-1}(y_{j(n), 0, N_1}')\notin \bigcup_{n<N_1}A_m \cup\bigcup_{m<N_1}B_m,\] 
and hence 
$\sigma_n \sigma_{N_0}^{-1}(y_{j(n), 0, N_1}')\in C$. 
Thinning out the sequence $(\sigma_n)_{n\in\N}$ 
we may assume there is a fixed $j$ with 
$j=j(n)$ all $n$. Thus for 
\[z'=\sigma_{N_0}^{-1}(y_{j, 0, N_1}')\] 
we have each $\sigma_n(z')\in C$ and hence in the limit $\rho_0(z')\in \overline{C}$. 
\hfill (Claim$\square$) 

An exactly similar claim gives the existence of some $z''$ with $\rho_0(z'')\in \overline{D}$, 
thereby completing the proof for this last remaining subsubcase, and hence the proof of the 
theorem. 
\end{proof} 




\begin{corollary} \label{cor1} 
Let $(X, d), G, \hat{G}$ be as above, 
again with $|G|>1$. 


Then there exists uniformly continuous 
\[f: \overline{\{(\pi\cdot x_0, \pi \cdot x_1):
\pi \in G\}}\rightarrow [0, 1]\] 
such that for any non-empty right ideal 
${\mathcal I}\subset\hat{G}$ there are
\[\rho, \gamma\in {\mathcal I}\]
with
\[f(\rho(x_0), \rho(x_1))=0,\]
\[f(\gamma(x_0), \gamma(x_1))=1.\] 
\end{corollary}

\begin{proof} 
We can assume that the ideal has the form 
${\mathcal I}=\j(\rho_0)$ for some $\rho_0\in \hat{G}$, 
and then \ref{cor1} follows directly from the definitions and 
the theorem. 
\end{proof} 




\begin{corollary} \label{cor2}
Let $(X, d), G, \hat{G}$ be as above,
again with $|G|>1$. Define $\hat{D}$ to be the usual 
metric 
on $\hat{G}$, with 
\[\hat{D}(\rho_0, \rho_1)= 
\sum_{n\in \N} 2^{-n}{\rm max}(1, d(\rho_0(z_n), \rho_1(z_n)))\] 
for some dense set $\{z_n: n\in \N\}\subset X$. 

Then there exists uniformly continuous 
\[\hat{f}: \hat{G}\rightarrow [0, 1]\] 
such that for  any non-empty right ideal
${\mathcal I}\subset\hat{G}$ there are
\[\rho, \gamma\in {\mathcal I}\]
with
\[\hat{f}(\rho)=0,\]
\[\hat{f}(\gamma)=1.\] 
\end{corollary} 

\begin{proof} Take $f$ and $x_0, x_1$ 
as in \ref{cor1} and let 
\[\hat{f}(\rho)= f(\rho(x_0), \rho(x_1)).\] 
It is immediate from the assumptions on $f$ that 
$\hat{f}$ has high oscillation on any non-trivial 
ideal and 
the definition of $\hat{D}$ implies that $\hat{f}$ is uniformly 
continuous. 
\end{proof} 


\begin{remark} 
In each of \ref{maintheorem}, \ref{cor1}, \ref{cor2} we 
can ask that $f$ not just be uniformly continuous but in fact 
Lipschitz. For instance in \ref{maintheorem} we 
can replace $f$ by 
\[f^*: X^2\rightarrow [0, 1]\] 
with 
\[f^*(z_0, z_1)={\rm min}(1, {\rm inf}\{d(z_0, y_0)+ d(z_1, y_1): 
f(y_0, y_1)=0)\}).\] 
$f^*$ is clearly Lipschitz and 
there will be some positive 
$\epsilon\leq 1$, depending on the 
uniformity of continuity of $f$, and 
fixed $x_0, x_1\in X$ 
such that $f^*$ will assume 
the value 0 and at least the value $\epsilon$ 
on 
\[\{(\rho_0(\pi(x_0)), \rho_0(\pi(x_1))): \pi\in G\}\] 
for any $\rho_0\in \hat{G}$. 
Then we can simply rescale and truncate $f^*$, 
replacing it by 
\[(z_0, z_1)\mapsto 
{\rm min}(1, \frac{f^*(z_0, z_1)}{\epsilon}).\] 
\end{remark} 

\subsection{Oscillation for Polish groups} 
\label{polishgroups} 

Theorem \ref{maintheorem} allows us to conclude with a kind of 
distortion theorem for Polish groups. 
If $G$ is a Polish group and $d_L$ is a left invariant metric, 
and $\hat{G}$ is the Cauchy completion of $G$ with respect to 
$d_L$, then there is a uniformly continuous 
\[f: \hat{G}\rightarrow [0,1]\] 
such that for any non-empty ideal ${\mathcal I}\subset \hat{G}$ we 
have $f$ attaining both 0 and 1 on ${\mathcal I}$. 

There are a couple of remarks and definitions before we can give the 
proof. 


\begin{theorem} (Birkhoff-Kakutani) 
Every Polish group has a compatible left invariant 
metric. 
\end{theorem} 

\begin{proof} See \cite{beckerkechris}. 
\end{proof} 

This is {\it not} to say that every Polish group admits a 
{\it complete} left invariant metric. 
$S_\infty$ for instance, the group of all permutations of 
the natural numbers with 
the topology of pointwise convergence, does not. 

\begin{lemma} \label{invariance} 
 Any two compatible left invariant metrics have the same Cauchy sequences. 
\end{lemma} 

\begin{proof} Note that a sequence $(g_n)_{n\in \N}$ is Cauchy with respect to one 
compatible left invariant metric (and hence all compatible left invariant metrics) 
if and only if for every open neighborhood $U$ of the identity there is an $N$ such that 
for all $m, n>N$
\[g^{-1}_mg_n\in U.\] 
\end{proof} 

Thus the {\it left completion} of a Polish group does not depend on the choice of 
a left invariant metric. This gives a way of exactly matching up the notions from 
$\S$\ref{biglentil} with the present situation. 

%Let $G$ be a Polish group, and $d_L$ a compatible left invariant metric, which, 
%after possibly truncating, we can assume is bounded by one. Let $\{x_n: n\in \N\}$ be a 
%dense subset of $G$ and define the left invariant metric $\hat{D}$ on 
%In$(G, d_L)$ as in $\S$\ref{lemmas} by 
%\[\hat{D}(\rho_0, \rho_1)=\sum_{n\in \N} 2^{-n}(d(\rho_0(x_n), \rho_0(x_n))).\] 
%Each $g\in G$ induces an element of Isom$(G, d_L)$ by considering the left 
%translation 
%\[\pi_g: h\mapsto gh,\] 
%and thus $\hat{D}$ provides a compatible left invariant metric on $G$, and thus, since 
%all left invariant metrics give the same notion of Cauchy completion, 
%we may identify the left completion of $G$ with the closure of 
%\[\{\pi_g: g\in G\}\] 
%in In$(G, d_L)$. 

\begin{notation} 
For $G$ a Polish group, let $\hat{G}$ be the Cauchy completion of $G$ with respect to 
some (equivalently, any) choice of a compatible left invariant 
metric $d_L$ on $G$. 
\end{notation} 



\begin{lemma} \label{cauchy}
If $(g_n)_{n\in \N}, (h_n)_{n\in \N}$ are Cauchy with respect to a left invariant metric 
on a Polish group $G$ then  
$(g_n h_n)_{n\in \N}$ is Cauchy with respect to the left invariant metric.
\end{lemma} 

\begin{proof} Nodding back to the proof of the last lemma, note that for any open neighborhood 
$U$ of the identity we can find an $N_1<N_2$ such that all bigger $n,m\geq N_2$ have 
\[g_m^{-1} g_n \in h_{N_1}Uh_{N_1}^{-1},\] 
and for all $m, n\geq N_1$
\[h_m^{-1}h_n\in U.\] 
Then at any $m, n\geq N_2$
\[(g_m h_m)^{-1} (g_n h_n) = (h_m^{-1} h_{N_1}) (g_m h_{N_1})^{-1}g_n h_{N_1} 
(h_{N_1}^{-1}h_n)\in U^3.\] 
Letting $U$ shrink to radius zero proves the lemma. 
\end{proof} 



Thus $G$ acts on its left completion by multiplication on the right. 

\begin{definition} A subset ${\mathcal I}\subset \hat{G}$ is said to be a 
{\it right ideal} 
if it is closed under composition and under multiplication on the right by elements 
in $G$. 
\end{definition} 

Note that we can identify each $g\in G$ with an isometry 
\[\pi_g: h\mapsto gh\] 
of $(G, d_L)$ for any left invariant metric. Since $G$ is dense in $\hat{G}$, any such 
isometry extends to an  isometry of $\hat{G}$. 

\begin{lemma} \label{careful} 
Under the natural identification of $G$ with a subgroup of 
{\rm Isom}$(\hat{G}, d_L)$ and $\hat{G}$ with a subset of 
{\rm In}$(\hat{G}, d_L)$, and for 
$\hat{D}$ defined as in $\S$\ref{lemmas}, we have that $\hat{G}$ equals the 
$\hat{D}$-closure of $G$ and the composition of elements in $\hat{G}$ in the sense of 
lemma \ref{lemmas} coincides with their 
composition as elements of ${\rm In}(\hat{G}, d_L)$ -- and in fact, for 
each $\epsilon>0$ there exist $\delta_1, \delta_2$ such that 
\[\hat{D}(\rho_0, \rho_1)<\delta_1\Rightarrow d_L(\rho_0, \rho_1)<\epsilon,\] 
\[d_L(\rho_0, \rho_1)<\delta_2\Rightarrow \hat{D}(\rho_0, \rho_1)<\epsilon.\] 
\end{lemma} 

\begin{proof} $\hat{D}$ provides a left invariant metric 
on $G$, and hence gives rise to the same completion by 
lemma \ref{invariance}. 
For checking the compositionality, note that the 
composition of the isometries associated 
to Cauchy $(g_n)_n, (h_n)_n$ sends each 
${h}\in{G}$ to the limit of $(g_nh_nh)_n$. 

It remains to show the equivalence of the two metrics defined on  
$\hat{G}$, as either the Cauchy completion of $G$ or the closure 
of $\{\pi_g: g\in G\}$ in In$(\hat{G}, d_L)$. 
We fix a dense subset $\{z_i: i\in \N\}$ of
$\hat{G}$ and define the metric $\hat{D}$ as usual. We can assume
that this dense subset lies inside $G$, since $G$ is dense in its
Cauchy completion, and we can also assume that $z_1=e$, the identity 
in $G$.  
It is clear that the topology obtained by viewing $\hat{G}$ as 
a subset of In$(\hat{G}, d_L)$ is at least as strong the topology 
given by viewing as the Cauchy completion, 
since for any $\epsilon >0$ we have that if 
\[\hat{D}(\rho_0, \rho_1)<\frac{\epsilon}{2}\] 
then 
\[d_L(\rho_0(z_1), \rho_1(z_1))=d_L(\rho_0z_1, \rho_1z_1) =d_L(\rho_0, \rho_1) 
<\epsilon.\] 
For the converse, consider $\epsilon>0$. We can 
find $\delta>0$ and   
an initial segment $z_1,..., z_N$ of our enumerated 
dense subset, 
such that for any $\rho_0, \rho_1\in \hat{G}$ with 
\[d_L(\rho_0(z_i), \rho_1(z_i))< \delta\] 
all $i\leq N$ we have 
\[\hat{D}(\rho_0, \rho_1)<\epsilon.\] 
Then we can find some $\delta_1$ such that 
\[d_L(g, e)<\delta_1\Rightarrow \forall i\leq N
(d_L(z_i^{-1}g z_i, e)<\delta,\] 
\[\therefore \forall g_0, g_1\in G\forall i\leq N
(d_L(g_0, g_1)<\delta_1\Rightarrow d_L(z_i^{-1}(g_1^{-1} g_0) z_i, e) 
=d_L(g_0z_i, g_1z_i)<\delta,\] 
\[\therefore\forall \rho_0, \rho_1\in \hat{G} \forall i\leq N 
(d_L(\rho_0, \rho_1)<\delta_1\Rightarrow d_L(\rho_0 z_i, \rho z_i)
<\delta),\] 
as required. 
\end{proof} 


Thus in particular 
the definition of {\it closed right ideal} in $\hat{G}$ given above 
agrees with the definition presented in $\S$\ref{lemmas}. 

\begin{theorem}  \label{polishdistort} 
Let $G$ be a Polish group of size bigger than one, and 
$\hat{G}$ its left completion with respect to some left invariant metric 
$d_L$. Then there is a uniformly continuous 
\[f: (\hat{G}, d_L)\rightarrow [0, 1]\] 
such that on any non-empty ideal ${\mathcal I}\subset \hat{G}$ we have 
$f$ obtaining both zero and one. 
\end{theorem} 

\begin{proof} As in the proof of theorem \ref{maintheorem} it suffices to 
show that we can obtain $f: \hat{G}\rightarrow [0, 3]$ such that on any non-empty 
ideal it hits values less than one and greater than 2. 
Following \ref{maintheorem} we can choose uniformly continuous 
$(\gamma_0, \gamma_1)\in \hat{G}$ and uniformly continuous 
\[f_0: \overline{\{(g\cdot \gamma_0, g\cdot \gamma_1): 
g\in G\}}\rightarrow [0, 3]\] 
such that for any $\rho_0\in \hat{G}$ and 
induced ideal $\j(\rho_0)\subset {\rm In}(\hat{G}, d_L)$ 
we attain values in both $[0, 1)$ and $(2, 3]$ on the set 
\[\{(\rho\cdot \gamma_0, \rho\cdot \gamma_1): \rho \in \j(\rho_0)\}.\]  
This induces a function 
\[f:\hat{G}\rightarrow [0,3]\] 
with the specification that 
\[f(\rho)=f_0(\rho\cdot \gamma_0, \rho\cdot \gamma_1).\] 
It follows immediately from \ref{careful} that 
$f$ is uniformly left continuous as a function on 
$(\hat{G}, d_L)$ and that $f$ obtains the required oscillation on 
any non-empty ideal. 
\end{proof} 

This theorem is in the form posed as a problem in 
\cite{kepeto}. 

\subsection{Connecting \ref{ultra} to the oscillation theorem} 
\label{derive} 

\begin{definition} $S_\infty$ is the group of all permutations 
of $\N$ with the topology of pointwise convergence. 
\end{definition} 

We specifically include permutations 
with infinite support in $S_\infty$. 
A basic open set in this topological group 
consists in all 
$\pi\in S_\infty$ such that 
\[\pi(i)=k_i\] 
all $i\leq N$, some choice of $k_1,..., k_N$. 

\begin{theorem} (Folklore) \label{closed} 
Any closed subgroup of 
$S_\infty$ equals Aut$(\m)$ for some ultrahomogeneous 
structure whose underlying set is $\N$. 
\end{theorem} 

\begin{proof} See \cite{beckerkechris}. 
\end{proof} 


\begin{notation} Given a countable 
ultrahomogeneous 
structure $\m$, 
let ${\rm In}(\m)$ be the set of all monomorphisms  
\[\rho: \m\hookrightarrow \m.\] 
We place a complete 
metric on ${\rm In}(\m)$ as follows. We first 
assume without loss that the underlying set of $\m$ is the 
set of natural numbers. Then for $\rho_0\neq \rho_1$ we 
let $\Delta(\rho_0, \rho_1)$ be the first $n$ with 
$\rho_0(n)\neq \rho_1(n)$, and then define   
\[d(\rho_0, \rho_1)=2^{-\Delta(\rho_0, \rho_1)}.\] 
\end{notation} 


Recall that in $\S$\ref{polishgroups} we defined 
$\hat{G}$ for $G$ a Polish group to be the Cauchy completion 
with respect to any compatible left invariant metric. 
We now have that $d$ provides a left invariant metric on 
${\rm In}(\m)$ and hence Aut$(\m)$, and so for $G={\rm Aut}(\m)$ 
we can identify $\hat{G}$ with the closure of 
Aut$(\m)$ with respect to $d$. 

Using ultrahomogeneity 
this $\hat{G}$ actually turns out to be the full 
semi-group of monomorphisms from $\m$ to itself. 

\begin{lemma} \label{derive1}
For $\m$ ultrahomogeneous,  $G={\rm Aut}(\m)$, 
we have that $\hat{G}$ equals ${\rm In}(\m)$. 
\end{lemma} 

\begin{proof}
Immediately from the definition of ultrahomogeneous we have that 
for any $\rho\in {\rm In}(\m)$ and $\vec a, \vec b\in \m^{<\infty}$, 
if $\rho(\vec a)=\vec b$ then there is an automorphism 
$\pi$ with $\pi(\vec a)=\vec b$. Since this holds for all 
$\vec a, \vec b$ we have $\rho\in \hat{G}$. 
\end{proof} 

\begin{lemma} \label{derive2} 
Let $\rho_0\in {\rm In}(\m)$. Then the ideal 
\[\j(\rho_0)=
\overline{\{\rho_0\circ \pi: \pi \in {\rm Aut}(\m)\}}\] 
equals 
all $\gamma\in {\rm In}(\m)$ with 
{\rm range}($\gamma)\subset$ {\rm range}$(\rho_0)$. 
\end{lemma} 

\begin{proof} It is clear 
from the definitions 
that if $\gamma\in \j(\rho_0))$ then 
its image is included in that of $\rho_0$. Conversely, 
if the range($\gamma)\subset$ range$(\rho_0)$ then at each 
$\vec a\in \m^{<\infty}$ we can first find $\vec c$ with 
$\rho_0(\vec c)=\gamma(\vec a)$ and then, since $\vec a, \vec c$ 
have the 
same type, we can find $\pi\in {\rm Aut}(\m)$ with 
$\pi(\vec a)=\vec c$ to obtain 
\[\rho_0\circ \pi(\vec a) =\gamma(\vec a).\] 
Since this holds for all such $\vec a$ we have that 
$\gamma$ is in the closure of $\{\rho_0 \circ \pi: 
\pi\in {\rm Aut}(\m)\}$. 
\end{proof} 

With all this background out of the way, 
we can obtain theorem \ref{polishdistort} for $G$ a 
closed subgroup of $S_\infty$ as follows.  

Applying \ref{closed} we can assume $G={\rm Aut}(\m)$ 
for some ultrahomogeneous $\m$. 
We use \ref{ultra} to find some $(a_0, a_1)\in \m^2$ 
and 
\[f:\m^2\rightarrow \{0, 1\}\] such that for  
$S$ the set of all pairs with the same type as $(a_0, a_1)$, 
we have $f$ obtaining both possible values on 
$S\cap \rho_0[\m]\times \rho_0[\m]$ for any monomorphism 
$\rho_0:\m\rightarrow \m$. 
Then we can define 
\[\hat{f}:\hat{G}\rightarrow \{0, 1\}\]
by $\hat{f}(\gamma)=f(\gamma(a_0), \gamma(a_1)).$ 
It is trivial to verify that this 
is left uniformly continuous 
Given 
any $\rho_0\in \hat{G}$ 
we choose $\vec b=(b_0, b_1), \vec c=(c_0, c_1)$ in 
range$(\rho)\cap S$ with $f(\vec b)=0, f(\vec c)=1$. 
We let $b_0', b_1', c_0', c_1'$ be the preimages under 
$\rho$ of $b_0, b_1, c_0, c_1$ respectively. 
Appealing to ultrahomogeneity we can find $\pi, \sigma\in 
{\rm Aut}(\m)$ with 
\[\pi(a_0)=b_0', \: \: \pi(a_1)=b_1',\] 
\[\sigma(a_0)=c_0', \: \:  \sigma(a_1)=c_1',\] 
and hence $\rho\circ \pi(\vec a)=\vec b$, 
$\rho\circ\sigma(\vec a)=\vec b$, and thus
\[\hat{f}(\rho\circ \pi)=f(\vec b)=0,\] 
\[\hat{f}(\rho\circ \sigma)=f(\vec c)=1,\] 
as required. 





%This section is just to point out that theorem 
%\ref{maintheorem} really does subsume 
%theorem \ref{ultra}. 

%\begin{definition} Given a countable ultrahomogeneous 
%structure $\m$ we associate a complete 
%{\it countable} metric space $(X_\m, d_\m)$ as 
%follows. 

%First we choose a one-to-one assignment of real numbers in the 
%open unit from $\frac{1}{4}$ to $\frac{1}{2}$ to quantifier 
%free types. This can be thought of as a function 
%\[\alpha: \m^{<\infty}\rightarrow (\frac{1}{4}, \frac{1}{2})\] 
%such that $\alpha(\vec a)=\alpha(\vec b)$ if and only if 
%they have the same quantifier free type. 

%We then take as the underlying set of $X_\m$ the 
%{\it disjoint} union of $\m$, $\m^1$, $\m^2$, $\m^3$,... 
%Then given any $\vec a=(a_0,..., a_n)\in \m^n$ we 
%set the distance in $X_\m$ of $\vec a$ to any $a_i$ to 
%be exactly $c(\vec a)$. For $a, b\in \m$, we let 
%$d_\m(a, b)=1$. For all other points in $X_\m$ we take the 
%induced path metric. Thus for 
%$\vec a\notequal \vec b\in \m^{<\infty}$ we let 
%\[d_\m(\vec a, \vec b)= c(\vec a) + c(\vec b)\] 
%if the sequences overlap and 
%\[d_\m(\vec a, \vec b)=c(\vec a) + c(\vec b) +1\] 
%if they are disjoint. 
%\end{definition} 

%\begin{lemma} \label{func1}
%Any onto isometry of $(X_\m, d_\m)$ fixes 
%$\m$ set wise and moves finite 
%sequences inside $\m$ to other finite sequences with the same 
%quantifier free type. 
%\end{lemma} 

%\begin{proof} $\m$ is recognizable inside $X_\m$ as the set of 
%points which are distance exactly one away from some other point, 
%and hence this set is preserved by any invertible 
%isometry. Inside $\m$ we similarly recognize the finite sequences 
%with the same type as some given $\vec a$, since these are exactly 
%the sequences of the same length as $\vec a$ for which there exists a 
%point sitting simultaneously 
%exactly distance $c(\vec a)$ away from them all. 
%\end{proof} 

%\begin{lemma} \label{func2} 
%Any automorphism of $\m$ extends uniquely to 
%an onto isometry of $X_\m$. 
%\end{lemma} 

%\begin{proof} The {\it existence} of an extension 
%follows directly from the invariant manner in which 
%$X_\m$ was defined. Uniqueness follows from 
%\ref{funct1}. 
%\end{proof} 

%\begin{lemma} \label{func3} 
%If $G$ is the isometry group of $X_\m$ and 
%$\rho\in \hat{G}$, then $\vec a$ is in the image of 
%$\rho$ if and only if each $a_i$ in the sequence is 
%in the image of $a$. 
%\end{lemma} 

%\begin{proof} Let $\sigma_n\rightarrow \rho$. 
%$\vec a$ is in the image of $\rho$ if and only if there 
%is single $\vec b$ such that for all sufficiently large 
%$n$ we have $\sigma_n(b_i)=a_i$ at each suitable $i$. This 
%in turn is equivalent to saying that each $a_i$ is in the 
%image of $\rho$. 
%\end{lemma} 








\subsection{$n=1$ and $n=2$ spaces} 
\label{tomandjerry}

\begin{definition} Let $(X, d)$ be a complete separable metric 
space and let $G$ be the isometry group of all onto isometries 
from $X$ to $X$. As previously defined, let $\hat{G}$ be its closure 
in the space of all isometries from $X$ into $X$. 
We say that $X$ is an {\it $n=1$ space} if there is a single point 
$x_0\in X$ and  
uniformly continuous 
\[f: \overline{\{\pi(x_0): \pi\in G\} }\rightarrow [0,1]\] 
such that for any $\rho_0\in \hat{G}$ we 
have $f$ achieving both 
extreme values on 
\[ \overline{\{\rho_0(\pi(x_0)): \pi\in G\} }.\] 
Otherwise we say that $X$ is an {\it $n=2$ space}. 
\end{definition} 

Thus this terminology is motivated by 
\ref{maintheorem} which says that $X$ is either an $n=1$ space 
or there is an 
\[f: \overline{\{(\pi(x_0), \pi(x_1)): 
\pi\in G\} }\rightarrow [0,1]\] 
which picks up both extreme values on 
any 
\[\overline{\{(\rho_0(\pi(x_0)), \rho_0(\pi(x_1))): \pi\in G\} }.\] 

While there are some $n=2$ spaces, most of the 
examples I know are trivial. 

\begin{example} Let $(X, d)$ be a countably infinite set equipped with 
the discrete metric. This is an $n=2$ space. 
\end{example} 

\begin{example} Recall from say \cite{gaokechris} that the 
{\it Urysohn space} ${\mathbb U}$ 
is the complete, separable, universal metric 
space which is ultrahomogeneous, in the 
sense that any isometry between finite subsets extends to an 
isometry of the whole space onto itself. 
 
The Urysohn space is an $n=1$ space: 
Fix some arbitrary $z_0$ in the space. 
We choose rapidly increasing $(k_n)_{n\in \N}$ with 
\[k_n >\sum_{m<n} 2k_m,\] 
and we let $B_m=\{z: k_m< d(z, z_0)< 2 k_m\}$. We can certainly 
find a uniformly continuous 
function $f$ which is 0 on each $B_{2m}$ and 
1 on each $B_{2m+1}$, and then any isometric embedding 
\[\rho:{\mathbb U}\rightarrow {\mathbb U}\] 
must pick up all but finitely many of the $B_m$'s in its 
image. 
\end{example} 

This argument arose in the course of conversations with 
Vladimir Pestov, and it is clear that it modifies to show 
any unbounded metric space to be an $n=1$ space. 

\def\U{{\mathbb U}}

\begin{example} A more mysterious example is the 
{\it bounded Urysohn space},$\U_{\leq 1}$ -- the 
complete separable ultrahomogeneous 
metric space which is universal for 
metric spaces bounded by 1 ($d(x, x')\leq 1$ all 
$x, x'$). 
\end{example} 

\begin{question} Is the bounded Urysohn space an 
$n=1$ space? 
\end{question} 

The next example might give some hope. 

\begin{example} Let $\U_{\{0, 1, 2\} }$ be the shaved 
down version $\U$ allowing only the distances 
$0, 1, 2$. This can be characterized as follows: It is the 
{\it countable} metric space where $d(x, x')$ is always 
either 0, 1, or 2, and which satisfies the 
following {\it extension} axiom: 
For any disjoint finite sequences 
$x_1, ..., x_N$ and $y_1,..., y_M$ we can find some $z$ with 
\[d(x_i, z)=1,\] 
\[d(y_j, z)=2,\] 
all $i\leq N, j\leq M$. 
(It is a routine back-and-forth argument to show that 
 $\U_{\{0, 1, 2\} }$ is the unique up to isometry countable 
ultrahomogeneous 
metric space universal among countable spaces 
realizing only distances 0, 1, 2. 
Compare for instance chapter 2 of \cite{beckerkechris}.) 

$\U_{\{0, 1, 2\} }$ is an $n=2$ space. The main point is that 
given any $x_1,..., x_N, y_1,..., y_M$ as above, the set of 
$z$ which satisfy $d(x_i, z)=1,$  $d(y_j, z)=2,$
all $i\leq N, j\leq M$ is already isometric to $\U_{\{ 0, 1, 2\} }$. 
Thus given $A\subset \U$ we either have that $A$ satisfies the 
extension axiom, or there is some  $x_1,..., x_N, y_1,..., y_M$ 
in $A$ such that the set of $z$'s with 
$d(x_i, z)=1,$  $d(y_j, z)=2,$
all $i\leq N, j\leq M$ forms a copy of  $\U_{\{ 0, 1, 2\} }$ 
disjoint from $A$. 
\end{example} 

Thus in trying to understand the situation with the 
bounded Urysohn space one might want to ask about the 
{\it rational bounded Urysohn space}. 

\begin{example} Let $\U_{\Q, \leq 1}$ be the 
{\it countable} ultrahomogeneous metric space which is universal among all 
countable metric spaces realizing only rational distances bounded by 
one. 
Although this space is 
not complete, it is natural from the model theoretic point of view. 
If we introduce a relation $R_q$ for each $q\in\Q\cap [0,1]$ with 
\[R_q(x, x')\] 
exactly when $d(x, x')=q$ then we obtain a 
countable ultrahomogeneous structure. 
\end{example} 

In the first draft of this paper the following question was posed: 

\begin{question} Does any partition of $\U_{\Q, \leq 1}$ into 
finitely many pieces have one piece in which there is an isometric 
copy of the whole space? 
\end{question} 


This has since been answered negatively 
by \cite{delaposa}. Their argument seems to leave unresolved the 
earlier problem of whether the bounded Urysohn space is an 
$n=1$ space. 

\def\B{{\mathbb B}} 

\subsection{The ragged end: 
Unclear 
connections with Banach space theory} 
\label{ragged} 

\begin{definition} Let $\B$ a Banach space and 
$(\B_1, d)$ the unit ball, with the obvious metric 
$d(x, x')=\nh x-x'\nh$. 
Let $G$ be the group of all isometries of 
$(\B_1,d)$ which are the restrictions of 
linear isometries of $\B$ onto itself. 
In the usual way we form a metric $\hat{D}$ on 
the space In($X, d$) of isometries from 
$X$ into itself with 
\[\hat{D}(\rho_0, \rho_1)=\sum_{n\in \N} 
2^{-n} d(\rho_0(x_n), \rho_1(x_n))\] 
for some dense subset $\{x_n: n\in \N\}$ 
and let $\hat{G}$ be the $\hat{D}$-closure of 
$G$. 

I will say that $\B$ is an {\it $n=1$ Banach space} 
if there exists  point $x_0\in \B_1$ and 
a uniformly continuous 
\[f: \overline{G\cdot x_0}\rightarrow [0,1]\] 
such that for any $\rho\in\hat{G}$ there are 
$y_0, z_0\in \rho[G\cdot x_0]$ with 
\[f(y_0)=0,\] 
\[f(z_0)=1.\] 
\end{definition} 

In the case of Hilbert space 
$\ell^2$ we can find for 
each infinite dimensional subspace $H\subset \ell^2$ 
a sequence of isometries $\sigma_n: \ell^2\rightarrow \ell^2$ 
along with $x(\xi)_{\xi\in H}$ such that  
\[\sigma_n(x(\xi))\rightarrow \xi\] 
and 
\[\{x(\xi):\xi\in H\}=\ell^2.\] 
Thus we obtain $\rho\in \hat{G}$ as the 
limit of the $\sigma_n|_{\B_1}$ with 
$\rho[\B_1]=H$. Thus the assertion that Hilbert space is 
an $n=1$ Banach space amounts to the assertion there is a 
uniformly continuous 
function from the unit ball to $[0,1]$  which realizes both 
extreme points on any infinite dimensional subspace, an 
assertion which has indeed been proved in Odell and Schlumprecht's 
celebrated solution to the {\it distortion problem}. 


\begin{theorem} (Odell-Schlumprecht; see \cite{odellschlumprecht}) 
If $\B=\ell^p$ for $1<0<\infty$ then there is a uniformly 
continuous function from the unit ball 
\[f:\B_1\rightarrow [0,1]\] 
which assumes both 0 and 1 on every infinite dimensional 
subspace. 
\end{theorem} 


This should compared with: 

\begin{theorem} (Gowers; see \cite{gowers})
If $\B=c_0$, then for any uniformly continuous function from 
the unit ball 
\[f:\B_1\rightarrow [0,1]\]
and any $\epsilon>0$ there is an isometric copy of $c_0$ appearing 
as a subspace on which $|f(x)-f(x')|$ is bounded by 
$\epsilon$. 
\end{theorem} 

However in contrast to $\ell^2$ there are isometric embeddings of 
$c_0$ into itself which do not appear in $\hat{G}$, the closure of the 
{\it onto} isometries. It is not hard to see that any onto isometry of 
$c_0$ must move take the canonical basis back to itself with possible 
multiplications by $-1$, and thus for $c_0$ we actually have that 
the onto isometries are {\it already} closed in the space of 
isometries, and thus, for almost trivial reasons,  
$c_0$ is an $n=1$ Banach space. 
It is therefore natural to ask: 

\begin{question} Does there exist a separable Banach space which is 
not an $n=1$ Banach space? 
\end{question} 





\begin{thebibliography}{99}
 




\bibitem{beckerkechris} H. Becker, A.S. Kechris, 
{\bf The descriptive set theory of Polish group actions,} 
London Mathematical Society Lecture Note Series, 
232, Cambridge University Press, Cambridge, 1996

\bibitem{delaposa} C. Delhomme, C. Laflamme, M. Pouzet, N. Sauer, 
{\it The indivisiability of countable metric spaces,} 
preprint, http://arxiv.org/abs/math.CO/0510254. 

\bibitem{gaokechris} S. Gao, A.S. Kechris, 
{\bf On the classification of Polish metric spaces up to isometry, }
Memoirs of the American Mathematical Society, vol. 766, 2003. 


\bibitem{gowers} T. Gowers,  {\it Lipschitz functions on classical spaces,} 
{\bf European Journal of Combinatorics,} vol 13 (1992), pp. 141--151. 


\bibitem{hodges} W. Hodges, {\bf Model theory,}   
Encyclopedia of Mathematics and its Applications, vol 42,  
Cambridge University Press, Cambridge, 1993.

%\bibitem{classical} A.S. Kechris, {\bf Classical descriptive set 
%theory,} Graduate Texts in Mathematics 156, Springer-Verlag, 
%New York, 1995. 

\bibitem{kepeto}A.S. Kechris, 
V.G. Pestov, S. T{o}dorcevic, 
{\it Fraiss\"{e} li\"{m}its, Ramsey theory and 
topological dynamics of automorphism groups}, 
to appear in {\bf Geometric and Functional Analysis}.  

\bibitem{keisler} H.J. Keisler, 
{\bf Model theory for infinitary logic,}  
Studies in Logic and the Foundations of Mathematics, Vol. 62,  
North-Holland Publishing Co., Amsterdam-London, 1971.


\bibitem{odellschlumprecht} E. Odell, T. Schlumprecht, 
{\it The distortion problem,} 
{\bf Acta Mathematica}, vol. 173 (1994), pp. 259--281.

\end{thebibliography} 


\leftskip 0.4in 

\noindent {\texttt greg@math.ucla.edu} 

\medskip

\noindent http://www.math.ucla.edu/\~{}greg/

\medskip 

\noindent 7340 MSB 

\noindent Mathematics 

\noindent UCLA 

\noindent 405 Hilgard Avenue

\noindent Los Angeles

\noindent CA 90095-1555

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