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\title{\huge Cardinalities in the projective hierarchy}         
\author{Greg Hjorth}        % Enter your name between curly braces
\date{2000}          % Enter your date or \today between curly braces
\maketitle

\tableofcontents   


\section{Introduction} 
\label{introduction} 

We show that the ``effective cardinality" of the 
collection of $\Ubf{\Pi}^1_{n+1}$ sets is strictly bigger than the 
effective cardinality of the $\Ubf{\Pi}^1_n$. The phrase 
{\it effective cardinality} is vague but 
can be made exact in the usual ways. For instance: 

\begin{theorem} Assume ${\rm AD}^{L(\R)}$. Then in $L(\R)$ there is 
no injection 
\[i: \Ubf{\Pi}^1_{n+1}\hookrightarrow \Ubf{\Pi}^1_n.\] 
\end{theorem} 

A few years ago Tony Martin showed a similar result, 
establishing the non-existence of an injection 
from $\Ubf{\Pi}^1_m$ to $\Ubf{\Pi}^1_n$ for $m$ sufficiently larger 
than $n$. His method 
did not seem to 
work for $m=n+1$. 


This present paper gives level 
by level calculations for the projective hierarchy, but it too 
falls short of a complete analysis, in as much as it leaves the position of 
the effective cardinals in the Wadge degrees largely obscure. At the low levels 
it takes some time for any new cardinals to appear. 
Whenever $\Gamma_1, \Gamma_2$ are non-trivial Wadge degrees strictly included in 
$\Ubf{\Delta}^0_2$ one has 
\[|\Gamma_1|_{L(\R)} = |\Gamma_2|_{L(\R)}.\] 
Beyond $\Ubf{\Delta}^0_2$ it is known from \cite{hjorth} that  
the different levels of the Borel hierarchy have distinct 
effective cardinalities. It is unclear whether 
there might be cardinals lying strictly 
between say $\Ubf{\Pi}^0_2$ and 
$\Ubf{\Pi}^0_3$, though Itay Neeman has proved that there 
is no $L(\R)$ 
injection from $\Ubf{\Pi}^0_{\alpha+2}$ to 
$\Ubf{\Delta}^0_{\alpha+2}$. Thus as a landmark of    
our ignorance: 

\begin{question} 
Is $|\Ubf{\Pi}^0_2|_{L(\R)}<|\Ubf{\Delta}^0_3|_{L(\R)}$? 
\end{question} 

\cite{andretta} provides a reference for virtually every known 
fact about the Wadge degrees, as well as all the general theory 
of $L(\R)$ required below. \cite{moschovakis} is the established  
text book for this kind of descriptive set theory. 
One can also refer to \cite{mamost} 
and \cite{martinsteel} for basic facts about scales in 
$L(\R)$ as well as the basis theorem for $\Ubf{\Sigma}^2_1$. 
$\S$9.2 of \cite{classification} overviews some of this material 
and along with \cite{dichotomy} presents an introduction to the 
theory of cardinality for non-wellorderable sets in $L(\R)$. 


\newpage 

\section{Outline} 

The plan is something like this. 
We suppose for a contradiction that 
there is an $L(\R)$ injection from $\Ubf{\Pi}^1_2$ 
into $\Ubf{\Pi}^1_1$. Formally speaking we are only working 
with the case $n=1$, but it will be clear that the 
proof generalizes. 

We then want to 
find some suitable forcing notion for introducing 
codes for $\Ubf{\Pi}^1_2$ sets, 
$\p_\h$, and some suitable Polish space of 
codes 
for $\Ubf{\Pi}^1_1$ sets, call it $\c_1$, and we argue 
from uniformization$(\Ubf{\Sigma}^2_1)$ that there 
must be an actual $L(\R)$ function 
\[\theta: {\rm Filters \: for\: } \p_\h\rightarrow  \c_1\] 
such that for all sufficiently generic 
$G_1, G_2\subset \p_\h$ we have 
that $G_1$ and $G_2$ give rise to the same 
$\Ubf{\Pi}^1_2$ sets if and only if 
$\theta(G_1)$ and $\theta(G_2)$ code the same
$\Ubf{\Pi}^1_1$ sets\footnote{In $\S$\ref{finishing} when the 
details are finally played out, we will use the Polish space of all 
downward closed subsets of $\{0,1\}^{<\omega}\times \omega^{<\omega}$ 
as our space $\c_1$ of codes; in the usual way this can be identified 
with a closed subset of $\{0,1\}^{(\{0,1\}^{<\omega}\times \omega^{<\omega})}$. 
$\p_\h$ is given in section \ref{hairy}, and 
will analogously consist of finite partial attempts to build a 
downward closed subset of 
$\{0,1\}^{<\omega}\times (\omega\times\omega)^{<\omega}$.}. 
Since all functions in $L(\R)$ are continuous 
on a comeager set we obtain that 
\[G\mapsto \theta(G)\] 
will be continuous for any sufficiently generic $G$. 

Given codes $C_1, C_2$ for $\Ubf{\Pi}^1_2$ 
we want to {\it continuously} assign generics 
\[G_1(C_1, C_2), G_2(C_1, C_2)\subset\p_\h\] 
such that $G_1$ and $G_2$ give 
rise to the same $\Ubf{\Pi}^1_2$ sets if and only if 
$C_1$ and $C_2$ give
rise to the same $\Ubf{\Pi}^1_2$. Then by composing 
with $\theta$ we have that 
\[(C_1, C_2)\mapsto (\theta(G_1(C_1, C_2)), 
\theta(G_2(C_1, C_2)))\] 
continuously reduces equality on pairs of codes 
for $\Ubf{\Pi}^1_2$ sets into equality on pairs of codes
for $\Ubf{\Pi}^1_1$ sets. Since the former is 
$\Ubf{\Pi}^1_3$ complete and the later is only $\Ubf{\Pi}^1_2$ 
we have a contradiction. 

In $\S$\ref{hairy} we present the notion of 
forcing $\p_\h$ for introducing generic $\Ubf{\Sigma}^1_1$ 
subsets of $\oo\times\oo$, which will then in turn give us a 
$\Ubf{\Pi}^1_2$ set by universally quantifying out the 
second coordinate. 
The point here is that the forcing gives rise to 
such ``hairy" and unruly $\Ubf{\Sigma}^1_1$ sets that for  
any generic $G$ and $\Ubf{\Sigma}^1_1$ set $A_S$ we will 
have room to find many ways to hide a copy of 
$A_S$ inside the $\Ubf{\Sigma}^1_1$ set given by $G$. 

$\S$\ref{amalgam} defines a method to ``amalgamate" two copies of 
$\p_\h$ relative to some given $\Ubf{\Sigma}^1_1$ sets. 
Given $A_{S_1}, A_{S_2}\subset 2^\o\times \oo$, 
both $\Ubf{\Sigma}^1_1$, we can continuously from some space of 
codes find 
a forcing 
\[\p_\h\otimes_{S_1, S_2}\p_\h;\] 
the generic object on this amalgamated forcing will provide 
a pair of generic (but not necessarily mutually generic) 
$G_1, G_2\subset \p_\h$, giving rise to 
\[A_{G_1}, A_{G_2}\subset \oo\times\oo\] 
\[\Phi: 2^\o\rightarrow \oo\] 
such that for all $(x, y)\in 2^\o\times\oo$ we have 
\[(x, y)\in A_{S_i}\Leftrightarrow (\Phi(x), y)\in A_{G_i}\] 
$i=1, 2,$  
and for all $z$ not in the range of $\Phi$ 
\[(z, y)\in A_{S_1}\Leftrightarrow (z, y)\in A_{S_2}.\] 

The technical part of the proof is to verify in 
$\S$\ref{dense} that the various open 
dense sets in $\p_\h$ 
appropriately lift up to open dense sets in 
$\p_\h\otimes_{S_1, S_2}\p_\h$ and that the generic object 
on this amalgamated forcing will have the various properties 
claimed in the paragraph before. The details in this 
technical part of the proof are numerous. Perhaps there are too many. 







\newpage 



\section{Notational issues} 

\label{notation} 

\begin{notation} For us $\omega$ denotes the set of natural numbers, and we 
begin counting at 0:- $\omega=\{0,1,2,...\}$. In this controversial 
issue we follow the usual practice of set theorists. 

We use $\lo$ to denote 
\[\bigcup_{k\in\o}\omega^k,\] 
where 
\[\omega^k=\{f:\{0, 1,...,k-1\}\rightarrow \omega\},\]
$2^{<\o}$ for 
\[\bigcup_{k\in\o}\{0,1 \}^k,\] 

and $(\o\times \o)^{<\omega}$ for 
\[\bigcup_{k\in\o}\omega^k\times \omega^k.\] 
We view $\lo$ as ordered by inclusion in the usual way. 
For $u, v\in \lo$ we write $u^\smallfrown v$ for the concatenation, 
obtained by letting the second 
sequence be adjoined onto the end of the first. For $j\in\o$ we 
write $u^\smallfrown j$, rather than $u^\smallfrown \langle j\rangle$, for 
the result of adjoining the sequence $\langle j \rangle $ of length one 
to $u$. 
 
Cheating ever so slightly, we shall describe the natural ordering 
on $(\o\times\o)^{<\omega}$ as inclusion and accordingly 
write 
\[(u, v)\subset (u', v')\] 
for $(u, v), (u', v')\in (\o\times\o)^{<\omega}$ when 
$u\subset u'$, $v\subset v'$. 

Very often we will use $s$ for an arbitrary element 
$s\in (\o\times\o)^{<\o}$ of the form $s=(u, v)$. 
For $s=(u, v)$, $t=(w, r)\in (\o\times \o)^{<\o}$ we let 
\[s^\smallfrown t=_{\rm df} (u^\smallfrown w, v^\smallfrown r).\] 

$\ell(u)$ denotes the length of $u$, so that 
$u\in \o^{\ell (u)}$. 


For $u\in \lo$ of non-zero length, 
we let $u^-$ denote its predecessor: So that 
$u^-\subset u$ and $\ell (u^-)=\ell(u)-1$. 
If $s=(u, v)\in (\o\times\o)^{<\o}$ then 
$s^-=(u^-, v^-)$. 


\end{notation} 


\begin{definition} A set $T\subset \lo\times (\o\times\o)^{<\o}$ 
is said to be {\it downwards closed} if whenever \[(u, (v, w))\in T\] 
and $u'\subset u$, $(v', w')\subset (v, w)$ then 
\[(u', (v', w'))\in T.\] 

Given such a $T$ we let $A_T$ be the $\Ubf{\Sigma}^1_1$ set of 
$(x, y)\in \oo\times \oo$ for which there exists $f\in \oo$ such that 
at each $n, m\in \o$ 
\[(x | _n, (y| _m, f| _m))\in S.\] 
\end{definition} 

Note that by the assumption of being downward closed, we in fact 
obtain that 
\[A_T=\{(x, y): \exists f\forall n(x| _n, 
(y| _n, f| _n))\in T\}.\]

\def\t{{\mathcal T}}

\begin{notation} 
For $(u,s)\in \lo\times (\o\times\o)^{<\o}$ we let 
\[\t_{(u,s)}=\{(u',s')\in\lo\times (\o\times\o)^{<\o}| 
u\subset u', s\subset s'\}.\] 
\end{notation} 

\newpage 


\section{Hairy $\Ubf{\Sigma}^1_1$ set forcing: $\p_\h$} 

\label{hairy}  

We describe a forcing notion which 
produces a complicated $\Ubf{\Sigma}^1_1$ set. 





\begin{definition} We let $\p_\h$ denote the set of pairs $(T, p)$ 
such that  
\begin{enumerate} 
\item  
\[T\subset \lo\times (\o\times \o)^{<\o}\] 
is finite and closed downwards; 

\item  
\[p: T\rightarrow \{0, 1\}\] 
has the property that if 
\[p(u, s)=0\]  and $u\subset u',$ 
$ s\subset s',$ 
then 
\[p(u', s')=0;\] 

\item $p(u,(0,0))\neq 0$ all $u\in\lo$. 
\end{enumerate} 


We order $\p_\h$ by extension, so that 
\[(T', p')\leq (T, p)\] 
if $T'\supset T$ and $p'\supset p$. 
\end{definition} 

\begin{lemma} \label{compatible} 
$(T, p), (T', p')\in \p_\h$ are such that 
\[p|_{T\cap T'}=p'|_{T \cap T'}\] 
then $(T, p), (T',p')$ are compatible, in the sense that there is 
some $(T^*, p^*) \in \p_\h$ with 
\[(T^*, p^*)\leq (T, p), (T',p').\] 
\end{lemma} 


\begin{proof} We let $T^{*}=T'\cup T$ and 
$p^{*}=p'\cup p$. We need to see that this pair $(T^{*}, p^{*})$ 
is an element of $\p_\h$. 

The first and last conditions in the definition of $\p_\h$ immediately 
hold, and for the second we see that if $(u, s)\in T^{*}$ with $p^{*}(u,s)=1$ 
then we may 
assume without loss that $(u, s)\in T$, and so for all $(u', s')\subset (u, s)$ we 
have $p(u',s')=1$ by $(T, p)\in \p_\h$, 
and hence $p^{*}(u',s')=1$. 
\end{proof}

\begin{lemma} \label{extend}
Suppose $(T', p')\leq (T, p)$, $(T',p')\neq (T, p)$. Then there is a condition 
$(T^*, p^*)$ with 
\[(T',p')\leq (T^*, p^*)\leq (T, p)\] 
and 
\[|T^*|=|T|+1.\]
\end{lemma} 

\begin{proof} 
Choose some $(u, s)\in T'\setminus T$. Choose $(u_0, s_0)$ with $lh(u_0)+lh(s_0)$ 
minimal subject to the constraint that 
\[u_0\subset u\]
\[s_0\subset s\]
\[(u_0, s_0)\notin T.\] 
Thus for any other $u_1\subset u_0$, $s_1\subset s_0$ with 
$(u_1, s_1)\neq (u_0, s_0)$ we have $(u_1, s_1)\in T$. 

Then let $T^*=T\cup\{(u_0, s_0)\}$ and $p^*=p'|_{T^*}$. 
\end{proof} 

\begin{definition} 

For $G\subset \p_\h$ a sufficiently generic filter we obtain a 
downward closed set $T_G\subset \lo\times (\o\times \o)^{<\o}$ by 
\[T_G=\{(u, s): \exists p\in G(p(u, s)=1)\}\] 
and a corresponding $\Ubf{\Sigma}^1_1$ set $A_G$ by taking 
the definition of $A_{T_G}$ from the previous section: 
\[A_G=_{\rm df}A_{T_G}=\{(x, y): \exists f\forall n(x| _n,
(y| _n, f| _n))\in T_G\}.\]

\end{definition} 

\newpage 




\section{And its amalgam: $\p_\h\otimes_{S_1, S_2}\p_\h$} 

\label{amalgam} 
\label{amalgamation} 

\begin{definition} Let $S_1, S_2\subset 2^{<\o}\times(\o\times\o)^{<\o}$ 
be two sets that are closed downwards. We then define 
$\p_\h\otimes_{S_1, S_2}\p_\h$ to be the set of sequences 
\[q=(T, p_1, p_2, U, A, \varphi, V, B, \psi, 
\sigma)\] 
satisfying the following conditions (for which we parenthetically 
provide some brief motivation along the way): 

\leftskip 0.4in 

\noindent (1) $(T, p_i)\in\p_\h$ for $i=1, 2$; (ultimately 
the forcing will try to generate two generics in $\p_\h$); 

\noindent (2) $U\subset 2^{<\o}$ is a finite non-empty subtree and 
\[\varphi: U\rightarrow \lo\]
is an ordering preserving injection; 

\noindent (3) $V\subset (\o\times\o)^{<\o}$ is a finite non-empty subtree 
and 
\[\psi: V\rightarrow (\o\times \o)^{<\o}\] is an order and length preserving 
injection, with the property that each 
\[\psi(v, w)=(v, \psi_0(v, w)),\] 
some $\psi_0(v, w)\in \o^{<\o}$; ($\varphi, \psi$ are partial 
attempts to describe a copy of $2^{<\o}\times (\o\times\o)^{<\o}$ 
inside $\o^{<\o}\times (\o\times\o)^{<\o}$); 

\noindent (4) if we define 
\[\hat{U}(q)=\{u|\exists u'\in U(u\subset \varphi(u'))\},\] 
\[\hat{V}(q)=\{s|\exists s'\in V(s\subset \psi(s'))\},\] 
then 

\leftskip 0.6in 

\noindent (a) $\hat{U}(q)\cap A=0;$ 

\noindent (b) $\hat{V}(q)\cap B=0$; 

\noindent (c) $\hat{U}(q)\times \hat{V}(q)\subset 
T\subset (\hat{U}(q)\cup A)\times (\hat{V}(q)\cup B)$, and in turn 
\[(\hat{U}(q)\cup A)\times (\hat{V}(q)\cup B)
\subset \{(u, s)|\exists u', s' ((u, s')\in T, (u', s)\in T)\},\] so 
that in particular $A$ and $B$ are finite; 

\leftskip 0.4in 

\no note that (3) actually implies $\hat{V}(q)=\psi[V]$; 
($A$ and $B$ describe ``forbidden" sets on which the copying 
functions $\varphi$ and $\psi$ can never extend; $T$ in turn at (c) 
includes everything necessitated by $A$, $B$, and the ranges of 
$\varphi$ and $\psi$); 

\noindent (5) for all $u\in U$, $s\in V$ we have $p_i(\varphi(u), \psi(s))=1$ 
if and only if $(u, s)\in S_i$, $i=1, 2$; 

\noindent (6) if we let $E_1(q)$ be the set of $(u, s)$ such that 

\leftskip 0.9in 

\noindent (i) $u\in A$

\noindent (ii) $u^-\in \hat{U}(q)$ 

\noindent (iii) $p_1(u, s^-)=p_2(u, s^-)=1$

\noindent (iv) $p_1(u, s)=1$ 

\noindent (v) $p_2(u, s)=0$ 

\noindent (so that $E_1(q)$ refers to certain nodes on which the 
generic below $(T, p_2)$ is unable to produce any infinite 
branches, while the generic below $(T,p_1)$ is still open to 
do so; ultimately we want that the generic objects should give 
rise to $\Ubf{\Sigma}^1_1$ sets which agree outside the ranges 
of $\psi$, and hence the function $\sigma$ below at (a)-(g) is 
intended to busily run around and even up this possible 
imbalance); 

\leftskip 0.4in 

and if we similarly let 
$E_2(q)$ be the set of $(u, s)$ such that 

\leftskip 0.9in 

\noindent (i) $u\in A$ 

\noindent (ii) $u^-\in \hat{U}(q)$ 

\noindent (iii) $p_1(u, s^-)=p_2(u, s^-)=1$

\noindent (iv) $p_1(u, s)=0$ 

\noindent (v) $p_2(u, s)=1$ 

\leftskip 0.4in 

\noindent then we obtain that 

\leftskip 0.6in 

\noindent (a) Domain$(\sigma)=\bigcup\{\t_{(u, s)}| (u, s)\in E_1(q)\}\cap T$; 


\noindent (b) Range$(\sigma)=\bigcup\{\t_{(u, s)}| (u, s)\in E_2(q)\}\cap T$; 

\noindent (c) if $(u, s)\in E_1(q)$ with $s=(v, w^\smallfrown j)$, then 
\[\sigma(u, s)=(u, (v, w^\smallfrown k))\] 
some $k$ with $(u, (v, w^\smallfrown k))\in E_2(q)$; 

\noindent (d) for each $u\in A$ with $u^-\in \hat{U}(q)$ there is a corresponding 
function 
\[\sigma_u: \bigcup\{\t_{(u, s)}| (u, s)\in E_1(q)\}\rightarrow \lo\] 
such that for all $(u, (v, w))\in E_1(q)$ and 
$(u', (v', w^\smallfrown r))\in \t_{(u, (v, w))}$ we 
have 
\[\sigma(u', (v', w^\smallfrown r))=(u', (v', w^\smallfrown \sigma_u(v', w^\smallfrown r)));\] 
so the $\sigma$ acting on some point $(u', s')$ above $(u, s)\in E_1(q)$ leaves $u'$ untouched, 
leaves the 
first coordinate of $s'$ untouched, and its value on the second coordinate of $s'$ is a function 
solely of $u$ and $s'$; 

\noindent (e) for each $(u, s)\in E_1(q)$ 
\[\sigma|_{\t_{(u, s)}}: \t_{(u, s)}\cong \t_{\sigma(u, s)}\] 
is an isomorphism of partially ordered sets; 

\noindent (f) for each $(u, s)\in E_1(q)$ and $u'\supset u$, $s'\supset s$ 
we have 
\[p_1(u',s')=p_2(\sigma(u',s'));\] 

\noindent (g) if $u\in A$ and 
\[p_1(u, s)\neq p_2(u, s)\] 
then there are $u'\subset u, s'\subset s$ with 
\[(u', s')\in E_1(q)\cup E_2(q).\] 

\leftskip 0in 

 







We order $\p_\h\otimes_{S_1, S_2}\p_\h$ by coordinate wise extension, 
except on the $\varphi$ coordinate where we permit  on  
terminal nodes of $U$ lengthening of the value of $\varphi(u)$. 
In other words, 
\[(T', p_1', p_2', U', A', \varphi', V', B', \psi', 
\sigma')\leq 
(T, p_1, p_2, U, A, \varphi, V, B, \psi,
\sigma)\] in 
$\p_\h\otimes_{S_1, S_2}\p_\h$ if 

\begin{enumerate} 

\item $T'\supset T$, 

\item $p_1'\supset p_1$, 

\item $p_2'\supset p_2$, 

\item $U'\supset U$, 

\item $A'\supset A$, 

\item $V'\supset V$, 

\item $B'\supset B$, 

\item $\sigma'\supset \sigma$, 

\item $\psi'\supset \psi$, 

\item for $u\in U$ not terminal we have 
\[\varphi'(u)=\varphi(u)\] while for $u\in U$ terminal in 
$U$ we have 
\[\varphi'(u)\supset \varphi(u){\rm .}\] 

\end{enumerate} 

The idea of the forcing is this: On $\hat{U}(q)\times \hat{V}(q)$ 
we have $p_1$ and $p_2$ constrained by the sets $S_1$ and $S_2$ 
respectively. Outside of $\hat{U}(q)\times \hat{V}(q)$ we try to 
make $p_1$ and $p_2$ match up as far as possible. Sometimes the 
commitments previously made on $\hat{U}(q)\times \hat{V}(q)$ 
prevent us, and this is tracked by the sets $E_1(q)$ and 
$E_2(q)$; and so below these bad points we use the $\sigma_u$ isomorphisms 
to at least ensure that the resulting $\Ubf{\Sigma}^1_1$ sets 
defined by the generic object $G$ will be the same outside the 
tree defined by projecting $\bigcup_{q\in G}\hat{U}(q)\times \hat{V}(q)$ 
to the set 
\[\{(u, v): \exists w\exists q\in G 
((u, (v, w))\in  \hat{U}(q)\times \hat{V}(q))\}.\] 

There is an assymetry between the requirements on $\varphi$ at (2) 
and $\psi$ at (3). $\psi$ is as a matter of law length preserving; $\varphi$ 
need not be. This shows up at \ref{claim3} below; it is technically 
convenient to have the ability to extend $\varphi$ so that 
\[p_1(\varphi(u), s)=p_2(\varphi(u), s)=0\] 
for any $s \in B$ and $u$ terminal in $U$. 

 

\end{definition} 

\newpage 

\section{Some dense sets for $\p_\h\otimes_{S_1, S_2}\p_\h$} 

\label{dense} 

For this section fix $S_1, S_2\subset 2^{<\o}\times (\o\times\o)^{<\o}$ 
downward closed. 

\begin{lemma} \label{5.0} Suppose 
\[q=(T, p_1, p_2, U, A, \varphi, V, B, \psi, 
\sigma)\] 
is in $\p_\h\otimes_{S_1, S_2}\p_\h$ and that we have conditions 
\[(T'',p'')\leq (T', p')\leq (T, p_1)\] 
with $|T'|=|T|+1$. 

Then there is 
\[q^*=(T^*, p_1^*, p_2^*, U^*, A^*, \varphi^*, V^*, B^*, \psi^*, 
\sigma^*)\] 
with 
\[(T^*, p_1^*)\leq (T',p')\] 
and $(T^*, p_1^*)$ still compatible with $(T'', p'')$. 
\end{lemma} 

\begin{proof} 
Let $(u, s)$ be such that $T'=T\cup{(u, s)}$. 

\smallskip 

Case(1) $u\in \hat{U}(q)$. 

\smallskip 

In this case it follows from $(u, s)\notin T$ that we have 
$s\notin \hat{V}(q)$. So we let 
\[T^*=T'=T\cup\{(u, s)\}\]
\[p_1^*(u,s)=p''(u, s)=p'(u,s)\] 
\[p_2^*(u, s)=0\]  
\[p_1^*|_T=p_1\]
\[p_2^*|_T=p_2\]
\[U^*=U\]
\[\varphi^*=\varphi\]
\[V^*=V\]
\[\psi^*=\psi\]
\[A=A^*\]
\[B^*=B\cup\{s\}\] 
\[\sigma^*=\sigma\]
\[q^*=(T^*, p_1^*, p_2^*, U^*, A^*, \varphi^*, V^*, B^*, \psi^*, 
\sigma^*).\]


Since $u\in \hat{U}(q)$ and $s\notin \hat{V}(q)$ we have no possibility of 
violating (5) and (6) in the definition of 
$\p_\h\otimes_{S_1, S_2}\p_\h$. $\hat{U}(q^*)=\hat{U}(q)$ and 
$\hat{V}(q^*)=\hat{V}(q)$, and so we maintain (4). 
$\varphi^*=\varphi$ and $\psi^*=\psi$ easily give (2) and (3). We have 
(1) by assumption of $(T', p')\in \p_\h$ and choice of 
$p_2^*$; here it is important that 
$s\neq (0,0)$ by case assumption and the part (3) of 
the amalgam forcing stating $V\neq \emptyset$, and so $p_2^*(u, s)=0$ 
creates no conflict with the definition of $\p_\h$.  
Finally since $(T^*, p_1^*)=(T',p')$ we obtain 
\[(T'', p'')\leq (T^*, p_1^*),\] 
which is certainly enough to show these two conditions compatible. 

\smallskip 

Case(2) $u^-\in \hat{U}(q)$, $u\notin \hat{U}(q)$. 

\smallskip 

Here we may as well assume that $s\neq (0, 0)$, since otherwise we would 
have $p'(u, s)=1$, and we could extend $q$ in a care free manner, with 
$p_1^*(u,s)=p_2^*(u, s)=1$, $A^*=A\cup\{u\}$, $B=B^*$, $\varphi^*=\varphi$, 
$\psi^*=\psi$, and as in case(1) the various conditions in the definition of 
the forcing follow easily. So we make this assumption. In particular then 
since $T'=T\cup\{(u, s)\}$ is downward closed, it follows that $(u,s^-)\in T$, 
and thus from the case assumption that $u\in A$. 
Similarly $(u^-, s)\in T$ so $s\in B\cup\hat{V}(q)$. 

\smallskip 

Subcase(2a) $p_1(u^-, s)=p_2(u^-, s)$ and for all $s^*\subset s$ 
we have $p_1(u, s^*)=p_2(u, s^*)$. 

\smallskip 
 
Then just let 
\[T^*=T'=T\cup\{(u, s)\}\]
\[p_1^*(u,s)=p_2^*(u,s)=p''(u, s)=p'(u,s)\] 
\[p_1^*|_T=p_1|_T\]
\[p_2^*|_T=p_2|_T\]
\[U^*=U\]
\[\varphi^*=\varphi\]
\[V^*=V\]
\[\psi^*=\psi\]
\[A^*=A\] 
\[B^*=B\] 
\[\sigma^*=\sigma\]
\[q^*=(T^*, p_1^*, p_2^*, U^*, A^*, \varphi^*, V^*, B^*, \psi^*, 
\sigma^*).\]


Since $(u^-, s)\in T$ we have no trouble with clause (c) of part 
(4) of the definition of 
$\p_\h\otimes_{S_1, S_2}\p_\h$, while (4)(a) and (4)(b) follow as 
in the last case. 
(6) is no problem, since we have added no new elements to 
$E_1(q^*)$, $E_2(q^*)$ nor any new elements to $T^*$ extending 
a point in $E_1(q)\cup E_2(q)$. (5) is for free since 
$U, \varphi, V, \psi$ are left unchanged. Similarly (2) and (3) 
continue to hold almost vacuously. We obtain (1) by assumption on 
$(T', p')$ along with our case hypothesis. As in the first case of 
the proof, we have $(T^*, p_1^*)\leq (T'',p'')$. 

\smallskip 

Subcase(2b) $p_1(u^-, s)\neq p_2(u^-, s)$ but for all $s^*\subset s,$ 
$s^*\neq s$, 
we have $p_1(u, s^*)=p_2(u, s^*)$. 

\smallskip 

If $p'(u,s)=0$ then we can go ahead and let 
\[p^*_1(u,s)=p^*_2(u,s)=0,\]
and the whole thing finishes up as in the earlier cases. 

So suppose instead that $p'(u,s)=1$. Then the definition of 
$\p_\h$ and the assumptions 
of this subcase entail  
\[p_1(u^-, s)=1,\]
\[p_2(u^-, s)=0.\] 

Here we have $s\neq(0,0)$. Thus we may 
write 
\[s=(v, w^\smallfrown j)\] 
for some $j\in\o$. Choose $k$ with 
\[(\hat{u}, (v, w^\smallfrown k))\notin T''\] 
all $\hat{u}\in\lo$. We define $\sigma^*$ to be the 
extension of $\sigma$ obtained by adding $(u, (v, w^\smallfrown j))$ 
to its domain and setting 
\[\sigma^*(u, (v, w^\smallfrown j))=(u, (v, w^\smallfrown k)).\] 
We let 
\[T^*=T\cup\{(u,s)\}\cup \{(\hat{u},(v, w^\smallfrown k))|\hat{u}\subset u\}.\] 
We extend $p_1$ and $p_2$ to $p_1^*$ and $p_2^*$ both having domain $T^*$ with 
\[p_1^*(u, s)=1\]
\[p_2^*(u,s)=0\]
and for all $\hat{u}\subset u$ 
\[p_1^*(\hat{u}, (v, w^\smallfrown k))=0\]
\[p_2^*(\hat{u}, (v, w^\smallfrown k))=1.\]
This clearly guarantees our $q^*$ will satisfy (1) from the 
definition of the forcing. 
As before we can let 
\[\varphi^*=\varphi\]
\[U^*=U\]
\[\psi^*=\psi\] 
\[V^*=V\] 
and obtain (2), (3), and (5)  
without any work. Letting $A^*=A$ and $B^*=B\cup\{(v, w^\smallfrown k)\}$ 
we obtain (4). Our choice of $\sigma^*$ maintains (6). We have 
$(T^*, p_1^*)$ compatible with $(T'', p'')$ by \ref{compatible}. 

\smallskip 

Subcase(2c) There is some $s^*\subset s$, $s^*\neq s$, with $p_1(u, s^*)
\neq p_2(u, s^*)$. 

\smallskip 

Then by (6)(g) of the definition of $\p_\h\otimes_{S_1, S_2}\p_\h$ 
we have some $s^*\subset s$, $s^*\neq s$ with 
\[(u, s^*)\in E_1(q)\cup E_2(q).\] 
We us assume the $(u,s^*)$ in $E_1(q)$, since the case of $(u,s^*)$ in $E_2(q)$ 
is similar, just 
with $\sigma^{-1}$ instead of $\sigma$. 

Let $s=(v, w^\smallfrown j)$. Fix $\hat{w}$ with 
\[\sigma(u, (v^-, w))=(u, (v^-, \hat{w})).\]
We then choose $k$ with 
\[(\hat{u}, (\hat{v}, \hat{w}^\smallfrown k))\notin T''\] 
for any $\hat{u}, \hat{v}$. By assumption on $\sigma$ we must 
have 
\[p_1(u, s^-)=p_2(\sigma (u, s^-))=p_2(u, (v^-, \hat{w}))=1\]
\[p_2(u, s^-)=0=p_1(\sigma(u, s^-)).\] 

We then let $T^*=T\cup \{(u, s)\}\cup \{(\hat{u}, (v, \hat{w}^\smallfrown k))| 
\hat{u}\subset u\}$ and we define $p_1^*, p_2^*$ to have domain 
$T^*$, extending $p_1, p_2$ by  
\[p^*_1(u, s)=p_2^*(u, (v, \hat{w}^\smallfrown k))=p'(u, s)=1\]
\[p_2^*(u, s)=p_1^*(u, (v, \hat{w}^\smallfrown k))=0\] 
and for all $\hat{u}\subset u$  
\[p_1^*(\hat{u}, (v, \hat{w}^\smallfrown k))=0\] 
\[p_2^*(\hat{u}, (v, \hat{w}^\smallfrown k))=p'(\hat{u}, s)=1.\] 

It is pretty clear that $(T^*, p_1^*)\in \p_\h$ for $i=1, 2$ and 
\[(T^*, p_1^*)\leq (T', p'),\] 
and by \ref{compatible} we have $(T^*, p^*_1)$ and $(T'', p'')$ compatible 
in $\p_\h$. 

Now we can continue on in the usual way, letting $\psi^*=\psi$, $\varphi^*=\varphi$, 
$U^*=U$, $V^*=V$, $A^*=A$, $B^*=B\cup \{(v, \hat{w}^\smallfrown k) \}$, 
and define $\sigma^*$ to be the extension of $\sigma$ to the 
domain$(\sigma)\cup \{(u,s)\}$ with 
\[\sigma^*(u,s)=(u, (v, \hat{w}^\smallfrown k)).\] 
Then take 
\[q^*=(T^*, p_1^*, p_2^*, U^*, A^*, \varphi^*, V^*, B^*, \psi^*, 
\sigma^*).\] 
As remarked above we have (1) in the definition of 
$\p_\h\otimes_{S_1, S_2}\p_\h$ since we still have 
each $(T^*, p_i^*)\in \p_\h$. (2), (3), (4), and (5) follow 
more or less for free. (6) follows from our exact choice of 
$\sigma^*$ and the construction above. 

\smallskip 

Case(3) $u, u^-\notin \hat{U}(q)$. 

\smallskip 

Subcase (3a) For all $\hat{u}\subset u$, $\hat{s}\subset s$, if $\hat{u}\in A$ 
and $(\hat{u}, \hat{s})\in T$ then 
\[p_1(\hat{u}, \hat{s})=p_2(\hat{u}, \hat{s}).\] 

\smallskip 

In particular this gives first of all that 
\[p_1(u^-, s)=p_2(u^-, s)\] 
\[p_1(u, s^-)=p_2(u, s^-)\] 
and secondly that no $(\hat{u}, \hat{s})\subset (u, s)$ is in the 
domain of $\sigma$. Then we proceed as in subcase(2a). 

\smallskip 

Subcase(3b) There exists $(\hat{u}, \hat{s})$ with 
\[\hat{u}\subset u\]
\[\hat{s}\subset s\]
\[\hat{u}\in A\] 
\[p_1(\hat{u}, \hat{s})\neq p_2(\hat{u}, \hat{s}).\] 

\smallskip 

It follows then from the definition of $q\in \p_\h\otimes_{S_1, S_2}\p_\h$ 
that for $u_0\subset u$ minimal with $u_0\in A$ we must have some $s_0\subset s$ 
with 
\[(u_0, s_0)\in E_1(q)\cup E_2(q);\] 
let us assume $(u_0, s_0)\in E_1(q)$, since the situation with 
$(u_0, s_0)\in E_2(q)$ is a mirror image of this one. 

Then $(u^-, s)\in $ domain$(\sigma)$, and so there is some 
$\hat{w}$ with 
\[\sigma(u^-, s)=(u^-, (v, \hat{w})),\] 
where $s=(v, w)$. This forces onto us the definition of 
$\sigma^*$: Its domain is domain$(\sigma)\cup \{(u, s)\}$ and 
$\sigma^*(u, s)=(u, (v, \hat{w}))$. The assumption on $\sigma$, that it 
accords with (6), implies that for all 
$\hat{u}$ with $u_0\subset \hat{u}\subset u$, $\hat{u}\neq u$ 
\[p_1(\hat{u}, s)=p_2(\hat{u}, (v, \hat{w}));\] 
similarly 
\[p_1(u, s^-)=p_2(u, \sigma(s^-))=p_2(u, (\sigma(s))^-)
=p_2(u, (v^-, \hat{w}^-)).\] 

Therefore we can let $T^*=T\cup\{(u, s), (u, (v, \hat{w}))\}$ and let $p_1^*, p^*_2$ be the 
extensions of $p_1, p_2$ to $T^*$ with 
\[p_1^*(u, s)=p'(u,s)\]
\[p_2^*(u, (v, \hat{w}))=_{\rm df} p_2^*(\sigma^*(u, s))=p'(u, s)\] 
\[p_1^*(u, (v, \hat{w}))=_{\rm df} p_1^*(\sigma^*(u, s))=p_2(u, s)=0.\]
We let $\varphi^*=\varphi$, $\psi^*=\psi$, $U^*=U$, $V^*=V$, $A^*=A$, $B^*=B$, and 
\[q^*=(T^*, p_1^*, p_2^*, U^*, A^*, \varphi^*, V^*, B^*, \psi^*, 
\sigma^*).\] 

We obtain (1) in the definition of the amalgamated forcing notion by 
assumption on $p'$ and from $\sigma$ satisfying (6) for $q$. (2), (3), (4), 
and (5) are again automatic. (6) follows from the assumption on $\sigma$ 
and our definition of $\sigma^*$, $p_i^*(u, s)$, $p_i(\sigma^*(u, s))$, $i=1, 2$. 

Finally we need to see that $(T^*, p_1^*)$ is still compatible with $(T'', p'')$. 
For this we need to check that they agree on their common domain. We first obtain 
that $p_1^*(u, s)=p'(u, s)=p''(u,s)$ by assumption on $p'$. On the other new element 
of $T^*$ we have $p_1^*(u, (v, \hat{w}))=0$, and we need to see that this does not differ from 
$p''(u, (v, \hat{w}))$. But by assumption $\sigma(u_0, s_0)\subset 
(u, (v, \hat{w}))$ we must have some $w_0\subset \hat{w}, v_0\subset v$ with 
\[p_1(u_0, (v_0, w_0))=0\]
and hence by our assumption $p''\leq p_1$ 
\[p''(u_0, (v_0, w_0))=0\]
hence 
\[p''(u, (v, \hat{w}))=0.\]




\end{proof}

\begin{corollary} \label{claim1} If $\co\subset \p_\h$ is open dense, then 
the set $(\co)^*$ of 
\[(T^*, p_1^*, p_2^*, U^*, A', \varphi^*,  V^*, B^*, \psi^*, 
\sigma^*)\in \p_\h\otimes_{S_1, S_2}\p_\h\] 
with $(T^*, p_1^*)\in \co$ is again open dense in 
$\p_\h\otimes_{S_1, S_2}\p_\h$. 
\end{corollary} 

\begin{proof} 
We successively apply \ref{5.0} and \ref{extend} 
to extend $(T_1, p_1)$ until 
we at last have a condition 
\[(T^*, p_1^*, p_2^*, U^*, A', \varphi^*,  V^*, B^*, \psi^*, \sigma^*)\] 
with $(T^*, p_1^*)$ actually extending some 
$(T'', p'')\in \co$. 
\end{proof} 





\begin{corollary} 
\label{corollary1} For each $u\in\lo$ and 
$s\in (\o\times\o)^{<\o}$ we have that 
the set of 
\[(T^*, p_1^*, p_2^*, U^*, A^*, \varphi^*, V^*, B^*, \psi^*, 
\sigma^*)\in \p_\h\otimes_{S_1, S_2}\p_\h\] 
with 
\[(u, s)\in T^*\] 
is open dense. 
\end{corollary} 

\begin{lemma} \label{claim1a} If $\co\subset \p_\h$ is open dense, then
the set $(\co)_*$ of
\[(T^*, p_1^*, p_2^*, U^*, A^*, \varphi^*, V^*, B^*, \psi^*, 
\sigma^*)\in \p_\h\otimes_{S_1, S_2}\p_\h\]
with $(T^*, p_2^*)\in \co$ is again open dense in
$\p_\h\otimes_{S_1, S_2}\p_\h$.
\end{lemma}

\begin{proof} 

The proof is exactly symmetrical to \ref{claim1}. 
\end{proof} 

\def\amalg{\p_\h\otimes_{S_1, S_2}\p_\h}

\begin{lemma} \label{claim2a} 
For each $u\in 2^{<\o}$ and $j\in \{0,1\}$, the set of 
\[(T', p_1', p_2', U', A', \varphi', V', B', \psi', 
\sigma')\in \p_\h\otimes_{S_1, S_2}\p_\h\] 
such that if 
$u\in U'$ then $u^\smallfrown j\in U'$ 
is dense. 
\end{lemma} 

\begin{proof} 


Choose some
\[q=(T, p_1, p_2, U, A, \varphi, V, B, \psi, 
\sigma)\in \p_\h\otimes_{S_1, S_2}\p_\h.\]
If $u\not\in U$ then we are already done, so suppose instead 
$u\in U$ and $u^\smallfrown j\notin U$. Then we can simply extend $\varphi$ to 
$\varphi'$ with 
\[\varphi'(u^\smallfrown j)=\varphi(u)^\smallfrown k\] 
for some 
\[\varphi(u)^\smallfrown k\not\in A\cup \hat{U}(q).\] 
For $s\in V$ we can set $p_i'(\varphi(u)^\smallfrown k, \psi(s))=1$ 
if $(u^\smallfrown j, s)\in S_i$ and $=0$ if  
$(u^\smallfrown j, s)\not\in S_i$. 
We let $\psi'=\psi$, $V'=V$, $A'=A$, $B'=B$, $U'=U\cup\{(u^\smallfrown j)\}$. 
We let $\hat{U}'=\{\hat{u}| 
\exists u\in U'(\hat{u}\subset \varphi'(u)\}$, 
$T'=T\cup \hat{U}'\times \hat{V}(q)$. 









\end{proof} 

\begin{corollary} \label{corollary2a} 
For each $u\in 2^{<\o}$, the set of
\[(T', p_1', p_2', U', A', \varphi', V', B', \psi', 
\sigma')\in \p_\h\otimes_{S_1, S_2}\p_\h\]
with 
$u\in U'$ 
is open dense.
\end{corollary} 

\begin{lemma} \label{claim2b} 
For each $(v, w)\in(\o\times\o)^{<\o}$ and $j, k\in\o$, the set of
\[(T', p_1', p_2', U', A', \varphi', V', B', \psi', \hat{V'}
\sigma')\in \p_\h\otimes_{S_1, S_2}\p_\h\]
such that if
$(v, w)\in V'$ then $(v^\smallfrown j, 
w^\smallfrown k)\in V'$
is dense.
\end{lemma} 

\begin{proof} 
This is similar to \ref{claim2a}. 
\end{proof} 



\begin{corollary} \label{corollary2b}
For each $s\in (\o\times\o)^{<\o}$, the set of
\[(T', p_1', p_2', U', A', \varphi', V', B', \psi', 
\sigma')\in \p_\h\otimes_{S_1, S_2}\p_\h\]
with
$s\in V'$
is open dense.
\end{corollary} 

\begin{lemma} 
\label{help} 
For all 
\[q=(T, p_1, p_2, U, A, \varphi, V, B, \psi, 
\sigma)\in \p_\h\otimes_{S_1, S_2}\p_\h\] 
and $W\supset U$ a subtree of $2^{<\o}$, there 
is 
\[(T', p_1', p_2', U', A', \varphi', V', B', \psi',
\sigma')\leq 
(T, p_1, p_2, U, A, \varphi, V, B, \psi, 
\sigma)\] 
with $U'=W$. 
\end{lemma} 

\begin{proof} 
We let $U'=W$, $V'=V$, $\psi'=\psi$,  
$A'=A$, $B'=B$, and then let 
\[\varphi': W\rightarrow \lo\] 
be any one to one order preserving function 
extending $\varphi$ such that for all $u\in W\setminus  U$ 
we have 
\[\varphi'(u) \notin \hat{U}(q)\cup A.\] 
We define 
\[\hat{U}'=\{\hat{u}|\exists u\in W(\varphi(u)\supset \hat{u})\}\] 
and 
\[T'=T\bigcup \hat{U'}\times \hat{V}(q),\] 
and we require that on each $(\varphi'(u), \psi(s))$ for $u\in W, 
s\in V$, and $i=1, 2$ 
\[p_i'(\varphi'(u), \psi(s))=1\Leftrightarrow (u, s)\in S_i,\]
\[p_i'(\varphi'(u), \psi(s))=0\Leftrightarrow (u, s)\not\in S_i.\] 
For other arguments $(u,s)\in \hat{U}'\times \hat{V}(q)$, not in $T\cup 
(\varphi'[W]\times \psi[V])$, we go to the maximal $\hat{u}\subset u$ 
with $\hat{u}\in \varphi[W]$ and let 
\[p_i'(u, s)=p_i'(\varphi'^{-1}(\hat{u}), \psi^{-1}(s)).\] 
(The properties of $\psi$ given at (3) in the definition of the forcing notion 
are already enough to imply $s$ is in the range of $\psi$). 
\end{proof} 




\begin{lemma} \label{claim3} The set of 
\[(T', p_1', p_2', U', A', \varphi',  V', B', \psi', 
\sigma')\in \p_\h\otimes_{S_1, S_2}\p_\h\] 
such that 

\leftskip 0.5in 

\noindent (1) for some $\ell $, 
\[U'=\{0,1\}^\ell =_{\rm df}\{u\mid u: \{0, 1, ...\ell -1\}\rightarrow 
\{0, 1\}\}{\rm ,}\]  
and 

\noindent (2) for all $u\in \{0,1\}^\ell $ and $s\in B'$ 
\[p_1'(\varphi'(u), s)=p_2'(\varphi'(u), s)=0\] 

\leftskip 0in 

\noindent is dense. 
\end{lemma} 

\begin{proof} 
Fix 
\[q=(T, p_1, p_2, U, A, \varphi,  V, B, \psi,  
\sigma)\in \p_\h\otimes_{S_1, S_2}\p_\h.\] 
Applying \ref{help} we can assume $U$ has the indicated 
form $U=\{0,1\}^\ell $ for some $\ell $.   
Let $u_1, u_2,...u_k$ ($k=2^\ell $) enumerate the terminal vertices in 
$U$. For each $i\leq k$ choose some $n(i)$ with 
\[\varphi(u_i)^\smallfrown n(i)\not\in A.\] 
Then let 
\[\varphi'(u_i) =\varphi(u_i)^\smallfrown n(i)\] 
and 
\[\varphi'(u)=\varphi(u)\] 
for $u\in U$ not terminal. Let 
\[T'=T\cup\{(\varphi(u_i)^\smallfrown n(i), s)\mid 
i\leq k, s\in B\cup \hat{V}(q)\}.\] 
We let $A'=A$ and $B'=B$. For $(u, s)\in T$ we of course 
let $p_i'(u, s)=p_i(u, s)$, $i=1, 2$. For 
\[(\varphi(u_i)^\smallfrown n(i), s)\in T'\setminus T\] 
we let 
\[p_i'(\varphi(u_i)^\smallfrown n(i), s)=0\] 
if $s\in B$ and 
\[p_i'(\varphi(u_i)^\smallfrown n(i), t)=p_i(\varphi(u_i), t)\] 
for $t=\hat{V}(q)$. 
\end{proof} 

\begin{corollary} 
\label{corollary3} 
For $s\in (\o\times\o)^{<\o}$, 
the set of 
\[q'=(T', p_1', p_2', U', A', \varphi',  V', B', \psi', 
\sigma')\in \p_\h\otimes_{S_1, S_2}\p_\h\] 
for which there exists $l\in \o$ with 

\leftskip 0.5in 

\noindent (1) $s\in \hat{V'}\cup B'$, and 

\noindent (2) $U'\supset \{0,1\}^l$, and 

\noindent (3) and we have either $s\in \hat{V}(q')$ or for all 
$u\in \{0, 1\}^l$ 
\[p_1'(\varphi(u), s)=p_2'(\varphi(u), s)=0\] 

\leftskip 0in 

\noindent is open dense. 
\end{corollary} 


\newpage 

\section{Some things to do with a generic} 

\label{things}

\begin{definition} For $G\subset 
\p_\h\otimes_{S_1, S_2}\p_\h$ meeting all the dense 
open sets from corollaries 
\ref{corollary2a} and \ref{corollary2b} we define 
\[\varphi_G: 2^{<\o}\rightarrow \lo\] by 
\[\varphi_G(u)=\hat{u}\] if there exists 
\[(T, p_1, p_2, U, A, \varphi, V, B, \psi, \sigma)\in G\] 
with $\varphi(u)=\hat{u}$ and $u$ not terminal in $U$.  
Similarly, 
\[\psi_G: (\o\times\o)^{<\o}\rightarrow (\o\times\o)^{<\o}\] 
is given by 
\[\psi_G(s)=\hat{s}\] if 
\[\exists
(T, p_1, p_2, U, A, \varphi, V, B, \psi, \sigma)\in G
[\psi(s)=\hat{s}]{\rm .}\] 
We then define 
\[\Phi_G:2^\omega\rightarrow \oo\] 
\[\Phi_G:x\mapsto \bigcup_{n\in\o}  \varphi_G(x|_n)\] 
and 
\[\Psi_G:\oo\times\oo\rightarrow \oo\times\oo\]  
\[\Psi_G: (y,f)\mapsto \bigcup_{m\in\o} \psi_G(y|_m,f|_m).\] 

We let 
\[\a_G=\bigcup\{x\in \oo\mid \exists
(T, p_1, p_2, U, A, \varphi, V, B, \psi, \sigma)\in G
\exists n\in\o
(x|_n\in A)\}{\rm .}\] 

We then let $\pi_1(G)\subset \p_\h$ 
be the set 
\[\{(T', p')\in \p_\h\mid  \exists
(T, p_1, p_2, U', A,  V, B, \psi, \sigma)\in G 
(T'\subset T, p'\subset p_1)\}{\rm ,}\] 
and we let $\pi_2(G)\subset \p_\h$
be the set
\[\{(T', p')\in\p_\h\mid  \exists
(T, p_1, p_2, U, A, \varphi, V, B, \psi, \sigma)\in G
(T'\subset T, p'\subset p_2)\}{\rm .}\] 
We let 
\[T_{\pi_1(G)}, T_{\pi_2(G)}\subset 
\lo\times(\o\times \o)^{<\o}\] 
be the subtrees indicated 
in section \ref{hairy}, from which we then 
obtain the corresponding $\Ubf{\Sigma}^1_1$ sets 
\[A_{\pi_i(G)}=_{\rm df} A_{T_{\pi_i(G)}}=\{(x, y)\mid 
\exists f\forall n \forall (T, p)\in \pi_i(G)(p(x|_n , 
(y|_n, f|_n))\neq 0)\}.\] 

\end{definition} 

\begin{lemma} 
\label{claim4} 
For all $G\subset \p_\h\otimes_{S_1, S_2}\p_\h$ 
meeting the dense sets of 
\ref{corollary1}, 
\ref{corollary2a}, 
\ref{corollary2b}, 
\ref{corollary3}, and all 
$x\in\a_G$,  and all $y\in \oo$, if 
\[(x, y)\in A_{\pi_1(G)}\] 
then 
\[(x, y)\in A_{\pi_2(G)}.\] 
\end{lemma} 

\begin{proof} 
Choose $f\in \oo$ with 
\[(x|_n, (y|_m, f|_m))\in T_{\pi_1(G)}\] 
all $n, m$. If every such 
$(x|_n, (y|_m, f|_m))\in T_{\pi_2(G)}$ then 
we are done, so instead suppose that 
there is some 
\[(T, p_1, p_2, U, A, \varphi, V, B, \psi, 
\sigma)\in G\] 
with $p_2(x|_n, (y|_m, f|_m))=0$ 
while 
that $p_1(x|_n, (y|_m, f|_m))=1$ by 
assumption on $T_{\pi_1(G)}$. 
By clause (6)(g) of the definition of 
$\amalg$ we may assume 
that $n$ is least with $x|_n\in A$ and $m$ is 
then least with 
$p_2(x|_n, (y|_m, f|_m))=0$. 
 

It then follows from clauses (6)(a) and (6)(b) of 
the definition of 
$\p_\h\otimes_{S_1, S_2}\p_\h$ 
that $(x|_n, (y|_m, f|_m))$ is in the 
domain of $\sigma$, 
\[\sigma(x|_n, (y|_m, f|_m))=(x|_n,(y|_m, w_0))\] 
for some $w_0\in \o^m$, and for all 
\[(T', p_1', p_2', U', A', \varphi', V', B', \psi', 
\sigma')\leq
(T, p_1, p_2, U, A, \varphi, V, B, \psi, 
\sigma)\] 
and $(x|_n^\smallfrown u_1, (y|_m^\smallfrown v_1, 
f|_m^\smallfrown w_1))\in T'$ we have 
\[p_1'(x|_n^\smallfrown u_1, (y|_m^\smallfrown v_1,
f|_m^\smallfrown w_1))
=p_2'(  \sigma'(x|_n^\smallfrown u_1, (y|_m^\smallfrown v_1,
f|_m^\smallfrown w_1)   )\] 
\[=p_2'(x|_n^\smallfrown u_1, (y|_m^\smallfrown v_1, \sigma'_{x|_n}
( y|_m^\smallfrown v_1,
f|_m^\smallfrown w_1))).\] 
In particular if we define $h\in\oo$ by 
\[h|_k=\hat{w}\] 
if and only if there is some 
\[(T', p_1', p_2', U', A', \varphi', V', B', \psi', 
\sigma')\in G\] with 
\[\sigma'_{x|_n}(y|_k, f|_k)=\hat{w}\] 
then 
\[(x|_n, (y|_m, h|_m))\in T_{\pi_2(G)}\]
all $n, m$.

\end{proof} 



\begin{lemma}
\label{claim5}
For all $G\subset \p_\h\otimes_{S_1, S_2}\p_\h$
meeting the dense sets of
\ref{corollary1},
\ref{corollary2a},
\ref{corollary2b},
\ref{corollary3}, and all
$x\in\a_G$,  and all $y\in \oo$, if
\[(x, y)\in A_{\pi_2(G)}\]
then
\[(x, y)\in A_{\pi_1(G)}.\]   
\end{lemma}

\begin{proof}  
Exactly symmetrical to the proof of 
\ref{claim4}, but with $\sigma^{-1}$ 
taking up the part of $\sigma$. 
\end{proof} 

\begin{lemma} 
\label{7.3} 
For all $G\subset \p_\h\otimes_{S_1, S_2}\p_\h$
meeting the dense sets of
\ref{corollary1},
\ref{corollary2a},
\ref{corollary2b},
\ref{corollary3}, 
and for all $x\in \oo$, 
\[x\in \a_G\] 
if and only if 
\[x\notin \Phi_G[2^\o].\] 
\end{lemma} 

\begin{proof} 
%This follows from the fact that for any 
%\[q=(T, p_1, p_2, U, A, \varphi,  V, B, \psi,  
%\sigma)\in \p_\h\otimes_{S_1, S_2}\p_\h\] 
%we must have $\varphi$ one to one. 
If $x\notin \a_G$ then by \ref{corollary1} 
we may successively find 
\[q=(T, p_1, p_2, U, A, \varphi,  V, B, \psi,  
\sigma)\in G\] and longer and 
longer $u \in U$ such that for any 
\[q'=(T', p_1', p_2', U', A', \varphi',  V', B', \psi',  
\sigma')\leq p\] 
we must have $\varphi'(u)\subset x$. Since any such $\varphi'$ 
must be one to one, we can obtain a unique $\hat{x}\in 2^\o$ which is the 
union of all the $u$'s as above, and hence satisfies $\Phi_G(\hat{x})=x$. 

Conversely if $x\in \a_G$ then from (4)(a) of the definition of the 
amalgamated forcing we have $x\notin \Phi_G[2^\o]$. 
\end{proof} 


\begin{lemma} 
\label{claim6} 
For $i=1, 2$, and  $G\subset \p_\h\otimes_{S_1, S_2}\p_\h$
meeting the dense sets of
\ref{corollary1},
\ref{corollary2a},
\ref{corollary2b},
\ref{corollary3}, and 
$x, y, f\in\oo$, 
if 
\[\forall n, m\in \o((\Phi(x)|_n, (y|_m, f|_m))\in T_{\pi_i(G)})\] 
then 
\[(y, f)=\Psi(y, f')\] 
for some $f'\in \oo$. 
\end{lemma} 

\begin{proof} For this we need an observation. 

\noindent{\bf Claim:} $f$ is in the image of $\Psi$ if and only if 
whenever $m\in\o$ and  
\[(T, p_1, p_2, U, A, \varphi, V, B, \psi, 
\sigma)\in G\] 
then 
\[(y|_m, f|_m) \notin B.\] 

\smallskip 

\noindent{\bf Proof of claim:} This equivalence follows the 
argument of \ref{7.3}. \hfill (Claim$\square$) 

\smallskip 

This claim granted, we apply \ref{corollary3} to see that 
if $s=(y|_m, f|_m)\in B$ for some 
\[(T, p_1, p_2, U, A, \varphi, V, B, \psi, 
\sigma)\in G\] 
then there 
would be some $l\in \o$ and 
\[(T', p_1', p_2', U', A', \varphi', V', B', \psi', 
\sigma')\in G\] 
with 
\[p_i'(\varphi(x|_l), s)=0.\] 

\end{proof} 



Putting together the last few claims we finally have:- 

\begin{proposition} \label{finally} 
There is a countable collection $({\mathcal D}_i)_i$ of open dense subsets of 
$\amalg$ such that for any filter $G$ meeting all these dense open sets we 
have:- 



(1) For all $z\in\oo\setminus \Phi_G[2^\o]$ and for all $y\in\oo$ 
\[(z, y)\in A_{\pi_1(G)}\Leftrightarrow (z, y)\in A_{\pi_2(G)}.\] 

(2) If $z=\Phi_G(x)\in \Phi_G[2^\o]$ then for all $y\in\oo$ we have 
\[(\Phi_G(x), y)\in A_{\pi_1(G)}\Leftrightarrow (x, y)\in A_{S_1}\]
\[(\Phi_G(x), y)\in A_{\pi_2(G)}\Leftrightarrow (x, y)\in A_{S_2}.\]

 

\end{proposition} 

\newpage 

\section{Generalities on finding generics} 

There are no proofs in this section.  The various 
assertions should not need them. 

\begin{notation} Let PO be the set of $x\in 2^{\o\times\o}$ 
such that $\leq_x$ defined by 
\[n\leq_x m\] 
if and only if 
\[x(n,m)=1\] 
provides a partial order on $\o$. For $y\in 2^{\o\times \o}$ and 
$n\in\o$ let 
\[D_{y,n}=\{m| y(n, m)=1\}.\] 
Then let $DPO$ be the set of $(x,y)\in $ PO$\times 2^{\o\times\o}$ 
such that for all $n$ we have $D_{y,n}$ dense in the partially 
ordered set 
\[(\o; \leq_x).\] 

\end{notation} 

\begin{lemma} \label{A} 
{\rm DPO} is a $\Ubf{\Pi}^0_2$ subset of $2^{\o\times\o}\times 
2^{\o\times\o}$. 
\end{lemma} 

\begin{lemma} \label{B} 
There is a continuous function 
\[\gamma_\o: {\rm DPO}\rightarrow 2^\o\] 
such that for all $(x, y)\in {\rm DPO}$ the set 
\[\{n\in\o| (\gamma(x,y))(n)=1\}\] 
provides a filter on $(\o;\leq_x)$ meeting each $A_{y,n}$. 
\end{lemma} 

\def\s{{\mathcal S}}
\def\t{{\mathcal T}} 
\def\f{{\mathcal F}}


\begin{notation} Let $\s(\p_\h)$ be the set 
of $y\in 2^{2^{<\o}\times (\o\times\o)^{<\o}}$ 
such that the set 
\[S(y)=_{\rm df}\{(u,s)| y(u,s)=1\}\] 
is closed downwards. Let $\f(\p_\h)$ be the set of $z\in 2^{\p_\h}$ such that 
\[G(z)=\{(T,p)| z(T,p)=1\}\] 
provides a filter on $\p_\h$. 
We view $2^{\p_\h}$ as having the product topology, under which it 
is homeomorphic to the Cantor space $2^\o$. 
\end{notation} 

$\s(\p_\h)$ can be viewed as a closed subset of the space 
\[(\{0,1\})^{2^{<\o}\times (\o\times\o)^{<\o}}.\] 
$\f(\p_\h)$ is a $\Ubf{\Pi}^0_2$ subset of $2^{\p_\h}$. Hence 
both these subspaces are Polish in their own right. 

To each $y_1, y_2\in\s(\p_\h)$ we can 
continuously associate a partial order on $\o$ isomorphic 
to $\p_\h\otimes_{S(y_1), S(y_2)}\p_\h$, and thus by lemma \ref{B}:  

\begin{lemma} 
\label{C} Let 
$(\co_i)_{i\in\o}$ be a sequence of dense open subsets of $\p_\h$. Then we may find a 
continuous function 
\[\gamma_{\p_\h\times\p_\h}: \s(\p_\h)\times\s(\p_\h)\rightarrow \f(\p_\h)\times\f(\p_\h)\] 
such that for all $(y_1, y_2)\in\s(\p_\h)\times\s(\p_\h)$, 
if 
\[(z_1, z_2)=\gamma_{\p_\h\times\p_\h}(y_1, y_2)\] 
then 

\leftskip 0.4in 

\noindent (1) $G(z_1)\cap \co_i\neq 0$ all $i\in\o$; 

\noindent (2) $G(z_2)\cap \co_i\neq 0$ all $i\in\o$; 

\noindent (3) there is a corresponding filter $G_{y_1, y_2}
\subset \p_\h\otimes_{S(y_1), S(y_2)}\p_\h$ meeting all the dense 
open sets of section \ref{dense} with 
\[\pi_1(G_{y_1, y_2})=G(z_1)\]
\[\pi_2(G_{y_1,y_2})=G(z_2).\] 

\end{lemma} 


Observe that we can reorganize 
$\gamma_{\p_\h\times\p_\h}$ into its coordinate functions 
and find a pair 
\[\gamma_1:  \s(\p_\h)\times\s(\p_\h)\rightarrow \f(\p_\h),\] 
\[\gamma_2:  \s(\p_\h)\times\s(\p_\h)\rightarrow \f(\p_\h),\] 
such that if 
\[(z_1, z_2)=\gamma_{\p_\h\times\p_\h}(y_1, y_2)\] 
then for $i\in\{1, 2\}$ we have $z_i=\gamma_i(y_1, y_2)$. 
We implicitly use this reorganization in the next section. 







\newpage 

\section{Finishing up} 

\label{finishing} 

I understand the next lemma to be folklore. 

\begin{lemma} Equality of $\Ubf{\Pi}^1_n$ sets is 
$\Ubf{\Pi}^1_{n+1}$ complete in the codes. 
\end{lemma} 

\begin{proof} For notational simplicity we assume $n=1$. 
Then given any $\Ubf{\Pi}^1_2$ set $P\subset 2^\omega$ 
we can appeal to the 
tree representation for $\Ubf{\Sigma}^1_1$ sets and 
continuously assign 
\[x \mapsto T_x\] 
to each point in $2^\o$ a tree $T_x\subset (\o\times\o)^{<\o}$ 
such that $x\in P$ if and only if 
\[\forall y\in\o^\o\exists f\in \o^\o\forall n
((y|_n, f|_n)\in T_x).\] 
Thus if we assign to each $x$ the $\Ubf{\Pi}^1_1$ set 
\[A_x=\{y\in\o^\o: \forall f\in \o^\o\exists n 
((y|_n, f|_n)\notin T_x)\}\] 
then we have 
\[x\in P\] 
if and only if 
\[A_x=\emptyset.\]
 
Thus we have reduced membership in $P$ to equality of a corresponding 
$\Ubf{\Pi}^1_1$ set with the empty set. It is easily checked that the 
assignment 
\[x \mapsto A_x\] 
is continuous in the codes. 
\end{proof} 



\begin{theorem} Assume {\rm AD}$^{L(\R)}$. Then there is no injection in 
$L(\R)$ from $\Ubf{\Pi}^1_{n+1}$ into $\Ubf{\Pi}^1_n$. 
\end{theorem} 

\begin{proof} Suppose instead we have 
\[{\Theta}: \Ubf{\Pi}^1_{n+1}\hookrightarrow \Ubf{\Pi}^1_n\] 
inside $L(\R)$. For the sake of concreteness let us assume 
$n=1$. 

Then by the basis theorem for $\Ubf{\Sigma}^2_1$ 
we may assume that $\Theta$ is itself $\Ubf{\Sigma}^2_1$, and 
hence by the uniformization theorem that there is a function 
in $L(\R)$ 
\[\theta: \f(\p_\h)\rightarrow 2^{2^{<\o}\times \lo}\] 
such that for all $z_1, z_2\in \f(\p_\h)$ we have 
\[\{x\in 2^\o: \forall y((x, y)\in A_{G(z_1)})\}=\{x\in 2^\o: \forall y((x, y)\in A_{G(z_2)})\}\]
if and only if 
\[\{x\in 2^\o|\forall y\in\oo\exists n ((x|_n, y|_n)\notin \theta(z_1))\}
=\{x\in 2^\o|\forall y\in\oo\exists n ((x|_n, y|_n)\notin \theta(z_2))\}.\] 

Since AD implies that all functions on Polish spaces 
are Baire measurable, we may find 
a comeager set on which $\theta$ is continuous. This means that there is a 
countable collection $(\co_i)_{i\in\o}$ of open dense subsets of $\p_\h$ 
such that $\theta$ is continuous on the $\Ubf{\Pi}^0_2$ subset $D$ of $\f(\p_\h)$ 
consisting of $\{z\in \f(\p_\h)\}$ such that 
\[\forall i(G(z)\cap \co_i\neq 0).\] 

But now applying \ref{C} we may find  {\it continuous} 
functions
\[\Gamma_1, \Gamma_2: \s(\p_\h)\times \s(\p_\h)\rightarrow D\] 
such that for all $(y_1, y_2)\in \s(\p_\h)\times \s(\p_\h)$ we have 
that if $\Gamma_i(y_1, y_2)=z_i$ ($i=1, 2$) 
then there is $G_{y_1, y_2}\subset \amalg$ meeting all the dense sets 
of $\S$\ref{dense} with $\pi_1(G_{y_1, y_2})=G(z_1)$ and 
$\pi_2(G_{y_1, y_2})=G(z_2)$ (where $\pi_1$, $\pi_2$ are defined as 
at the start of section \ref{things}). 

\smallskip 

\noindent {\bf Claim:} If $\Gamma_1(y_1, y_2)=z_1$, $\Gamma_2(y_1, y_2)=z_2$, then 
\[\{x\in \o^\o: \forall r((x, r)\in A_{G(z_1)})\}
=\{x\in \o^\o: \forall r((x, r)\in A_{G(z_2)})\}\] 
if and only if 
\[\{x\in 2^\o: \forall r((x, r)\in A_{S(y_1)})\}=
\{x\in 2^\o: \forall r((x, r)\in A_{S(y_2)})\}.\]

\smallskip 

\noindent {\bf Proof of claim:} 
We will use \ref{finally} several times in this proof. For this we 
need to keep in mind that 
\[A_{\pi_1(G_{y_1, y_2})}=A_{G(z_1)},\]
\[A_{\pi_2(G_{y_1, y_2})}=A_{G(z_2)}.\]


Suppose first that 
\[\{x\in \o^\o: \forall r((x, r)\in A_{G(z_1)})\}
\neq\{x\in \o^\o: \forall r((x, r)\in A_{G(z_2)})\}.\] 
Without loss we may assume there is some $x_0, r_0$ with 
$x_0\in 
\{x\in \o^\o: \forall r((x, r)\in A_{G(z_1)})\}$ but 
\[(x, r_0)\notin A_{G(z_2)}.\] 
Then by the statement of \ref{finally}(1) we must have some $\hat{x}$ 
with 
\[\Phi_{G_{y_1, y_2}}(\hat{x})=x_0.\] 
By \ref{finally}(2) we have for all $r\in\o^\o$
\[(x_0, r)\in A_{G(z_1)}\Leftrightarrow (\hat{x}, r)\in A_{S(y_1)}\] 
\[(x_0, r)\in A_{G(z_2)}\Leftrightarrow (\hat{x}, r)\in A_{S(y_2)}\] 
and thus 
\[(\hat{x}, r_0)\notin A_{S(y_2)}\] 
while 
\[\forall r((\hat{x}, r)\in A_{S(y_1)}).\] 
Thus 
\[\{x\in 2^\o: \forall r((x, r)\in A_{S(y_1)})\}
\neq\{x\in 2^\o: \forall r((x, r)\in A_{S(y_2)})\}.\] 



So now for the other direction. Let us assume $x_0$, $r_0$ have been chosen so that 
$\forall r(({x}_0, r)\in A_{S(y_1)})$ but 
$(({x}_0, r_0)\notin A_{S(y_2)})$; the other case, with $y_1$ and 
$y_2$ interchanged, is essentially identical. 
Let $\hat{x}=\Phi_{G(y_1, y_2)}(x_0)$. Then by 7.5(2) we have for all 
$r$ that 
\[(\hat{x}, r)\in A_{G(z_1)}\] 
but 
\[(\hat{x}, r_0)\notin A_{G(z_2)}.\] 
\hfill (Claim$\Box$) 

\smallskip 

But then the composition 
\[(y_1, y_2)\mapsto (\theta(\Gamma_1(y_1, y_2)), 
\theta(\Gamma_2(y_1, y_2)))\] 
gives a continuous reduction of the equivalence relation of coding the 
same $\Ubf{\Pi}^1_2$ set to the equivalence relation of coding the 
same $\Ubf{\Pi}^1_1$ set. Since the former is  
$\Ubf{\Pi}^1_3$ complete while the later is only $\Ubf{\Pi}^1_2$, we have 
our desired contradiction. 

\end{proof} 








\newpage 

\begin{thebibliography}{99}

\bibitem{andretta} A. Andretta, {\bf Notes on descriptive set theory}, 
unpublished manuscript, 
University of Turino, 2000. 

\bibitem{dichotomy} G. Hjorth, {\it A dichotomy for the 
definable universe,} 
{\bf Journal of Symbolic Logic,} 
vol. 60(1995), pp. 1199-1207. 

\bibitem{hjorth} G. Hjorth, 
{\it An absoluteness principle for 
Borel sets,} {\bf Journal of Symbolic Logic,} 
vol. 63(1998), pp. 663--693. 

\bibitem{classification} G. Hjorth, 
{\bf Classification and Orbit 
Equivalence Relations,} 
American Mathematical Society Mathematical Surveys and 
Monographs Series, Rhode Island, 2000. 

\bibitem{mamost} D.A. Martin, Y.N. Moschovakis, 
J.R. Steel, {\it The extent of definable scales,} 
{\bf Bulletin of the American Mathematical Society,} 
vol. 6(1982), pp. 435--440. 

\bibitem{martinsteel} D.A. Martin, J.R. Steel, 
{\it The extent of scales in $L(\R)$,} 
{\bf Cabal Seminar 79-81,} (eds. A.S. Kechris, 
D.A. Martin, J.R. Steel), vol. 1019, pp.86--96. 



\bibitem{moschovakis} 
Y.N. Moschovakis, {\bf Descriptive Set Theory,} 
North-Holland, Amsterdam, 1980. 




\end{thebibliography}

\leftskip 0.5in







greg@math.ucla.edu

\bigskip



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UCLA

CA 90095-1555

\leftskip 0in

\end{document}




