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\begin{document}


\setbeamercovered{dynamic}



\title{Suslin Cardinals and Scales From AD}



\author{S.\ Jackson}




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






\date{{\color{red} Conference on the Core Model Induction and HOD Mice\\
July-August, 2010, Muenster}}






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








\begin{frame}
\section{Introduction}
\frametitle{Basic Facts}

We work throughout in $\zf+\ad+\dc$. 
\mys


Our goal is to give a complete as possible picture of the structure of the
Suslin cardinals and scaled pointclasses just assuming $\ad$. The methods  are 
``descriptive set theoretic.'' 
\mys

The methods here just use $\ad$ although $\ad^+$ is needed to get that 
the supremum of the Suslin cardinals is a Suslin cardinal. 
\mys


Will use work of Cabal in particular work of {\color{red}Kechris}, {\color{red}Steel}, and 
{\color{red}Woodin}, and in particular 
a method of {\color{red}Martin} for analyzing the next Suslin cardinal. 
\mys


\end{frame}




\begin{frame}

\begin{defn}
$A \subseteq \ww$ is $\kappa$-Suslin if there is a tree $T \subseteq (\omega \times \kappa)^{<\omega}$
with \[A=p[T]=\{ x \colon \exists f \in \kappa^\omega \ \forall n \ (x \res n, f\res n) \in T\}.\]
\end{defn}


We let $S(\kappa)$ denote the pointclass of $\kappa$-Suslin sets. $S({< \kappa})=
\bigcup_{\lambda < \kappa} S(\lambda)$. 
\mys

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

\end{frame}





\begin{frame}
Suslin representations are closely related to scales.
\mys


\begin{defn}
An $\alpha$-semiscale on $A \subseteq \ww$ is a sequence of norms 
$\varphi_n \colon A \to \alpha$ such that if $\{ x_m\}_{m \in \omega} 
\subseteq A$, $x_m \to x \in \ww$,  and for each $n$, $\varphi_n(x_m)$ is eventually equal to some 
$\lambda_n \in \on$, then $x \in A$.
\end{defn}
\mys

We have the following easy and basic fact:
\mys

\begin{fact}
$A =p[T]$ for a tree $T$ on $\omega \times \alpha$ iff $A$ admits an $\alpha$-semiscale.
\end{fact}
\end{frame}


\begin{frame}
If $A=p[T]$, let $\varphi_n(x)=$ the $n$th coordinate of the leftmost branch of $T_x$.
$\vec \varphi=\vec\varphi_T$ is the {\color{red}semiscale from $T$}.
\mys


If $\{ \varphi_n\}$ is a semiscale on $A$, define $T$ by 
$(s,\vec \alpha) \in T $ iff $\exists x \in A\ \exists n \ [
s=x \res n \wedge \vec\alpha=(\varphi_0(x),\dots,\varphi_{n-1}(x)) ]$.
$T=T_{\vec\varphi}$ is the {\color{red}tree from $\vec\varphi$}.
\mys

To get scales, we add the lower semicontinuity property:
\mys


\begin{defn}
$\vec \varphi$ is a {\color{red}scale} on $A$ if it is a semiscale and whenever 
$x_m \to x$ and $\varphi_n(x_m)\to \lambda_n$ then $\varphi_n(x)\leq \lambda_n$.
\end{defn}

\end{frame}





\begin{frame}
If $A=p[T]$ and we let \[ \varphi'_n=|\langle \varphi_0(x), \dots,\varphi_n(x)\rangle|_\lex,\]
then $\vec\varphi'$ is a scale on $A$ (with norms into $\alpha^n$).




\begin{itemize}
\item
If $\vec\varphi$ is a scale on $A$, then $\vec \varphi_{T_{\vec \varphi}}=\vec \varphi$.
\item
If $T=T_{\vec \varphi}$ is the tree of a scale $\vec \varphi$, then 
$T_{\vec \varphi_T}=T_{\vec \varphi_{T_{\vec \varphi}}}=T_{\vec\varphi}=T$.
\end{itemize}
\mys


Not every tree $T$, however, is the tree of a scale. For a general tree $T$ we just have 
$T_{\vec \varphi_T} \subseteq T$ and these trees have the same projection (if 
$x \in p[T]$ and $\vec \beta$ is the left-most branch of $T_x$ then $(x,\vec \beta) \in 
[T_{\varphi_T}]$).
\mys

\begin{itemize}
\item
For any tree $T$ we have $\varphi_{T}=\varphi_{T_{\varphi_{T}}}$.
\end{itemize}
\mys

In particular, both maps $T \mapsto T_{\varphi_T}$ and $\varphi\mapsto 
\varphi_{T_\varphi}$ are idempotent.
\end{frame}



\begin{frame}
The scale $\vec \varphi'$ corresponding to a semiscale $\vec \varphi$ may be on a 
slightly bigger ordinal, but we have the following.
\mys




\begin{fact}
For all cardinals $\kappa$, the following are equivalent.
\begin{enumerate}
\item
$A$ is $\kappa$-Suslin.
\item
$A$ admits a $\kappa$-semiscale.
\item
$A$ admits a $\kappa$-scale.
\item
$A$ admits a $\kappa$-very good scale.
\end{enumerate}
\end{fact}
\end{frame}

\begin{frame}
proof of (1) $\Rightarrow$ (3): Let $A=p[T]$ where $T$ is a tree on $\omega \times \kappa$.
\mys

\underline{Case 1}. $\cof(\kappa)>\omega$. 
\mys


For $x \in p[T]$, let $\psi_0(x)=$ least $\alpha<\kappa$ such that $T_x \res \alpha$
is illfounded. Let \[\psi_n(x)=|(\psi_0(x), \varphi^{\psi_0(x)}_0(x),\dots, \varphi^{\psi_0(x)}_{n-1}(x)
)|_{\psi_0(x)}  \]


where $\varphi^\alpha_n$ is the semiscale corresponding to $T\res \alpha$ and 
$| (\alpha,\beta_0, \dots,\beta_{n-1})|_\gamma$ denotes the rank of 
$(\beta_0, \dots,\beta_{n-1})$ in lexicographic ordering on $(\gamma+1)^{n+1}$.
\mys

\underline{Case 2}. $\cof(\kappa)=\omega$. 
\mys

Let $\kappa=\sup_n \kappa_n$. 
\mys

Can easily get a $T'$ with $p[T']=p[T]=A$ and such that if $(s,\vec \alpha)\in T'$
then $\alpha_i < \kappa_i$. Let $\vec \psi$ be scale from $T'$, where $\psi_n$
uses lex ordering on $\kappa_n^{n+1}$. 

\end{frame}




\begin{frame}
\begin{lem}
Let $\kappa$ be a Suslin cardinal.

\begin{enumerate}
\item
There is a strictly increasing sequence $\{ A_\alpha\}_{\alpha<\kappa}$ of $\kappa$-Suslin sets.
\item
If $\cof(\kappa) >\omega$ then we may take the $A_\alpha$ to be $<\kappa$ Suslin.
\end{enumerate}
\end{lem}

{\color{blue}Proof.}
Suppose first $\cof(\kappa)=\omega$. Let $B\in S(\kappa)-S(<\kappa)$ and let 
$\vec \varphi$ be a scale on $B$ with $|\varphi_n|=\lambda_n<\kappa$. We must have 
$\sup_n \lambda_n=\kappa$. Let $A=\{ x \colon x' \in B\}$ where $x'(i)=x(i+1)$. 
Define $\vec \psi$ on $A$ by: $\psi_0(x)=\varphi_{x(0)} (x')$ and 
$\psi_{i+1}(x)=\psi_i(x')$. $\vec\psi$ is a $\kappa$-scale on $A$ and $|\psi_0|=\kappa$. 
Let $A_\alpha=\{ x \in A \colon \psi_0(x)\leq \alpha\}$. 
\end{frame}

\begin{frame}
Suppose next that $\cof(\kappa)>\omega$. Let $A$ and $\vec \psi$ be as above so 
$|\psi_0|=\kappa$. For $\alpha < \beta <\kappa$ let 
\[ A_{\alpha,\beta}=\{ x \colon \forall i\ \psi_i(x)<\beta\} \cup 
\{ x \colon \forall i \ \psi_i(x)\leq \beta \wedge \psi_0(x)\leq \alpha\}.\]
Each $A_{\alpha,\beta}$ is $\kappa$-Suslin. Order the indices by reverse lexicographic
ordering. 


For each $\alpha <\kappa$, there is an $x \in A$ with $\psi_0(x)=\alpha$. 
Let $\beta=\sup_i \psi_i(x)$. Then \[x  \in A_{\alpha,\beta}
-\bigcup_{(\alpha',\beta')<(\alpha,\beta)} A_{\alpha',\beta'}.\]

Thus we get an increasing $\kappa$-length subsequence of $\kappa$-Suslin sets.
\end{frame}


\begin{frame}
We have the following immediate fact.


\begin{lem}
For any $\kappa$, $S(\kappa)$ is closed under $\exists^\ww$, $\wedge_\omega$, $\vee_\omega$.
\end{lem}
\mys

We also have:
\mys


\begin{thm}[Kechris]
For any Suslin cardinal $\kappa$, $S(\kappa)$ is non-selfdual.
\end{thm}
\end{frame}


\begin{frame}
Recall a { \color{red}$\bg$-norm} on $A$ is a norm $\varphi \colon A \to \on$
such that the following relations are in $\bg$:

 
\begin{align*}
x<^* & y \leftrightarrow (x \in A) \wedge (y \notin A \vee (y \in A \wedge \varphi(x)<\varphi(y)))\\
x\leq^* & y \leftrightarrow (x \in A) \wedge (y \notin A \vee (y \in A \wedge \varphi(x)\leq\varphi(y)))
\end{align*}


A {\color{red}$\bg$-scale} on $A$  is a scale $\{ \varphi_n\}$ with each $\varphi_n$
a $\bg$-norm.


\begin{defn}
{\color{red}$\pwo(\bg)$} if every $A \in \bg$ admits a $\bg$-norm. {\color{red}$\sca(\bg)$} 
if every $A \in \bg$ admits a $\bg$-scale.
\end{defn}
\end{frame}


\begin{frame}
Note that if $\varphi$ is a $\bg$-norm on the $\bg$ set $A$, then 
each 

\begin{align*}
A^\varphi_{\leq \alpha}=&\{ x \in A \colon \varphi(x)\leq \alpha\} \in \bd\\
A^\varphi_{< \alpha}=&\{ x \in A \colon \varphi(x)< \alpha\} \in \bd\
\end{align*}
[If $\varphi(x_0)=\alpha$ then $x \in A^\varphi_{\leq \alpha} \leftrightarrow 
x \leq^* x_0 \leftrightarrow \neg (x_0 <^* x)$, and likewise for 
$A^\varphi_{<\alpha}$.]
\mys


So if $\bd$ is closed under $\wedge$, $\vee$ (e.g., if $\bg$ is closed under $\wedge$, $\vee$)
then each $A^\varphi_\alpha=\{ x \colon \varphi(x)=\alpha\}\in \bd$.
\end{frame}

\begin{frame}
\frametitle{Basic Facts: Pointclasses}

Recall a {\color{red}pointclass} $\bg$ is a collection $\bg \subseteq \sP(\ww)$
closed under Wadge reduction (continuous preimages). 
\mys

We say $\bg$ is selfdual if $\bg=\bgd=\{ \ww-A \colon A \in \bg\}$.
\mys

{ \color{blue}Wadge's Lemma}: For any $A, B$ either $A \leq_w B$ or $B \leq_w \ww-A$. 
\mys

This gives a strict linear order on the {\color{red}Wadge degrees}, the equivalence classes
of pairs $(A,\ww-A)$ under Wadge reduction.
\mys

{\color{blue}Martin-Monk Theorem}: The Wadge degrees are wellfounded. 
\mys

We let $o(A)$ denote the Wadge rank of $[(A,\ww-A)]$. 
We let $o(\bg)= \sup \{ o(A) \colon A \in \bg\}$.
\end{frame}


\begin{frame}
$A$ is {\color{red}selfdual} if $A \equiv_w \ww-A$. 
\mys

The pointclasses are essentially the initial segment of the Wadge degrees.
Nonselfdual pointclasses correspond to nonselfdual degrees
(a selfdual pointclass $\bd$ can occur in one of $4$ ways).
\mys


\begin{fact}[Steel, VanWesep]
The selfdual and the nonselfdual Wadge degrees alternate. At limit ordinals of cofinality 
$\omega$ there is a selfdual degree and at ordinals of uncountable cofinality 
a nonselfdual degree.
\begin{itemize}
\item
If $o(A)$ is a limit of cofinality $\omega$, $A$ is the degree of a countable
join of sets of lower degree.
\item
If $A$ is selfdual and $o(A)$ a successor, then $A$ is the degree of the join
of the two sets in the previous nonselfdual pair.
\end{itemize}
\end{fact}

\end{frame}



\begin{frame}

\begin{defn}
$\sep(\bg)$ if for every $A, B \in \bg$ with $A \cap B=\emptyset$ there is a 
$C \in \bd=\bg \cap \bgd$ with $A \subseteq C \subseteq \ww-B$.
\end{defn}


\begin{thm}
For every nonselfdual $\bg$, exactly one of $\sep(\bg)$, $\sep(\bgd)$
holds.
\end{thm}



\begin{fact}
For any nonselfdual $\bg$ we have $\pwo(\bg)\Rightarrow \neg \sep (\bg)$
(and so $\sep (\bgd)$).
\end{fact}
\mys


So, at most one side can have the $\pwo$ property.
\end{frame}



\begin{frame}
We are mainly interested in {\color{red}Lev\'y} classes:


\begin{defn}
A nonselfdual pointclass $\bg$ is a Lev\'y pointclass if it is closed under 
$\exists^\ww$ or $\forall^\ww$. 
\end{defn}
\mys

\begin{rem}
A nonselfdual pointclass closed under $\exists^\ww$ is closed under 
$\vee^\omega$. Likewise, closure under $\forall^\omega$ implies closure under 
$\wedge^\omega$.
\end{rem}
\mys


Let $\bs^1_\alpha$  ($\alpha <\Theta$) enumerate the Lev\'y classes closed under 
$\exists^\ww$, and $\bp^1_\alpha$ their duals.
\end{frame}


\begin{frame}
The Lev\'y classes fall into {\color{red}projective-like hierarchies}
analyzed as follows.
\mys



\begin{defn}
For $\bg$ a pointclass, $\delta(\bg)$ is the supremum of the lengths of the 
$\bg$ prewellorderings of $\ww$.
\end{defn}


\begin{fact}
If $\Lambda$ is a selfdual pointclass closed under $\exists^\ww$ and $\wedge$,
then $o(\bd)=\delta(\bd)$.
\end{fact}


\begin{rem}
In this case $o(\Lambda)$ is also the supremum of the lengths of the $\Lambda$
wellfounded relations.
\end{rem}
\end{frame}




\begin{frame}
\begin{defn}[Steel]
For $\bg$ a pointclass, let 
\[\Lambda(\bg)=\bigcup \{ \Lambda \subseteq \bg\colon 
\Lambda \text{ is selfdual, closed under }
\exists^\ww, \wedge \}.\]
\end{defn}



So, $\Lambda(\bg)$ is selfdual, closed under $\exists^\ww$, $\wedge$, and 
$\lambda=o(\Lambda(\bg))$ is a limit ordinal.
\mys



\underline{Type 1}: If $\cof(\lambda)=\omega$ and $o(B)=\lambda$, then $B$ is selfdual and 
$\exists^\ww B$ is equal to $\bigcup_\omega \Lambda$. Let $\bs^\lambda_0=
\bigcup_\omega\Lambda$, and then define $\bs^\lambda_n$, $\bp^\lambda_n$ over this as usual.
\end{frame}



\begin{frame}
\underline{Types 2 and 3}:
If $\cof(\lambda)>\omega$ and $o(B)=\lambda$, then $(B, \ww-B)$ is a nonselfdual pair. 
We assume these classes are not closed under quantifiers in these cases.
Let $\bs^\lambda_{-1}$ be the side with $\sep(\bs^\lambda_{-1})$, and 
apply quantifiers to get the $\bs^\lambda_n$, $\bp^\lambda_n$ as usual. 
\mys

[Type 3 means $\bs^\lambda_{-1}$ is closed under $\wedge$, type~2 if it is not.]
\mys



\underline{Type 4}: If $\cof(\lambda)>\omega$ and both the $B$ and $\ww-B$
sides are closed under quantifiers, let $\bg=\bg(\lambda)$ be the side such that  $\sep(\bgd(\lambda))$.
Let $\bs^\lambda_1=\bg \vee \bgd$, $\bp^\lambda_1=\bg \wedge \bgd$. Apply quantifiers
to get  the $\bs^\lambda_n$, $\bp^\lambda_n$.
\mys


We call the classes $\bp^\lambda_{-1}$ the {\color{red}Steel Pointclasses}.
\mys


The $\bs^\lambda_{n}$, $\bp^\lambda_{n}$ for $n \geq -1$ enumerate all of the 
Lev\'y classes.
\end{frame}

\begin{frame}
\frametitle{A quick proof}
Quick proof that the Steel class $\bg=\bp^\lambda_{-1}$ is closed under $\forall^\ww$:
\mys

Recall $\sep(\bgd)$, we show $\exists^\ww \bgd \subseteq \bgd$. 
\mys

$\exists^\ww \bgd$
is easily a pointclass, so if this fails then $\bg \subseteq \exists^\ww \bgd$. 
\mys

Let $A \in \bg-\bd$ and write $A=p[B]$ with $B\in \bgd$. Let $C=(\ww-A)\times \ww$, so 
$C \in \bgd$. Let $D\in \bd$ separate $B$, $C$. Then $A=p[D] \in \bd$, a contradiction.
\end{frame}


\begin{frame}
\begin{rem}
All of the Lev\'y classes are closed under $\wedge$, $\vee$
except for the Steel classes in type 2 and the $\bs^\lambda_0$, $\bp^\lambda_0$
in type~4.
\mys

All of the Lev\'y classes are closed under $\wedge^\omega$, $\vee^\omega$
except for the above and $\bs^\lambda_0$, $\bp^\lambda_0$ in type~1.


\end{rem}
\end{frame}


\section{Prewellorderings}

\begin{frame}
\frametitle{Prewellordering Property}


Type~1: Let $A=\bigcup_n A_n$, each $A_n \in \bd$. For $x \in A$ let $\varphi(x)=\mu n\ (x \in A_n)$.
Then $x <^*y \leftrightarrow \exists n\ (x \in A_n \wedge y \notin A_n)$. Likewise for $\leq^*$. 
Then propagate by periodicity.
\mys




Types 2 and 3:
We show the  Steel class $\bg= \bp^\lambda_{-1}$ has the prewellordering property. Let $\rho$ be least such that 
$\bd$ is not closed under $\rho$-unions ($\ad^+$). $\rho$ is regular. 
$\rho \leq \lambda$ as otherwise there is a $\bd$
prewellordering of length $\lambda$, a contradiction.
\mys



In fact, $\rho=\cof(\lambda)$: If $\rho <\lambda$ and $\{ A_\alpha\}_{\alpha <\rho} \subseteq \bd$
is increasing with union $A\notin \bd$, then we must have $\sup_{\alpha <\rho} o(A_\alpha)=\lambda$.
Otherwise let $B\in \Lambda$ with $o(B)>o(A_\alpha)$ for all $\alpha$. By the coding lemma 
there is a $\bd$ set $C$ of reals $z$ each of which reduces some $A_\alpha$ to $B$, and every $A_\alpha$
is reduced to $B$ by some $z \in C$. Then $x \in A \leftrightarrow 
\exists z \in C\ (z(x)\in B)$. So, $A \in \bd$, a contradiction.
\end{frame}


\begin{frame}
If $\rho >\cof(\lambda)$ then for some $\beta<\lambda$, for cofinally many $\alpha<\rho$
$o(A_\alpha)<\beta$. This is a contradiction as above.
\mys


Following Steel, let \[ \bg^*=\{ \bigcup_{\alpha <\rho} A_\alpha \colon 
\forall \alpha\ (A_\alpha \in \bd) \wedge \{ A_\alpha\}_{\alpha <\rho}
\text{ is } \bs^1_1 \text{ bounded } \}.\] 


\begin{claim} $\bg= \bg^*$.
\end{claim}

{\color{blue}Proof}
If $\bigcup_\rho \bd \subseteq \bgd$ (and so equal), then $\pwo(\bgd)$, a contradiction. 
So, $\bigcup_\rho \bd \subseteq \bg$. Note that $\bigcup_\rho \bd$ is closed under $\exists^\ww$. 
So, $\bigcup_\rho \bd \supseteq \exists^\ww \bg=\bs^\lambda_0$. 
\mys

Next, $\bg \subseteq \bg^*$. For let $ A \in \bg$,  and let 
Let $C$ be the set of codes of $\bd^1_1$ subsets of $A$.
\end{frame}


\begin{frame}
More precisely, let
$S\subseteq \ww \times \ww$ be $\bs^1_1$. Say $S=p[F]$
where $F$ is closed. 
\mys

Then,



\begin{align*}
x \in C&
\leftrightarrow S_x \subseteq A \\ & 
\leftrightarrow \forall y \ (y \in S_x \rightarrow y \in A) \\ &
\leftrightarrow \forall y, z\ ((y,z)\in F \rightarrow y\in A)
\end{align*}

So, $C \in \bg$ ($\bg$ is closed under unions with open sets). 
\mys

Write $C=\bigcup_{\alpha<\rho} C_\alpha$, an increasing union with each $C_\alpha \in \bd$.
\mys


Let $A_\alpha(x)\leftrightarrow \exists y\ (C_\alpha(y) \wedge S_y(x))$, so $A_\alpha \in \bd$.
This works.
\end{frame}


\begin{frame}
We show that  $\bigcup_\rho \bd \subseteq \exists^\ww \bg$. Let $A=\bigcup_{\alpha < \rho}A_\alpha$
with each $A_\alpha \in \bd$. Fix a $B \in \bg-\bd$ and write it as a $\bs^1_1$ bounded union:
$B=\bigcup_{\alpha<\rho} B_\alpha$. 
\mys

By $\bs^1_1$ boundedness, II has a winning strategy $\sigma$ such that if 
$x \in B$ then $\sigma(x)$ is a $\bg$ code for some $A_\alpha$ where $\alpha > 
\varphi(x)=\mu \beta\ (x \in B_\beta)$. 
\mys

Then $x \in A \leftrightarrow \exists y\ (y \in B \wedge (\sigma(y))(x)\in U)$,
where $U \in \bg-\bd$. 
\mys


So, $\bg \subseteq \bg^* \subseteq \exists^\ww \bg$.
\end{frame}

\begin{frame}
Finally, $\bg^*$ is closed under $\forall^\ww$:
\mys


If $A\subseteq \ww \times \ww$ is in $\bg^*$, $A=\bigcup_{\alpha<\rho} A_\alpha$ (a 
$\bs^1_1$ bounded union), then $B=\forall^\ww A = \bigcup_{\alpha<\rho}
B_\alpha$ where $B_\alpha=\forall^\ww A_\alpha$. This is also a $\bs^1_1$ bounded union. 
\mys


It now follows that $\bg=\bg^*$ and $\bigcup_\rho \bd=\bs^\lambda_0$. 
\mys

Exactly same arguments work in type~4 except $\bg=\bg^*=\exists^\ww\bg$.
\end{frame}



\begin{frame}
We now easily have $\pwo(\bp^\lambda_{-1})$. For let $A \in \bp^\lambda_{-1}$, 
\[ A=\bigcup_{\alpha<\rho} A_\alpha\]
an increasing $\bs^1_1$-bounded union, where $A_\alpha \in \bd$. 
\mys

For $x \in A$ let $\varphi(x)=\mu \alpha\ (x \in A_\alpha)$.
\mys

This is $\bp^\lambda_{-1}$-norm on $A$. For example 
\[ x <^*_\varphi y \leftrightarrow \exists \alpha <\rho\ (x \in A_\alpha \wedge 
y \notin A_\alpha).\]
\mys

This is easily a $\bs^1_1$-bounded union.
\end{frame}


\begin{frame}
Type~4: We have $\bg=\bg^*=\exists^\ww \bg=\bigcup_\lambda \bd$. Note here that 
$\rho=\lambda$: if $\rho=\cof(\lambda)<\lambda$ then there is a $\bd$ pwo of length $\rho$.
Let $A\in \bg-\bd$ and write $A=\bigcup_{\alpha <\rho}A_\alpha$, $A_\alpha \in \bd$. 
By coding lemma there in a $\bd$ set of codes $C$ which reduce the $A_\alpha$ to a $\bgd$ set. 
This computes $A \in \bgd$, a contradiction. 

\[ x \in A \leftrightarrow \exists \sigma\ (\sigma \in C \wedge \sigma(x) \in U)\]
where $U \in \bgd-\bd$. 
\mys

We use here that $\bs^\lambda_{-1}$ is closed under $\wedge$. More generally, if 
$\bs^\lambda_{-1}$ is closed under intersections with $\bd$, then $\lambda$ is regular.
\end{frame}


\begin{frame}
Note in this case that 

\begin{align*}
\bg&= \bs^1_1\text{-bounded unions of } \bd \text{ sets  }\\ &
=\bd\text{-bounded unions of } \bd \text{ sets (usual boundedness argument) }\\ &
= \bigcup_\lambda \bd \text{ (since  $\bg=\exists^\ww \bg$)}
\end{align*}
\mys





Recall $\bp^\lambda_1=\bg(\lambda) \wedge \bgd(\lambda)$. We show $\pwo(\bp^\lambda_1)$. 
\mys


Let $A \in \bg \cap \bgd $ (where $\bg=\bg(\lambda)$. 
\mys

Say $A= B \cap C$ where 
$B \in \bg$, $C\in \bgd$. 
\end{frame}


\begin{frame}
Write $B=\bigcup_{\alpha<\lambda} B_\alpha$, an increasing $\bd$-bounded union. 
\mys


Write $D=\ww-C=\bigcup_{\alpha<\lambda}D_\alpha$, an increasing $\bd$-bounded union.
\mys

For $x \in A=B\cap (\ww- D)$, let $\varphi(x)=\mu \alpha\ (x \in B_\alpha)$. 
\mys

Then \[ x <^*_\varphi y \leftrightarrow 
 (x \in C) \wedge \exists \alpha <\lambda\ \exists \beta \leq \alpha\ 
(x \in B_\beta) \wedge (y \notin B_\beta \vee y \in D_\alpha).\]





This is an intersection of a $\bgd$ set ($\{ (x,y)\colon x \in C\}$) with a 
$\bg$ set (a $\lambda$-union of $\bd$ sets). Similarly for $\leq^*_\varphi$.
\end{frame}

\begin{frame}
\frametitle{Useful Fact}

\begin{lem}[Martin]
Let $\bg$ be nonselfdual, closed under $\forall^\ww$, $\vee$, and assume $\pwo(\bg)$.
Then $\bd$ is closed under $<\delta(\bd)$ unions and intersections.
\end{lem}


\begin{proof}
Otherwise by the coding lemma $\bgd=\bigcup_\kappa \bd$ for some least $\kappa<\delta$.
But then $\pwo(\bgd)$, a contradiction.
\end{proof}

\end{frame}










\section{Wellordered Unions}






\begin{frame}
\frametitle{Wellordered Unions}


\begin{thm}
Suppose $\bg$ is nonselfdual, $\pwo(\bg)$, and $\exists^\ww \bg \subseteq \bg$. 
Then $\bg$ is closed under wellordered  unions.
\end{thm}



\underline{Case I}: $\bg$ closed under $\wedge^\omega$, $\vee^\omega$ but not $\forall^\ww$.
\mys

Let 

\begin{align*}
\delta_1& =\sup \{ |\prec| \colon \prec\, \in \bg \ \wedge \ (\prec \text{ is wellfounded })\}\\ 
\delta_2& =\sup  \{ |\prec| \colon \prec\, \in \exists^\ww \bgd \ \wedge \ (\prec \text{ is wellfounded })\}
\end{align*}

\end{frame}


\begin{frame}
Let $\rho$ be least so that some $\bigcup_{\alpha<\rho} A_\alpha \notin \bg$, with each $A_\alpha 
\in \bg$. So, $\rho$ is regular, uncountable. Easily $\delta_1<\delta_2$ (use closure of 
$\bg$ under $\wedge^\omega$). We must have $\bigcup_\rho \bg \supseteq \exists^\ww \bgd$. 
We must have $\rho \geq \delta_2$ as otherwise there is a least $\rho' \leq \rho$
such that $\bd_1=\bd(\exists^\ww \bgd)$  is not closed under wellordered unions, and by the coding lemma 
$\bigcup_{\rho'} \bd_1= \exists^\ww \bgd$ which shows $\pwo(\exists^\ww \bgd)$, a contradiction 
to periodicity (which shows $\pwo(\forall^\ww \bg)$). So, $\delta_1<\delta_2 \leq \rho$. 
\mys

Let $\prec \in \exists^\ww \bgd$ be wellfounded of length $<\delta_1$, and write 
$\prec=\bigcup_{\alpha<\rho} A_\alpha$, each $A_\alpha \in \bg$. For $x \in \dom(\prec)$, 
let $\zeta(x)=$ the eventual rank of $x$ in $A_\alpha$ (for $\alpha$ large enough). This is
an order-preserving map from $\prec$ to $\delta_1$, a contradiction. 
\end{frame}


\begin{frame}
\underline{Case II}: $\bg$ is closed under $\exists^\ww$, $\forall^\ww$.
\mys

So, $\bg$ is closed under $\wedge^\omega$, $\vee^\omega$.
\mys


By $\pwo(\bg)$, clearly $\bgd$ is also not closed under wellordered unions. 
Let $\rho_1$ be least so $\bg$ is not closed under $\rho$-unions. Let $\rho_2$ be same for $\bgd$.
$\rho_1$, $\rho_2$ are regular.
\mys

Then $\bigcup_{\rho_1} \bg \supseteq \bgd$. So argument from before shows every $A\in \bgd$
is a $\bs^1_1$-bounded union of length $\rho_1$ of $\bg$ sets. Let $\varphi$ be corresponding norm
on $A$. 
\mys



Play game: I plays $x$, II plays $y,z$, and II wins iff $(x \in A)\rightarrow  
(y \text{ codes } A_\alpha) \wedge (z \in A_\alpha- \bigcup_{\beta<\alpha}A_\beta)$
for some $\alpha \geq \varphi(x)$. 
\end{frame}



\begin{frame}
By $\bs^1_1$-boundedness II has a winning strategy $\tau$. 
\mys


Define a relation $\prec$ by $ x \prec y$ iff $(x,y \in A) \wedge (\tau(y)_1 \notin A_{\tau(x)_0})$. 
\mys

Then $\prec$ is a $\bgd$ wellfounded relation of length $\rho_2$ (since $\rho_2$ is regular). 
By the coding lemma, $\rho_2 >\rho_1$.
\mys


A symmetrical argument shows $\rho_1>\rho_2$, a contradiction.
\end{frame}


\begin{frame}
\underline{Case III}: $\bg$ not closed under $\forall^\omega$ or $\wedge^\omega$.
\mys



We must have that $\lambda$ is of Type~1, so $\cof(\lambda)=\omega$. 
\mys


Also, $\bg=\bs^\lambda_0=\bigcup_\omega \Lambda$. 
\mys


Let $\rho$ be least such that $\bigcup_\rho \bg \nsubseteq \bg$. So, $\rho$ is regular.
By coding lemma, $\rho >\lambda$. 
\mys


By Wadge, $\bigcup_\rho \bg \supseteq \bs^\lambda_1$. Let $A\in \bgd-\bg$ and write 
$A=\bigcup_{\alpha <\rho} A_\alpha$ with each $A_\alpha \in \bg$.
\end{frame}



\begin{frame}
As before, let $S$ be universal $\bs^1_1$ and let $C(z)\leftrightarrow 
\forall x\ (S_z(x) \rightarrow x \in A)$. So, $C \in \bp^\lambda_1$. Write 
$C=\bigcup_{\alpha<\rho}C_\alpha$ with $C_\alpha \in \bg$. Let 
$A'_\alpha(x)\leftrightarrow \exists z \in C_\alpha \ (S_z(x))$, so $A'_\alpha \in \bg$
and $A$ is the $\bs^1_1$ bounded union of the $A'_\alpha$. So, we may assume
the $\{ A_\alpha\}_{\alpha <\rho}$ is a $\bs^1_1$ union.
\mys

Let $U$ be universal $\bs^\lambda_0$ set. 
\mys


Play the game where I plays $x$, II plays $y,z$ and II wins iff 
\[ (x \in A) \rightarrow \exists \alpha <\rho \ (U_y=A_\alpha \wedge z \in A_\alpha -\bigcup_{\beta <\alpha}
A_\beta)
\]
\end{frame}


\begin{frame}
By boundedness, II has a winning strategy $\tau$. This gives a $\bp^\lambda_0$
wellfounded relation (in fact prewellordering) of length $\rho$:




\[ x_1 \prec x_2 \leftrightarrow (x_1, x_2 \in A) \wedge (\tau(x_2)_1 \notin U_{\tau(x_1)_0})
\]


By the coding lemma we then have $\bigcup_\rho \bg \subseteq \bs^\lambda_1$, and so 
$\bigcup_\rho \bg =\bs^\lambda_1$. 
\mys



Let $\rho'\leq \rho$ be least such that $\bigcup_{\rho'}\bd^\lambda_1 \nsubseteq \bd^\lambda_1$. 
By the coding lemma and the above, $\bigcup_{\rho'}\bd^\lambda_1 \subseteq \bs^\lambda_1$
and so $\bigcup_{\rho'}\bd^\lambda_1 = \bs^\lambda_1$. This show $\pwo(\bs^\lambda_1)$,
a contradiction. 
\end{frame}






\begin{frame}
\frametitle{Lengths of Wellordered Unions}

\begin{thm}[Chuang]
Let $\bg$ be nonselfdual, closed under $\forall^\ww$ and $\vee$. Then there is no
strictly increasing or decreasing sequence of sets in $\bg$ of length $(\delta(\bg)^+)$.
\end{thm}


{\color{blue}Proof.} Let $\delta=\delta(\bg)$ be the supremum of the lengths of the $\bd=
\bg \cap \bgd$ prewellorderings. 
\mys

Let $\{A_\alpha\}_{\alpha <\delta^+}$ be a strictly increasing sequence of $\bg$ sets. 
We may assume $\bigcup_{\beta <\alpha} A_\beta \subsetneq A_\alpha$. 
\end{frame}


\begin{frame}
Note that $\pwo(\exists^\ww \bg)$ and $\exists^\ww \bg$ is closed under $\wedge^\omega$, $\vee^\omega$,
so is closed under wellordered unions. 
\mys


\underline{Case I}: $\bg$ not closed under $\exists^\ww$.
\mys


Let $\varphi$ be the norm corresponding to $\{ A_\alpha\}_{\alpha <\delta^+}$.
\mys

Let $x \prec y \leftrightarrow (x,y \in A) \wedge (\varphi(x)<\varphi(y))$, so $\prec \in \exists^\ww \bg$.
\mys

Let $U$ be universal for $\bg$.
\mys


Define

\[
C(x,y,z)\leftrightarrow \exists \alpha <\delta^+\ [
(U_x=A_\alpha) \wedge (y,z \in A) \wedge (\varphi(y)=\alpha)\wedge (\varphi(z)=\alpha+1)]
\]
\end{frame}



\begin{frame}
By coding lemma, let $S\subseteq C$ be in $\exists^\ww \bg$ and such that for all $\alpha<\delta^+$
there is an $(x,y,z)\in S$ with $U_x=A_\alpha$.
\mys


Every set in $\bg$ is a $\delta$ union of sets in $\bd$, so $S$ is a $\delta$-union 
of sets in $\exists^\ww \bd \subseteq \bgd$. Say $S=\bigcup_{\beta<\delta} S_\beta$,
where $S_\beta \in \bgd$. 
\mys

Note that every $\bgd$ wellfounded relation has length $<\delta$ (otherwise, by the coding lemma,
there is a $\bgd$ unbounded subset of a $\bg$ complete norm, a contradiction; we use here the closure
of $\bgd$ under $\wedge$). 
\end{frame}


\begin{frame}
Consider the following prewellordering on $S_\beta$:
\mys



\begin{align*}
(x_1,y_1,z_1)\leq_\beta (x_2,y_2,z_2) &\leftrightarrow (x_1,y_1,z_1),  (x_2,y_2,z_2) \in S_\beta
\wedge \varphi(y_1)\leq \varphi(y_2) \\ &
\leftrightarrow  (x_1,y_1,z_1),  (x_2,y_2,z_2) \in S_\beta \wedge y_1 \in U_{x_2} \\ &
\leftrightarrow  (x_1,y_1,z_1),  (x_2,y_2,z_2) \in S_\beta \wedge z_2 \notin U_{x_1}
\end{align*}


So, $\leq_\beta$ in the intersection of $S_\beta \times S_\beta$ with a $\bg$ or a $\bgd$ set.
So, $\leq\beta \in \bgd$.
\mys


This gives a one-to-one map $\alpha \mapsto (\beta,\gamma)$ 
of $\delta^+$ into $\delta \times \delta$, a contradiction:
For $\alpha <\delta^+$, let $\beta<\delta$ be least such that some $(x,y,z)\in S_\beta$
has $\varphi(y)=\alpha$, and let $\gamma= |(x,y,z)|_{\leq_\beta}$. 
\end{frame}



\begin{frame}
\underline{Case II}: $\bg$ closed under $\exists^\ww$.
\mys



Let $\{ A_\alpha\}_{\alpha<\delta^+}$ and $\prec$ be as before. Then $\prec \in \exists^\ww 
(\bg \wedge \bgd)$ (this class is closed under wellordered unions). 
\mys


Use coding lemma to get $S \in \exists^\ww (\bg \wedge \bgd)$ as before. Note that 
$\exists^\ww (\bg \wedge \bgd)=\bigcup_{\delta} \bgd$. So write $S=\bigcup_{\beta<\delta}
S_\beta$ with each $S_\beta \in \bgd$. 
\mys

Every  $\bgd$ wellfounded relation has length $<\delta$ as before, and this gives a contradiction
as before.
 

\begin{rem}
In case~II we can avoid appeal to closure theorem by considering cases as to whether there is a $\bg$
wellfounded relation of length $\delta^+$.
\end{rem}
\end{frame}

\begin{frame}
\begin{ques}
Can we remove the assumption that $\bg$ is closed under $\vee$ in Chuang's theorem?
\end{ques}


\begin{rem}
There are two cases where $\forall^\ww \bg \subseteq \bg$, $\pwo(\bg)$, and $\bg$ is not closed under 
$\vee$. Namely, at the base of a Type~2 hierarchy ($\bg$ is the Steel class $\bp^\lambda_{-1}$),
and for the class $\bg \wedge \bgd$ where $\bg$ is closed under quantifiers. 
\mys

In the second case similar arguments show that there is no strictly increasing or decreasing sequence
of $\bg \wedge \bgd$ sets of length $\lambda^+$ ($\lambda=o(\Lambda)$, $\Lambda=\bg \cap \bgd$).
\mys

For the Steel pointclass we don't know.
\end{rem}
\end{frame}


\section{Scales and Suslin Cardinals}


\begin{frame}
\frametitle{Two Background Results}
We will need the following two fundamental results.


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



\begin{thm} [Martin, Woodin]
Every tree $T$ on $\omega \times \kappa$, where $\kappa$ is less than the supremum of the Suslin
cardinals, is weakly homogeneous.
\end{thm}
\end{frame}


\begin{frame}
\frametitle{Type 1 Case}
Suppose $\kappa$ is a limit Suslin cardinal (i.e., $\kappa$ is a limit of Suslin cardinals). 
\mys


First consider the case $\cof(\kappa)=\omega$. Let $\Lambda=S(<\kappa)$. 
So, $\Lambda$ is closed under quantifiers,
negation. Let $\lambda=o(\Lambda)$.
\mys


{\color{blue}Claim}. $\lambda=\kappa$.
\mys


{\color{blue}Proof}. If $\lambda>\kappa$ then there is a $\Lambda$ prewellordering of length $\kappa$. 
By the coding lemma, $S(\kappa) \in \Lambda$, a contradiction. So, $\lambda \leq \kappa$. 
To show $\lambda \geq \kappa$ it suffices to show that there are $\Lambda$ prewellorderings 
of length $\kappa'<\kappa$, where $\kappa$ is a Suslin cardinal. 
\mys

There is a $\kappa'$ increasing sequence $\{ B_\beta\}_{\beta <\kappa'}$ of sets in $S(\kappa')
\subseteq \Lambda$. We can find a pointclass $\bg' \subseteq \Lambda$ with 
$\exists^\ww \bg' =\bg'$, $\pwo(\bg')$. By closure theorem this gives a 
$\bg'$ prewellordering of length $\kappa$. 
\qedsymbol
\end{frame}



\begin{frame}
Since $S(\kappa)$ is closed under $\exists^\ww$ and $\wedge^\omega$, $\vee^\omega$, 
$S(\kappa)\supseteq \bs^\kappa_1$. By coding lemma, $S(\kappa)\subseteq \bs^\kappa_1$
(there is a prewellordering of length $\lambda$ in $\bs^\kappa_0$). 
So, $S(\kappa)=\bs^\kappa_1$. 
\mys


We show $\sca(\bs^\kappa_0)$. 
\mys


Let $\kappa=\sup_n \kappa_n$, each $\kappa_n$ a Suslin cardinal. Let $A \in \bs^\kappa_0$,
say $A=\bigcup_n A_n$ where $A_n \in S(\kappa_n)$. Let $A_n=p[T_n]$ and let $\psi^n$ be the 
corresponding scale. For $x \in A$ let $\varphi_0 (x)=\mu n\ (x \in A_n)$, and 
\[ \varphi_{i+1}(x)= \langle \varphi_0(x), \psi^{\varphi_0(x)}_i(x)\rangle.\]

This is easily a $\bs^\kappa_0$ scale. 
Then propagate by periodicity. 
\end{frame}

\begin{frame}
So, $\sca(\bp^\lambda_1)$, $\sca(\bs^\lambda_2)$, $\sca(\bp^\lambda_3), \dots$.
\mys


By Kunen-Martin, a $\bp^\lambda_1$ scale has norms of length $\leq \lambda^+$ (all
initial segments are in $\bd^\lambda_1 \subseteq S(\lambda)=\bs^\lambda_1$). 
\mys

By the coding lemma, there are $\bs^\lambda_1$ prewellorderings of length any $\alpha <\lambda^+$.
So, $\lambda^+$ is the supremum of the lengths of the $\bs^\lambda_1$ wellfounded relations and
so $\lambda^+$ is regular. The $\bp^\lambda_1$-norms on a $\bp^\lambda_1$-complete set
must have length $\lambda^+$ (otherwise we violate boundedness of $\bs^\lambda_1$
sets in the norm). 
\mys


So, $\lambda^+$ is a regular Suslin cardinal. From the coding lemma,
$S(\lambda^+)=\bs^\lambda_2$. 
\end{frame}


\begin{frame}
\frametitle{Quick Summary of Projective Arguments}
The remaining arguments are just as in the projective hierarchy.
\mys

A $\bp^\lambda_{2n+3}$ norm on a $\bp^\lambda_{2n+3}$ set has length 
$\delta^\lambda_{2n+3}=\delta(\bd^\lambda_{2n+3})$. Such a set cannot be
$\alpha$-Suslin for $\alpha <\delta^\lambda_{2n+3}$ (coding lemma). So, 
$\delta^\lambda_{2n+3}$ is a Suslin cardinal and easily $S(\delta^\lambda_{2n+3})=
\bs^\lambda_{2n+4}$. From coding lemma and closure of $\bp^\lambda_{2n+3}$
under $\vee$ we have that $\delta^\lambda_{2n+3}$ is regular. 
\mys

A $\bp^\lambda_{2n+2}$ set has a scale with norms in $\bd^\lambda_{2n+3}$ and so of 
length $<\delta^\lambda_{2n+3}$. So, the set is $\lambda_{2n+3}$-Suslin for some 
\[\delta_{2n+1}<\lambda_{2n+3}<\delta_{2n+3}.\]
\end{frame}



\begin{frame}
Clearly $S(\lambda_{2n+3})\supseteq \bs^\lambda_{2n+3}$, and we have equality.
\mys


\begin{claim}[Kechris]
$\delta_{2n+3}=(\lambda_{2n+3})^+$ and $\cof(\lambda_{2n+3})=\omega$. 
\end{claim}

{\color{blue}Proof}.
By Kunen-Martin every $\bs^\lambda_{2n+3}$ wellfounded relation has length $\leq (\lambda_{2n+3})^+$,
so $\delta_{2n+3}\leq (\lambda_{2n+3})^+$, and so $\delta_{2n+3}= (\lambda_{2n+3})^+$.
\mys

If $\cof(\lambda_{2n+3})>\omega$, then $\bs^\lambda_{2n+3}=\bigcup_{\lambda_{2n+3}} S(<\lambda_{2n+3})$
(using the coding lemma for $\supseteq$). 
So,  $\bs^\lambda_{2n+3}=\bigcup_\rho \bd^\lambda_{2n+3}$ where $\rho\leq \lambda_{2n+3}$
is least so that $\bd^\lambda_{2n+3}$ is not closed under $\rho$-unions.
\mys

Then we have $\pwo(\bs^\lambda_{2n+3})$, a contradiction.
 























\end{frame}
















\begin{frame}
\frametitle{Type 2, 3 cases}

Suppose now $\kappa$ is a limit Suslin cardinal with $\cof(\kappa)>\omega$. 
Let $\Lambda =S(<\kappa)$. We assume we are in Type~2 or 3, that is, the Steel
class $\bp^\lambda_{-1}$ is not closed under $\exists^\ww$. 
\mys


As above we have $\lambda=\kappa$.
\mys



$\bs^\kappa_0$ is closed under wellordered unions, so $S(\kappa)=\bigcup_\kappa S(<\kappa)
\subseteq \bs^\kappa_0$. 
\mys

So, either $S(\kappa)=\bs^\kappa_0$ or $S(\kappa)=\bs^\kappa_{-1}$. We show the latter
case does not occur.
\end{frame}


\begin{frame}
Let $\bg$ be the collection of $A=p[T]$ where $T$ is a homogeneous tree on $\kappa$. 
$\bg$ is easily a pointclass. 
\mys

We have $S(\kappa)= \exists^\ww \bg$ as every tree on $\omega \times \kappa$ is weakly homogeneous
(we are assuming $\kappa$ is not the largest Suslin cardinal). So, $\bg \nsubseteq \Lambda$.
\mys


It suffices to show the following:

\begin{claim}
$\forall^\ww \bg \subseteq S(\kappa)$.
\end{claim}


{\color{blue}Proof}.
Let $A(x) \leftrightarrow \forall y\ B(x,y)$, where $B=p[T]$, $T$ a homogeneous tree on 
$\omega \times \omega \times \kappa$. 
\end{frame}

\begin{frame}
For $x \in A$ consider the game:
\mys

I plays out $y$, II plays out $\vec \alpha$, and II wins iff 
$(x,y, \vec \alpha) \in [T]$. 
\mys

This is a closed game for II, and II has a winning strategy by the homogeneity of $T$. 
\mys


So define $T'$ on $\omega \times \kappa$ by:

$(s, \vec \alpha)\in T'$ iff for all $u=u_m \in \omega^{<\omega}$ with $m< (s)$
we have \[(s \res \lh(u), u, (\alpha_{i_1}, \alpha_{i_2},\dots, \alpha_{i_\lh(u)}))\in T \]

where $i_k$ is such that $u_{i_k}= u\res k$.
\end{frame}



\begin{frame}
\begin{rem}
It follows that $\bg=\bp^\lambda_{-1}$. This gives another characterization of the Steel
pointclasses as the projections of $\kappa$-homogeneously Suslin trees. 
\end{rem}



\begin{rem}
One can also rule out the case $S(\kappa)=\bs^\lambda_{-1}$ by using Woodin's
proof of the closure of the Suslin cardinals which shows that there is a $\kappa$
prewellordering in $S(\kappa)$ (and so $S(\kappa)\nsubseteq \bs^\lambda_{-1}$).
\end{rem}
\end{frame}


\begin{frame}
So, $S(\kappa)=\exists^\ww \bg$, $\bg$ the Steel pointclass,
\mys

First we show $\sca(S(\kappa))$. Since $\cof(\kappa)>\omega$, $S(\kappa)=\bigcup_\kappa S(<\kappa)
=\bigcup_\kappa \Lambda$.
\mys


Let $A\in S(\kappa)$, $A=p[T]$, $T$ a tree on $\omega \times \kappa$. 
\mys


For $x \in A$, let $\varphi_0(x)=\mu \alpha\ (x \in p[T\res \alpha])$. 
Let \[\varphi_{n+1}(x)=\langle \varphi_0(x), \psi^{\varphi_0(x)}_0(x),
\dots, \psi^{\varphi_0(x)}_n(x) \rangle,\] 
where $\psi^\alpha$ is the scale from $T \res \alpha$. Each $\varphi_n \in \bigcup_\kappa \Lambda
=S(\kappa)$.
\end{frame}


\begin{frame}
We now show $\sca(\bg)$ where $\bg=\bp^\lambda_{-1}$ is the Steel pointclass.
\mys



Let $\rho \leq \kappa$ be least such that $\bigcup_\rho \bd \nsubseteq \bd$ (so actually $\rho=\cof(\kappa)$).
Recall $\bg=\bs^1_1$-bounded $\rho$-unions of $\bd$ sets.
\mys



Let $A \in \bg$. Let $U$ be universal $\bs^1_1$. Let $C=\{ y \colon U_y \subseteq A\}$. 
Then, as before, $ C\in \bg$. Let $C=p[T]$, $T$ a tree on $\omega \times \kappa$. 
Let $S$ be a tree on $(\omega)^3$ with $U(y,x)\leftrightarrow (x,y)\in p[S]$. 
\mys

Define the tree $V$ on $(\omega)^3 \times \kappa$ by:

\[
(s,t,u,\vec \alpha) \in S \leftrightarrow
(s,t,u)\in S \wedge (t,\vec \alpha)\in V.\]
\end{frame}


\begin{frame}
Clearly $A=p[V]$.
\mys





Identify $V$ with a tree $V'$ on $\omega \times \kappa$ by reverse lexicographic order on 
$\omega \times \omega \times \kappa$ (i.e., order by ordinal first). 
\mys


Modifying $V'$ slightly to $V''$ we may assume that for $(s,\vec \alpha)\in V''$ we have $\alpha_0> 
\max \{ \alpha_1,\dots,\alpha_k\}$.
\mys



Let $\{ \varphi_i\}$ be the scale corresponding to $V''$. 
\mys


Each $\varphi_i$ maps into $\kappa$. We show that $<^*_i$, $\leq^*_i$ are in $\bg$.
\end{frame}



\begin{frame}
For example, $<^*_i=\bigcup_{\alpha <\rho} B_\alpha$ where:

\[
(x ,y)\in B_\alpha \leftrightarrow 
\exists \beta \leq f(\alpha)\ [((x,y)\in A \wedge \varphi_i(x)=\beta) \wedge 
\neg (y \in A \wedge \varphi_i(y)\leq \beta) \]


where $f \colon \rho \to \kappa$ is cofinal.
\mys



To see this is a $\bs^1_1$-bounded union, let $E\subseteq <^*_i$ be $\bs^1_1$. 
Let $E'=\{ x \colon \exists y\ (x,y) \in E\}$, so $E' \in \bs^1_1$ and $E' \subseteq A$.
\mys

Let $E'=U_y$. Fix $\vec \alpha$ such that $(y,\vec \alpha)\in [T]$. 
This gives an ordinal $\gamma<\kappa$ such that $E' \subseteq p[V'' \res \gamma]$. 
\mys


This shows $E'$ is bounded in the $\{ B_\alpha\}$ union.
\end{frame}

\begin{frame}
We propagate scales from $\bs^\lambda_0$ by periodicity to 
$\bp^\lambda_1$, $\bs^\lambda_2$, $\bp^\lambda_3, \ldots$. 
\mys


As before, all the $\delta^\lambda_{2n+1}$ are regular Suslin cardinals,
and $S(\delta^\lambda_{2n+1})=\bs^\lambda_{2n+2}$. 
\mys


Also as before, $\delta^\lambda_{2n+1}=(\lambda_{2n+1})^+$ where $\cof(\lambda_{2n+1})=\omega$. 
Also, $S(\lambda_{2n+1})=\bs^\lambda_{2n+1}$. 

\end{frame}


\begin{frame}
Consider now limit Suslin $\kappa$ with $\Lambda=S(<\kappa)$ of type~4.
\mys



We previously showed $\lambda=o(\Lambda)=\kappa$. Recall $\lambda$ is regular in this case. 
Previous arguments also showed $\sca(\bg)$, where 
$\bg$ is at the base (so $\pwo(\bg)$, $\bg$ closed under quantifiers).
\mys

However, we cannot propagate scales by periodicity.
\mys


\begin{ques}
How do we get the next Suslin cardinal?
\end{ques}

We will present a method of {\color{red}Martin} for constructing the next Suslin cardinal.
\end{frame}




\begin{frame}
\frametitle{The Largest Suslin Cardinal}
Assume there is a largest Suslin cardinal $\kappa$. 
\mys


\begin{claim}
$\kappa$ is a regular limit Suslin cardinal. $\bg=S(\kappa$) and $\sca(\bg)$. Also, $S(\kappa)$
is closed under quantifiers.
\end{claim}


{\color{blue}Proof}.
$S(\kappa)$ is closed under $\forall^\ww$ as otherwise $\forall^\ww S(\kappa)$
would be a larger pointclass admitting scales (by periodicity). So,
$S(\kappa)$ is closed under quantifiers.
\mys


So, $\bd=\bg \cap \bgd$ is closed under quantifiers, $\wedge$, $\vee$. Let 
$\lambda=o(\bd)$. 
\end{frame}


\begin{frame}
We must have that $\lambda$ is regular as otherwise $\bg$ would not be closed under 
$\wedge^\omega$, $\vee^\omega$. 
\mys

Suppose $\lambda <\kappa$. There is a $\kappa$ increasing sequence $\{ A_\alpha\}_{\alpha <\kappa}$
of sets in $S(\kappa)$. Let $\bg=S(\kappa)$ or $\bg=\check{S(\kappa)}$, with $\pwo(\bg)$. 
This gives either an increasing or a decreasing $\kappa$-length sequence of sets in $\bg$,
contradicting Chuang's theorem. 
\mys

So, $\kappa=\lambda$. In particular, $\kappa$ is regular. Since $\cof(\kappa)>\omega$,
$S(\kappa)=\bigcup_\kappa \bd$. 
\mys


$\kappa$ must be a limit of Suslin cardinals as if there were a largest 
Suslin $\kappa'<\kappa$ then $S(\kappa)=\bigcup_\kappa S(\kappa')$, a contradiction as we can find
a $\bg_0 \supseteq S(\kappa')$ closed under wellordered unions. 

Previous arguments show $\sca(S(\kappa))$.
\end{frame}

\section{The Next Suslin Cardinal}

\begin{frame}
\frametitle{The Envelope}

Let $\bg$ be a pointclass, and $\kappa \in \on$. We define the $\bg, \kappa$ {\color{red}envelope}
as follows.
\mys


\begin{defn}[Martin]
Let $\sA=\{ A_\alpha\}_{\alpha<\kappa}$, each $A_\alpha \subseteq \ww$. 
Then $\overline{\sA}$ is the set of $A\subseteq \ww$ such that for all countable
$S\subseteq \ww$, there is an $\alpha <\kappa$ such that $S\cap A=S\cap A_\alpha$.

We let 
\[\Lambda(\bg,\kappa)=\{ \overline{\sA} \colon  \sA \subseteq \bg \wedge || \sA||\leq \kappa\}.\]
\end{defn}

\end{frame}


\begin{frame}
\begin{lem}
Let $\bg$ be nonselfdual, closed under $\forall^\ww$, and $\pwo(\bg)$ (if $\bd$ not closed under 
$\exists^\ww$ the assume $\sca(\exists^\ww \bg$ with norms into $\kappa$).
Then $\Lambda(\bd,\kappa)=\Lambda(\bg,\kappa)=\Lambda(\exists^\ww\bg,\kappa)$ where 
$\kappa=\delta(\bd)$.
\end{lem}



{\color{blue}Proof}.
There is an $\exists^\ww\bg$ prewellordering $(W,\varphi)$ of length $\kappa$ such that for all 
$\alpha<\kappa$, $\{ x \in W\colon \varphi(x)=\alpha\} \in \bd$. 
\mys

Let $\{ A_\alpha\}_{\alpha<\kappa}$ be given with $A_\alpha \in \exists^\ww \bg$. 


By coding lemma, $C=\{(x,y)\colon x \in W \wedge y \in A_{\varphi(x)} \} \in \exists^\ww \bg$. 
Recall $\exists^\ww \bg = \bigcup_\rho \bd$ where $\rho=\cof(\kappa)$. 
Write $C=\bigcup_{\beta<\rho}C_\beta$ with $C_\beta \in \bd$.
\mys


For $\alpha <\kappa$, $\beta <\rho$, let 

\[
y \in A_{\alpha,\beta} \leftrightarrow \exists x\ ( (x \in W \wedge \varphi(x)=\alpha)\wedge 
(x,y) \in C_\beta).
\]

This works if $\exists^\ww \bd =\bd$.

\end{frame}

\begin{frame}
Assume now $\bg=\forall^\ww \bd$ (sketch).
\mys

Let $\varphi_n$ be an $\exists^\ww$-scale on an $\exists^\ww \bg$ set $W$.
Let $C$ be as above using norm $\varphi_0$ (we may assume $\varphi_)$ is onto $\kappa$).
\mys

For $\alpha <\kappa$ let $\beta(\alpha)<\kappa$ be the least reliable ordinal $>\alpha$.
\mys


By Becker-Kechris there is uniformly in $\alpha<\kappa$ and $z \in \ww$ a closed game 
$G_\alpha(z)$ for II, where I plays ordinals $<\beta(\alpha)$, II plays ordinals $<\kappa$, 
such that II has a winning strategy iff $z \in A_\alpha$. 
\mys


Let $G_{\alpha,\gamma}(z)$ be the game where II's moves are restricted to ordinals $<\gamma$.
\mys


Let $A_{\alpha,\gamma,\delta}$ be the set of $z$ such that I doesn't win $G_{\alpha,\gamma}$
with rank less than $\delta$. Each $A_{\alpha,\gamma,\delta} \in \bd$ by the closure of $\bd$
under $<\kappa$ unions and intersections.


\end{frame}












\begin{frame}
The next lemma is a ``universality'' property.


\begin{lem}
Same hypotheses as previous lemma. then there is a single $\sA=\{ A_\alpha\}_{\alpha <\kappa}$
with each $A_\alpha \in \bd$ such that every set in $\Lambda(\bg,\kappa)$ is Wadge reducible
to a set in $\overline{\sA}$.
\end{lem}


{\color{blue}Proof}.
Let $(W,\varphi)$ again be an $\exists^\ww \bg$ pwo on an $\exists^\ww\bg$ complete set $W$.
Let $U_1, U_2$ be a universal $\bg$ sets for subsets of $\ww$ and $(\ww)^2$.
\mys


Define 
\[ (x,y)\in A_\alpha \leftrightarrow \exists z, u\ [(z \in W \wedge \varphi(z)=\alpha)\wedge 
U_2(x,z,u)\wedge U_1(u,y)].\]

Each $A_\alpha \in \bg$, Given any $\sA'=\{A'_{\alpha} \}_{\alpha<\kappa}$ with $A'_\alpha \in \bg$,
by the coding lemma there is an $x_0$ such that $A'_\alpha=(A_\alpha)_{x_0}$ for all $\alpha<\kappa$.
\end{frame}


\begin{frame}


Let $A' \in \overline{\sA'}$.
\mys


For each Turing degree $d$ let $\beta(d)<\kappa$ be least such that 
$A' \cap d= A'_{\beta(d)}\cap d= (A_{\beta(d)})_{x_0}$. 
\mys


Define $A \subseteq \ww \times \ww$ by

\[ (x,y)\in A\leftrightarrow \forall^* d\ [ (x,y)\in A_{\beta(d)}].\]
\mys


Then $A \in \overline \sA$ and $A'=A_{x_0}$ so $A' \leq_W A$.

\end{frame}



\begin{frame}
\begin{cor}
Under same hypotheses, $\Lambda(\bg,\kappa)$ is closed under $\wedge$, $\vee$, $\neg$.
\end{cor}


This follows immediately using $\Lambda(\bg,\kappa)=\Lambda(\bd,\kappa)$.
\mys


The next lemma says that if $\bg$ is closed under quantifiers then so is $\Lambda(\bg,\kappa)$.
\mys


\begin{lem}
Suppose $\bg$ is nonselfdual, closed under $\exists^\ww$, $\forall^\ww$, and $\pwo(\bg)$.
Let $\kappa=o(\bd)$. Then $\Lambda(\bg,\kappa)$ is closed under $\exists^\ww$, $\forall^\ww$.
\end{lem}


{\color{blue}Proof}.


Let $A\subseteq \ww \times \ww$ be in $\overline{\sA}$ with $\sA=\{A_\alpha\}_{\alpha<\kappa}$,
and each $A_\alpha \in \bd$.
Let $B(x)\leftrightarrow \exists y \ A(x,y)$.
\end{frame}


\begin{frame}



View every $z \in \ww$ as coding a countable set $s_z \subseteq \ww$ and a function 
$f_z \colon s_z \to \{ 0,1\}$. For every degree $d$ let $\alpha_z(d)<\kappa$ be least (if it exists)
such that \[\forall x \in s_z\ [ f_z(x)=1 \leftrightarrow \exists y \leq d \ 
(x,y)\in A_{\alpha_z(d)}].\]

Let $C=\{ z \colon \forall^* d\ (\alpha_z(d) \text{ is defined }) \}$. Then $C \in \bg$.
\mys

Let $\varphi$ be the norm on $C$ corresponding to the pwo 
\[ z_1 \preceq z_2 \leftrightarrow 
\forall^* d\ [\alpha_{z_1}(d) \leq \alpha_{z_2}(d)].\]


This is a $\bg$ pwo on $C$ and so has length $\kappa$. 
\mys


For $z \in C$ define $B_{\varphi(z)}$ by 

\[
z \in B_{\varphi(z)}\leftrightarrow \forall^* d\ [ \exists y \leq d\ ((x,y) \in A_{\alpha_z(d)})].\]
\end{frame}


\begin{frame}
\frametitle{Coding Measures}
Let $\bg$ and $\kappa=o(\bd)$ be as above.
Fix an $\exists^\ww \bg$ norm $(W,\varphi)$ of length $\kappa$ (with each $W_\alpha\in \bd$). 
\mys


Let $U$ be universal $\exists^\ww \bg$. For $z \in \ww$ let $B_z=\{ \alpha <\kappa 
\colon \exists x \in W\ (\varphi(x)=\alpha\}$. By the coding lemma, every 
subset of $\kappa$ is of the form $B_z$.
\mys



 Let $\mu$ be a measure on $\kappa$. The {\color{red}code set} is defined by
\[ C_\mu=\{ z \colon \mu(B_z)=1\}.\]
\end{frame}

\begin{frame}
\begin{lem}
Let $\bg$, $\kappa$ be as above. Then $A \in \Lambda(\bg,\kappa)$ iff 
there is a measure $\mu$ on $\kappa$ such that $A \leq_W C_\mu$.
\end{lem}


{\color{blue}Proof}.
($\Leftarrow$) Let $\mu$ be a measure on $\kappa$. Define $\{ A_{\alpha}\}_{\alpha <\kappa}$
by $z\in A_\alpha \leftrightarrow \alpha \in B_z$. Then $C_\mu \in \overline{\sA}$
(by countable additivity of the measure).
\mys



($\Rightarrow$) Let $A \in \overline{\sA}$, $\sA=\{ A_\alpha\}_{\alpha<\kappa}$. Define 
$f \colon \sD \to \alpha$ by $f(d)=\mu \alpha<\kappa \ (A \cap d= A_\alpha \cap d)$.
Let $\mu=f(\nu)$ where $\nu$ is the Martin measure on the degrees. 
\end{frame}


\begin{frame}
Let $R(x,y)\leftrightarrow ( x \in W \wedge y \in A_{\varphi(x)}$, so $R \in \exists^\ww \bg$
by the coding lemma.
\mys


By $s$-$m$-$n$ theorem there is a continuous function $h$ such that 
$U_{h(y)}=\{ x \in W \colon y \in A_{\varphi(x)}\}$. That is, 
$B_{h(y)}=\{ \alpha <\kappa \colon y \in A_\alpha\}$. 
\mys


Then 

\begin{align*}
y \in A & \leftrightarrow \forall^*_\nu d\ (y \in A_{f(d)} ) \\ & 
\leftrightarrow \forall^*_\mu \alpha \ (y \in A_\alpha) \\ & 
\leftrightarrow h(y) \in C_\mu.
\end{align*}
\end{frame}


\begin{frame}
\frametitle{Upper bound for the next (semi) scale}

\begin{thm}
Let $\bg$ be nonselfdual, closed under $\forall^\ww$, and $\pwo(\bg)$. Assume every 
$\exists^\ww \bg$ set admits a  $\exists^\ww \bg$ scale with norms into $\kappa=\delta(\bd)$. 
Assume also that there is a Suslin cardinal greater than $\kappa$. Then every set in 
$\forall^\ww \bgd$ admits a semi-scale with all norms in $\Lambda(\bg,\kappa)$.
\end{thm}


{\color{blue}Proof}.
In all cases, $\cof(\kappa)>\omega$. 
[This follows by usual boundedness arguments if $\bg$ is closed under $\vee$. $\bg$
cannot be at the base of a type~1 hierarchy. The remaining problematic case is where 
$\bg=\bg_0 \wedge \bgd_)$ where $\bg_0$ is closed under quantifiers. This case cannot occur 
as otherwise $\bgd_0$ would admit scales with a fixed projective class over 
$\bg_0$, contradicting the next result.]
\end{frame}



\begin{frame}
Let $T$ be a tree on $\omega \times \kappa$ with $A=p[T]$ a complete $\exists^\ww \bg$ set. 
Since $\cof(\kappa)>\omega$ we may assume that for all $(s,\vec \alpha)\in T$ that 
$\alpha_0 >\max \{ \alpha_1,\dots,\alpha_k\}$. 
\mys

$T$ is weakly homogeneous (there is a larger Suslin cardinal.
\mys


The homogeneous tree construction produces a tree $T'$ with $p[T']=B=\ww-A$, so $B \in \forall^\ww \bgd$.
\mys


For $x \in B$, let $f \colon T_x \to \kappa$ be the rank function (using KB order on $T$; view $T$
as a homogeneous tree on $\omega \times \omega \times \kappa$ with measures 
$\mu_{s,t}$). 
\end{frame}


\begin{frame}
Let \[\varphi_n(x)=[ f \res T_{x\res \lh(t_n),t_n}]_{\mu_{x \res \lh(t_n),t_n}}.\]

Then $\vec \varphi$ is a semiscale on $B$. We show the norms are in $\Lambda(\bg,\kappa)$.
\mys



We have (assume $x\res \lh(t_n)=y\res\lh(t_n)=s_n$) 

\[
x <^*_n y \leftrightarrow \forall_{\mu_{s_nt_n}} \vec \alpha\ 
[\exists \beta <\kappa \ ( |(s_n,\vec \alpha)|_{T_x} \leq \beta 
\wedge \neg  |(s_n,\vec \alpha)|_{T_y} \leq \beta) ].
\]

The relation 

\begin{align*}
R(\vec w,x,y)& \leftrightarrow (\vec w \in W)\wedge \exists \vec \alpha , \beta <\kappa\ 
[ (\varphi(\vec w)=\vec \alpha)\\ & \wedge ( |(s_n,\vec \alpha)|_{T_x} \leq \beta 
\wedge \neg  |(s_n,\vec \alpha)|_{T_y} \leq \beta) ].
\end{align*}

is in $\exists^\ww \bg$.  This shows $<^*_n \in \Lambda(\bg,\kappa)$.
\end{frame}


\begin{frame}

\begin{rem}
It is not clear if we can get a scale whose norms are in $\Lambda(\bg,\kappa)$.
The norms from the Suslin representation from the homogeneous tree construction
are of the form $\psi_n(x)=[ f' \res {x\res \lh(t_n),t_n}]_{\mu_{x \res \lh(t_n),t_n}}$
where $f'$ is the minimal ``almost everywhere  rank function.'' We don't know if these norms
are in $\Lambda(\bg,\kappa)$.
\mys

If this is true then we can fill the remaining gap in the final analysis.
\end{rem}
\end{frame}


\begin{frame}
\frametitle{Lower bound for next scale}


\begin{lem}
Let $\bg$ be nonselfdual, closed under $\forall^\ww$, and $\pwo(\bg)$. 
Let $A$ be $\forall^\ww \bgd$-complete. Then $A$ does not admit a scale all of whose norms
are Wadge reducible to some $B\in \Lambda(\bg,\kappa)$.
\end{lem}



{\color{blue}Proof}.
Fix $\sA=\{A_\alpha\}_{\alpha<\kappa}$ with $A_\alpha \in \bd$ which is universal for $\Lambda(\bg,\kappa)$.
\mys

Define $D \subseteq \ww \times \ww$ by:



\begin{align*}
D(x,y)& \leftrightarrow \exists r \leq_T x\ \forall^*_\nu d\ \exists \alpha <\kappa\ 
\forall n,m \in \omega\ \\ &
[ y(n)=m \leftrightarrow \exists z \leq_T d\ \forall w \leq_T d\ 
r(n,m,x,z(w),w)\in A_\alpha]. 
\end{align*}
\end{frame}


\begin{frame}
$D \in \forall^\ww \exists^\ww \bg$, and all sections of $D$ are countable.
\mys


Consider $D^c \in \exists^\ww \forall^\ww \bgd$. 
\mys

Since we are assuming every $\forall^\ww \bgd$ set admits a scale with norms reducible to $B$, $D^c$
has a uniformizing function $f$ of the form 

\[
f(x)(n)=m \leftrightarrow \Game \langle z,w \rangle \ r(n,m,x,z,w)\in B
\]

for some continuous $r \colon \omega^2 \times (\ww)^3 \to \ww$.
\end{frame}



\begin{frame}
Then 

\[ f(x)(n)=m  \leftrightarrow \forall^*_\nu d\ \exists z \leq_T d\ \forall w \leq_T d\ 
r(n,m, x, z(w),w) \in B.\]

Let $x \geq_T r$  and $y=f(x)$, so $\neg D(x,y)$. 
\mys

For $d \in \sD$, let $\alpha(d)<\kappa$  be such that $A_{\alpha(d)}\cap d=B\cap d$. 
\mys


Then 
\[ y(n)=m \leftrightarrow \forall^*_\nu d\ \exists z \leq_T d\ \forall w \leq_T d\ 
r(n,m, x, z(w),w) \in A_{\alpha(d)}\]

which shows $D(x,y)$, a contradiction.
\end{frame}


\begin{frame}
\begin{rem}
A semi-scale can be converted to a scale within the next projective class over 
a class containing the norms of a semiscale. So, if $\bg$ (and hence $\Lambda$)
is closed under quantifiers, then $\Lambda$ is the least class containing the norms
of a semiscale on a $\bgd$ complete set.
\end{rem}


\begin{ques}
In this case do we get a scale on $\bgd$ with norms in $\Lambda$?
\end{ques}

\begin{cor}
If $\bg$ is nonselfdual, closed under quantifiers, and $\sca(\bg)$, then 
$\lambda=o(\Lambda)$ is the next Suslin cardinal after $\kappa=o(\bd)$ 
(assuming $\kappa$ is not the largest Suslin cardinal).
\end{cor}
\end{frame}

\begin{frame}
\begin{lem}
Suppose  $\bg$ is nonselfdual, closed under quantifiers, and $\sca(\bg)$ 
(and  $\kappa=o(\bd)$ is not the largest Suslin cardinal). Then 
every set in $\Lambda$ is $\lambda$-Suslin.
\end{lem}


{\color{blue}Proof}.
Let $B\in \Lambda$ and $A \in \bg-\bgd$. Let $A=p[T]$ with $T$ a tree on $\omega \times \lambda$. 
Let $\{\psi_n\}$ be the scale on $A$ from $T$. Fix nonselfdual $\bg_0\subsetneq \bg_1 \subseteq \Lambda$
with $B, \ww-B \in \bg_0$ and $\pwo(\bg_1)$, $\exists^\ww \bg_1=\bg_1$. Fix $n$ so that $|\psi_n|>$ 
sup of the lengths of the $\bg_1$ prewellorderings. 
\mys

For $\alpha<|\psi_n|$, let $A_\alpha=\{ x \in A \colon \psi_n(x) \leq \alpha \}$. We cannot
have $A_\alpha \in \bg_0$ for all $\alpha$, so $B\leq_W A_\alpha$ for some $\alpha$. 
\mys


But $A_\alpha$ and hence $B$ is $\lambda$-Suslin.
\end{frame}


\subsection{More on $\Lambda$}

\begin{frame}
Assume $\bg=S(\kappa)$ is closed under quantifiers, $\kappa=o(\bd)$ is not the largest Suslin cardinal,
and $\Lambda=\Lambda(\bg,\kappa)$. Let $\lambda=o(\Lambda)$. So, $\cof(\lambda)=\omega$. 
\mys


Let $\bs_0=\bs^\lambda_0=\bigcup_\omega S(<\kappa)$, etc. Recall $\pwo(\bs_0)$, $\pwo(\bp_1)$, etc.
\mys


\begin{lem}
$S(\lambda)=\bs_1$.
\end{lem}


{\color{blue}Proof}.
$S(\lambda)$ is closed under $\wedge^\omega$, $\vee^\omega$, and $\exists^\ww$ so $\bs_1 \subseteq S(\lambda)$.
From the coding lemma, $S(\lambda)\subseteq \bs_1$.
\end{frame}


\begin{frame}
\begin{lem}
$\delta_1=\delta(\Delta_1)=\lambda^+$ and $\lambda^+$ is regular.
\end{lem}


{\color{blue}Proof}.
$\delta_1$ is the supremum of the lengths of the $\bs_1$ wellfounded relations, and so is regular.
Since $\bd_1=S(\lambda)$ we have $\delta_1=\lambda^+$ by Kunen-Martin and the coding lemma.
\mys




\begin{lem}
Let $B \in \Lambda$, $\rho <\lambda$, and $\sB=\{ B_\beta\}_{\beta <\rho}$ be such that 
$B_\beta \leq_W B$ for each $\beta$. Then $\overline{\sB}\subseteq \Lambda$.
\end{lem}
\end{frame}


\begin{frame}

{\color{blue}Proof}.


For $\rho_0,\rho_1<\lambda$ let $\Lambda_{\rho_0,\rho_1}$ be the union of all $\overline{\sB}$
where $\sB=\{ B_\beta\}_{\beta <\rho_0}$ and $o(B_\beta)<\rho_1$. 
\mys

If the lemma fails, we first show that some $\Lambda_{\rho_0,\rho_1}$ contains $\bs_2$.
\mys


Each $\Lambda_{\rho_0,\rho_1}$ is selfdual and is closed under $\exists^\omega$. So if lemma fails then 
$\Lambda_{\rho_0,\rho_1} \supseteq \bp_1$ for some $\rho_0, \rho_1$.
\mys


Now we run the previous argument for the closure of $\Lambda$ under quantifiers. Let 
$B \in \overline{\sB}$ where $B=\{ B_\beta\}_{\beta <\rho_0}$ and let $A=\exists^\ww B$.
\end{frame}



\begin{frame}

View every real $z$ as coding a countable set $s_z$ and a function $f_z \colon s_z \to \{ 0,1\}$.
Define
\[ z \in C \leftrightarrow 
\forall^* d\ \exists \alpha<\rho_0\ \forall x \in s_z\ ( f_z(x)=1 \leftrightarrow 
\exists y \leq_T d\ (x,y)\in B_\alpha).
\]

Easily $C\in \Lambda$. Let $\prec$ be the prewellordering on $C$ as before. So, $\prec \in \Lambda$
and so $\rho'_0=|\prec| <\lambda$. 
\mys


Define for $\alpha =|z|$ ($z \in C$) 
\[
x \in A'_\alpha\leftrightarrow \forall^* d\ \exists y \leq_T d\ (x,y)\in B_{\alpha_z(d)}.
\]
where $\alpha_z(d)$ is the least $\alpha<\rho_0$ satisfying the above.
Then $A \in \overline{ \{A'_\alpha\}}_{\alpha<\rho'_0} $.
\end{frame}



\begin{frame}
So, we may assume $\Lambda_{\rho_0,\rho_1}\supseteq \bs_2$. Fix $A\in \overline{\sA}$,
$\sA=\{ A_\alpha\}_{\alpha<\rho_0}$ with $A$ a complete $\bs_2$ set. 
\mys


Now we run the previous non-uniformization argument.
\mys

Define

\begin{align*}
D(x,y)& \leftrightarrow 
\exists \tau \leq_T x\ \forall^* d\ \exists \alpha <\rho_0\ 
\forall m,n \in \omega\ \\ & [y(m)=n \leftrightarrow 
\exists z \leq_T d\ \tau(x,z,m,n)\in A_\alpha].
\end{align*}


Easily $D \in \Lambda$ and so is $\lambda$-Suslin. So $D^c$ admits a $\bs_2$ uniformization,
and thus a uniformization $f$ reducible to $A$, say by $\tau$. For $x \geq_T \tau$ we have
$D(x,f(x))$, a contradiction.
\end{frame}


\begin{frame}
\begin{lem}
$\lambda$ is closed under ultrapowers.
\end{lem}

{\color{blue}Proof}.
Let $\mu$ be a measure on $\alpha<\lambda$ and $\beta<\lambda$. Fix a prewellordering $\prec$
of length $>\max\{\alpha,\beta\}$ and let $\bg_0\subseteq \Lambda$ be nonselfdual containing $\prec$
and closed under $\exists^\ww$, $\wedge$, $\vee$. Let $U$ be universal $\bg_0$.
\mys


Then $C_\mu \in \Lambda(\bg_0,\alpha)\subseteq \Lambda$ as witnessed by the sequence 
$A_\alpha=\{ x \colon \exists y\ |y|_\prec =\alpha \wedge U(x,y)\}$ as before,
\mys


From this and the coding lemma we easily have that $j_\mu(\beta)<\lambda$.
\end{frame}


\begin{frame}

\begin{lem}
$\delta_1=\lambda^+$ is closed under ultrapowers.
\end{lem}


{\color{blue}Proof}.
Follows immediately from previous lemma and $\cof(\lambda)=\omega$.
\mys

\begin{lem}
$\delta_1$ is a Suslin cardinal, $S(\delta_1)=\bs_2$, and $\sca(\bs_2)$.
\end{lem}


{\color{blue}Proof}.
Let $A=p[T]$ be a complete $\bs_1$ set, where $T$ is a tree on $\omega\times \lambda$. 
$T$ is weakly homogeneous, and the homogeneous tree construction gives a tree 
$T'$ with $\ww-A=p[T']$. and $T'$ is a tree on $\omega \times \sup_\mu j_\mu(\delta_1)$
where $mu$ ranges over measures on $\lambda$. So, $T'$ is on $\omega \times \delta_1$. 
\mys

This show $\delta_1$ is a Suslin cardinal and $\bs_2 \subseteq S(\delta_1)$. Since 
$\pwo(\bp_1)$, the coding lemma gives $S(\delta_1)\subseteq \bs_2$. 
\end{frame}


\begin{frame}


Since $\delta_1$ is regular and $\bs_2$ is closed under wellordered unions, previous arguments
give $\sca(\bs_2)$.
\mys


\begin{rem}
We can show that $\bd_1$ (and $\bs_1$, $\bp_1$) is closed under measure quantification by 
measures on $\lambda$. Using this can show that every $\bp_1$ set admits a semi-scale whose norms are $\bp_1$.
\end{rem}


\begin{ques}
Do we have $\sca(\bs_0)$? Do we have $\sca(\bp_1)$?
\end{ques}
\end{frame}


\begin{frame}

\begin{defn}
A tree $T$ on $\omega \times \kappa$ is {\color{red}strongly homogeneous}
if there are measures $\mu_s$ on $T_s$ such that:
\begin{itemize}
\item
$\vec \mu$ witnesses the homogeneity of $T$.
\item
There are $\mu_s$ measure one sets $A_s$ such that for all $x$ with $T_x$ wellfounded,
The ranking function $f$ on $T_x \res \vec A_s$ has minimal values for $[f_s]_{\mu_s}$
where $f_s$ is the function on $T_s$ induced by $f$. 
\end{itemize}
\end{defn}


\begin{fact}
If every $\kappa$-homogeneously Suslin set is $\kappa$-strongly homogeneously Suslin
(where $\kappa=o(\bd)$ in the type~4 case) then every set in $\Lambda$ admits a scale all of whose norms
are in $\Lambda$.
\end{fact}
\end{frame}


\begin{frame}

\begin{ques}
Is every $\kappa$-homogeneously Suslin set a $\kappa$-strongly homogeneously Suslin set?
\end{ques}


\end{frame}

\subsection{Other Characterizations of the Envelope}
\begin{frame}

Let $\bg$ be nonselfdual, closed under quantifiers, $\bg=S(\kappa)$ where $\kappa=o(\bd)$.
Let $A\in\bg-\bgd$ and let $A=p[T]$ where $T$ is a tree on $\omega\times \kappa$.
\mys



\begin{defn}[Steel]
$\env(\bg)$ is the set of $A \subseteq \ww$ such that for some $z_0 \in \ww$, for any countable set
of reals $z$ containing $z_0$ we have $A\cap z \in L(T,z)$.
\vspace{10pt}

$\env'(\bg)$ is the set of $A \subseteq \ww$ such that for some $z_0 \in \ww$, for any countable set
of reals $z$ containing $z_0$ we have $A\cap z$ is definable in $ L(T,z)$ from finitely
many ordinals, $T$,  $z$.
\end{defn}
\end{frame}

\begin{frame}

\begin{rem}
We can also consider the variations $\tenv$, $\tenv'$ which are defined like 
$\env$ and $\env'$ except we replace ``for all countable $z$ containing $z_0$''
with ``for all  $d \geq_T z_0$.''
\end{rem}

Clearly $\env \subseteq \tenv$ and $\env' \subseteq \tenv'$.
\mys



\begin{thm}
For $\bg$ as above, $\Lambda(\bg,\kappa)=\env(\bg)=\env'(\bg)=\tenv(\bg)$.
\end{thm}



{\color{blue}Proof}.
Clearly $\env' \subseteq \env \subseteq \tenv$ and $\tenv' \subseteq \tenv$.
\end{frame}


\begin{frame}
We first show that $\Lambda(\bg,\kappa)\subseteq \env'(\bg)$.
\mys

Let $A \in \overline{\sA}$ where $\sA=\{ A_\alpha\}_{\alpha<\kappa}$, each $A_\alpha \in \bg$. 
\mys


Let $W=p[T] \in \bg-\bgd$ and $\varphi=\varphi_0$ a $\bg$-norm on $W$.
\mys


From the coding lemma, the relation

\[
R(x,y)\leftrightarrow (x \in W)\wedge (y \in A_{\varphi(x)})
\]

is in $\bg$. Let $z_0$ be a $\bg$ code for $R$. Let $z$ be a countable set containing $z_0$.
We claim that $A\cap z$ is ordinal definable in $L(T,z)$ from $T$, $z$ and $z_0$. 
\end{frame}

\begin{frame}
Let $\alpha <\kappa$ be such that $A\cap z=A_\alpha \cap z$. It suffices to show that 
$A_\alpha \cap z$ is definable in $L(T,z)$ from $T$, $z$, $\alpha$, and $z_0$. 
\mys
l

This follows as in Becker-Kechris. For $y \in z$, consider the game $G_{\alpha,T,z}(y)$:
\mys


I \qquad $x'$, $\vec \beta$
\mys

II \qquad $x$, $\vec \gamma$, $\vec \delta$
\mys


Rules: $(x'\res n, \vec \beta\res n)\in T$, $(x \res n, \vec \gamma\res n)\in T$, 
$\beta_0=\gamma_0=\alpha$, $(z_0\res n, x\res n, y\res n, \vec \delta \res n)\in T$
(we assume $W$ is universal $\bg$ set). 
\mys

In addition, II must make the additional moves as in Becker-Kechris to prove that 
$\varphi(x')\leq \varphi(x)$.
\end{frame}

\begin{frame}
These games are (uniformly) closed for II and are in $L(T,z)$ (for $z$ containing $z_0$).
\mys

Winning the game is absolute to $L(T,z)$.
\mys

\begin{align*}
y \in A_\alpha & \leftrightarrow \exists x\ [(x \in W \wedge \varphi(x)=\alpha) \wedge W(z_0,x,y)]
\\ &
\leftrightarrow \text{II has a winning strategy in } G_{\alpha,T,z}.
\end{align*}

So, $z \cap A=z \cap A_\alpha$ is definable from $\alpha$ and $z_9$ in $L(T,z)$. 

\end{frame}



\begin{frame}
We next show that $\tenv \subseteq \tenv'$. Let $A \in \tenv$. So, $\forall^* d\ (A\cap d \in 
L(T,d))$.
\mys



Consider the following game:
\mys



I \qquad x
\mys

II \qquad y, z
\mys

II wins iff $y\geq_T x$, $z \leq_T y$ and the following holds: 
\mys


Let $d=\{ y' \colon y'\leq_T y\}$. 
Let $\alpha$, $\varphi$ be the least ordinal and formula such that 
$A \cap d$ is definable in $L(T, d)$ by the formula $\varphi$ using 
parameters $\alpha$, $T$, $d$, and some $w \in d$. Then $A \cap d$ is 
definable in $L(T,d)$ using $\varphi$ and the parameters $\alpha$, $T$, $d$, and $z$. 
\end{frame}



\begin{frame}
If I had a winning strategy $\sigma$, let $y \geq_T \sigma$ be such that $A \cap d 
\in L(T,d)$ ($d$ as above). Let $\alpha$,$\varphi$ be the least ordinal and formula such that 
$A \cap d$ is definable in $L(T,d)$ from $\alpha$, $T$, $d$, and some $w \in d$. 
II can then play $y$, $w$ to defeat $\sigma$.
\mys


Fix a strategy $\tau$ for II. Then for any $d \geq _T \tau$, $A\cap d$ is definable in 
$L(T,d)$ from an ordinal, $T$, $d$, and $\tau$. 
\mys



Namely, let $\alpha$, $\varphi$ be least such that $A \cap d$ is definable in $L(S,d)$
from $\alpha$, $S$, $d$ and some $w \in d$ by $\varphi$. Then 
$ u \in A \cap d$ iff $L(S,d)\models \exists x \in d \ \exists v\ (\varphi(\alpha,S,d,\tau(x)_1)
\wedge u \in v)$.
\end{frame}



\begin{frame}
Finally, we show that $\tenv'(\bg)\subseteq \Lambda (\bg,\kappa)$.
\mys


As before, view every $z$ as coding a countable set $s_z$ and an $f_z \colon z \to \{0,1\}$.
\mys


For degree $d$, let $\alpha_z(d)=\langle \alpha,\beta,\gamma,\varphi\rangle <\kappa$ be least, 
if it exists, such that 
$\beta,\gamma <\alpha$ and for all $u \in s_z$ we have
\[
f_z(u)=1 \
\leftrightarrow L_\alpha(T\cap\beta, d)\models \varphi(\gamma, T\cap \beta, d,u)
\]

Let $C=\{ z \colon \forall^* z\ \alpha_z(d) \text{ is defined}\}$. 
(we suppress the fixed real parameter). As before, $C\in \bg$ and the natural norm $\psi$
on $C$ has length $\kappa$. 
\end{frame}



\begin{frame}
For $z \in C$ with $\psi(z)=\alpha$ define $A_\alpha$ by:




\begin{align*}
x \in A_\alpha & \leftrightarrow 
\forall^* d\ \exists \alpha, \beta,\gamma,\varphi<\kappa \ 
[\alpha_z(d)=\langle \alpha,\beta,\gamma,\varphi\rangle \\ & \wedge 
L_\alpha(T\cap \beta,d)\models \varphi(\gamma,T\cap\beta,d,x)].
\end{align*}


Then $A \in \overline{ \{A_\alpha\}}_{\alpha<\kappa}$.













\end{frame}












































\end{document}
