19

inputs/lecture_13.tex

@@ -130,7 +130,7 @@
 The next goal is to show the following:
 (However the method might be more interesting than the result)
 \begin{theorem}[Silver]
-    \yalabel{Silver's Theorem}{Silver}{thm:silver}
+    \yalabel{Silver's Theorem}{Silver}{thm:silver1}
     If $2^{\aleph_\alpha} = \aleph_{\alpha + 1}$
     for all $\alpha < \omega_1$,
     then $2^{\aleph_{\omega_1}} = \aleph_{\omega_1 + 1}$.

inputs/lecture_18.tex

@@ -100,7 +100,7 @@
                 $j(\alpha)$ is an ordinal such that $j(\alpha) \ge \alpha$.
             \end{claim}
             \begin{subproof}
-                $\alpha \in \Ord$
+                $\alpha \in \OR$
                 can be written as
                 \[\forall  x \in \alpha .~\forall y \in  x.~y \in \alpha
                     \land \forall x \in \alpha .~ \forall y \in  \alpha.~(x \in y \lor x = y \lor y \in x).

inputs/lecture_19.tex

@@ -0,0 +1,222 @@
+\lecture{19}{2024-01-11}{Forcing}
+
+Recall that $\exists x \in y.~ \phi$ abbreviates
+$\exists x.~ x \in y \land \phi$
+and $\forall x \in y.~\phi$ abbreviates $\forall x.~x \in y \to \phi$.
+
+\begin{definition}[Arithmetical Hierarchy]
+    Let $\phi$ be a $\cL_{\in}$-formula.
+    We say that $\phi$ is  \vocab{$\Delta_0$} (or \vocab{$\Sigma_0$}
+    or \vocab{$\Pi_0$})
+    iff it is in the smallest set $\Gamma$ of formulas
+    such that
+    \begin{enumerate}[(1)]
+        \item $\Gamma$ contains all \vocab[Formula!atomic]{atomic} formulas ($x \in y$,$x=y$).
+        \item If $\phi, \psi \in \Gamma$,
+            then so are $\lnot \phi$
+            and $\phi \land \psi$.%
+            \footnote{It follows that $\phi \lor \psi$, $\phi \to \psi$ and $\phi \leftrightarrow \psi$
+                are also in $\Gamma$.}
+        \item If $\phi \in \Gamma$,
+            then $(\exists x \in y.~\phi), (\forall x \in y.~\phi) \in \Gamma$.
+    \end{enumerate}
+
+    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 $\phi(x_0,\ldots,x_n) \in \Pi_n$,
+    then $(\exists x_0.~\ldots\exists x_n.~\phi(x_0,\ldots,x_n) \in \Sigma_{n+1}$.
+
+    $\Delta_n \coloneqq \Sigma_n \cap \Pi_n$.
+\end{definition}
+
+\begin{notation}
+    Assume that $M$ is transitive
+    and $\phi$ is sentence.
+    Then
+    \[
+    M \models \phi
+    \]
+    means that $(M, \in\defon{M}) \models \phi$.
+
+    If $ a_0,\ldots,a_n \in M$
+    and $\phi(x_0,\ldots,x_n)$ is a  $\cL_\in$-formula,
+    then we say $M \models \phi(a_0,\ldots,a_n)$
+    iff $M$ satisfies $\phi(x_0,\ldots,x_n)$
+    for the assignment $x_i \mapsto a_i$.
+\end{notation}
+
+\begin{lemma}
+    Let $M$ be transitive, $\phi \in \Delta_0$
+    and $a_0,\ldots, a_n \in M$.
+    Then $M \models\phi(a_0,\ldots,a_n)$
+    iff $V \models\phi(a_0,\ldots,a_n)$.
+\end{lemma}
+\begin{proof}
+    Clearly $M \models a_i \in a_j \iff V \models a_i \in a_j$
+    and $M \models a_i = a_j \iff V \models a_i = a_j$,
+    i.e.~the lemma holds for atomic $\phi$.
+    % TODO transitivity needed?
+
+    It is clear that if $M \models \phi_i \iff V \models \phi_i, i = 1,2$,
+    then also $M \models \lnot \phi_i \iff V \models \lnot \phi_i$
+    and $M \models \phi_1 \land \phi_2 \iff V \models \phi_1 \land \phi_2$.
+
+    Assume that the lemma holds for $\phi$.
+    Then it also holds for $\exists a_i \in a_j.~\phi$:
+    We have that $a_i \in a_j$ is atomic and by the assumption that the lemma holds for $\phi$
+    so since $M$ is transitive, a witness can be transferred from $V$ to $M$
+    and vice versa.
+    The case of $\forall a_i \in a_j.~\phi$ can be treated similarly.
+\end{proof}
+A similar arguments yields \vocab{upwards absoluteness} for $\Sigma_1$-formulas
+and \vocab{downwards absoluteness} for $\Pi_1$-formulas:
+\begin{lemma}
+    Let $M$ be transitive.
+    Let $\phi(x_0,\ldots,x_n) \in \cL_\in$ and $a_0,\ldots,a_n \in M$.
+    Then
+    \begin{itemize}
+        \item If  $\phi$ is $\Sigma_1$, then
+            \[
+            M \models\phi(a_0,\ldots,a_n) \implies V \models \phi(a_0,\ldots,a_n).
+            \]
+        \item If $\phi$ is $\Pi_1$, then
+            \[
+            V \models\phi(a_0,\ldots,a_n) \implies M \models \phi(a_0,\ldots,a_n).
+            \]
+    \end{itemize}
+\end{lemma}
+\begin{definition}
+    Assume that $T$ is a theory and $\phi \in \cL_\in $ a formula
+    We say that $\phi$ is  \vocab{$\Delta_1^T$}
+    iff there are formulas $\psi$, $\tau$
+    such that $\psi \in \Sigma_1$, $\tau \in \Pi_1$
+    and
+    \[
+    T \vdash\phi \leftrightarrow \psi \leftrightarrow\tau.
+    \]
+\end{definition}
+Again by a similar argument we get:
+\begin{lemma}
+    Let $M$ be a transitive model of a theory $T$.
+    Let $\phi$ be a $\Delta_1^T$ formula
+    and $a_0,\ldots,a_n \in M$.
+    Then $M \models \phi(a_0,\ldots,a_n) \iff V \models \phi(a_0,\ldots,a_n)$.
+\end{lemma}
+\begin{lemma}
+    Let $\phi$ denote the statement ``$R$ is a well-founded relation''.
+    Then $\phi \in \Delta_1^{\ZFC^-}$.
+\end{lemma}
+\begin{proof}
+    $\phi$ is equivalent to
+     \begin{itemize}
+        \item $R$ is a relation ($\Delta_0$) and
+        \item $\forall b.~b \cap \ran(R) = \emptyset \lor \exists x \in b.~\text{``$x$ is $R$-minimal''}$.
+    \end{itemize}
+    We only need to care about the second point.
+    This is equivalent (using \AxC!) to
+    the statement that there is no
+    \[
+    f\colon \omega \to \dom(R) \cup \ran(R) \text{ such that } \forall n < \omega.~f(n+1)Rf(n),
+    \]
+    which can be written as a $\Pi_1$-formula.
+    With the help of ranks, we can also write it as a $\Sigma_1$-formula:
+    \[
+    \exists r\colon \OR \to \dom(R) \cup \ran(R).~
+    \forall x \in \dom(R) \cup \ran(R).~ r(x) = \{\sup(r(y) +1) : y R x\}.
+    \]
+
+    So $\phi \in \Delta_1^{\ZFC^-}$.
+\end{proof}
+\begin{lemma}
+    Assume that $M$ is transitive. Then
+    \begin{enumerate}[(1)]
+        \item $M \models \AxExt$.
+        \item $M \models \AxFund$.
+        \item If $\omega \in M$, then $M \models \AxInf$.
+        \item If $M$ is closed under $(x,y) \mapsto \{x,y\}$,
+            then $M \models \AxPair$.
+        \item If $M$ is closed under $x \mapsto \bigcup x$,
+            then $M \models \AxUnion$.
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    \begin{enumerate}[(1)]
+        \item Let $x, y \in M$
+            such that $M \models \forall t .~t \in  x \iff t \in y$.
+            Since $M$ is transitive $V \models \forall t.~t  \in x \iff t \in y$.
+            Since $V \models \Ext$, we can apply
+            $V \models x = y \iff M \models x = y$.
+        \item We need to show $M \models \forall y \neq \emptyset.~\exists x \in y.~x \cap y = \emptyset$.
+            Let $y \in M$. Since $V \models \AxFund$,
+            $V \models \exists x \in y .~x \cap y = \emptyset$.
+            Note that this is a $\Delta_0$-formula,
+            hence $M \models \exists x \in y.~x \cap y = \emptyset$.
+        \item By assumption $\omega \in M$.
+            Since $M$ is transitive, we get $\omega \subseteq M$.
+            Hence $\omega$ is a witness for $\AxInf$.
+        \item Trivial.
+        \item Trivial.
+    \end{enumerate}
+\end{proof}
+
+
+\section{Forcing}
+Recall that a structure $\bP = (P, \le )$
+is a partially ordered set (\vocab{poset})
+if $\le $ is reflexive, symmetric nd transitive.
+
+\begin{definition}
+    A non-empty poset $\bP = (P, \le )$ is called a  \vocab{forcing notion}.
+    The elements of $P$ are called \vocab{conditions}.
+    If $q \le p$ we say that $q$ is \vocab{stronger} than $p$.%
+    \footnote{i.e.~it carries more information.}
+    $D \subseteq P$ is called \vocab{dense}
+    iff $\forall p \in P.~\exists q \in D.~q \le p$.
+
+    Let $p \in P, D \subseteq P$. Then $D$ is \vocab{dense below $p$}
+    iff $\forall  P \ni q \le p.~\exists r \in D.~r \le q$.
+
+    $G \subseteq  P$ is called a \vocab{filter}
+    iff
+    \begin{enumerate}[(1)]
+        \item $\forall p,q \in G.~\exists r \in G.~r \le p \land r \le q$.
+        \item $(p \in G \land p \le q) \implies q \in G$.
+    \end{enumerate}
+
+    For $p,q \in P$ we say that $p$ and $q$ are \vocab{compatible},
+    $p || q$,
+    iff $\exists r \in P .~r \le p \land r \le  q$.
+    Otherwise they are \vocab{incompatible}, $p \perp q$.
+
+    Let $\cD$ be a family of dense subsets of $P$
+    and $G$ a filter.
+    We say that $G$ is \vocab{$\cD$}-generic
+    iff $\forall  D \in \cD.~G \cap D \neq \emptyset$.
+\end{definition}
+
+\begin{lemma}
+    Let $\cP = (P, \le )$ be a poset,
+    $\cD$ a countable family of dense subsets of $P$
+    and $p \in P$.
+    Then there exists a $\cD$-generic filter $G \subseteq  P$
+    such that $p \in G$.
+\end{lemma}
+\begin{proof}
+    Fix $p$ as above.
+    Let $\langle D_n : n < \omega \rangle$ be an enumeration of $\cD$.
+    Let $p_0 \le  p$ be such that $p_0 \in D_0$.
+    If $p_n$ is given, let $p_{n+1} \le p_n$ be such that
+    $p_{n+1} \in  D_{n+1}$.
+    This is possible since $\cD$ is a collection of dense sets.
+    Define $ G \coloneqq \{ q \in P : \exists n.~p_n \le q\}$.
+
+    $G$ is a filter:
+    Let $r,q \in G$. Let $n_r, n_q < \omega$ such that $p_{n_r} \le r$
+    and $p_{n_q} \le q$.
+    Let $m = \max \{n_r, n_q\}$.
+    Then $p_m$ is a common extension.
+
+    Clearly $G$ is $\cD$-generic.
+\end{proof}
+

logic.sty

@@ -129,7 +129,7 @@
 \DeclareSimpleMathOperator{CH}
 \DeclareSimpleMathOperator{GCH}
 \DeclareSimpleMathOperator{DC}
-\DeclareSimpleMathOperator{Ord}
+%\DeclareSimpleMathOperator{Ord}
 \DeclareSimpleMathOperator{OR} % Ordinals
 \DeclareSimpleMathOperator{trcl}
 \DeclareSimpleMathOperator{tcl}

logic2.tex

@@ -42,6 +42,7 @@
 \input{inputs/lecture_16}
 \input{inputs/lecture_17}
 \input{inputs/lecture_18}
+\input{inputs/lecture_19}
 
 
 \cleardoublepage