\documentclass{amsart} \usepackage{amssymb} \usepackage{amsmath} \usepackage{comment} \theoremstyle{plain} \newtheorem{thm}{Theorem}[section] \newtheorem{cor}[thm]{Corollary} \newtheorem{lem}[thm]{Lemma} \theoremstyle{definition} \newtheorem{defn}[thm]{Definition} \newtheorem{exer}{Exercise} \theoremstyle{remark} \newtheorem{rem}[thm]{Remark} \newtheorem{fact}[thm]{Fact} \newtheorem{hyp}[thm]{Hypothesis} \newcommand{\Q}{\mathbb{Q}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\R}{\mathbb{R}} \newcommand{\res}{\restriction} \newcommand{\dom}{\operatorname{dom}} \newcommand{\ran}{\operatorname{ran}} \newcommand{\cof}{\operatorname{cof}} \newcommand{\ot}{\text{o.t.}} \newcommand{\on}{\text{ON}} \newcommand{\card}{\text{CARD}} \newcommand{\ac}{\text{AC}} \newcommand{\zf}{\text{ZF}} \newcommand{\zfc}{\text{ZFC}} \newcommand{\sP}{\mathcal{P}} \newcommand{\sI}{\mathcal{I}} \newcommand{\sF}{\mathcal{F}} \newcommand{\ch}{\text{CH}} \newcommand{\gch}{\text{GCH}} \newcommand{\cub}{\text{c.u.b.}} \newcommand{\Cub}{\text{Cub}} \newcommand{\ns}{\text{NS}} \newcommand{\fin}{\text{FIN}} \newcommand{\sB}{\mathcal{B}} \newcommand{\lub}{\text{l.u.b.}} \newcommand{\glb}{\text{g.l.b.}} \newcommand{\mbar}[1]{\overline{#1}} \newcommand{\sat}{\text{sat}} \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}