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