lecture 23

inputs/intro.tex

@@ -21,3 +21,5 @@ please send me a message:\\
 These notes follow the way the material was presented in the lecture rather
 closely. Additions (e.g.~from exercise sheets)
 and slight modifications have been marked with $\dagger$.
+
+Cut off for the exam is Christmas.

inputs/lecture_23.tex

@@ -0,0 +1,130 @@
+\lecture{23}{2024-01-25}{Negation of CH is consistent to ZFC}
+
+\begin{goal}
+    We want to construct a model of $\ZFC$
+    such that  $2^{\aleph_0} \ge  \aleph_2$.
+    % = ?
+\end{goal}
+
+Let $M$ be a countable transitive model of $\ZFC$.
+Suppose that  $M \models \CH$
+(otherwise we are done).
+
+Let $\alpha = \omega_2^M$.
+
+Let $\bC(\alpha) \coloneqq  \{p : p\colon  \alpha \to  \bC \text{ is a function such that } \{\xi < \alpha : p(\xi) \neq \emptyset\} \text{ is finite} \}$,
+ordered by $p \le_{\bC(\alpha)} q$ iff
+$p(\xi) \le_{\bC} q(\xi)$ for all $\xi < \alpha$.
+
+Recall that $\bC$ is the set of finite sequences of natural numbers
+ordered by  $p \le_{\bC} q$ iff $p \supseteq q$.
+
+Let $g $ be $\bC(\alpha)$-generic over $M$.
+For $\xi < \alpha$ let $x_\xi = \bigcup \{ p(\xi) : p \in g\}$.
+We have already seen that $x_\xi\colon \omega \to  \omega$
+is a function
+and $x_\xi \neq  x_\eta$ for $\xi \neq \eta$.
+
+We have $M[g] \models\ZFC$.\footnote{We only handwaved this step.}
+As $g \in M[g]$,
+we have $\langle x_\xi : \xi < \alpha\rangle \in M[g]$.
+Therefore $M[g] \models \text{``$2^{\aleph_0} \ge \alpha$''}$.
+Also $\alpha = \omega_2^M$.
+However the proof is not finished yet,
+since we need to make sure,
+that $M[g]$ does not collapse cardinals.
+
+We only have $M[g] \models 2^{\aleph_0} \ge \aleph_2^M$,
+i.e.~we need to see $\aleph_2^{M[g]} = \aleph_2^M$.
+
+\begin{claim}
+    Every cardinal of $M$ is still
+    a cardinal of $M[g]$.
+\end{claim}
+This suffices, because
+then $\aleph_0^M = \aleph_0^{M[g]}$,
+$\aleph_1^M = \aleph_1^{M[g]}$,
+$\aleph_2^{M}= \aleph_2^{M[g]}$, $\ldots$
+
+\begin{definition}
+    Let $(\bP, \le )$ be a partial order.
+    We say that $\bP$ has the
+    \vocab{countable chain condition} 
+    (\vocab{c.c.c.})%
+    \footnote{it should really be the ``countable antichain condition''}
+    iff there is no uncountable
+    antichain,
+    i.e.~every uncountable $V \subseteq \bP$
+    contains compatible $p \neq q$.
+\end{definition}
+
+We shall prove:
+\begin{claim}
+    \label{l23:c:1}
+    For all $\beta$, $\bC(\beta)$ has the c.c.c.
+\end{claim}
+\begin{claim}
+    \label{l23:c:2}
+    If $\bP \in M$ and
+        $M \models \text{`` $\bP$ has the c.c.c.''}$
+        and $h$ is generic over $M$,
+        then all $M$-cardinals are still
+        $M[h]$ cardinals.
+        \footnote{Being a cardinal is  $\Pi_1$,
+            so $M[h]$ cardinals are always $M$ cardinals.}
+\end{claim}
+\begin{refproof}{l23:c:2}
+    Suppose not.
+    Let $\kappa$ be minimal such that $M \models \text{``$\kappa$ is a cardinal''}$,
+    but $M[h] \models \text{``$\kappa$ is not a cardinal''}$.
+    Then $\kappa = (\lambda^+)^M$ for some unique $M$-cardinal $\lambda < \kappa$.
+    By minimality, $\lambda$ is also an $M[h]$-cardinal.
+
+    Let $f \in M[h]$ be such that $M[h] \models \text{``$f$ is a surjection from $\lambda$ onto $\kappa$''}$.
+    There is a name $\tau \in M^{\bP}$ with $\tau^h = f$.
+
+    We then have some $p \in h$
+    with $p  \Vdash_M^\bP \text{``$\tau$ is a surjection from $\check{\lambda}$ onto $\check{\kappa}$''}$.
+
+    Let $\xi < \lambda$.
+    Consider $X_\xi \coloneqq  \{\eta < \kappa: \exists  q \le_{\bP} p .~q  \Vdash \tau(\check{\xi}) = \check{\eta}\} \in M$.
+
+    $X_\xi$ is countable in $M$ by the following argument (in $M$):
+    For every $\eta \in X_\xi$,
+    let  $q_\eta \le p$ be such that $q_\xi  \Vdash^{\bP}_M \tau(\check{\xi}) = \check{\eta}$.
+    The set $\{q_\eta : \eta \in X_\xi\}$ is an antichain
+    as for $\eta_1 \neq \eta_2$ we have that $q_{\eta_i}  \Vdash \tau(\check{\xi}) = \check{\eta_i}$,
+    so they are not compatible.
+    So $\{q_\eta : \eta \in X_\xi\}$ is countable by the c.c.c.
+    Thus $X_\xi$ is countable.
+
+
+
+    Therefore we may define a function in $M$
+    \begin{IEEEeqnarray*}{rCl}
+        F\colon \lambda \times \omega &\longrightarrow & \kappa
+    \end{IEEEeqnarray*}
+    such that for all $\xi < \lambda$
+    \[
+    \{F(\xi,n) : n < \omega\}  = X_\xi.
+    \]
+    
+    $F$ is surjective since $f$ is surjective:
+    For $\eta < \kappa$,
+    there is some $\xi < \lambda$ such that $M[h] \models \text{``$f(\xi) = \eta$''}$,
+    there is some $\overline{q} \in h$ with $\overline{q}  \Vdash ^{\bP}_M \tau(\check{\xi}) = \check{\eta}$.
+    Pick $q \le \overline{q},p$.
+    This shows $\eta \in X_\xi$
+    hence $\eta = F(\xi, n)$ for some $n$.
+    But $|\lambda \times \omega| = |\lambda| = \lambda$,
+    so in $M$ there is a surjection $F' \colon  \lambda \to  \kappa$,
+    but $\kappa$ is a cardinal in $M$ $\lightning$.
+\end{refproof}
+
+\begin{refproof}{l23:c:1}
+    Omitted.
+    % TODO combinatorial argument
+\end{refproof}
+
+
+

logic2.tex

@@ -46,6 +46,7 @@
 \input{inputs/lecture_20}
 \input{inputs/lecture_21}
 \input{inputs/lecture_22}
+\input{inputs/lecture_23}
 
 
 \cleardoublepage