w23-logic-2/inputs/lecture_15.tex

242 lines
8.2 KiB
TeX
Raw Normal View History

2023-12-07 15:54:40 +01:00
\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}
2024-02-13 02:00:58 +01:00
\gist{%
2023-12-07 15:54:40 +01:00
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 $C = \diagi_{\nu < \alpha} C_\nu$.
2024-01-17 11:51:59 +01:00
By \yaref{lem:diagiclub} $C$ is a club.
2023-12-07 15:54:40 +01:00
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 $\lightning$
2024-02-13 02:00:58 +01:00
}{%
\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.~ $\forall \nu < \kappa.~\exists C_\nu \text{ club}.~C_\nu \cap S_\nu = \neq$.
\item $\diagi_{\nu < \alpha} C_\nu$ 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}
}
2023-12-07 15:54:40 +01:00
\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
\[
2024-02-13 02:00:58 +01:00
f\colon \leftindex^k V_\theta \to V_\theta,
2023-12-07 15:54:40 +01:00
\]
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}
2024-02-13 02:00:58 +01:00
% TODO ANKI-MARKER
2023-12-07 15:54:40 +01:00
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}