\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 $\ZFC \models \phi \leftrightarrow \psi$ 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} \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} \label{lem:d0absolute} Let $M$ be transitive, $\phi \in \Delta_0$ and $a_0,\ldots, a_n \in M$. Then $M \models\phi(a_0,\ldots,a_n)$ 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 $\phi$. 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} \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 $p$.% \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} \label{lem:egenfilter} 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}