\documentstyle[12pt, oldlfont]{article}

\input{amssymb.def}
\input{amssymb.tex}





% Set left margin - The default is 1 inch, so the following 
% command sets a 1.25-inch left margin.
\setlength{\oddsidemargin}{-0.25in}

% Set width of the text - What is left will be the right margin.
% In this case, right margin is 8.5in - 1.25in - 6in = 1.25in.
\setlength{\textwidth}{7in}

% Set top margin - The default is 1 inch, so the following 
% command sets a 0.75-inch top margin.
\setlength{\topmargin}{-0.7in}

% Set height of the header
\setlength{\headheight}{0.5in}

% Set vertical distance between the header and the text
\setlength{\headsep}{0.2in}

% Set height of the text
\setlength{\textheight}{8.8in}

% Set vertical distance between the text and the
% bottom of footer
\setlength{\footskip}{0.4in}

\pagestyle{myheadings}
%\markright{\empty}



% Set the beginning of a LaTeX document
\begin{document}


\def\Ubf#1{{\baselineskip=0pt\vtop{\hbox{$#1$}\hbox{$\sim$}}}{}}
\def\ubf#1{{\baselineskip=0pt\vtop{\hbox{$#1$}\hbox{$\scriptscriptstyle\sim$}}}{}}
\def\R{{\Bbb R}}
\def\V{{\Bbb V}}
\def\N{{\Bbb N}}
\def\n{{\Bbb N}}
\def\Q{{\Bbb Q}}
\def\O{{\cal O}}
\def\e{\epsilon}
\def\h{{\hat{X}}}
\def\B{{\cal B}}

\title{{\bf Variations on Scott} (preliminary report)}   
      % Enter your title between curly braces
\author{Greg Hjorth}        % Enter your name between curly braces
\date{\today}          % Enter your date or \today between curly braces
\maketitle

{\bf $\S$0. Introduction} 

Recent work in the general theory of Polish groups has 
looked for inspiration from the study of countable models, 
where a major tool in analyzing isomorphism types is provided by 
the Scott analysis. One has many results that were first 
proved in the case of the ``logic action" on countable models 
by an appeal to the Scott analysis, and only later generalized to 
general Polish group actions -- usually in some {\it ad hoc} manner, 
and occasionally in a somewhat diluted form. On top of this there were  
even a small number of results known for closed subgroups of 
$S_{\infty}$ by use of the 
Scott analysis but open for more general Polish groups. 

In this note we present one attempt to generalize the Scott analysis. 
For each $\alpha<\omega_1$ one provides an equivalence relation 
$\approx_{\alpha}$ that is coarser than the original orbit equivalence 
relation, but such that at each point $x$ 
in the space there will be 
an $\alpha_x$ such that orbit of $x$ is equal to the set of points 
$\approx_{\alpha_x}$-equivalent to $x$. 
The essential new idea in this -- as against simply trying to 
rewrite what was known specifically for $S_{\infty}$ -- is 
to define the $\approx_{\alpha}$ in terms of {\it transitive 
non-symmetric relations}, $\leq_{\beta}$, $\beta<\alpha$. 

This presents a departure from the analysis for closed subgroups of 
$S_{\infty}$, where the existence of a neighbourhood basis of clopen 
subgroups around the identity allows us to define the equivalence relation 
``the models ${\cal M}$ and ${\cal N}$ have the same $\alpha$th approximation 
to their Scott sentence" in terms of equivalence relations obtained by 
looking at the subgroups of $S_{\infty}$ consisting of all permutations 
fixing some finite subset of ${\Bbb N}$. In some sense, there is a 
canonical way to approximate $S_{\infty}$ by a countable collection of 
clopen subgroups. This method of approximation lifts to the orbit 
equivalence relation, and allows us to define 
``the models $\langle {\cal M},\vec a \rangle$ and $\langle{\cal N}, \vec b\rangle$ 
have the same $\alpha$th approximation 
to their Scott sentence" in terms of the set of $\beta$th approximations for 
$\{\langle {\cal M},\vec a c\rangle: c\in\N\}$ and 
$\{\langle {\cal N},\vec b d\rangle: d\in\N\}$ as $\beta$ ranges over $\alpha$. 

Here any naive attempt to extend this to arbitrary groups runs afoul if there is 
no such basis of open subgroups around the identity. More generally, it is known 
(see \cite{friedman}, \cite{hjorth}) that there are Polish groups whose actions do not 
admit objects in HC as complete invariants, and thus even more 
sophisticated attempts must necessarily have a structurally very different form 
to the original Scott analysis. 

The chief results obtained in this new analysis are as follows: 

\medskip 

{\bf 0.1 Theorem} Let $G$ be a Polish group and $X$ a Polish $G$-space. 
If every orbit is $\Ubf{\Pi}^0_{\alpha}$ then the orbit equivalence relation 
is $\Ubf{\Pi}^0_{\alpha+\omega}$. 

\medskip 

This result was known in the case that $G$ is a closed subgroup of the infinite 
symmetric group. Previously Ramez Sami had used a fairly abstract argument to 
show that 

\medskip 

{\bf 0.2 Theorem} Let $G$ be a Polish group and $X$ a Polish $G$-space. 
If every orbit is $\Ubf{\Pi}^0_{\alpha}$ some $\alpha<\omega_1$ 
then the orbit equivalence relation 
is Borel. (See \cite{sami}.) 

\medskip 

We also show that for each $\alpha$ we may assign to each $x\in X$ a $G$-invariant 
$\Ubf{\Pi}^0_{\alpha+1}$ (namely: the $\approx_{\alpha+1}$ equivalence class of 
$x$) which {\it determines} the invariant $\Ubf{\Pi}^0_{\alpha}$ sets met by 
$[x]_G$. The point is that this assignment is highly effective. So for instance 
one obtains results such as 

\medskip 

{\bf 0.3 Theorem} ($AD^{L({\Bbb R})}$)  
Let $G$ be a Polish group and $X$ a Polish $G$-space. 
Then in $L({\Bbb R})$ there is an injection 
\[i:X/G\hookrightarrow \Ubf{\Pi}^0_{\alpha+1}\] 
such that orbits $[x]_G$ and $[y]_G$ with the same image under 
$i$ will meet exactly the same invariant $\Ubf{\Pi}^0_{\alpha}$ sets. 

\medskip 

which the author at least had previously wondered about. Hitherto this was 
known for $G$ a closed subgroup of $S_{\infty}$. I also have a vague memory 
(this is just a preliminary report) that Slawek Solecki had verified 
0.3 in some further special cases. 

We also obtain by these methods an entirely new proof of the well known fact that:  

\medskip 

{\bf 0.4 Theorem}(folkore) 
Let $G$ be a Polish group and $X$ a Polish $G$-space. 
Then every orbit is Borel. (See \cite{beckerkechris}.)  

\medskip 

The point is that the new proof makes no appeal to descriptive set 
theoretic generalities, but is a kind of ``bottom up" proof. In this 
respect it closely resembles the proof using the Scott analysis that 
every countable model has Borel isomorphism type. 

Some discussion should be made of further results that are known for 
$S_{\infty}$ but have escaped proof in general: 

\medskip 

{\bf 0.5 Conjecture} The set of $x\in X$ such that 
\[x\in \bigcup_{n\in\N}\Ubf{\Pi}^0_{\alpha+n}\] 
is Borel for all $\alpha<\omega_1$. 

\medskip 

{\bf 0.6 Conjecture} Let $E$ be a Borel equivalence relation with 
$E\leq_B E^X_G$, $X$ some Polish $G$-space. Then there is a Polish  
$G$-space $Y$ with $E\leq_B E^Y_G$ {\it and} $E^Y_G$ Borel. 

\medskip 

Harvey Friedman has observed 0.6 for the 
case that $G=S_{\infty}$. (See \cite{friedman}.) I suspect that there 
is a fundamental obstacle to proving 0.6 and that a further new 
idea is needed. 

\bigskip 

{\bf $\S$1. Some proofs} 

From now on let $G$ be a Polish group and $X$ a Polish $G$-space. 

\medskip 

{\bf 1.1 Definition} For $V_0, V_1\subset G$ open and non-empty, 
$x_0, x_1\in X$, write 
\[(x_0, V_0)\leq_1(x_1, V_1)\] 
if 
\[\overline{V_0\cdot x_0}\subset \overline{V_1\cdot x_1}.\]
Write 
\[(x_0, V_0)\leq_{\alpha+1}(x_1, V_1)\] 
if for all $W_0\subset V_0$ open non-empty there is $W_1\subset V_0$ 
non-empty and open with 
\[(x_1, W_1)\leq_{\alpha}(x_0, W_0).\] 
For $\lambda$ a limit 
\[(x_0, V_0)\leq_{\lambda}(x_1, V_1)\] 
expresses that at every $\alpha<\lambda$ 
\[(x_0, V_0)\leq_{\alpha}(x_1, V_1).\] 

\medskip 

{\bf 1.2 Lemma} For $W_0\subset V_0$, $W_1\supset V_1$, 
all open and non-empty, 
\[(x_0, V_0)\leq_{\alpha}(x_1, V_1)\] 
implies 
\[(x_0, W_0)\leq_{\alpha}(x_1, W_1).\] 

Proof. Immediate from the definitions. \hfill $\Box$ 

\medskip 

From now on let $\B$ be a countable basis for $G$ and let 
$\B_0$ be the non-empty elements of $\B$. 
Assume $G\in\B$. 

\medskip 

{\bf 1.3 Lemma} 
\[(x_0, V_0)\leq_{\alpha+1}(x_1, V_1)\] 
if for all $W_0\subset V_0$in $\B_0$ there is $W_1\subset V_0$ 
in $\B_0$ with  
\[(x_1, W_1)\leq_{\alpha}(x_0, W_0).\] 

Proof. From 1.2. \hfill $\Box$ 

\medskip 

{\bf 1.4 Corollary} Let $(\hat{V}_n)_{n\in\N}$ enumerate $\B_0$. 
Then for all $\alpha<\omega_1$ 
\[\{(x_0, x_1, n,m): (x_1, \hat{W}_n)\leq_{\alpha}(x_0, \hat{W}_m)\}\]
is a $\Ubf{\Pi}^0_{\alpha+k}$ set for some $k\in\N$. 

\medskip 

{\bf 1.5 Lemma} (a) Each $\leq_{\alpha}$ is transitive. 

(b) $\alpha<\beta$ implies $\leq_{\alpha}\supset \leq_{\beta}$ (in the 
sense that $(x_1, W_1)\leq_{\beta}(x_0, W_0)$ implies 
$(x_1, W_1)\leq_{\alpha}(x_0, W_0)$. 

Proof. (a) is almost immediate from the definitions. (b) routinely by transfinite 
induction once we observe the slightly tricky fact that 
$\leq_{1}\supset \leq_{2}$. \hfill $\Box$ 

\medskip 

{\bf 1.6 Definition} Set $x_0\approx_{\gamma}x_1$ if for all $i\in\{0,1\}$, 
$\beta<\gamma$, 
$W_i\in \B_0$,  there exists 
$\hat{W}_i\subset W_i $ with $\hat{W}_i\in\B_0$ and 
$W_{1-i}$ with 
\[(x_i, \hat{W}_i)\leq_\beta (x_{1-i}, W_{1-i})\leq_\beta (x_i, W_i).\]

\medskip 

Note that $x_0E^X_G x_1$ implies $x_0\approx_{\gamma}x_1$. 

\medskip 

{\bf 1.7 Lemma} For $V_0, V_1\in \B_0$ and $\alpha<\omega_1$, $\alpha\neq 0$, 
there is an 
``essentially $\Ubf{\Pi}^0_{\alpha}$ relation" (in the sense of \cite{hjkelo}) 
$R$ such that if 
\[x_0\approx_{\gamma}x_1\] 
then $(x_0, V_0)\leq_{\alpha}(x_1, V_1)$ if and only if $R(x_0, x_1)$. 

Proof. (I say that $R$ {\it essentially} has a certain complexity if there is a 
Polish topology on $X$ compatible with the Borel structure in which this 
complexity is achieved.) 

For the base case of $\alpha=1$, we assign to $x_0, x_1$ the closures 
$\overline{V_0\cdot x_0}, \overline{V_1\cdot x_1}$ in ${\cal F}(X)$ 
(the standard Effros Borel space of closed sets), and note that it is 
essentially closed to assert in ${\cal F}(X)$ that one set includes 
another. At limit $\alpha$ the lemma resolves into triviality in light of 
1.4. 

So let us assume the lemma is true at $\alpha$ and attempt to step up to 
$\alpha+1$. Then  $(x_0, V_0)\leq_{\alpha+1}(x_1, V_1)$ if and only if 
for all $\hat{W}_0\subset W_0\subset 
V_0$ with $\hat{W}_0, W_0\in \B_0$, for all $W_1\in \B_0$ 
\[ [(x_0, \hat{W}_0)\leq_{\alpha}(x_1, W_1)\leq_{\alpha}
\leq_{\alpha}(x_0, W_0)]\Rightarrow \exists \hat{W}_1\subset V_1
[\hat{W}_1\in \B_0\wedge (x_1,\hat{W}_1)\leq_{\alpha}
(x_1, W_1)].\] 
Since the relation $\exists \hat{W}_1\subset V_1
[\hat{W}_1\in \B_0\wedge (x_1,\hat{W}_1)\leq_{\alpha}
(x_1, W_1)]$ is unary and Borel, it is essentially closed, and the lemma for 
$\alpha$ implies it for $\alpha+1$. \hfill $\Box$ 

\medskip 

{\bf 1.8 Lemma} For $\alpha<\omega_1$, $\alpha\neq 0$, the relation 
\[x_0\approx_{\alpha} x_1\]
is essentially $\Ubf{\Pi}^0_{\alpha+1}$. 

Proof. We prove this by transfinite induction on $\alpha$, with 
the base and limit stages again trivial. So assume true at $\beta$ and 
try to wench up to $\alpha=\beta+1$. The relation 
$x_0\approx_{\beta} x_1$ is essentially $\Ubf{\Pi}^0_{\beta+1}=
\Ubf{\Pi}^0_{\alpha}$ by inductive assumption. Then we can apply 1.7 to 
just read off that the definition of $x_0\approx_{\alpha} x_1$ 
is essentially $\Ubf{\Pi}^0_{\beta+2}$, as required. \hfill $\Box$ 

\medskip 

{\bf 1.9 Lemma} Let $B\subset X$ be $\Ubf{\Pi}^0_{\alpha}$ 
($\alpha\in [1,\omega_1)$). Suppose $(x_0, V_0)\leq_{\alpha} 
(x_1, V_1)$ and $x_1\in B^{*V_1}$. 

Then  $x_0\in B^{*V_0}$. 

Proof. This is trivial at the base case of $\alpha=1$, since it amounts 
to calculating $\overline{V_0\cdot x_0}$ and $\overline{V_1\cdot x_1}$. 
Now suppose it is true at all $\beta<\alpha$ and that 
\[B=\bigcap_{i\in\N}B_i\]
with each $B_i\in \Ubf{\Sigma}_{\beta(i)}^0$ some $\beta(i)<\alpha$. 
Then assuming 
\[(x_0, V_0)\leq_{\alpha} (x_1, V_1)\] 
and $i\in\N$ 
and $W_0\subset V_0$ with $W_0\in\B_0$. We just need to show that 
$x_0\in (B_i)^{\Delta W_0}$. 

So choose $W_1\subset V_1$ with $W_1\in \B_0$ and 
\[(x_0, W_0)\geq_{\beta(i)} (x_1, W_1).\] 
Now if $x_0$ is {\it not} in $(B_i)^{\Delta W_0}$, then for 
$C=X\setminus B_i$ we have 
\[x_0\in C^{*W_0}\]
and so by assumption on $\beta(i)<\alpha$ we have 
\[x_1\in C^{*W_1},\]
with a contradiction to $x_1\in (B_i)^{\Delta W_1}$. \hfill $\Box$ 

\medskip 

Thus sharpening Sami's theorem: 

\medskip 

{\bf 1.10 Corollary} If every orbit is $\Ubf{\Pi}^0_{\alpha}$ then 
$E_G^X$ is $\Ubf{\Pi}^0_{\alpha+\omega}$. 

Proof. Since $x_0 E^X_G$ implies $x_0\approx_{\alpha+\omega} x_1$ 
implies  $x_0\in([x_1]_G)^{*G}$ by 1.9 and the assumption on the 
Borel complexity of $[x_1]_G$, and hence $x_0 E^X_G x_1$. 
Thus $x_0 E^X_G$ if and only if $x_0\approx_{\alpha+\omega} x_1$, 
and so the corollary follows 1.4. \hfill $\Box$ 

\medskip 

The rest of this note may seem like a heck of a lot of puff and furry to 
reprove a well known result -- that every orbit is Borel. I believe that 
the techniques here may have further applications for the ``fine analysis 
of the Borel complexity of orbits", and moreover I simply find it 
philosophically interesting to give a proof resembling the argument 
for the logic action that appeals to the Scott analysis, 
in as much as we may see the Borel complexity of the orbit $[x]_G$ 
as being a direct result (1.18 below) of a countable length 
analysis of the stabilizer of the point $x$. 
It seems that these calculations would also play an essential 
role in any attempt to complete the conjectures 0.5 and 0.6. 

From now on assume that $d$ is a bounded right invariant metric on $G$. 
(I do {\it not} ask that it also be complete -- so that such a metric 
always exists). Let $G_0$ be a countable dense subgroup of $G$ and let 
$\B_0$ consist of all sets of the form $B_q(g)=_{df}\{d(h, g)<q\}$ 
for $g\in G_0$ and $q$ a positive rational. 

\medskip 

{\bf 1.11 Definition} For $x\in X$ let $\delta(x)$ be least such that 
for all $g_0, g_1\in G_0$, $n\in\N$, $q_0, q_1\in\Q$ with $q_0, q_1>1/n$, 
if 
\[(x, B_{q_0}(g_0))\leq_{\alpha}(x, B_{q_1}(g_1))\]
then 
\[(x, B_{q_0-1/n}(g_0))\leq_{\alpha+1}(x, B_{q_1}(g_1+1/n)).\] 

\medskip 

By 1.5(b) and countability of $\B_0$, such a $\delta(x)<\omega_1$ must 
exist. Note that $x_0E^X_G x_1$ implies $\delta(x_0)=\delta(x_1)$;  
the fact that $\B_0$ is not necessarily translation invariant requires us 
to give a little ground in 1.11 so as to make $x\mapsto \delta(x)$ 
$G$-invariant. 

\medskip 

{\bf 1.12 Lemma} Suppose $\delta(x)=\delta$ and $(x, V_0)\leq_{\delta+1}
(x, V_1)$. Then $(x,V_0)\leq_{\delta+2} (x, V_1)$. 

Proof. Fix $W_0\subset V_0$ with $W_0=B_q(g)$ some $q\in\Q^+$, $g\in G_0$. 
We need $W_1\subset V_1$ with $(x, W_1)\leq_{\delta+1} (x, W_0)$. 

Choose $\bar{q}<q$ with $\bar{q}\in \Q^+$ and let 
$\hat{W}_0=B_{\bar{q}}(g)$. Then by definition of $(x, V_0)\leq_{\delta+1}
(x, V_1)$ we can find $\hat{W}_1\in \B_0$ with 
\[(x, \hat{W}_0)\geq_{\delta} (x, \hat{W}_1).\] 
By assumption on $\delta$ we can find $W_1\subset \hat{W}_1$ with 
\[(x, {W}_0)\geq_{\delta+1} (x, {W}_1).\]
\hfill $\Box$ 

\medskip 


{\bf 1.13 Corollary} If $\delta(x)\leq\delta$ and $(x, V_0)\leq_{\delta+1}
(x, V_1)$, then $(x,V_0)\leq_{\delta+2} (x, V_1)$. 

Proof. Induction on $\delta$, starting with the base case 
$\delta(x)=\delta$ discussed at 1.12. \hfill $\Box$ 


\medskip 

{\bf 1.14 Lemma} Suppose $\delta(x_1)\leq \delta$, $x_0\approx_{\delta+1}
x_1$, $V_0, V_1\in \B$ with 
\[(x_0, V_0)\delta_{\delta+1}(x_1, V_1).\] 

Then $(x_0, V_0)\leq_{\delta+2} (x_1, V_1)$. 

Proof. Fix $W_0\subset V_0$ in $\B_0$. By $x_0\approx_{\delta+1} x_1$ 
we may find $\hat{W}_1\in \B_0$ with $(x_1, \hat{W}_1)\leq_{\delta+1} 
(x_0, W_0)$. Transitivity of $\leq_{\delta+1}$ yields  
$(x_1, \hat{W}_1)\leq_{\delta+1} 
(x_1, V_1)$.  
Assumption on $\delta$ and 1.13 implies $(x_1, \hat{W}_1)\leq_{\delta+2} 
(x_1, V_1)$. 

Thus we may find $W_1\in \B_0$ with $W_1\subset V_1$ and 
$(x_1, \hat{W}_1)\geq_{\delta+1} (x_1, W_1)$. Then transitivity 
of $\leq_{\delta+1}$ gives $(x_0, W_0)\geq_{\delta+1} (x_1, W_1)$ 
as required. \hfill $\Box$ 


\medskip 

{\bf 1.15 Corollary} Suppose that $\delta(x_0)=\delta(x_1)\leq\delta$ and 
that \[x_0\approx_{\delta+1}x_1.\] 
Then for all ordinals $\alpha$
\[x_0\approx_{\alpha}x_1.\] 

Proof. A routine induction on $\alpha$ shows that for $i\in\{0,1\}$ and 
$V_i, V_{1-i}\in\B_0$ 
\[(x_i, V_i)\leq_{\delta+1} (x_{1-i}, V_{1-i})\] 
implies 
\[(x_i, V_i)\leq_{\alpha} (x_{1-i}, V_{1-i}).\] 
 \hfill $\Box$ 

\medskip 

{\bf 1.16 Lemma} Suppose $\delta(x_1)\leq \delta$ and $x_0\approx_{\delta+2} x_1$. 
Then $\delta(x_0)\leq\delta+2$. 

Proof. Suppose $(x_0, W_0)\leq_{\delta+2} (x_0, V_0)$. It suffices to produce 
$\hat{W}_0\subset V_0$ in $\B_0$ with 
\[(x_0, W_0)\geq_{\delta+2} (x_0, \hat{W}_0).\]
Since $x_0\approx_{\delta+2} x_1$ find $W_1$ with 
\[(x_1, W_1)\leq_{\delta+2} (x_0, W_0).\]
By transitivity of $\leq_{\delta+2}$ 
\[(x_1, W_1)\leq_{\delta+2} (x_0, V_0).\]
So there exists $\hat{W}_0\subset V_0$ with 
\[(x_0, \hat{W}_0)\leq_{\delta+2}(x_1, W_1).\] 
Then by lemma 1.14, 
\[(x_0, \hat{W}_0)\leq_{\delta+2}(x_1, W_1)\leq_{\delta+2} (x_0, W_0),\]
and we finish by the transitivity of $\leq_{\delta+2}$. 
 \hfill $\Box$ 

\medskip 

{\bf 1.17 Corollary} If $\delta(x_1)\leq \delta$ and $x_0\approx_{\delta+3} x_1$, then 
for all $\alpha$ 
\[x_0\approx_{\alpha} x_1.\] 

Proof. Let $\bar{\delta}=\delta+2$. By lemma 1.16, 
$\delta(x_0)\leq \bar{\delta}$. 
Since $x_0\approx_{\bar{\delta}}x_1$ we have the conclusion by 
lemma 1.14. \hfill $\Box$ 

\medskip 

{\bf 1.18 Proposition} Suppose $\delta(x_0), \delta(x_1)\leq \delta$ 
and $V_0, V_1\in\B_0$ with 
\[(x_0, V_0)\leq_{\delta+1} (x_1, V_1).\] 
Then there exist $g\in\overline{V_0}$ and 
$h\in\overline{V_1}$ with 
\[g\cdot x_0=h\cdot x-1.\] 

Proof. In what follows note that if $Wgh\subset V$ then for all $x\in X$ 
and $\alpha\in\omega_1$ 
\[(Wg, h\cdot x)\leq_{\alpha} (V,x).\] 
We already fixed a compatible right invariant metric $d$ for $G$, and 
let us know add to this a compatiable {\it complete} metric 
$\hat{d}$. We will now build group elements $g_{0,0}, g_{0,1}, g_{0,2}, ..., 
h_{0,0}, h_{0,1}, h_{0,2},...g_{1,0}, g_{1,1}, g_{1,2}, ..., 
h_{1,0}, h_{1,1}, h_{1,2},...$ and open neighbourhoods of the identity 
$V_{0,0}, V_{0,1}, V_{0,2},... V_{1,0}, V_{1,1}, V_{1,2},...$ such that 

\leftskip 0.5in 

\noindent (I) $h_{i, j}=g_{i, j}g_{i, j-1}...g_{i,o}$ for $i=0,1$, $j\in\N$; 
and so $h_{i, j+1}=g_{i, j+1}h_{i,j}$; 

\noindent (II) for all $g, g'\in V_{i, j}$ ($i=0,1$, $j\in\N$) 
\[\hat{d}(gh_{i,j}, h_{i,j})<2^{-j}),\]
\[\hat{d}(g, g')<2^{-j};\]

\noindent (III) each $g_{i,j+1}\in V_{i,j}$ ($i=0,1$, $j\in\N$); 

\noindent (IV) each $V_{i,j}h_{i,j}\subset V_i$ ($i=0,1$, $j\in\N$); 

\noindent (V) each $(h_{0,j}\cdot x_0, V_{0,j})
\leq_{\delta+1} (h_{1,j}\cdot x_1, V_{1,j}$. 

\leftskip 0in 

If we suceed with all this, then we can find elements 
\[g={\rm lim}_{j\rightarrow\infty} h_{0,j},\]
\[h={\rm lim}_{j\rightarrow\infty} h_{1,j}.\] 
From (V) we may obtain group elements $\delta_i, \epsilon_i\rightarrow 1_G$ 
with 
\[(\delta_ig\cdot x_0, V_{0,j})\leq_1 (\epsilon_ih\cdot x_1, V_{1, j}),\]
\[\therefore \overline{V_{0,j}\delta_i g x_0}\subset
\overline{V_{1,j}\epsilon_ih\cdot x_1}\]
and thus (by (II) $(V_{0,j}\delta_i)_{i\in\N}$ and 
$(V_{1,j}\epsilon_i)_{i\in\N}$ are neighbourhood bases for the 
identity) we have $g\cdot x_0=h\cdot x_1$. 

So suppose we have produced the above elements and open sets for $j\leq N$. 
Then 
\[(h_{0,N}\cdot x_0, V_{0,N})\leq_{\delta+1} (h_{1, N}\cdot x_1, V_{1,N}).\]
Then we may find $W\subset V_{1, N}$ with 
\[(h_{0,N}\cdot x_0, V_{0,N})\geq_{\delta+1} (h_{1, N}\cdot x_1, W)\]
by lemma 1.14. Choose $g_{1, N+1}\in W$ and let $h_{1, N+1}
=h_{1, N+1}g_{1, N+1}$. We can then let $V_{1, N+1}$ be a small enough 
neighbourhood of the identity to ensure (II) and (IV) with 
$V_{1, N+1}g_{1, N+1}\subset W$. Repeating exactly the same step on the 
other side we pass from 
\[(h_{0,N}\cdot x_0, V_{0,N})\geq_{\delta+1} (h_{1, N+1}\cdot x_1, V_{1, N+1})\]
to some suitable 
\[(h_{0,N+1}\cdot x_0, V_{0,N+1})\leq_{\delta+1} (h_{1, N+1}\cdot x_1, V_{1, N+1}).\]
and continue. \hfill $\Box$ 

\medskip 

In particular: 

\medskip 

{\bf 1.19 Corollary} $y\approx_{\delta(x)+1} x\Rightarrow y E^X_G x$ for 
all $x, y\in X$. 

\medskip 

And thus a different proof of the well known fact: 

\medskip 

{\bf 1.20 Corollary} For all $x\in X$, $[x]_G$ is Borel. 

\medskip 

Here seems as good as place as any to collect together some calculations 
regarding the $\leq_{\alpha}$ relations. 

\medskip 

{\bf 1.21 Lemma} (i) 
\[(Wg,g^{-1}\cdot x)\leq_{\alpha} (W, x)\] all $W\in\B_0$, $x\in X$, 
$g\in G$, $\alpha\in\omega_1$. 

(ii) If $Wgh\subset V$, $W, V$ open, non-empty, then for all $x\in X$, $\alpha\in 
[1,\omega_1)$ 
\[(Wg, h\cdot x)\leq_{\alpha} (V, x).\] 

(iii) 
\[(V,  y)\leq_{\alpha} (W, x)\]
if and only if 
\[(gV,  y)\leq_{\alpha} (gW,  x).\] 

%(iv) If $W_0, W_1, V_0, V_1$ are open neighbourhoods of the identity, $g, h\in G$, 
%$(W_0)^{-1}=W_0$, $W_1\subset W_0$, $V_1\subset V_0$, $x, y\in X$, 
%with 
%\[(hV_1, y)\leq_{\alpha} (gW_1, x)\]
%and 
%\[W_0(h^{-1}V_1h)\subset V_0\]
%then 
%\[(gW_0, x)\leq_{\alpha} (hV_0, y).\] 

%(v) For all $V_0, W\in \B_0$ there are $W_1\subset W_0\subset W$ and 
%$V_1\subset V_0$ in $\B_0$ such that for all $x, y\in X$ and $\alpha\in[1\,omega_1)$ 
%\[(V_1, y)\leq_{\alpha}(W_1, x)\]
%implies 
%\[(W_0, x)\leq_{\alpha}(V_0, y).\]

Proof. (i) is trivial and (ii) follows (i). (iii) is proved by transfinite 
induction on $\alpha$ (starting with the base case $\alpha=1$, where 
one notes that $\overline{gV\cdot y}=g\cdot\overline{V\cdot y}$ and 
$\overline{gW\cdot x}=g\cdot\overline{W\cdot x}$). \hfill $\Box$. 

%The trickiest of 
%all these statements is (iv), 
%and note here that proving it at a given $\alpha$ implies the statement 
%for that $\alpha$ at (v). We prove the two by simultaneous induction. 

%For this purpose 
%let $\beta<\alpha$  and $\hat{W}\in \B_0$ be a subset of $W_0$. 
%Using $(hV_1, y)\leq_{\alpha} (gW_1, x)$ let $W_2\subset gW_1$ 
%be in $\B_0$ 
%be such that $(hV_1, y)\geq (W_2, x)$. 
%Then choose $\hat{h}\in G_0\cap W_0$ with 
%$\hat{h}\hat{W}\cap W_2\neq\emptyset$. After possibly shrinking 
%$\hat{W}$ we may assume that 
%$\hat{h}\hat{W}\subset W_2$. 
%Thus we have 
%\[(hV_1, y)\geq_{\beta} (\hat{h}\hat{W}, x)\]
%\[\therefore (\hat{h}^{-1}hV_1, y)\geq_{\beta} (\hat{W}, x)\]
%and since $\hat{h}^{-1}hV_1=h(h^{-1}\hat{h}^{-1}h V_1)\subset V_0$ by 
%assumption on the respective open sets


















\begin{thebibliography}{99}
%\bibitem{becker}H. Becker, {\it Vaught's conjecture for complete left invariant 
%groups}, handwritten notes, University of North Carolina at Columbia, 1996. 

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

\bibitem{friedman} H. Friedman, {\it Baire and Borel reducibility}, in preparation. 

\bibitem{hjorth} G. Hjorth, {\bf Classification and orbit equivalence relations,} manuscript 
(see the web cite below). 

\bibitem{hjkelo} G. Hjorth, A.S. Kechris, A. Louveau, {\it Borel equivalence 
relations induced by actions of the symmetric group,} to appear in the 
{\bf Annals of Pure and Applied Logic.} 

\bibitem{sami} R. Sami, {\it Polish group actions and the topological 
conjecture}, 
{\bf Transactions of the American Mathematical Society,} 
vol. 341(1994), pp. 335-353. 



\end{thebibliography} 


6363 MSB

Mathematics

UCLA

CA90095-1555

greg@math.ucla.edu

www.math.ucla.edu/\~{}greg


% Set the ending of a LaTeX document
\end{document}