lecture 21

inputs/lecture_14.tex

@@ -225,7 +225,7 @@ We have shown (assuming \AxC to choose contained clubs):
     $\lambda \not\in D_\alpha$.
     Since $D_\alpha$ is closed,
     we get $\sup(D_{\alpha} \cap \lambda) < \lambda$.
-    In particular $\sup (\lambda \cap\diagi_{\beta < \kappa} D_{\beta}) \le \max (\alpha ,\sup(D_\alpha \cap \lambda) < \lambda$.
+    In particular $\sup (\lambda \cap\diagi_{\beta < \kappa} D_{\beta}) \le \alpha \cup \sup(D_\alpha \cap \lambda) < \lambda$.
 \end{remark}
 
 \begin{definition}

inputs/lecture_19.tex

@@ -23,9 +23,13 @@ and $\forall x \in y.~\phi$ abbreviates $\forall x.~x \in y \to \phi$.
 
     If $\phi(x_0,\ldots,x_m) \in \Sigma_n$,
     then $(\forall x_0.~\ldots\forall x_m.~\phi(x_0,\ldots,x_m)) \in \Pi_{n+1}$.
+    If $\ZFC \models \phi \leftrightarrow \psi$
+    and $\phi \in \Sigma_n$, then $\psi \in \Sigma_n$.
 
     If $\phi(x_0,\ldots,x_m) \in \Pi_n$,
     then $(\exists x_0.~\ldots\exists x_m.~\phi(x_0,\ldots,x_m) \in \Sigma_{n+1}$.
+    If $\ZFC \models \phi \leftrightarrow \psi$
+    and $\phi \in \Pi_n$, then $\psi \in \Pi_n$.
 
     $\Delta_n \coloneqq \Sigma_n \cap \Pi_n$.
 \end{definition}

inputs/lecture_21.tex

@@ -0,0 +1,161 @@
+\lecture{21}{2024-01-18}{}
+
+\begin{goal}
+    We want to show that certain statements are consistent with $\ZFC$
+    (or  $\ZF$), for instance  $\CH$.
+
+    We start with a model $M$ of $\ZFC$.
+    Usually we want $M$ to be transitive.
+
+    We want to enlarge $M$ to get a bigger model,
+    where our desired statement holds,
+    i.e.~add more reals to violate $\CH$.
+
+    However we need to do this in a somewhat controlled way,
+    so we can't just do it the way one builds field extensions.
+    In particular, when trying to violate $\CH$ we need to make sure that
+    we don't collapse cardinals.
+\end{goal}
+\begin{remark}
+    The idea behind forcing is clever.
+    Unfortunately an easy ``how could I have come up with this myself''-approach
+    does not seem to exist.
+\end{remark}
+\begin{remark}
+    How can a countable transitive model $M$ even exist?
+
+    $M$ believes some statements that are wrong from the outside perspective.
+    For example there exists $\aleph_1^M \in M$
+    such that $M \models x = \aleph_1$.
+    $\aleph_1^M$ is indeed an ordinal (since  being an ordinal is a  $\Sigma_0$-statement).
+    However $\aleph_1^M$ is countable, since $M$ is countable
+    and transitive.
+    This is fine.
+    (Note that ``$\aleph_1^M$ is uncountable'' is a $\Pi_1$-statement.)
+\end{remark}
+
+\begin{idea}[The method of \vocab{forcing}]
+    Start with $M$, a countable transitive model of $\ZFC$
+    and let $\mathbb{P} \in M$ be a partial order,
+    where $p \le q$ means that $p$ has ``more information''
+    than $q$.
+
+    A filter $g \subseteq \mathbb{P}$ is $\mathbb{P}$-generic over $M$
+    iff $g \cap D \neq  \emptyset$ for all dense $D \subseteq \mathbb{P}$,
+    $D \in M$.
+
+    Next steps:
+    \begin{enumerate}[(1)]
+        \item Define the \vocab{forcing extension} $M[g]$.
+        \item Show that $M[g] \models \ZFC$.
+        \item Determine other facts about (the theory of) $M[g]$.
+            This depends on the partial order  $\mathbb{P}$ we chose
+            in the beginning (and maybe $M$).
+    \end{enumerate}
+\end{idea}
+
+\begin{example}[Prototypical example]
+    Let $\bP = 2^{< \omega}, p \le q \mathop{:\iff} p \supseteq q$ be Cohen forcing,
+    often denoted $\bC$.
+
+    Let $M$ be a countable transitive model of $\ZFC$.
+    Since the definition of $\bC$ is simple enough,
+    $\bC \in M$.
+    Let $g$ be $\bC$-generic over $M$.
+    
+    \begin{claim}
+        \label{ex:cohen:c1}
+        For each $n \in \omega$,
+        the set $D_n \coloneqq  \{ p \in \bC : n \in  \dom(p)\}$
+        is dense.
+    \end{claim}
+    \begin{subproof}
+        This is trivial.
+    \end{subproof}
+    \begin{claim}
+        \label{ex:cohen:c2}
+        $D_n \in M$.
+    \end{claim}
+    \begin{subproof}
+        The definition of $D_n$ is absolute.
+    \end{subproof}
+    \begin{claim}
+        \label{ex:cohen:c3}
+        If $p,q \in g \cap D_n$,
+        then $p(n) = q(n)$.
+    \end{claim}
+    \begin{subproof}
+        $g$ is a filter, so $p$ and $q$ are compatible.
+        $p,q \in D_n$ makes sure that $p(n)$ and $q(n)$ are defined.
+    \end{subproof}
+    Let $x = \bigcup g$.
+    By \yaref{ex:cohen:c3}, $x \in 2^{\le \omega}$.
+    By \yaref{ex:cohen:c1} and \yaref{ex:cohen:c2},
+    we have $g \cap D_n \neq \emptyset$ for all $n < \omega$,
+    hence $n \in \dom(x)$ for all $n < \omega$.
+    So $x \in 2^{\omega}$.
+
+    \begin{claim}
+        Let $z \in 2^{\omega}$, $z \in M$.
+        Then $D^z = \{p \in \bC : \exists n \in \dom(p) .~p(n) \neq z(n)\} $
+        is dense.
+    \end{claim}
+    \begin{subproof}
+        Trivial.
+    \end{subproof}
+    \begin{claim}
+        $D^z \in M$ for all $z \in 2^{< \omega}$ with $z \in M$.
+        Therefore, $g \cap D^z \neq \emptyset$ for all $z \in M$,
+        $z\colon  2^{<\omega}$.
+        Hence $x \neq z$ for all $z \in M$, $z \in 2^{< \omega}$.
+        In other words $x \not\in M$.
+    \end{claim}
+
+    The new real $x$ does not do too much damage to $M$
+    when adding it.\footnote{We still need to make this precise.}
+    (Some reals would completely kill the model.)
+
+    Now let $\alpha$ be an ordinal in $M$.
+    Let
+    \begin{IEEEeqnarray*}{rCll}
+        \bC(\alpha) &\coloneqq& \{p \colon &\text{$p$ is a function with domain $\alpha$,}\\
+                    &&&\text{$p(\xi) \in \bC$ for all $\xi < \alpha$,}\\
+                    &&&\text{$\{\xi < \alpha : p(\xi) \neq \emptyset\}$ is finite}\}
+    \end{IEEEeqnarray*}
+    ($\alpha$ many copies of $\bC$ with \vocab{finite support}).
+    
+    For $p, q \in \bC(\alpha)$ define $p \le q :\iff \forall \xi < \alpha .~p(\xi) \supseteq q(\xi)$.
+    We have $\bC(\alpha) \in M$
+    
+    Let $g$ be $\bC(\alpha)$-generic over $M$.
+    Let $x_\xi = \bigcup \{p(\xi) : p \in g\}$
+    for $\xi < \alpha$.
+    $x_\xi \in 2^{ \omega}$:
+    For each $n < \omega$ and $\xi < \alpha$,
+    \[
+    D_{n,\xi} \coloneqq  \{ p \in \bC(\alpha) : n \in \dom(p(\xi))\}  \in M
+    \]
+    and $D_{n,\xi}$ is dense.
+
+    \begin{claim}
+        For all $\xi, \eta < \alpha$, $\xi \neq \eta$,
+        \[
+        D^{\xi, \eta} \coloneqq  \{ p \in \bC(\alpha) : \exists  n \in \dom(p(\xi)) \cap \dom(p(\eta)) .~
+        p(\xi)(n) \neq  p(\eta)(n)\}
+        \]
+        we have that $D^{\xi, \eta} \in M$
+        and is $D^{\xi, \eta}$ dense.
+    \end{claim}
+    Therefore if $\xi \neq \eta$, $x_\xi \neq x_\eta$.
+
+    Currently this is not very exciting,
+    since we only showed that for a countable transitive model $M$,
+    there is a countable set of reals not contained in $M$.
+    The interesting point will be, that we can actually add these reals
+    to $M$.
+\end{example}
+
+Next steps:
+\begin{itemize}
+    \item Make sense of $M[g]$.
+\end{itemize}

logic2.tex

@@ -44,6 +44,7 @@
 \input{inputs/lecture_18}
 \input{inputs/lecture_19}
 \input{inputs/lecture_20}
+\input{inputs/lecture_21}
 
 
 \cleardoublepage