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\newcommand{\dom}{\operatorname{dom}}
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\newcommand{\cof}{\operatorname{cof}}
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\newcommand{\card}{\text{CARD}}
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\newcommand{\cub}{\text{c.u.b.}}
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\newcommand{\ns}{\text{NS}}
\newcommand{\fin}{\text{FIN}}
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\newcommand{\lub}{\text{l.u.b.}}
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\begin{document}
\begin{center}
\bf The c.u.b.\ Filter and Silver's Theorem
\end{center}
\bigskip

\section{Ideals and Filters}



We first recall the standard notions of ideal and filter.


\begin{defn}
An \emph{ideal} on a set $X$ is a collection $\sI \subseteq \sP(X)$
of subsets of $X$ satisfying:
\begin{enumerate}
\item
If $A \in \sI$ and $B \subseteq A$, then $B \in \sI$.
\item
If $A,B \in \sI$, then $A \cup B \in \sI$. 
\end{enumerate}
We say the ideal $\sI$ is proper if $X \notin \sI$ (equivalently
$\sI \neq \sP(X)$). 
\end{defn}

We think of an ideal as a notion of smallness for the subsets of $X$;
those subsets of $X$ which are in $\sI$ are the small ones. 

The ``dual'' notion is the concept of a filter:


\begin{defn}
A \emph{filter} on a set $X$ is a collection $\sF \subseteq \sP(X)$
of subsets of $X$ satisfying:
\begin{enumerate}
\item
If $A \in \sF$ and $B \supseteq A$, then $B \in \sF$.
\item
If $A,B \in \sF$, then $A \cap B \in \sF$. 
\end{enumerate}
We say the filter $\sF$ is proper if $\emptyset \notin \sF$ (equivalently
$\sF \neq \sP(X)$). 
\end{defn}


Recall than an \emph{ultrafilter} on a set $X$ is a maximal filter. 
Equivalently, an ultrafilter is a filter $\sF$ with the
property that for every $A \in \sP(X)$, either 
$A \in \sF$ or $X-A \in \sF$. It is a standard fact that from the
axiom of choice one may extend any filter on a set $X$ to an 
ultrafilter. 



\begin{exer}
Show that $\sI \subseteq \sP(X)$ is an ideal iff 
$\sF=\{ A \colon X-A \in \sI \}$ is a filter. 
\end{exer}

We say an ideal $\sI$ is $\kappa$-additive if whenever 
$\alpha < \kappa$ and $\{ A_\beta\}_{\beta < \alpha}$ is an 
$\alpha$ sequence of members of $\sI$, then $\bigcup_{\beta< \alpha}
A_\beta \in \sI$. The dual notion would be: a filter $\sF$ 
is $\kappa$-additive if whenever 
$\alpha < \kappa$ and $\{ A_\beta\}_{\beta < \alpha}$ is an 
$\alpha$ sequence of members of $\sF$, then $\bigcap_{\beta< \alpha}
A_\beta \in \sI$. Note that $\kappa$-additive refers to closure under 
\emph{less than } $\kappa$ unions (or intersections). 


The notions of ideal and filter are thus interchangeable, and we
will pass back and forth between the two. For $\sI$ an ideal 
(or $\sF$ a filter), we sometimes call the sets $A \in \sI$ 
(or sets $A$ such that $X-A \in \sF$) ``measure zero.'' 
We call the $A$ such that $X-A \in \sI$ (or $A \in \sF$) 
``measure one.'' If neither $A \in \sI$ nor $X-A \in \sI$, we 
say $A$ is ``positive.'' 





\begin{exer}
Show that for any ideal (or filter) there is a largest $\lambda 
\in \card$ such that $\sI$ is $\lambda$-additive. We call
this the \emph{additivity} of the ideal (or filter).
\end{exer}



\begin{exer}
Let $\kappa$ be a cardinal and let $\sI$ be the ideal of 
subsets of $\kappa$ which have size $< \kappa$. Identify the 
additivity of this ideal.
\end{exer}

If $\sI$ is an ideal (or filter) on a set $X$, an \emph{antichain} is a 
collection $\{ A_\alpha \}$ of $\sI$-positive subsets of 
$X$ such that $A_\alpha \cap A_\beta \in \sI$ for all $\alpha \neq \beta$. 
We say the ideal is $\lambda$-saturated if all anti-chains have size
$< \lambda$. The saturation of the ideal, $\sat(\sI)$ is the 
largest $\lambda$ such that $\sI$ is $\lambda$-saturated (which
is easily well-defined). 



If $\sI$ is an ideal (or $\sF$ a filter) on a set $X$, and $S \subseteq 
X$ is positive, then define the notion of the ideal (or filter) 
restricted to $S$, which we denote by $\sI_{| S}$ (or $\sF_{| S}$),
and defined by $\sI_{| S}= (\sI \cap \sP(S)) \cup (X-S)$ (that is we
declare complement of $S$ to be in the restricted ideal, i.e, the 
restricted ideal ``lives'' on $S$). Equivalently, 
$\sF_{| S}= \{ A \cap S \colon A \in \sF\}$. 




\section{Boolean Algebras}




\begin{defn}
A Boolean algebra is a set $\sB$ with two distinguished 
elements $0$ and $1$ and two binary operations
$+$, $\cdot$, and one unary operations $A \mapsto \mbar{A}$. 
The axioms are:

(commutative laws) $a+b=b+a$, $a\cdot b = b \cdot a$.

(associative laws) $a+(b+c)=(a+b)+c$, 
$a \cdot (b \cdot c)=(a \cdot b) \cdot c$.

(distributive laws) $a \cdot(b+c)=a \cdot b + a \cdot c$,
$a+(b \cdot c)= (a+b) \cdot (a+c)$. 

(identity laws) $a+a=a$, $a \cdot a=a$. 

(de Morgan's laws) $\mbar{a+b}=\mbar{a} \cdot \mbar{b}$,
$\mbar{a \cdot b}= \mbar{a} + \mbar{b}$. 

(negation laws) $a+\mbar{a}=1$, $a \cdot \mbar{a}=0$. 

($0$, $1$ laws) $0+a=a$, $0 \cdot a=0$, $1+a=1$, $1 \cdot a=a$. 
\end{defn}


In analogy with set operations, we sometimes write $\vee$ for $+$
and $\wedge$ for $\cdot$ in a Boolean algebra. We also sometimes 
write $a^c$ for $\mbar{a}$. 
The axioms imply all of the usual set identities. 


\begin{exer}
Show that in any Boolean algebra $a= \mbar{\mbar{a}}$. 
Show that $a+a \cdot b=a$ and $a \cdot(a+b)=a$. 
Show that $a \cdot b= a $ iff $a+ b = b$ iff $a \cdot (\mbar{b})=0$.
\end{exer}

We write $a \leq b$ in a Boolean algebra to denote $a \cdot b =a$,
or equivalently, $a + b=b$. 
We also write $a-b$ for $a \cdot(\mbar{b})$. 
We have $a \leq b $ iff $\mbar{b} \leq 
\mbar{a}$. 


The concepts of ideal, filter, ultrafilter generalize naturally
from $\sP(X)$ to any Boolean algebra. 

\begin{defn}
An ideal on the Boolean algebra $\sB$ is a collection 
$\sI \subseteq \sB$ satisfying:
\begin{enumerate}
\item
If $a \in \sI$, and $b \leq a$ then $b \in sI$. 
\item
If $a$, $b$ are in $\sI$, then $a+b \in \sI$.
\end{enumerate}
The ideal $\sI$ is proper if $1 \notin \sI$. 

A filter on the Boolean algebra $\sB$ is a collection 
$\sF \subseteq \sB$ satisfying:
\begin{enumerate}
\item
If $a \in \sF$, and $a \leq b$ then $b \in sF$. 
\item
If $a$, $b$ are in $\sF$, then $a\cdot b \in \sF$.
\end{enumerate}
The filter $\sF$ is proper if $0 \notin \sF$.
An ultrafilter on $\sB$ is a maximal filter. 
\end{defn}


It is straightforward to check that a filter $\sF$ on a Boolean 
algebra $\sB$ is an ultrafilter iff for ever $a \in \sB$
either $a \in \sF$ or $\mbar{a} \in \sF$. With $\ac$, every filter
on a Boolean algebra can be extended to an ultrafilter (the proof
is the same as that for filters on $\sP(X)$. 




If $X$ is any set, then all $\sB \subseteq \sP(X)$ which 
contains $\emptyset$, $X$, and is closed under finite unions, 
finite intersections, and complements is a Boolean algebra
under the operations of union, intersection, and complement. We 
call such a $\sB$ an algebra of subsets of $X$. 
Conversely, Stone's theorem says 
any boolean algebra is isomorphic to an algebra of subsets of some 
set $X$: 


\begin{thm} ($\zfc$)
Every Boolean algebra is isomorphic to an algebra of subsets of
some set $X$.
\end{thm}



\begin{proof}
Let $\sB$ be a Boolean algebra. Let $X=\{ u \colon u$ is an ultrafilter
on $\sB \}$. 
Define $\pi \colon \sB \to \sP(X)$ by 
$\pi(a)= \{ u \in X \colon a \in u \}$. Let $S=\ran(\pi)$. 
We claim that $\pi$ is a Boolean algebra isomorphism between 
$\sB$ and the algebra of sets $S$. Clearly $\pi(0)=\emptyset$
and $\pi(1)=X$. 
That $S$ is a field of sets
and $\pi$ is a homomorphism (i.e., preserves the Boolean operations)
follows from the equations: $\pi(a \cdot b)= \pi(a) \cap \pi(b)$,
$\pi(a+b)=\pi(a) \cup \pi(b)$, $\pi(\mbar{a})=X-\pi(a)$. 
For example, the first equation says $a \cdot b$ is in an ultrafilter
iff $a$ and $b$ are. It is immediate from the definition that this
in fact holds for all filters. For the second equation, note that 
$\pi(a), \pi(b) \subseteq \pi(a+b)$ as $a, b \leq a+b$. 
If $u \in \pi(a+b)$ but $u \notin \pi(a)$ and $u \notin \pi(b)$, then 
since $u$ is an ultrafilter, $u \in \pi(\mbar{a})$ and 
$u \in \pi(\mbar{b})$. Since $u$ is a filter, 
$u \in \pi(\mbar{a} \cdot \mbar{b})$, and so 
$u \in \pi((\mbar{a} \cdot \mbar{b} ) \cdot (a+b))= \pi(0)=\emptyset$.
The third equation follows from the fact that any ultrafilter 
must contain either $a$ or $\mbar{a}$. It remains to show that 
$\pi$ is one-to-one. Suppose $a \neq b$. Without loss of 
generality $a \nleq b$ (since if $a \leq b$ and $b \leq a$ then 
$a=a\cdot b=b$). So, $a-b \neq 0$. Let $u$ be an ultrafilter on 
$\sB$ with $a-b \in u$. Then $u \in \pi(a)$ but $u \notin \pi(b)$. 
\end{proof}







\begin{defn}
A Boolean algebra $\sB$ is \emph{complete} is for every 
$A \subseteq \sB$, a least upper bound, $\lub(A)$ for 
the elements of $A$ under $\leq$ exists. Equivalently, 
for every $A$ the greatest lower bound $\glb(A)$ exists. We 
also write $\Sigma(A)$, $\sup(A)$ for $\lub(A)$ and 
$\Pi(A)$, $\inf(A)$ for $\glb(A)$. A Boolean algebra is 
said to be $\kappa$-complete if $\Sigma(A)$, $\Pi(A)$
exists for all $A$ of size $< \kappa$. 
\end{defn}



For example, $\sP(X)$ is a complete Boolean algebra. 
On the other hand, $\sP(\omega)/\fin$ is not complete, where
$\fin$ denotes the ideal of finite subsets of $\omega$. 
To see this, let $A_n$ be disjoint infinite subsets of $\omega$
whose union is $\omega$. Then $\{ [A_n] \}_{n \in \omega}$ does
not have a least upper bound. 


The notions of $\kappa$-additive and $\kappa$-saturated generalize
from ideals on a set $X$ (i.e., the Boolean algebra $\sP(X)$)
to arbitrary Boolean algebras:


\begin{defn}
An ideal $\sI$ on a $\kappa$-complete
Boolean algebra $\sB$ is said to be $\kappa$-additive if 
$\Sigma(A) \in \sI$ whenever $A \subseteq \sI$ and $|A| < \kappa$. 
A Boolean algebra is $\kappa$-saturated if every antichain in 
$\sB$ has size $< \kappa$. $\sat(\sB)$ is the largest $\kappa$
such that $\sB$ is $\kappa$-saturated. 
\end{defn}

Thus, an ideall $\sI$ on $\kappa$ is $\lambda$-saturated iff 
the Boolean algebra $\sP(\kappa)/ \sI$ is $\lambda$-saturated. 

We will be mainly interested in complete Boolean algebras. 
For complete Boolean algebras it is a theorem that 
$\sat(\sB)$ is a regular cardinal. 



\section{The \cub\ Filter}




We introduce now a specific filter of basic importance called the
\cub\ filter (the corresponding ideal is called the 
non-stationary ideal). 


\begin{defn}
If $A \subseteq \on$, then the \emph{closure} of $A$, $\mbar{A}$, 
is the set of 
all $\alpha \in \on$ such that $\forall \beta < \alpha\ 
\exists \gamma \ (\beta < \gamma \leq \alpha \wedge \gamma \in A)$. 
We cay $A$ is \emph{closed} if $A= \mbar{A}$. 
\end{defn}

It is easy to see that $\mbar{A}$ consists of $A$ together with 
the ordinals $\alpha$ which are limit points of $A$, that is, 
$\alpha$ is a limit ordinal and $A$ is unbounded in $\alpha$. 
Thus, $A$ is closed iff it contains all its limit points. 

The topological terminology is justified. We can put a 
topology on the ordinals (order topology) by defining the basic open
sets to be of the form $(\alpha, \beta)= \{ \gamma \colon 
\alpha < \gamma < \beta\}$ (together with $\{ 0\}$ since $0$
is the least element in the ordering of ordinals). It is easily
checked that is a base for a topology (in fact, this is true for 
any linear ordering on any set). A neighborhood base at $\alpha$
consists of sets of the form $(\beta, \alpha]$, where $\beta < \alpha$. 
In this topology, the closure operation defined
above is just topological closure in the order topology. 



\begin{defn}
Let $\alpha$ be a limit ordinal. We say $C \subseteq \alpha$ is 
\cub\ is $C$ is closed and unbounded in $\alpha$. 
If $\cof(\alpha)> \omega$, then the \cub\ filter on $\alpha$,
$\Cub(\alpha)$ is defined to be the collection of subsets of $\alpha$
which contain a \cub\ set. The corresponding ideal is denoted $\ns(\alpha)$;
the ideal of non-stationary subsets of $\alpha$ (terminology 
explained below). 
\end{defn}



\begin{exer}
Show that if $\cof(\alpha)=\omega$ then $\Cub(\alpha)$ is not 
a filter.
\end{exer}



We let $\sP(\kappa)/\ns$ denote the set of equivalence classes
$[A]$, for $A \in \sP(\kappa)$, under the equivalence relation 
$A \sim B$ iff $A \triangle B \in \ns(\kappa)$. In fact, for 
any ideal $\sI$ on $\kappa$ we may consider 
$\sP(\kappa)/\sI$. This forms a Boolean algebra. 




\begin{lem}
Suppose $\cof(\alpha) > \omega$. Then $\Cub(\alpha)$
is a filter, and is $\cof(\alpha)$-additive. In fact, the
intersection of $< \cof(\alpha)$ many \cub\ subsets of $\alpha$ is
\cub. 
\end{lem}


\begin{proof}
By definition if $A \in \Cub(\alpha)$ and $B \supseteq A$ then 
$B \in \Cub(\alpha)$. Suppose $\delta < \cof(\alpha)$ 
and $\{ A_ \beta \}_{\beta < \delta}$ is a sequence of 
sets in $\Cub(\alpha)$. Using $\ac$, we may assume that all of the
$A_\beta$ are actually \cub\ subsets of $\alpha$, and 
show their intersection is \cub. 
Clearly $\bigcap_{\beta< \delta} A_\beta$ is closed. We must show it
is unbounded in $\alpha$. For each $\beta < \delta$, let 
$f_\beta \colon \alpha \to \alpha$ be given by $f_\beta(\gamma)=
$ the least element of $A_\beta$ which is $> \gamma$. 
Fix $\eta < \alpha$. Let $\eta_0=\eta$, and let 
$\eta_{n+1}= \sup_{\beta< \delta} f_\beta(\eta_n)$. Note that 
$\eta_{n+1} < \alpha$ as $\cof(\alpha)> \delta$. Since also 
$\cof(\alpha)> \omega$, $\eta_\omega= \sup_n(\eta_n) < \alpha$. 
Each $A_\beta$ is unbounded in $\eta_\omega$ (as there is a point of 
$A_\beta$ between $\eta_n$ and $\eta_{n+1}$ for any $n$), and thus 
$\eta_\omega \in A_\beta$ for all $\beta< \delta$. 
\end{proof}




In discussing the \cub\ filter $\Cub(\alpha)$, there is actually no loss
of generality is assuming $\alpha$ is a regular cardinal. 
For assume $\cof(\alpha)=\kappa$. Let $\{ \gamma_\eta \}_{\eta < \kappa}$
be a continuous, increasing, cofinal sequence in $\alpha$. 
By continuous we mean that for $\eta$ limit that 
$\gamma_\eta=\sup_{\eta' < \eta} \gamma_{\eta'}$. Let $C=\{ \gamma_\eta 
\colon \eta < \kappa\}$. Then $C$ is \cub\ in $\alpha$. 
The map $A \mapsto A'=\{ \gamma_\eta \colon \eta \in A\}$ is a bijection
between $\sP(\kappa)$ and $\sP(C)$ which preserves the notion of 
\cub\ since the $\gamma_\eta$ are continuous. Thus we have an isomorphism 
between $\sP(\kappa)/\Cub$ and $\sP(C)/\Cub$. Finally, the map 
$A \mapsto A \cap C$ is a Boolean algebra isomprphism between 
$\sP(\alpha)/\Cub$ and $\sP(C)/\Cub$ (the map is one-to-one since 
$C$ is \cub). Thus, $\sP(\kappa)/\Cub \cong \sP(\alpha)/\Cub$ as 
Boolean algebras. Thus, as far as discussions concerning the \cub\ 
filter are concerned, we may replace $\alpha$ by the set $C$ of 
size $\kappa=\cof(\alpha)$. 



\begin{defn}
Let $\kappa$ be a cardinal and $A_\alpha \subseteq \kappa$
for $\alpha < \kappa$. The \emph{diagonal intersection} of the 
$A_\alpha$ is defined by $\triangledown A_\alpha= 
\{ \beta < \kappa \colon \forall \alpha < \beta\ (\beta \in A_\alpha \}$. 
The \emph{diagonal union} is defined by 
$\triangle A_\alpha = \{ \beta < \kappa \colon \exists \alpha < \beta\ 
(\beta \in A_\alpha)\}$.
\end{defn}





\begin{defn}
A filter $\sF$ (or ideal $\sI$) is said to be \emph{normal}
if whenever $A_\alpha$, $\alpha< \kappa$, are in $\sF$ (or $\sI$), then 
$\triangledown A_\alpha \in \sF$ (resp.\ $\triangle A_\alpha \in \sI$).
\end{defn}


\begin{lem} \label{normal}
For every regular cardinal $\kappa$, the filter $\Cub(\kappa)$
is normal. 
\end{lem}



\begin{proof}
Assume $A_\alpha \in \cub(\kappa)$ for all $\alpha < \kappa$. Using 
$\ac$, let $C_\alpha \subseteq A_\alpha$ be \cub. It suffices to show that 
$\triangledown C_\alpha$ is \cub\ in $\kappa$. The 
diagonal intersection is easily closed, we show it is also unbounded.
For each $\alpha < \kappa$, let $f_\alpha \colon \kappa \to \kappa$ 
be given by $f_\alpha(\eta)=$ least element of $C_\alpha$ greater than 
$\eta$. Let $\eta_0< \kappa$. Define $\eta_{n+1}= 
\sup_{\alpha < \eta_n} f_\alpha(\eta_n)$. Let $\eta_\omega=
\sup_n \eta_n$. Note that if $\alpha < \eta_\omega$, then for 
all $n$ such that $\eta_n > \alpha$, there is a point of $C_\alpha$
between $\eta_n$ and $\eta_{n+1}$, and hence $\eta_\omega \in C_\alpha$. 
Thus, $\eta_\omega \in \triangledown A_\alpha$. 
\end{proof}




An immediate but important consequence of this lemma is Fodor's 
theorem. To state it, we introduce the important notion of 
stationarity. 


\begin{defn}
Let $\kappa$ be a regular cardinal. Then $S \subseteq \kappa$ is 
\emph{stationary} if $S \cap C \neq \emptyset$ for every \cub\ 
$C \subseteq \kappa$. 
\end{defn}


Note that $S$ being stationary is just saying that $S$ is positive
with respect to the $\Cub$ filter on $\kappa$. That is, $S$ is not
in the corresponding ideal (which is why we called this
ideal the non-statioary ideal).  


\begin{thm} (Fodor's Theorem) \label{fodor}
Let $\kappa$ be a regular cardinal, $S \subseteq \kappa$ 
be stationary, and $f \colon S \to \kappa$
be \emph{pressing down}, that is, $f(\alpha) < \alpha$ for all 
$\alpha  \in S$. Then there is a stationary set $S' \subseteq S$ on which $f$ 
is constant.
\end{thm}


\begin{proof}
If not, then for all $\alpha  \in S$ there is a set $A_\alpha \in 
\Cub(\kappa)$ such that $f(\beta) \neq \alpha$ for all 
$\beta \in A_\alpha \cap S$. From lemma~\ref{normal}, 
$\triangledown A_\alpha \in \Cub(\kappa)$ (for $\alpha \notin S$ we may
take $A_\alpha=\kappa$), and thus there is some 
$\beta \in (\triangledown A_\alpha) \cap S$ as $S$ is stationary. 
Then $f(\alpha)<\alpha$ and so $\alpha \in A_{f(\alpha)} \cap S$, 
a contradiction to the definition of $A_{f(\alpha)}$. 
\end{proof}


\begin{exer}
Let $\kappa$ be regular and $f_\alpha \colon \kappa \to \kappa$ for all 
$\alpha < \kappa$. Show that $C=\{ \beta< \kappa \colon 
\forall \alpha < \beta\ (\beta$ is closed under $f_\alpha) \}$ is 
\cub\ in $\kappa$. 
\end{exer}





If $\lambda < \kappa$ are regular cardinals, then $S^\kappa_\lambda=
\{ \alpha < \kappa \colon \cof(\alpha)=\lambda \}$ is stationary in 
$\kappa$. For example, for $\kappa=\aleph_2$ this gives two disjoint
stationary subsets of $\aleph_2$, namely $S_{\omega}$ and $S_{\omega_1}$.
We will show now more generally that any stationary subset 
$S \subseteq \kappa$ of a regular cardinal $\kappa$ can 
be split into $\kappa$ many disjoint stationary subsets. For 
successor $\kappa$ this is due to Ulam, and for limit $\kappa$
to Solovay. 

We consider first the successor case and prove a slightly
more general result. 


\begin{thm} (Ulam) \label{ulam}
Let $\kappa$ be a successor cardinal and $\sI$ a $\kappa$-additive 
ideal on $\kappa$ containing all the singletons. Then there is a $\kappa$
size family of pairwise disjoint $\sI$-positive subsets of $\kappa$.
\end{thm}



\begin{proof}
Let $\kappa=\lambda^+$. For each $\rho < \kappa$ let $f_\rho
\colon \lambda \to \kappa$ be a bijection. For each $\alpha < \lambda$
and $\beta < \kappa$ let $X^\alpha_\beta= 
\{ \rho > \beta \colon f_\rho(\alpha)=\beta \}$. 
For each $\beta < \kappa$ there is an $\alpha(\beta)< \lambda$
such that $X^{\alpha(\beta)}_{\beta} \notin \sI$ since $\sI$ is 
$\kappa$-additive and $\bigcup_{\alpha< \lambda} X^\alpha_\beta
=\kappa- (\beta+1)$, which is not in $\sI$. For some $\alpha_0 < \lambda$
we must have $|\{ \beta \colon \alpha(\beta)=\alpha_0 \}|=\kappa$. 
If $S=\{ \beta \colon \alpha(\beta)=\alpha_0 \}$, then for 
$\beta_1 \neq \beta_2 \in S$ we have $X^{\alpha_0}_{\beta_1} \cap 
X^{\alpha_0}_{\beta_2}=\emptyset$. 
\end{proof}



\begin{cor} \label{corulam}
If $\kappa$ is a successor cardinal and $S \subseteq \kappa$
is stationary, then $S$ can be split into $\kappa$ many 
paiwise disjoint stationary subsets.
\end{cor}

\begin{proof}
Consider $\sI_{| S}$, 
the non-stationary ideal 
restricted to $S$. This is a $\kappa$-additive, proper ideal containing
all the singletons. From theorem~\ref{ulam}, let 
$A_\alpha$, $\alpha < \kappa$, be a $\kappa$ sequence of pairwise
disjoint $\sI$-positive subsets of $\kappa$. Then $A'_\alpha= 
A_\alpha \cap S$ form a $\kappa$-sequence of pairwise disjoint $\sI$-positive
subsets of $S$. We can enlarge one, if necessary, so they
union to $S$.
\end{proof}



If $\kappa$ is a regular cardinal, and $S \subseteq \kappa$ is 
stationary, we define the set of \emph{thin points} $\tilde{s} \subseteq
S$ by $\alpha \in \tilde{S}$ iff $S \cap \alpha$ is not stationary
in $\alpha$. 

\begin{lem}
Let $\kappa$ be regular and $S \subseteq \kappa$ be stationary and
consist of limit ordinals. Then $\tilde{S}$ is stationary.
\end{lem}

\begin{proof}
Let $C \subseteq \kappa$ be \cub. Let $\alpha$ be the least limit point of
$C$ which is in $S$ (which exists as $C'$ is also \cub). If $\cof(\alpha)
> \omega$, then $C' \cap \alpha$ is \cub\ in $\alpha$ and is 
disjoint from $S$, and so $\alpha \in S'$. If $\cof(\alpha)=
\omega$, then there is an $\omega$ sequence of successor ordinals
cofinal in $\alpha$, which also gives a \cub\ subset of 
$\alpha$ missing $S$. 
\end{proof}



We now prove the limit case of theorem~\ref{corulam}. Actually, the
proof (due to Solovay) works for both limit and successor cardinals,
and provides a different proof of theorem~\ref{ulam}.

\begin{thm}
Let $\kappa$ be a regular cardinal. Then every stationary $S \subseteq \kappa$
can be split into $\kappa$ many pairwise disjoint stationary subsets.
\end{thm}


\begin{proof}
Let $\kappa$ be regular, and $S \subseteq \kappa$ be stationary. Without
loss of generality we may assume $S$ consists of limit ordinals. 
Let $\tilde{S}$ be the thin points of $S$. For each $\alpha \in \tilde{S}$,
let $\eta^\alpha_\xi$, $\xi < \cof(\alpha)$, be an increasing
continuous sequence with supremum $\alpha$ which misses $S$. 
We claim that there is $\xi$ such that for all $\delta < \kappa$ the 
set $\{ \alpha \in \tilde{S} \colon \eta^\alpha_\xi > \delta\}$ is stationary. 
If not, then for each $\xi$ there is a $\rho(\xi) < \kappa$ and a 
\cub\ set $C_\xi$ such that for all $\alpha \in \tilde{S} \cap C_\xi$
we have $\eta^\alpha_\xi < \rho(\xi)$ (if $\eta^\alpha_\xi$ is defined). 
Let $C = \triangledown C_\xi$, and
let $D \subseteq C$ be \cub\ and closed under the function 
$\xi \mapsto \rho(\xi)$. Since $\tilde{S}$ is stationary, let 
$\alpha < \beta$ be two elements of $\tilde{S} \cap D$. 
Then for each $\xi \in \cof(\beta) \cap \alpha$ we have 
$\eta^\beta_\xi < \alpha$. This shows that $\cof(\beta) \geq \alpha$
and that $\eta^\beta_\alpha$ is defined and equal to $\alpha$ (since 
the sequence $\eta^\beta_\xi$ is continuous). By definition of the
$\eta^\beta_\xi$, this shows $\alpha \notin \tilde{S}$, a contradiction. 

Fix now $\xi$ as in the claim. Note that $\alpha \mapsto 
\eta^\alpha_\xi$ is pressing down. For each $\gamma< \kappa$, by the claim 
and Fodor's theorem there is a $\tau(\gamma)> \gamma$ and a stationary
set $S_\gamma \subseteq \tilde{S}$ such that $\eta^\alpha_\xi=\tau(\gamma)$
for all $\alpha \in S_\gamma$. Since $\kappa$ is regular, 
there is a $\kappa$ size set $A \subseteq \kappa$ such that 
$\tau(\alpha) \neq \tau(\beta)$ for $\alpha\neq \beta \in A$. 
Then the sets $S_\delta=\{ \alpha \in \tilde{S} \colon 
\eta^\alpha_\xi= \tau(\delta) \}$, for $\delta \in A$,  
are pairwise disjoint and stationary.
\end{proof}






\section{Silver's Theorem}


We prove a theorem of Silver which shows a significant restriction
on the continuum function at singular cardinals of uncountable 
cofinality. 


\begin{thm}
Let $\kappa$ be a singular cardinal of uncountable cofinality. 
If the $\gch$ hols below $\kappa$ (i.e., $\forall  \lambda < \kappa\ 
(2^\lambda= \lambda^+)$), then it holds at $\kappa$ as well. 
\end{thm}

\begin{proof}
Let $\kappa_\alpha$, $\alpha< \cof(\kappa)$ be an increasing, continuous
sequence of cardinals cofinal in $\kappa$. For $A \subseteq \kappa$
consider the function $f_A$ with domain $\cof(\kappa)$ where 
$F_A(\alpha)=A \cap \kappa_{\alpha}$. Since 
$2^{\kappa_\alpha}=\kappa_{\alpha+1}$, we may identify $f_A$ with a function
satisfying $f(\alpha) \in \kappa_{\alpha+1}$. Consider the collection
$F=\{ f_A \colon A \subseteq \kappa \}$ of all such functions. 
Note that this forms an \emph{almost disjoint} family of functions, that is,
if $A \neq B$ then $\exists \alpha < \cof(\kappa)\ \forall \beta > \alpha\ 
(f_A(\beta) \neq f_B(\beta))$. 

Let $g \colon \cof(\kappa) \to \kappa$ with $g(\alpha) < \kappa_{\alpha
+1}$ for all $\alpha$. Let $F_g$ denote those $f \in F$ such that 
$\{ \alpha < \cof(\kappa) \colon f(\alpha) \leq  g(\alpha)\}$ is stationary. 
We claim that for any such $g$, $|F_g| \leq \kappa$. To see this, 
let $\pi_\alpha \colon g(\alpha)+1 \to \kappa_\alpha$ be a bijection. 
If $f \in F_g$
then there is a stationary set $S_f \subseteq \cof(\kappa)$ and an ordinal 
$\delta_f < \kappa$ such that for all $\alpha \in S_f$, 
$\pi_\alpha(f(\alpha)) < \delta_f$ (by Fodor's theorem). 
Let $h_f \colon S_f \to \delta_f$ be the function $h_f(\alpha)=
\pi_\alpha(f(\alpha))$.  The map $f \mapsto (S_f, \delta_f, h_f)$
is one-to-one on $F$ since $(S_f, \delta_f, h_f)$ determines $f$ on $S$, which 
determines $f \in F$ as $F$ is an almost disjoint family. 
There are at most $2^{\cof(\kappa)}< \kappa$ choices for $S_f$, 
$\kappa$ many choices for $\delta_f$, and $\sup_{\delta < \kappa} \delta^
{\cof(\kappa)}< \kappa$ many choices for $h_f$. Thus, $|F_g|
\leq \kappa$. 


We now show that $|F| \leq \kappa^+$. We define  a sequence
$f_\alpha \in F$ recursively so that for $\beta < \alpha$
we have $\{ \xi \colon 
f_\beta(\xi) < f_{\alpha}(\xi) \}$ is stationary. 
Assume $f_\beta$ has been defined for 
$\beta < \alpha$. If for every $f \in F$ there is a $\beta < \alpha$
such that $\{ \xi \colon f(\xi) \leq  f_\beta(\xi)\}$ is stationary, 
then stop the construction. Otherwise, let $f_{\alpha} \in F$ be such
that $\forall \beta < \alpha\ \{ \xi \colon 
f_\beta(\xi) < f_{\alpha}(\xi) \} \in \Cub(\cof(\kappa))$. 
In particular, for all $\beta < \alpha$, 
$\{ \xi \colon 
f_\beta(\xi) < f_{\alpha}(\xi) \}$ is stationary. $f_{\kappa^+}$
cannot be defined by the claim. Thus, we end with a collection
$\{ f_\alpha\}_{\alpha < \lambda}$, where $\lambda \leq \kappa^+$. 
Every $f \in F$ is then in some $F_{f_{\alpha}}$, and from the claim
it follows that $|F| \leq \kappa \cdot \lambda \leq \kappa^+$. 
\end{proof}












\end{document}
