\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}
    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 $C = \diagi_{\nu < \alpha} C_\nu$.
    By \yaref{lem:diagiclub} $C$ is a club.\todo{Show that it \emph{is} a club not just contains one}
    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 $\lightning$
\end{proof}

\subsection{Some model theory and a second proof of Fodor's Theorem}


Recall the following:

\begin{definition}
    A substructure $X \subseteq  V_\theta$\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
    \[
    f\colon {}^k V_\theta \to V_\theta,
    \]
    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{refproof}{thm:fodor}
    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 is increased 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{refproof}{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$
        a 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{refproof}{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{refproof}
    \end{refproof}
    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}
\end{refproof}


