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\newcommand{\ww}{{\omega^\omega}}



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\begin{document}


\setbeamercovered{dynamic}



\title{Partition Triples Under AD}



\author{S.\ Jackson\\
(joint with A.\ Apter and B.\ L\"{o}we)}



\institute{
Department of Mathematics\\
University of North Texas\\
jackson@unt.edu
}






\date{{\color{red} BEST 18}\\
March, 2009\\
Boise, ID}






\begin{frame}
\titlepage
\end{frame}


\section{Overview}




\begin{frame}
We assume the {\em axiom of determinacy} $\ad$  throughout. 
We work in the theory $\zf+\ad+\dc$. 
\bigskip

We prove a polarized partition result for certain triples of cardinals
and give applications to forcing over models of $\ad$. 
\bigskip


Main result is a strengthening/generalization of a result proved by Kechris.
\end{frame}



\subsection{Partition Properties}


\begin{frame}
To state the main $\ad$ result, we first recall some partition notation.
\medskip

$(\kappa)^\lambda$ denotes the increasing functions from $\lambda$ to $\kappa$. 
\bigskip



\begin{defn}[Erd\"{o}s-Rado]
$\kappa \rightarrow (\kappa)^\lambda$ means for every partition $\sP
\colon (\kappa)^\lambda \to \{ 0,1\}$ there is a \color{red}
homogeneous \color{black} $H \subseteq \kappa$ of size $\kappa$.
\medskip

$H$ is homogeneous if $\sP \res (H)^\lambda$ is constant.
\end{defn}

\end{frame}









\begin{frame}
\begin{defn}
$\kappa$ has the \color{red}weak \color{black} partition property 
if $\kappa \rightarrow (\kappa)^\lambda$ for all $\lambda <\kappa$.
\medskip


$\kappa$ has the \color{red}strong \color{black} partition property if $\kappa \rightarrow (\kappa)^\kappa$.
\end{defn}
\bigskip


In $\mathsf{ZFC}$ no cardinal has an infinite exponent parition relation.
\end{frame}







\begin{frame}
Let $\vec \kappa= (\kappa_\alpha)_{\alpha< \rho}$ be a sequence of regualr cardinals. 
\medskip
Let $\vec \lambda=(\lambda_\alpha)_{\alpha<\rho}$ be a sequence or ordinals, 
$\lambda_\alpha \leq \kappa_\alpha$.
\bigskip


A \color{red} block function \color{black} from $\vec \lambda$ to 
$\vec \kappa$ is a function with domain  $\oplus_{\alpha <\rho} \lambda_\alpha$ which 
sends the copy of $\lambda_\alpha$ into $\kappa_\alpha$. 
\bigskip





\begin{defn}[Polarized Partition Property]
$\vec \kappa  \rightarrow (\vec \kappa)^{\vec \lambda}$
if for every partition $\sP$ of the block functions from $\vec \lambda$ to $\vec \kappa$ there is an
$H= \oplus_{\alpha<\rho} H_\alpha$, where $H_\alpha \subseteq \kappa_\alpha$, which is
homogeneous for $\sP$. 
\end{defn}
\end{frame}


\subsection{Suslin Cardinals}


\begin{frame}
If $T$ is a tree on a set $Z$, $[T]=\{ f \in Z^\omega \colon \forall n\ f\res n \in T\}$
(set of infinite branches through $T$).



\begin{defn}
We say $A \subseteq \ww$ is $\kappa$-Suslin if there is a tree $T \subseteq \omega \times \kappa$
such that $A=p[T]=\{ x \colon \exists f \in \kappa^\omega \ (x,f) \in [T]\}$.
\end{defn}
\medskip

Let $S(\kappa)$ denote the collection of $\kappa$-Suslin sets. 
\medskip


\begin{defn}
$\kappa$ is a \color{red} Suslin cardinal \color{black} 
if $S(\kappa)-\bigcup_{\lambda < \kappa} S(\lambda) \neq \emptyset$.
\end{defn}

\end{frame}








\begin{frame}
The Suslin cardinals are closely related to the projective ordinals.
\medskip

In $\zf$:
\begin{itemize}
\item
$\omega$ is the first Suslin cardinal and $S(\omega)=\bs^1_1$.
\item
$\omega_1$ is the next Suslin cardinal and $S(\omega_1) \supseteq \bs^1_2$.
\end{itemize}
\medskip


Recall $\de 1{n}$ is the supremum of the lengths of the $\bd^1_n$ prewellorderings of the reals.
\medskip



With $\ad$:
\begin{itemize}
\item
$\de 1{2n+2}=(\de 1{2n+1})^+$
\item
$\de 1{2n+1}=\lambda_{2n+1}^+$, where $\cof(\lambda_{2n+1})=\omega$
\item
$\de 1{n}$ is regular.
\end{itemize}



\end{frame}





\begin{frame}
The first $\omega$ Suslin cardinals are:

$$
\lambda_1, \ \de 11, \ \lambda_3, \ \de 13, \ \lambda_5, \ \de 15, \dots$$
\medskip

where: $\lambda_1=\omega$, $\lambda_3=\aleph_{\omega}$, $\lambda_5=\aleph_{\omega^{\omega^\omega}}, \dots$.
\bigskip



Also: $S(\lambda_1)=\bs^1_1$, $S(\de 11)=\bs^1_2$, $S(\lambda_3)=\bs^1_3$, $S(\de 13)=\bs^1_4, \dots$




\end{frame}





\begin{frame}
\begin{thm}[Steel-Woodin]
$(\ad)$ The Suslin cardinals are closed below their supremum. $(\ad^+)$ The Suslin 
cardinals are closed below $\Theta$. 
\end{thm}
\medskip


We say a Suslin cardinal $\kappa$ is a \color{red} limit Suslin \color{black}
cardinal if $\kappa$ is a limit of Suslin cardinals, otherwise $\kappa$ is a 
\color{red} successor Suslin \color{black} cardinal.
\medskip


In $\lr$ there is a largest Suslin cardinal $\de 21$ (Martin-Steel), and 
$S(\de 21)=\bs^2_1$ is closed under real quantification.
\medskip




$\de 21$ is regular and a limit of Suslin cardinals, so $\de 21$ is an 
\color{red} inaccessible Suslin cardinal. \color{black}
\medskip


In fact, the inaccessible Suslin cardinals are stationary in $\de 21$.
\end{frame}



\subsection{Main Theorem}



\begin{frame}
We now state our main partition result.



\begin{thm}[$\ad$]
For all inaccessible Suslin cardinals $\kappa$ we have:

$$
(\kappa,\kappa^+,\kappa^{++}) \rightarrow (\kappa,\kappa^+,\kappa^{++})^{(\kappa,\kappa,\kappa)}$$
\end{thm}



\begin{cor}
$\kappa$, $\kappa^+$, $\kappa^{++}$ are all measurable. 
\end{cor}

\end{frame}


\subsection{Application}




\begin{frame}
We consider the possible cofinalities/measurability of $\aleph_1$, $\aleph_2$, $\aleph_3$. 
\bigskip


\begin{itemize}
\item
There are $3$ possibilities of $\aleph_1$ ($\cof=\omega$, $regular$, measurable).
\item
There are $4$ possibilities for $\aleph_2$, and $5$ possibilities for $\aleph_3$.
\item
$13$ of the $60$ total possibilities are ``trivially inconsistent.''
For example, $\aleph_1$ regular, $\cof(\aleph_2)=\aleph_1$, and $\cof(\aleph_3)=\aleph_2$.
\end{itemize}




\begin{thm}
Assuming suitable large cardinals, all of the remaining $47$ patterns are consistent
with $\zf$. 
\end{thm}
\end{frame}




\section{$\ad$ Background}


\begin{frame}
We review a few more more facts about pointclasses and Suslin cardinals.
\bigskip

\begin{itemize}
\item
By a Wadge degree we mean the equivalence class of a pair $(A, A^c)$ if
$A \nequiv A^c$ (non-selfdual) or of  $(A)$ if $A \equiv A^c$ (selfdual). 



\item
The selfdual and non-selfdual degrees alternate. At limit ordinals of cofinality $\omega$ 
there is a selfdual degree, and for uncountable cofinality a non-selfdual degree. 



\item
The separation property falls on exactly one of the sides of a non-selfdual pointclass. 
\end{itemize}

\end{frame}







\begin{frame}
By a \color{red} Levy class \color{black} we mean a non-selfdual pointclass 
closed under either $\exists^\ww$ or $\forall^\ww$ (or both).
\bigskip
Let $\bs^1_\alpha$, $\bp^1_\alpha$ enumerate these classes. 
\bigskip


\begin{fact}[Kechris-Solovay-Steel]
If $\bd$ is selfdual and closed under quantifiers, $\wedge$, $\vee$, then 
$o(\bd)=\delta(\bd)$
\end{fact}
\bigskip


$o(\bd)=\sup \{ |A|_w \colon A \in \bd \}$.
\bigskip

$\delta(\bd)= \sup \{ |{\prec}| \colon {\prec} \in \bd \}$.
\end{frame}




\begin{frame}
Every Levy pointclass lies in a \color{red} projective-like \color{black}
hierarchy. 
\bigskip



Let $C= \{ o(\bd) \colon \exists^\ww \bd \subseteq \bd, \wedge \bd \subseteq \bd\}$.
\bigskip




Given a levy class $\bg$. let $\alpha$ be the larget element of $C$ such that 
$\bd_\alpha =\{ A \colon o(A)<\alpha\} \subseteq \bg$.
\bigskip



$\bd_\alpha$ is the base of a projective-like hierarchy containing $\bg$. 
\bigskip


This hierarchy falls into one of four types.
\end{frame}






\begin{frame}
We specialize to the case of Suslin cardinals. Let $\alpha$ be a limit Suslin cardinal. 
Let $\bd=\bigcup_{\lambda <\kappa} S(\lambda)$. Then $\bd$ is at the base of a projective-like
hierarchy.





\begin{description}
\item[Type I] $\cof(\alpha)=\omega$. 
Let  $\bs^\alpha_0$ 
be the collection of countable unions
of sets in $\bd$. Then $\bs^\alpha_0$, $\bp^\alpha_1$, etc.\ have the scale 
property. $\alpha^+$ is a Suslin cardinal, and a 
$\bp^\alpha_1$ scale on a $\bp^\alpha_1$-complete set
has norms of length $\kappa^+$. We have $\mathbf{S}(\kappa)=\bs^\alpha_1$ and 
$\mathbf{S}(\kappa^+)=\bs^\alpha_2$. 




\item[Type II or III] $\cof(\alpha)>\omega$, $\bg$ not closed under real quantifiers. 
$\bg$ is the Steel classof Wadge rank $\alpha$. 
So, $\bd=\bg \cap \bgd$, and $\bg$ is closed under 
$\wedge$, $\forall^\R$. Then $\mathbf{S}(\kappa)=\exists^\R \bg$, and 
$\sca(\bg)$, $\sca(\mathbf{S}(\kappa))$ hold.



\item[Type IV] In this case, the pointclasses $\bg$, $\bgd$  of Wadge degree $\kappa$
are closed under real quantification. Let $\bg$ be such that $\bgd$ has the separation property. 
Then $\sca(\bg)$, and $\mathbf{S}(\kappa)= \bg$. 
\end{description}
\end{frame}





\begin{frame}
Of particular importance for the proof are the Type I hierarchies.
\bigskip


Fix now an inaccessible Suslin $\kappa$, with corresponding pointclass $\bg$
(of Wadge rank $\kappa$). 
\bigskip


$\kappa$ is a limit Suslin cardinal. Let $C$ be the c.u.b.\ set of limit Suslin cardinals
below $\kappa$.
\bigskip


We are interested in the points of $C$ of cofinality $\omega$ (Type I hierarchy).
\end{frame}









\begin{frame}
We have the following picture.
\bigskip





\begin{center}
\begin{picture}(250,150)(0,-125)
\put(0,0){\line(1,0){250}}

\put(175,-5){\line(0,1){10}}
\put(173,-12){$\kappa$}


\put(195,-5){\line(0,1){10}}
\put(193,-12){$\kappa^+$}


\put(215,-5){\line(0,1){10}}
\put(213,-12){$\kappa^{++}$}




\put(75,-5){\line(0,1){10}}
\put(85,-5){\line(0,1){10}}
\put(90,-5){\line(0,1){10}}
\put(93,-5){\line(0,1){10}}

\put(95,-5){\line(0,1){10}}
\put(93,-12){$\alpha$}

\put(110,-5){\line(0,1){10}}
\put(108,-12){$\alpha^+$}


\put(125,-5){\line(0,1){10}}
\put(123,-12){$\alpha^{++}$}


\put(150,-12){$\cdots$}

\put(95,0){\fcolorbox{black}{red}{\makebox(8,0.5){}}}



\end{picture}
\end{center}
\vspace{-100pt}

$$
\bs^\alpha_0=\bigcup_{\lambda<\alpha} S(\lambda)
$$

$$
S(\alpha)=\bs^\alpha_1, \qquad S(\alpha^+)=\bs^\alpha_2
$$



$$
\sca(\bs^\alpha_0,) \qquad \sca(\bp^\alpha_1)
$$
\end{frame}


\section{Plan of the Proof}




\begin{frame}
\frametitle{Plan of the proof}

Let $\mu$ be the $\omega$-cofinal normal measure on $\kappa$. ($\kappa$ has the strong partition 
property by Kechris-Kleinberg-Moschovakis-Woodin).
\medskip

Let $\mu_\alpha$ be the $\omega$-cofinal normal measure on $\alpha^+$. We have 
$j_{\mu_\alpha}(\alpha^+)=\alpha^{++}$. 
\bigskip


\begin{enumerate}
\item
We show $[\alpha \mapsto \alpha^+]_\mu=\kappa^+$.


\item
We show $\delta \doteq [\alpha \mapsto \alpha^{++} ]_\mu \leq \kappa^{++}$.


\item
We show for all $\theta < \omega_1$ that $(\kappa,\kappa^+,\delta)\rightarrow
(\kappa,\kappa^+,\kappa^{++})^\theta$.

\qquad It follows that $\delta$ is regular, so $\delta =\kappa^{++}$.


\item
We show $(\kappa,\kappa^+,\delta)\rightarrow
(\kappa,\kappa^+,\kappa^{++})^\theta$ for all $\theta <\kappa$.


\item
Finally, we show $(\kappa,\kappa^+,\delta)\rightarrow
(\kappa,\kappa^+,\kappa^{++})^\kappa$
\end{enumerate}


\end{frame}






\section{The Countable Exponent Case}


\subsection{Construction of the Trees}



\begin{frame}
We define two trees $T^+$ and $T^{++}$ on $\omega \times \kappa$ to analyze the ultrapowers 
$[\alpha \mapsto \alpha^+]_\mu$ and $[\alpha \mapsto \alpha^{++}]_\mu$.
\bigskip



Will have:


\begin{lem}
For any $f \colon \kappa \to \kappa$ with $f(\alpha)<\alpha^+$ there is a $x \in \ww$ with $T^+_x$
wellfounded and such that $\forall^*_\mu \alpha\ f(\alpha)<|T^+_x \res \alpha|$.
\end{lem}
\medskip


This immediately shows that $[\alpha \mapsto \alpha^+]_\mu \leq \kappa^+$, 
and the lower bound follows easily from the coding lemma. 




\end{frame}







\begin{frame}
For the tree $T^{++}$ we will have:



\begin{lem}
For any $f \colon \kappa \to \kappa$ with $f(\alpha)<\alpha^{++}$ there is a $x \in \ww$
with $T^{++}_x$ wellfounded such that 
$\forall^*_\mu \alpha\ (f(\alpha)< [\beta \mapsto |T^{++}_x \res \beta|]_{\mu_\alpha})$.
\end{lem}
\bigskip


The lemma and the normality of $\mu_\alpha$ show that 
$[\alpha \mapsto \alpha^{++}]_\mu \leq \kappa^{++}$.
\medskip


These lemmas will also give a coding of the functions into $\kappa^+$ or $\kappa^{++}$
we use to get the partition relations.
\end{frame}



\subsection{Construction of $T^+$}


\begin{frame}
\frametitle{Construction of $T^+$}
Fix a $\bg$-complete set $P$ and a $\bg$-scale $\{ \varphi_n \}_{n \in \omega}$ on $P$. 

\qquad -- we use $\varphi_0$ to code ordinals $< \kappa$. 
\bigskip



Say $\alpha$ is  {\em strongly reliable} if for all $\beta < \alpha$:


$$
\sup\{ \varphi_n(x) \colon x \in P \wedge \varphi_0(x) \leq \beta\} <\alpha
$$


The set of strongly reliable ordinals  is c.u.b.\ in $\kappa$. 
\bigskip




Let 

$(x,y) \in R \leftrightarrow (x,y \in P \wedge \varphi(x) < \varphi(y)\}$.
\medskip

and  

$(x,y) \in R^\alpha  \leftrightarrow (x,y \in P \wedge \varphi(x) < \varphi(y)<\alpha\}$.

\end{frame}










\begin{frame}
Note that $R^\alpha  \in \bs^\alpha_0- \bigcup_{\lambda > \alpha} S(\lambda)$.
\bigskip

Also, we uniformy have a $\bs^\alpha_0$ scale on $R^\alpha$ (essentially by restricting the
scale $\vec \varphi$ to ordinals below $\alpha$). 
\bigskip



Starting from this, we uniformy get $\bs^\alpha_1$ universal sets $B^\alpha$
and $\bp^\alpha_1$ sets $Q^\alpha$ and $\bp^\alpha_1$ scales $\vec \psi^\alpha$ on $Q^\alpha$. 
\bigskip


Let $W=\{ x \colon \forall n\ (x)_n \in P\}$. $x \in W$ will code the ordinal 
$|x|=\sup_n \varphi_0((x)_n)$. The scale on $P$ easily gives a scale on $W$.Let 
$V_W$ be the corresponding tree. 
\bigskip



\end{frame}








\begin{frame}
We first construct a tree $U$ on $\omega \times \omega \times \kappa$
with the following properties:

\begin{enumerate}
\item
If $x \in W$ and $|x| \in C$, then $U_{x,y}$ is wellfounded iff 
the $\bs^\alpha_1$ relation coded by $y$ is wellfounded. 

\item
For $x,y$ as above, $|U_{x,y} \res \alpha| \geq |B^{|x|}_y|$, the 
$\bs^\alpha_1$ ($\alpha=|x|$) coded by $y$. 
\end{enumerate}
\bigskip



\color{red} Key Point: \color{black} For $x,y$ as above, the entire tree 
$U_{x,y}$ is wellfounded (not just $U_{x,y} \res \alpha$).
\bigskip


\color{red} idea: \color{black}
$U$ is constructed as in the proof of the Kunen-Martin theorem, but we use the components of the real 
$x$ to verify the appropriate reals are in $B^{|x|}_y$.

\end{frame}







\begin{frame}
Suppose $f \colon \kappa \to \kappa$ and $f(\alpha)<\alpha^+$ for $\alpha \in C$. 
\bigskip



Consider the game $G_f$:
\bigskip


\begin{center}
\begin{tabular}{rl}
I & $r$ \\
II & $x,y$
\end{tabular}
\end{center}
\bigskip

II wins the run iff 

$$(r \in W) \rightarrow (x \in W \wedge B^{|x|}_y  \text{ is wellfounded
} \wedge |B^{|x|}_y| >f(|x|).$$
\end{frame}





\begin{frame}
A boundedness argument shows that II has a winning strategy. 
\bigskip

This suggests the following definition of the tree $T^+$ on 
$(\omega)^2 \times \kappa \times (\omega)^2 \times \kappa$:
\bigskip




$(\sigma, r,  \vec \alpha, x,y, \vec \beta) \in [T^+]$ iff:


\begin{enumerate}
\item
$(r, \vec \alpha) \in [T_W]$.
\item
$\sigma(r)=(x,y)$
\item
$(x,y,\vec \beta) \in [U]$.
\end{enumerate}
\bigskip


Then $T^+_\sigma$ is wellfounded and $|T^+_\sigma \res \alpha|>f(\alpha)$ 
for $\mu$ almost all $\alpha$. 
\end{frame}


\subsection{Construction of $T^{++}$}

\begin{frame}
\frametitle{Construction of $T^{++}$}


We first construct a tree $V$ on $(\omega)^2 \times \kappa$
with the following properties:


\begin{enumerate}
\item
$(x,y) \in [V]$ iff $x \in W$ and for all $n$, $(y)_n$ codes a $\bs^{|x|}_1$
wellfounded relation $B^{|x|}_{(y)_n}$.
\item
If $x \in W$, $|x| \in C$ then there is a c.u.b.\ $D \subseteq \alpha^+$ 
such that if $\gamma \in D$, $y \in \ww$  and for all $n$ $(y)_n$
codes a $\bs^{|x|}_1$ wellfounded relation of rank $<\gamma$, then
$V_{x,y} \res \gamma$ is illfounded.
\end{enumerate}
\bigskip


\color{red} main point: \color{black}
We can translate the $\bp^{|x|}_1$ statement asserting the wellfoudedness
of the $B^{|x|}_{(y)_n}$ into $\bp^\beta_1$ statements for any $\beta \geq |x|$
(use the $(x)_i$ as in the definition of $U$).


\end{frame}




\begin{frame}
Suppose $x \in W$, $|x| =\alpha\in C$, and $g \colon \alpha^+ \to \alpha^+$. 
\bigskip
Play the game $G_g$:


\begin{center}
\begin{tabular}{rl}
I & $z$ \\
II & $w$
\end{tabular}
\end{center}



II wins the run iff: 

$$
(\forall n\ B^{\alpha}_{(z)_n} \text{ is wellfounded }) \rightarrow 
(B^\alpha_w  \text{ is wellfounded } \wedge |B^\alpha_w| > g( \sup_n 
|B^\alpha_{(y)_n}|) )
$$

\end{frame}






\begin{frame}
By boundedness, II has a winning strategy $\tau$ for any $G_g$. 
\bigskip

Suppose now $f \colon \kappa \to \kappa$ with $f(\alpha)< \alpha^{++}$.
\bigskip









Play the game $G_f$:


\begin{center}
\begin{tabular}{rl}
I & $r$ \\
II & $x,\tau$
\end{tabular}
\end{center}
\bigskip



$r,x$ will be in $W$ and $\tau$ will be strategy  for a game $G_g$ where $[g]_{\mu_\alpha} >f(\alpha)$,
where $\alpha=|x|$.
\end{frame}












\begin{frame}
More precisely, II wins the run iff:


\begin{equation*}
\begin{split}
r \in W & \rightarrow (x \in W \wedge |x| =\alpha \geq |y|) \\ &\wedge \forall z\ 
[ \forall n\ B^\alpha_{(z)_n} \text{ is wellfounded } \rightarrow 
\\ & \qquad B^\alpha_{\tau(z)} \text{ is wellfounded }  \wedge 
|B^\alpha_{\tau(z)} | \geq g( \sup_n |B^\alpha_{(z)_n}|) ]
\end{split}
\end{equation*}


\noindent
for some $g \colon \alpha^+ \to \alpha^+$  with $[g]_{\mu_\alpha} \geq f(\alpha)$.

\end{frame}







\begin{frame}
II has a winning strategy $\sigma$ for any $f$, and this suggests the definition of $T^{++}$:
\bigskip



$(\sigma, r, \vec \alpha, x, \tau, y, z,\vec \beta, \vec \gamma) \in T^{++}$ iff:


\begin{enumerate}
\item
$(r, \vec \alpha) \in [T_W]$.
\item
$\sigma(r)=(x,\tau)$.
\item
$(x,y,\vec \beta)\in [V]$
\item
$\tau(y)=z$.
\item
$(x,a,\vec \gamma) \in [U]$
\end{enumerate}
\bigskip

The properties of $U$ and $V$ show that $T^{++}$ has the desired property.
\end{frame}


\section{Proof for the countable case}


\begin{frame}
\frametitle{The countable exponent $\theta$ case.}


Fix a bijection $\pi \colon \omega \cdot \theta \to \omega$.
\bigskip


We code cofinally in $\kappa^+$, $\kappa^{++}$ many ordinals using sections of our trees:
$T^+_x$, $T^{++}_x$.
\bigskip




Suppose $\sP$ is a partition of the block functions from $3 \times \theta$ to 
$(\kappa,\kappa^+,\kappa^{++})$.
\bigskip

Consider the game $G_\sP$:



\begin{center}
\begin{tabular}{rl}
I & $x,y,z$ \\
II & $x',y',z'$
\end{tabular}
\end{center}
\bigskip



\end{frame}







\begin{frame}





(1) If there is an $j<\omega \cdot \theta$ such that $(x)_{\pi(j)} \notin P_0$ or
$(x')_{\pi(j)} \notin P_0$, then player~I wins iff for the least such $j$, 
$(x)_{\pi(j)} \in P_0$.
\medskip



(2) Suppose next that there is an  $\alpha<\kappa$ such that one of the following holds. 

\begin{enumerate}
%\renewcommand{\labelenumi}{(\alph{enumi})}

\item[(a)]  There is a $j< \omega \cdot \theta$ such that either $T^+_{(y)_{\pi(j)}}
\res \alpha$ or $T^+_{(y')_{\pi(j)}}\res \alpha$ is
illfounded.



\item[(b)]  There is a $\beta < \alpha^+$ and a $j < \omega \cdot
\theta$ such that either $T^{++}_{(z)_{\pi(j)}} \res \beta$ 
or  $T^{++}_{(z')_{\pi(j)}} \res \beta $ is
illfounded.
\end{enumerate}



Let $\alpha<\kappa$ be least such that (a) or (b) above holds. 
If (a) holds, let $j$ be least such that (a) holds for $\alpha$ and this $j$. 
In this case, Player I wins provided $T^+_{(y)_{\pi(j)}}$ is wellfounded. 
If (a) does not hold at $\alpha$, but (b) does, let $(\beta,j)$ be lexicographically least
such that (b) holds. Player I wins in this case provided 
 $T^{++}_{(z)_{\pi(j)}} \res \beta$  is wellfounded. 







\end{frame}








\begin{frame}

Assume II has a winning strategy $\tau$.
\bigskip


We define c.u.b.\ sets $C_) \subseteq \kappa$, $C_1 \subseteq \kappa^+$, and $C_2 \subseteq \kappa^{++}$.
\bigskip


For example, to define $C_2$ we define for $\alpha \in C$, $\beta, \gamma < \alpha^+$
and $j <\omega \cdot \theta$:







\begin{equation*}
\begin{split}
A_{\alpha, \beta,\eta, j}=& \{ (x,y,z) \,;\, \forall j\ ( (x)_{\pi(j)}
\in P_0 \wedge \varphi_0((x)_{\pi(j)}) < \alpha) 
\\ & \wedge 
\forall \alpha' < \alpha\ \forall \beta \ <(\alpha')^+\ \forall j\ 
(T^+_{(y)_{\pi(j)}} \res \alpha \text{ and } 
T^{++}_{(z)_{\pi(j)}} \res \beta \text{ are wellfounded})
\\ & 
\wedge \forall j \
|T^+_{(y)_{\pi(j)}} \res \alpha| < \beta \wedge 
\forall (\beta',j') \leq_\lex (\beta,j) \ (
|T^{++}_{(z)_{\pi(j)}} \res \beta | \leq \eta ) \}.
\end{split}
\end{equation*}




\end{frame}







\begin{frame}
We have: $A_{\alpha,\beta,\gamma,j} \in \bd^\alpha_1$.
\bigskip





Since $\tau$ is winning for Player~II, for each $(x,y,z) \in
A_{\alpha, \beta, \eta,j}$, if $\tau(x,y,x)= (x',y',z')$ then
$\forall (\beta',j') \leq_\lex (\beta,j)\   
T^{++}_{(z')_{\pi(j')}}\res \beta$ is wellfounded. 
\medskip



By
boundedness, $$\rho_2(\alpha,\beta,\eta,j) := \sup \{ |
T^{++}_{(z')_{\pi(j')}} \res \beta| \,;\, (x',y',z') \in
\tau[A_{\alpha,\beta,\eta,j}] \wedge j' \leq j  \} <
\alpha^+.$$
\medskip


Let $C_2^\alpha \subseteq \alpha^+$ be c.u.b.\ closed under $\rho_2$. 
The $C^\alpha_2$ lift to $C_2 \subseteq \kappa^{++}$.















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\end{document}