lecture 20

inputs/lecture_19.tex

@@ -24,8 +24,8 @@ 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 $\phi(x_0,\ldots,x_n) \in \Pi_n$,
+    If $\phi(x_0,\ldots,x_m) \in \Pi_n$,
-    then $(\exists x_0.~\ldots\exists x_n.~\phi(x_0,\ldots,x_n) \in \Sigma_{n+1}$.
+    then $(\exists x_0.~\ldots\exists x_m.~\phi(x_0,\ldots,x_m) \in \Sigma_{n+1}$.
 
     $\Delta_n \coloneqq \Sigma_n \cap \Pi_n$.
 \end{definition}
@@ -47,6 +47,7 @@ and $\forall x \in y.~\phi$ abbreviates $\forall x.~x \in y \to \phi$.
 \end{notation}
 
 \begin{lemma}
+    \label{lem:d0absolute}
     Let $M$ be transitive, $\phi \in \Delta_0$
     and $a_0,\ldots, a_n \in M$.
     Then $M \models\phi(a_0,\ldots,a_n)$
@@ -56,7 +57,6 @@ and $\forall x \in y.~\phi$ abbreviates $\forall x.~x \in y \to \phi$.
     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 $M \models \lnot \phi_i \iff V \models \lnot \phi_i$
@@ -65,8 +65,8 @@ and $\forall x \in y.~\phi$ abbreviates $\forall x.~x \in y \to \phi$.
     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 $M$
+    so since $M$ is transitive,
-    and vice versa.
+    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
@@ -167,6 +167,7 @@ is a partially ordered set (\vocab{poset})
 if $\le $ is reflexive, symmetric nd transitive.
 
 \begin{definition}
+    \label{def:forcingwords}
     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 $p$.%
@@ -196,6 +197,7 @@ if $\le $ is reflexive, symmetric nd transitive.
 \end{definition}
 
 \begin{lemma}
+    \label{lem:egenfilter}
     Let $\cP = (P, \le )$ be a poset,
     $\cD$ a countable family of dense subsets of $P$
     and $p \in P$.

inputs/lecture_20.tex

@@ -0,0 +1,230 @@
+\lecture{20}{2024-01-15}{}
+
+\begin{idea}
+    We want to add a new object that satisfies certain condition.
+    The elements of the forcing notion correspond to approximations
+    of this object.
+
+    A filter picks some information which we want to be true.
+    Being a filter ensures that this information does not contradict itself.
+\end{idea}
+
+\begin{definition}
+    Assume that $M$ is a transitive model of $\ZFC$,
+    and $\bP \in M$ a poset.
+    $G \subseteq \bP$ is said to be \vocab{$M$-generic for $\bP$} 
+    if whenever $D \subseteq  \bP$ is dense
+    and in $M$, then $G \cap D \neq \emptyset$.
+\end{definition}
+\begin{remark}
+    That is the same as being $\{D \subseteq \bP \text{ dense } :  D \in M\}$-generic
+    with generic defined as in \yaref{def:forcingwords}.
+\end{remark}
+
+% \begin{remark}
+%     We can not prove that such a transitive model actually exists (we need inaccessible cardinals for that).
+%     However there are transitive models of finite fragments of $\ZFC$.
+%     % cf. Kuhnen, Chapter 7
+%     So proofs using tansitive models can be made more general by
+%     talking only about finite fragments, i.e.~a finite subset of the axioms of $\ZFC$ (since every proof can only
+%     use finitely many axioms).
+% \end{remark}
+
+\begin{definition}[Cohen Forcing]
+    \yalabel{Cohen Forcing}{Cohen Forcing}{def:cohenf}
+    Let $\bP$ be the set of finite partial function $p$ from $\omega$ to $2$,
+    i.e.~$\cP = 2^{<\omega}$.
+
+    The order on $\bP$ is described by $q \le  p \colon\iff q \supseteq p$.
+    $\bP$ is called the \vocab{Cohen forcing}.
+\end{definition}
+\begin{fact}
+    Assume $X \subseteq 2^\omega$ is countable,
+    Then there is $x \in 2^\omega \setminus X$.
+\end{fact}
+Of course we already know that, but let's use it
+to test our machinery:
+\begin{proof}
+    Assume that $X = \{x_n \colon  n \in \omega\}$
+    is an enumeration of $X$.
+    Let $D_n = \{ p \in \bP : \exists i \in \dom(\bP).~x_n(i) \neq p(i)\}$.
+    This makes sure that we get a ``new'' element
+    not belonging to $X$.
+
+    \begin{claim}
+        $D_n$ is dense in $\bP$.
+    \end{claim}
+    \begin{subproof}
+        Assume $q \in \bP$.
+        Let $i = 1 + \max(\dom(q))$.
+        Note that $i \not\in \dom(q)$.
+        Let $p = q \cup \{(i, 1-x_n(i))\}$.
+        Then $p \in D_n$.
+    \end{subproof}
+    Let $E_i = \{p \in \bP : i \in \dom(\bP)\}$.
+    This makes sure that our ``new'' element
+    is defined everywhere.
+    \begin{claim}
+        $\forall  i < \omega.~E_i \subseteq \bP \text{ is dense}$.
+    \end{claim}
+    \begin{subproof}
+        Assume $q \in \bP$.
+        If $ i \in \dom(q)$ pick $p = q \in E_i$.
+        If $i \not\in \dom(q)$,
+        let $p = q \cup \{(i,0)\} \in E_i$.
+    \end{subproof}
+    Let $\cD = \{D_n : n < \omega\} \cup \{E_i : i < \omega\}$.
+    This is a countable subset of dense sets.
+    By \yaref{lem:egenfilter}
+    there is a $\cD$-generic filter $G$.
+    Let $y = \bigcup G$.
+    Note that $y$ is a function, since any two elements of $G$ are compatible.
+\end{proof}
+Note that the ``new'' element did already exists, so we used forcing language
+to find it but didn't actually do anything.
+
+\begin{lemma}
+    Let $M$ be a transitive model of $\ZFC$
+    and let $\bP = (P, \le ) \in M$.
+
+    Let $D \subseteq  \bP$, $D \in M$, $p \in \bP$.
+    Then
+    \begin{enumerate}[(1)]
+        \item $\bP$ is a partial order iff $M \models \text{``$\bP$ is a partial order''}$.
+        \item $D$ is dense in  $\bP$ iff  $M \models \text{``$D$ is dense in $\bP$''}$.
+        \item $D$ is dense below  $p$ iff  $M \models \text{``$D$ is dense below $p$''}$
+            (this only makes sense if $p \in M$).
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    All the definitions are $\Delta_0$, so we can apply
+    \yaref{lem:d0absolute}.
+\end{proof}
+
+\begin{definition}
+    Assume that $M$ is a transitive model of $\ZFC$
+    and $\bP \in M$ is a poset.
+    $G \subseteq \bP$ is called a
+    \vocab{$\bP$-generic filter over $M$}
+    or  \vocab{$M$-generic filter for $\bP$} 
+    if
+    \[
+    \forall D \in M.~((D \subseteq \bP \text{ is dense}) \implies G \cap D \neq \emptyset).
+    \]
+\end{definition}
+\begin{corollary}
+    If $M$ is a countable transitive model of $\ZFC$,
+    $\bP \in M$ is a poset
+    and $p \in \bP$,
+    then there is an $M$-generic filter $G \subseteq \bP$
+    with $p \in G$.
+\end{corollary}
+\begin{remark}
+    The filter usually exists outside of $M$.
+    $M$ itself does not think that $M$ is countable,
+    since $M \models \ZFC$.
+    But from the outside, we see that $M$ is countable,
+    so we can find a filter.
+\end{remark}
+
+\begin{definition}
+    Assume that $\bP$ is a poset.
+    $\bP$ is said to be \vocab{atomless} 
+    if for all $p \in \bP$ there are $q,r \in \bP$
+    such that
+    \begin{enumerate}[(1)]
+        \item $q \le p$, $r \le  p$,
+        \item $q \bot r$.
+    \end{enumerate}
+\end{definition}
+\begin{example}
+    The \yaref{def:cohenf} is atomless.
+\end{example}
+Usually we are only interested in atomless partial orders.
+\begin{lemma}
+    Assume that $M$ is a transitive model of $\ZFC$,
+    $\bP \in M$ an atomless poset
+    and let $G \subseteq  \bP$ be $M$-generic for $\bP$.
+    Then $G \not\in M$.
+\end{lemma}
+\begin{proof}
+    Towards a contradiction assume $G \in M$.
+    Define $D\coloneqq  \bP \setminus G$.
+    We'll show that $D \subseteq \bP$ is dense,
+    which is a contradiction, since $G$ was assumed to be $M$-generic.
+    Let $q \in \bP$ and
+    let $r,s$ be two extensions of $q$
+    such that  $r \bot s$.
+    These exist because $\bP$ is atomless.
+    Since $G$ is a filter,
+    it can contain at most one of $\{r,s\}$,
+    wlog.~$s \not\in G$.
+    In particular, $s \in D$ and $s \le q$.
+    Hence $D$ is dense in $\bP$.
+\end{proof}
+
+\begin{lemma}
+    Assume that $M$ is a transitive model of $\ZFC$,
+    $\bP \in M$ a poset, $G \subseteq \bP$ an $M$-generic filter
+    and $p \in G$.
+
+    If $D$ is dense below $p$,
+    then $G \cap D \neq  \emptyset$.
+\end{lemma}
+\begin{proof}
+    Let $E = D \cup \{ q \in  \bP : q \bot p\}$.
+    $E \subseteq  \bP$ is dense:
+    Let $r \in  \bP$.
+    \begin{itemize}
+        \item If $r || p$ let $s \le r,p$.
+            Since $D$ is dense below $p$,
+            there exists $\overline{s} \in D$ such that $\overline{s} \le s$.
+            Since $D \subseteq E$, $\overline{s} \in E$.
+        \item If $r \bot p$, then it is obvious that $r \in E$.
+    \end{itemize}
+
+    Since $E \in M$, $G \cap E \neq \emptyset$.
+    \begin{IEEEeqnarray*}{rCl}
+        G \cap (D \cup \{q \in \bP : q \bot p\}) &\neq & \emptyset\\
+        \implies (G\cap D) \cup \underbrace{(G \cap \{q \in \bP : q \bot p\})}_{\emptyset} & \neq & \emptyset\\
+    \end{IEEEeqnarray*}
+\end{proof}
+
+\begin{definition}
+    Assume that $\bP$ is a poset.
+    \begin{enumerate}[(1)]
+        \item $A \subseteq \bP$ is said to be an \vocab{antichain} 
+            iff for all $p \neq q$ in $A$,
+            $p \bot q$.
+        \item  An antichain $A \subseteq \bP$ is a
+            \vocab{maximal antichain}
+            iff $\forall r \in \bP$, there exists $a \in \bP$
+            such that $p || r$.
+        \item $X \subseteq  \bP$ is said to be \vocab{open} 
+            if $\forall  p \in X.~\forall q \le p.~ q \in X$.
+    \end{enumerate}
+\end{definition}
+\begin{remark}
+    Note that if $A$ is a maximal antichain
+    in $\bP$, then it is maximal in $(\{A \subseteq \bP : A \text{ is an antichain}\}, \subseteq )$.
+    Using $\AxC$, every antichain can be extend to a maximal antichain.
+
+    The statement ``$A$ is an antichain'' is  $\Delta_0$.
+    
+    Note that ``every antichain of $\bP$ is countable'' is not necessarily
+    absolute between transitive models of $\ZFC$.
+    
+\end{remark}
+\begin{lemma}
+    Assume that $M$ is a transitive model of $\ZFC$,
+    $\bP \in M$ a poset and $G \subseteq \bP$ a filter.
+    Then the following are equivalent:
+    \begin{enumerate}[(1)]
+        \item $G$ is $\bP$-generic over $M$.
+        \item $G \cap A \neq \emptyset$ for every
+            maximal antichain $A \in M$.
+        \item $G \cap D \neq \emptyset$ for every dense
+            open $D \in M$ with $D \subseteq \bP$
+    \end{enumerate}
+\end{lemma}
+We'll prove this next time.