lecture 15

inputs/lecture_14.tex

@@ -130,11 +130,12 @@ We have shown (assuming \AxC to choose contained clubs):
     \] 
 \end{definition}
 \begin{lemma}
+    \label{lem:diagiclub}
     Let $\kappa$ be a regular, uncountable cardinal.
     If $\langle C_{\beta} : \beta < \kappa \rangle$
     is a sequence of club subsets of $\kappa$,
     then $\diagi_{\beta < \kappa} C_{\beta}$
-    contains a club.
+    contains a club. % TODO: contains or is?
 \end{lemma}
 \begin{proof}
     Let us fix $\langle C_{\beta} : \beta < \alpha \rangle$.

inputs/lecture_15.tex

@@ -0,0 +1,229 @@
+\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}
+
+

logic2.tex

@@ -38,6 +38,7 @@
 \input{inputs/lecture_12}
 \input{inputs/lecture_13}
 \input{inputs/lecture_14}
+\input{inputs/lecture_15}
 
 
 \cleardoublepage