<<<<<<< HEAD \lecture{16}{2023-12-11}{} Recall \yaref{thm:fodor}. \begin{question} What happens if $S$ is nonstationary? \end{question} Let $S \subseteq \kappa$ be nonstationary, $\kappa$ uncounable and regular. Then there is a club $C \subseteq \kappa$ with $C \cap S = \emptyset$. Let us define $f\colon S \to \kappa$ in the following way: If $\alpha \in S$ and $C \cap \alpha \neq \emptyset$, then $\max(C \cap \alpha) < \alpha$. Define \[ f(\alpha) \coloneqq \begin{cases} 0 &: C \cap \alpha = \emptyset,\\ \max(C \cap \alpha) &:C\cap \alpha \neq \emptyset. \end{cases} \] For all $\alpha > 0$, we have that $f(\alpha) < \alpha$. If $\gamma \in \ran(f)$ then $f(\alpha) = \gamma$ implies either $\gamma = 0$ and $\alpha < \min(C)$ or $\gamma \in C$ and $\gamma < \alpha < \gamma'$ where $\gamma' = \min(C \setminus (\gamma + 1))$. Thus for all $\gamma$, there is only an interval of ordinals $\alpha \in S$ where $f(\alpha) = \gamma$. \todo{Move this to the definition of filter} Recall that $F \subseteq \cP(\kappa)$ is a filter if $X,Y \in F \implies X \cap Y \in F$, $X \in X, X \subseteq Y \subseteq \kappa \implies Y \in F$ and $\emptyset \not\in F, \kappa \in F$. \begin{definition} A filter $F$ is an \vocab{ultrafilter} iff for all $X \subseteq \kappa$ either $X \in F$ or $\kappa \setminus X \in F$. \end{definition} \begin{example} Examples of filters: \begin{enumerate}[(a)] \item Let $\kappa \ge \aleph_0$ and let $F = \{X \subseteq \kappa: \kappa \setminus X \text{ is finite}\}$. This is called the \vocab{Fr\'echet filter} or \vocab{cofinal filter}. It is not an ultrafilter (consider for example the even and odd numbers\footnote{we consider limit ordinals to be even}). \item Let $\kappa$ be uncountable and regular. Then $\cF_\kappa \coloneqq \{X \subseteq \kappa: \exists C \subseteq \kappa \text{ club in $\kappa$}. C \subseteq X\}$. \end{enumerate} \end{example} \begin{question} Is $\cF_\kappa$ an ultrafilter? \end{question} This is certainly not the case if $\kappa \ge \aleph_2$, because then $S_0 \coloneqq \{\alpha < \kappa : \cf(\alpha) = \omega\}$ and $S_1 \coloneqq \{\alpha < \kappa : \cf(\alpha) = \omega_1\} $ are both stationary and clearly disjoint. So neither $S_0$ nor $S_1 \subseteq \kappa \setminus S_0$ contains a club. For $\kappa < \aleph_1$ this argument does not work, since there is only on cofinality. \begin{theorem}[Solovay] \yalabel{Solovay's Theorem}{Solovay}{thm:solovay} Let $\kappa$ be regular and uncountable. If $S \subseteq \kappa$ is stationary, there is a sequence $\langle S_i : i < \kappa \rangle$ of pairwise disjoint stationary sets of $\kappa$ such that $S = \bigcup S_i$. \end{theorem} \begin{corollary} $\cF_{\aleph_1}$ is not an ultrafilter. \end{corollary} \begin{proof} Apply \yaref{thm:solovay} to $S = \aleph_1$. Let $\aleph_1 = A \cup B$ where $A$ and $B$ are both stationary and disjoint. Then use the argument from above. \end{proof} \begin{refproof}{thm:solovay}% %\footnote{``This is one of the arguments where it is certainly % worth it to look at it again''} % TODO: Look at this again and think about it. We will only proof this for $\aleph_1$. Fix $S \subseteq \aleph_1$ stationary. For each $0 < \alpha < \omega_1$, either $\alpha$ is a successor ordinal or $\alpha$ is a limit ordinal and $\cf(\alpha) = \omega_1$. Let $S^\ast \coloneqq \{\alpha \in S \setminus \{0\} : \alpha \text{ is a limit ordinal}\} $. $S^\ast$ is still stationary: Let $C \subseteq \omega_1$ be a club, then $D = \{\alpha \in C \setminus \{0\} : \alpha \text{ is a limit ordinal}\} $ is still a club, so \[S^\ast \cap C = S^\ast \cap D = S \cap D \neq \emptyset.\] Let \[ \langle \langle \gamma_n^{\alpha} : n < \omega \rangle : \alpha \in S^\ast\rangle \] be such that $ \langle \gamma_n^\alpha : n < \omega \rangle$ is cofinal in $ \alpha$. \begin{claim} \label{thm:solovay:p:c1} There exists $n < \omega$ such that for all $\delta < \omega_1$ the set \[ \{\alpha \in S^\ast : \gamma_n^\alpha > \delta\} \] is stationary. \end{claim} \begin{subproof} Otherwise for all $n < \omega$, there is a $\delta$ such that $\{\alpha \in S^\ast : \gamma_n^{\alpha} > \delta\}$ is nonstationary. Let $\delta_n$ be the least such $\delta$. Let $C_n$ be a club disjoint from \[ \{\alpha \in S^\ast : \gamma_n^{\alpha} > \delta_n\}, \] i.e.~if $\alpha \in S^\ast \cap C_n$, then $\gamma_n^{\alpha} \le \delta_n$. Let $\delta^\ast \coloneqq \sup_{n< \omega}\delta_n$. Let $C = \bigcap_{n < \omega} C_n$. Then $C$ is a club. We must have that if $\alpha \in S^\ast \cap C$ then $\gamma_n^{\alpha} \le \delta^\ast$ for all $n$. % But now things get a bit fishy: Let $C' \coloneqq C \setminus (\delta^\ast + 1)$. $C'$ is still club. As $\delta^\ast$ is stationary, we may pick some $\alpha \in S^\ast \cap C'$. But then $\gamma_n^{\alpha} > \delta^\ast$ for $n$ large enough as $\langle \gamma_n^{\alpha} : n < \omega \rangle$ is cofinal in $\alpha$ $\lightning$. \end{subproof} Let $n < \omega$ be as in \yaref{thm:solovay:p:c1}. Consider \begin{IEEEeqnarray*}{rCl} f\colon S^\ast&\longrightarrow & \omega_1\\ \alpha&\longmapsto & \gamma^{\alpha}_n. \end{IEEEeqnarray*} Clearly this is regressive. We will now define a strictly increasing sequence $\langle \delta_i : i < \omega_1 \rangle$ as follows: Let $\delta_0 = 0$. For $0 < i < \omega_1$ suppose that $\delta_j, j < i$ have been defined. Let $\delta \coloneqq (\sup_{j < i} \delta_j) + 1$. By \yaref{thm:solovay:p:c1} (rather, by the choice of $n$), we have that $\{\alpha \in S^\ast : \gamma_n^{\alpha} > \delta\}$ is stationary. Hence by Fodor there is some stationary $T \subseteq S^\ast$ and some $\delta'$ such that for all $\alpha \in T$ we have $\gamma_n^{\alpha} = \delta'$. Write $\delta_i = \delta'$ and $T_i = T$. \begin{claim} \label{thm:solovay:p:c2} Each $T_i$ is stationary and if $i \neq j$, then $T_i \cap T_j = \emptyset$. \end{claim} \begin{subproof} The first part is true by construction. Let $j < i$. Then if $\alpha \in T_i$, $\alpha' \in T_j$, we get $\gamma_n^{\alpha'} = \delta_j < \delta_i = \gamma_n^{\alpha}$ hence $\alpha \neq \alpha'$. \end{subproof} Now let \[ S_i \coloneqq \begin{cases} T_i &: i > 0,\\ T_0 \cup (S \setminus \bigcup_{j > 0} T_j) &: i = 0. \end{cases} \] Then $\langle S_i : i < \omega_1 \rangle$ is as desired. \end{refproof} We now want to do another application of \yaref{thm:fodor}. Recall that $2^{\kappa} > \kappa$, in fact $\cf(2^{\kappa}) > \kappa$ by \yaref{thm:koenig}. Trivially, if $\kappa \le \lambda$ then $2^{\kappa} \le 2^{\lambda}$. This is in some sense the only thing we can prove about successor cardinals. However we can say something about singular cardinals: \begin{theorem}[Silver] \yaref{Silver's Theorem}{Silver}{thm:silver} Let $\kappa$ be a singular cardinal of uncountable cofinality. Assume that $2^{\lambda} = \lambda^+$ for all (infinite) cardinals $\lambda < \kappa$. Then $2^{\kappa} = \kappa^+$. \end{theorem} \begin{definition} $\GCH$, the \vocab{generalized continuum hypothesis} is the statement that $2^{\lambda} = \lambda^+$ holds for all infinite cardinals $\lambda$, \end{definition} Recall that $\CH$ says that $2^{\aleph_0} = \aleph_1$. So $\GCH \implies \CH$. \yalabel{thm:silver} says that if $\GCH$ is true below $\kappa$, then it is true at $\kappa$. The proof of \yalabel{thm:silver} is quite elementary, so we will do it now, but the statement can only be fully appreciated later. ||||||| ede97ee ======= \lecture{16}{2023-12-14}{Silver's Theorem} We now want to prove \yaref{thm:silver}. More generally, if $\kappa$ is a singular cardinal of uncountable cofinality such that $2^{\lambda} = \lambda^+$ for all $\lambda < \kappa$, then $2^{\kappa} = \kappa^+$. \begin{remark} The hypothesis of \yaref{thm:silver} is consistent with $\ZFC$. \end{remark} We will only proof \yaref{thm:silver} in the special case that $\kappa = \aleph_{\omega_1}$. The general proof differs only in notation. \begin{remark} It is important that the cofinality is uncountable. For example it is consistent with $\ZFC$ that $2^{\aleph_n} = \aleph_{n+1}$ for all $n < \omega$ but at the same time $2^{\aleph_{\omega}} = \aleph_{\omega + 2}$. \end{remark} \begin{refproof}{thm:silver} We need to count the number of $X \subseteq \aleph_{\omega_1}.$ Let us fix $\langle f_\lambda : \lambda < \kappa \text{ an infinite cardinal} \rangle$ such that $f_{\lambda}\colon \cP(\lambda) \to \lambda^+$ is bijective for each $\lambda < \kappa$. For $X \subseteq \aleph_{\omega_1}$ define \begin{IEEEeqnarray*}{rCl} f_X\colon \omega_1 &\longrightarrow & \aleph_{\omega_1} \\ \alpha &\longmapsto & f_{\aleph_\alpha}(X \cap \aleph_\alpha). \end{IEEEeqnarray*} \begin{claim} For $X,Y \subseteq \aleph_{\omega_1}$ it is $X \neq Y \iff f_X \neq f_Y$. \end{claim} \begin{subproof} $X \neq Y$ holds iff $X \cap \aleph_\alpha \neq Y \cap \aleph_\alpha$ for some $\alpha < \omega_1$. But then $f_X(\alpha) \neq f_Y(\alpha)$. \end{subproof} For $X, Y \subseteq \aleph_{\omega_1}$ write $X \le Y$ iff \[ \{\alpha < \omega_1 : f_X(\alpha) \le f_Y(\alpha)\} \] is stationary. \begin{claim} For all $X,Y \subseteq \aleph_{\omega_1}$, $X \le Y$ or $Y \le X$. \end{claim} \begin{subproof} Suppose that $X \nleq Y$ and $Y \nleq X$. Then there are clubs $C,D \subseteq \omega_1$ such that \[ C \cap \{\alpha < \omega_1 : f_X(\alpha) \le f_Y(\alpha)\} = \emptyset \] and \[ D \cap \{\alpha < \omega_1 : f_Y(\alpha) \le f_X(\alpha)\} = \emptyset. \] Note that $C \cap D$ is a club. Take some $\alpha \in C \cap D$. But then $f_X(\alpha) \le f_Y(\alpha)$ or $f_{Y}(\alpha) \le f_X(\alpha)$ $\lightning$ \end{subproof} \begin{claim} \label{thm:silver:p:c3}. Let $X \subseteq \aleph_{\omega_1}$. Then \[ |\{Y \subseteq \aleph_{\omega_1} : Y \le X\}| \le \aleph_{\omega_1}. \] \end{claim} \begin{subproof} Write $A \coloneqq \{Y \subseteq X_{\omega_1} : Y \le X\}$. Suppose $|A| \ge \aleph_{\omega_1 + 1}$. For each $Y \in A$ we have that \[ S_Y \coloneqq \{\alpha : f_Y(\alpha) \le f_X(\alpha)\} \] is a stationary subset of $\omega_1$. Since by assumption $2^{\aleph_1} = \aleph_2$, there are at most $\aleph_2$ such $S_Y$. Suppose that for each $S \subseteq \omega_1$, \[ |\{Y \in A : S_Y = S\}| < \aleph_{\omega_1 + 1}. \] Then $A$ is the union of $\le \aleph_2$ many sets of size $< \aleph_{\omega_1 + 1}$. Thus this is a contradiction since $\aleph_{\omega_1 + 1}$ is regular. So there exists a stationary $S \subseteq \omega_1$ such that \[ A_1 = \{Y \subseteq \aleph_{\omega_1} : S_Y = S\} \] has cardinality $\aleph_{\omega_1 + 1}$. We have \[f_Y(\alpha) \le f_X(\alpha) = f_{\aleph_\alpha}(X \cap \aleph_\alpha)< \aleph_{\alpha + 1}\] for all $Y \in A_1, \alpha \in S$. Let $\langle g_{\alpha} : \alpha \in S \rangle$ be such that $g_\alpha\colon \aleph_{\alpha} \twoheadrightarrow f_X(\alpha) + 1$ is a surjection for all $\alpha \in S$. Then for each $Y \in A_1$ define \begin{IEEEeqnarray*}{rCl} \overline{f}_Y\colon S &\longrightarrow & \aleph_{\omega_1} \\ \alpha &\longmapsto & \min \{\xi : g_\alpha(\xi) = f_Y(\alpha)\}. \end{IEEEeqnarray*} Let $D$ be the set of all limit ordinals $< \omega_1$. Then $S \cap D$ is a stationary set: If $C$ is a club, then $C \cap D$ is a club, hence $(S \cap D) \cap C = S \cap (D \cap C) \neq \emptyset$. Now to each $Y \in A$ we may associate a regressive function \begin{IEEEeqnarray*}{rCl} h_Y \colon S \cap D &\longrightarrow & \omega_1 \\ \alpha &\longmapsto & \min \{\beta < \alpha : \overline{f}_Y(\alpha) < \aleph_{\beta}\}. \end{IEEEeqnarray*} $h_Y$ is regressive, so by \yaref{thm:fodor} there is a stationary $T_Y \subseteq S \cap D$ on which $h_Y$ is constant. By an argument as before, there is a stationary $T \subseteq S \cap D$ such that \[ |A_2| = \aleph_{\omega_1 +1}, \] where $A_2 \coloneqq \{Y \in A_1 : T_Y = T\}$. Let $\beta < \omega_1$ be such that for all $Y \in A_2$ and for all $\alpha \in T$, $h_Y(\alpha) = \beta$. Then $\overline{f}_Y(\alpha) < \aleph_\beta$ for all $Y \in A_2$ and $\alpha \in T$. There are at most $\aleph_\beta^{\aleph_1}$ many functions $T \to \aleph_\beta$, but \begin{IEEEeqnarray*}{rCl} \aleph_\beta^{\aleph_1} &\le & 2^{\aleph_\beta \cdot \aleph_1}\\ &=& \aleph_{\beta+1} \cdot \aleph_2\\ &<& \aleph_{\omega_1}. \end{IEEEeqnarray*} Suppose that for each function $\tilde{f}\colon T \to \aleph_\beta$ there are $< \aleph_{\omega_1 + 1}$ many $Y \in A_2$ with $\overline{f}_Y \cap T = \tilde{f}$. Then $A_2$ is the union of $<\aleph_{\omega_1}$ many sets each of size $< \aleph_{\omega_1 + 1}$ $\lightning$. Hence for some $\tilde{f}\colon T \to \aleph_\beta$, \[ |A_3| = \aleph_{\omega_1 + 1}, \] where $A_3 = \{Y \in A_2 : \overline{f}_Y\defon{T} = \tilde{f}\}$. Let $Y, Y' \in A_3$ and $\alpha \in T$. Then \[ \overline{f}_Y(\alpha) = \overline{f}_{Y'}(\alpha), \] hence \[ f_{\aleph_\alpha}(Y \cap \aleph_\alpha) = f_Y(\alpha) = f_{Y'}(\alpha) = f_{\aleph_\alpha}(Y' \cap \aleph_\alpha), \] i.e.~$Y \cap \aleph_\alpha = Y' \cap \aleph_\alpha$. Since $T$ is cofinal in $\omega_1$, it follows that $Y = Y'$. So $|A_3| \le 1 \lightning$ \end{subproof} Let us now define a sequence $\langle X_i : i < \aleph_{\omega_1 + 1} \rangle$ of subsets of $\aleph_{\omega_1 + 1}$ as follows: Suppose $\langle X_j : j < i \rangle$ were already chosen. Consider \[ \{Y \subseteq \aleph_{\omega_1} : \exists j < i.~Y \le X_j\} = \bigcup_{j < i} \{Y \subseteq \aleph_{\omega_1} : Y \le X_j\}. \] This set has cardinality $\le \aleph_{\omega_1}$ by \yaref{thm:silver:p:c3}. Let $X_i \subseteq \aleph_{\omega_1}$ be such that $X_i \nleq X_j$ for all $j < i$. The set \[ \{Y \subseteq \aleph_{\omega_1} : \exists i < \aleph_{\omega_1 + 1} .~Y \le X_i\} = \bigcup_{i < \aleph_{\omega_1 + 1}} \{Y \subseteq \aleph_{\omega_1} : Y \le X_i\} \] has size $\le \aleph_{\omega_1 + 1}$ (in fact the size is exactly $\aleph_{\omega_1 + 1}$). But \[ \{Y \subseteq \aleph_{\omega_1} : \exists i < \aleph_{\omega_1 + 1} .~Y \le X_i\} = \cP(\aleph_{\omega_1 + 1}) \] because if $X \subseteq \aleph_{\omega_1 + 1}$ is such that $X \nleq X_i$ for all $i < \aleph_{\omega_1 + 1}$, then $X_i \le X$ for all $i < \aleph_{\omega_1 + 1}$, so such a set $X$ does not exist by \yaref{thm:silver:p:c3}. \end{refproof} >>>>>>> 5ee970194987e24c54bfc9c787cb219e15e71520