\lecture{15}{2023-12-07}{} % RECAP % Let $\kappa$ be uncountable and regular. % If $\alpha < \kappa$ and $\{C_{\xi} : \xi < \alpha\}$ % is a family of sets which are club in $\kappa$ % then $\bigcap_{\xi < \alpha} C_\xi$ is a club. % % If $\{C_\xi : \xi < \kappa\}$ is a family of sets which are club in $\kappa$, % then $\diagi_{\xi < \kappa} = \{\alpha < \kappa : \alpha \in \bigcap_{\xi < \alpha} C_\xi\}$ % is a club. % END RECAP \begin{theorem}[Fodor] \yalabel{Fodor's Theorem}{Fodor}{thm:fodor} Let $\kappa$ be a regular and uncountable cardinal. Let $S \subseteq \kappa$ be stationary and let $f\colon S \to \kappa$ be \vocab{regressive} in the following sense: $f(\alpha) < \alpha$ for all $\alpha \in S$. Then there exists a stationary subset $T \subseteq S$ and some $\nu < \kappa$ such that $f(\alpha) = \nu$ for all $\alpha \in T$. \end{theorem} \begin{proof} \gist{% Let $S, f$ be given. For $\nu < \kappa$ set $S_\nu \coloneqq \{\alpha \in S : f(\alpha) = \nu\}$. % f^{-1}(\nu)$. We aim to show that one of the $S_\nu$ is stationary. Suppose otherwise. Then for every $\nu$ there exists a club $C_\nu$ such that $S_\nu \cap C_\nu = \emptyset$.\footnote{Here we use \AxC to choose the $C_\nu$ uniformly.} Let $C = \diagi_{\nu < \kappa} C_\nu$. By \yaref{lem:diagiclub} $C$ is a club. So we may pick some $\alpha \in C \cap S$. In particular $\alpha \in C_\nu$ for all $\nu < \alpha$. Hence $f(\alpha) \neq \nu$ for all $\nu < \alpha$, so $f(\alpha) \ge \alpha$. But $f$ is regressive $\lightning$ }{% \begin{itemize} \item For $\nu < \kappa$ set $S_\nu \coloneqq \{\alpha \in S : f(\alpha) = \nu\}$. \item Suppose none of the $S_\nu$ is stationary, i.e.~ $\forall \nu < \kappa.~\exists C_\nu \text{ club}.~C_\nu \cap S_\nu = \emptyset$. \item $\diagi_{\nu < \kappa} C_\nu$ is club. \item Pick $\alpha \in C \cap S$. But then $\forall \nu < \alpha.~\alpha \in C_\nu$, i.e.~$f(\alpha) \ge \alpha$. \end{itemize} } \end{proof} \subsection{Some model theory and a second proof of Fodor's Theorem} Recall the following: \begin{definition} A substructure $X \subseteq V_\theta$\gist{\todo{make this more general. Explain why $V_\theta$ is a model}}{} is an \vocab{elementary substructure} of $V_\theta$, denoted $X \prec V_{\theta}$,\footnote{more formally $(X,\in ) \prec (V_{\theta})$} iff for all formulae $\phi$ of the language of set theory and for all $x_1,\ldots,x_k \in X$, \[ (X; \in\defon{X}) \models \phi(x_1,\ldots,x_k) \iff(V_\theta; \in\defon{V_\theta}) \models\phi(x_1,\ldots,x_k). \] \end{definition} \begin{remark} Löwenheim-Skolem allows us to find elementary substructures of arbitrary sizes. How do we do this? Let $\phi$ be a formula. A \vocab{Skolem-function} over $V_\theta$ for $\phi$ is a function \[ f\colon \leftindex^k V_\theta \to V_\theta, \] where $k$ is the number of free variables of $\exists v.~\phi$ and for all $x_1,\ldots,x_k \in V_\theta$, if $(V_\theta, \in) \models \exists v.~\phi(v,x_1,\ldots,x_k)$ then $(V_\theta, \in) \models \phi(f(x_1,\ldots,x_k),x_1,\ldots,x_k)$. Using \AxC such Skolem-functions can be easily found for all formulae. \end{remark} There is a sufficient criterion for $X \subseteq V_{\theta}$ to be an elementary substructure of $V_\theta$. \begin{lemma}[Tarski-Vaught Test] Let $X \subseteq V_\theta$. For each formula $\phi$, let $f_\phi$ be a Skolem function over $V_\theta$ for $\phi$. If for every $\phi$ and for all $x_1,\ldots, x_k \in X$ (where $k$ is the number of free variables of $\exists v.~\phi$) $f_\phi(x_1,\ldots,x_k) \in X$, then $X \prec V_{\theta}$. \end{lemma} Let's do a second proof of \yaref{thm:fodor}. \begin{yarefproof}{thm:fodor} \gist{% Fix $\theta > \kappa$ and look at $V_{\theta}$. Fix $S \subseteq \kappa$ stationary and $f\colon S \to \kappa$ regressive. For each formula $\phi$ fix a Skolem function $f_\phi$ over $V_\theta$ for $\phi$. Let $(X_\xi: \xi \le \kappa)$ be a sequence of elementary substructures of $V_\theta$ defined as follows: Let $X_0$ be the least $X$ such that $S, f \in X$ and $X$ is closed under $f_\phi$. Note that $X_0$ is countable. For $\xi < \kappa$ let $X_{\xi + 1}$ be the least $X \subseteq V_\theta$ such that $X_\xi \subseteq X$, $\min(\kappa \setminus X_\xi) \in X$ and $X$ is closed under all $f_\phi$. For limits $\lambda \le \kappa$ let \[ X_\lambda \coloneqq \bigcup_{\xi < \lambda} X_\xi. \] Note that $|X_{\xi}| = |X_{\xi + 1}|$ but the size may increase at limits. It is easy to see inductively that $|X_{\xi}| < \kappa$ for every $\xi < \kappa$, while $X_\xi \subsetneq X_{\xi'}$ for all $\xi < \xi' \le \kappa$. Also $\xi \subseteq X_\xi$ for all $\xi \le \kappa$. \begin{claim} \label{thm:fodor:p2:c1} There is a club $C \subseteq \kappa$ such that $X_\xi \cap \kappa = \xi$ for all $\xi \in C$. \end{claim} \begin{yarefproof}{thm:fodor:p2:c1} Write $C = \{\xi < \kappa:X_\xi \cap \kappa = \xi\}$. Trivially $C$ is closed. Let us show that $C$ is unbounded in $\kappa$. Let $\zeta < \kappa$. Let us define a strictly increasing sequence $ \langle \xi_n : n < \omega \rangle$ as follows. Set $\xi_0 \coloneqq \zeta$. Suppose $\xi_n$ has been chosen. Look at $X_{\xi_n} \cap \kappa$. Since $|X_{\xi_n} \cap \kappa| < \kappa$, $\sup (X_{\xi_n} \cap \kappa) < \kappa$. Set $\xi_{n+1} \coloneqq \sup(X_{\xi_n} \cap \kappa) + 1$. Set $\xi \coloneqq \sup_{n<\omega} \xi_n$. Clearly $\zeta < \xi$. \begin{claim} \label{thm:fodor:p2:c1.1} $\xi \in C$, i.e.~$X_\xi \cap \kappa = \xi$. \end{claim} \begin{yarefproof}{thm:fodor:p2:c1.1} If $\eta < \xi$, then $\eta < \xi_n$ for some $n$ and then $\eta \in \xi_n \subseteq X_{\xi_n} \subseteq X_{\xi}$. Now let $\eta \in X_\xi \cap \kappa$. Then $\eta \in X_{\xi_n}$ for some $n < \omega$, so $\eta < \xi_{n+1} < \xi$, hence $X_{\xi} \cap \kappa \subseteq \xi$. \end{yarefproof} \end{yarefproof} Now let $\alpha \in S \cap C$, i.e.~$X_\alpha \prec V_{\theta}$ and $\alpha = X_{\alpha} \cap \kappa$. $f \in X_{\alpha}$ and $f$ is regressive, so $f(\alpha) < \alpha$. Write $\nu = f(\alpha)$. Let $T = \{\xi \in S: f(\xi) = \nu\}$. We have $T \in X_{\alpha}$, as $T$ is definable from $S,f,\nu \in X_\alpha$. \begin{claim} $T$ is stationary. \end{claim} \begin{subproof} Otherwise there is a club $D \subseteq \kappa$ such that $D \cap T = \emptyset$, i.e.~ \[V_\theta \models \exists D .~ D\text{ club in $\kappa$} \land D \cap T = \emptyset\] hence \[X_\alpha\models \exists D .~ D\text{ club in $\kappa$} \land D \cap T = \emptyset.\] So there is $D \in X_\alpha$ such that \[ X_\alpha \models D \text{ is club in $\kappa$} \land D \cap T = \emptyset, \] hence \[ V_\theta \models D \text{ is club in $\kappa$} \land D \cap T = \emptyset. \] In other words, there is some club $D \in X_\alpha$ with $D \cap T = \emptyset$. We have $\alpha \in T$ as $\alpha \in S$ and $f(\alpha) = \nu$. Let us show that $\alpha \in D$, which gives a contradiction. For $\alpha \in D$ it suffices to show that $D \cap \alpha$ is unbounded in $\alpha$. Let $\xi < \alpha$. As $D$ is unbounded in $\kappa$, $\exists \eta > \xi .~ \eta \in D$, so \[ V_{\theta} \models \exists \eta > \xi .~ \eta \in D, \] hence \[ X_\alpha \models \exists \eta > \xi .~ \eta \in D. \] Hence there is some $\eta \in X_\alpha$ with $\eta \in D$. This means that $\xi < \underbrace{\eta}_{\in D} < \alpha$. \end{subproof} }{% \begin{itemize} \item Fix $S \subseteq \kappa$ stationary, $f\colon S \to \kappa$ regressive. \item Fix $\theta > \kappa$ look at $V_\theta$, fix Skolem functions $f_\phi$ over $V_{\theta}$ for all $\phi$. \item Define sequence of elementary substructures $\langle X_\xi, \xi < \kappa \rangle$: \begin{itemize} \item $X_0$ minimal st.~closed under $f_\phi$ and $S,f \in X$ (note: countable). \item $X_{\xi+1}$ minimal st.~$X_{\xi} \subseteq X_{\xi+1}$, closed under $f_\phi$ and $\min(\kappa \setminus X_\xi) \in X$. (note: $|X_\xi| = |X_{\xi + 1}|$) \item $X_{\lambda} \coloneqq \bigcup_{\xi < \lambda} X_\xi$ (note: cardinality may increase) \end{itemize} \item Note: strictly increasing, $|X_\xi| < \kappa$, $\xi \subseteq X_\xi$. \item $C\coloneqq \{\xi < \kappa: X_\xi \cap \kappa = \xi\}$ is club: \begin{itemize} \item clearly closed. \item unbounded in $\kappa$: \begin{itemize} \item Take $\zeta < \kappa$, Define $\langle \xi_n : n < \omega \rangle$ by $\xi_0 \coloneqq \zeta$, $\xi_{n+1} \coloneqq \sup(X_{\xi_n} \cap \kappa)$ ($ < \kappa$), $\xi \coloneqq \sup \xi_n < \kappa$. \item $X_\xi \cap \kappa = \xi$, i.e.~$\xi \in C$: \begin{itemize} \item $\eta < \xi \implies \exists n.~\eta < \xi_n \implies \eta \in \xi_n \subseteq X_{\xi_n} \subseteq X_\xi$ $\implies \xi \subseteq X_\xi \cap \kappa$. \item $\eta \in X_\xi \cap \kappa \implies \exists n.~\eta \in X_{\xi_n} \implies \eta < \xi_{n+1} < \xi \implies X_\xi \cap \kappa \subseteq \xi$. \end{itemize} \end{itemize} \end{itemize} \item Take $\alpha \in S \cap C$, $\nu \coloneqq f(\alpha) < \alpha$, $T \coloneqq \{\xi \in S : f(\xi) = \nu\}$. ($T \in X_\alpha$, since it can be defined from $S, f, \nu$) \item $T$ is stationary: Suppose $D \cap T = \emptyset$ for a club $D$. $X_\alpha \prec V_\theta \implies$ find such $D \in X_\alpha$. \item Clearly $\alpha \in T$. Want ($\lightning$) $\alpha \in D$: \begin{itemize} \item Suffices $\sup (D \cap \alpha) = \alpha$. Let $\xi < \alpha$. \item $D$ unbounded, so $V_\theta \models \exists \eta > \xi : \eta \in D$ \item $\implies X_\alpha \models \exists \eta > \xi . ~\eta \in D$ \item $\implies \exists \eta > \xi. ~\eta \in D \land \underbrace{\eta < \alpha}_{\in X_\alpha}$. \end{itemize} \end{itemize} } \end{yarefproof}