Josia Pietsch
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8.4 KiB
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233 lines
8.4 KiB
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\lecture{19}{2024-01-11}{Forcing}
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Beginning with this lecture,
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the material is no longer relevant for the exam.
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Recall that $\exists x \in y.~ \phi$ abbreviates
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$\exists x.~ x \in y \land \phi$
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and $\forall x \in y.~\phi$ abbreviates $\forall x.~x \in y \to \phi$.
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\begin{definition}[Arithmetical Hierarchy]
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Let $\phi$ be a $\cL_{\in}$-formula.
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We say that $\phi$ is \vocab{$\Delta_0$} (or \vocab{$\Sigma_0$}
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or \vocab{$\Pi_0$})
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iff it is in the smallest set $\Gamma$ of formulas
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such that
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\begin{enumerate}[(1)]
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\item $\Gamma$ contains all \vocab[Formula!atomic]{atomic} formulas ($x \in y$,$x=y$).
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\item If $\phi, \psi \in \Gamma$,
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then so are $\lnot \phi$
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and $\phi \land \psi$.%
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\footnote{It follows that $\phi \lor \psi$, $\phi \to \psi$ and $\phi \leftrightarrow \psi$
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are also in $\Gamma$.}
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\item If $\phi \in \Gamma$,
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then $(\exists x \in y.~\phi), (\forall x \in y.~\phi) \in \Gamma$.
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\end{enumerate}
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If $\phi(x_0,\ldots,x_m) \in \Sigma_n$,
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then $(\forall x_0.~\ldots\forall x_m.~\phi(x_0,\ldots,x_m)) \in \Pi_{n+1}$.
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If $\ZFC \models \phi \leftrightarrow \psi$
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and $\phi \in \Sigma_n$, then $\psi \in \Sigma_n$.
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If $\phi(x_0,\ldots,x_m) \in \Pi_n$,
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then $(\exists x_0.~\ldots\exists x_m.~\phi(x_0,\ldots,x_m) \in \Sigma_{n+1}$.
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If $\ZFC \models \phi \leftrightarrow \psi$
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and $\phi \in \Pi_n$, then $\psi \in \Pi_n$.
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$\Delta_n \coloneqq \Sigma_n \cap \Pi_n$.
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\end{definition}
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\begin{notation}
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Assume that $M$ is transitive
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and $\phi$ is sentence.
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Then
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\[
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M \models \phi
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\]
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means that $(M, \in\defon{M}) \models \phi$.
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If $ a_0,\ldots,a_n \in M$
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and $\phi(x_0,\ldots,x_n)$ is a $\cL_\in$-formula,
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then we say $M \models \phi(a_0,\ldots,a_n)$
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iff $M$ satisfies $\phi(x_0,\ldots,x_n)$
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for the assignment $x_i \mapsto a_i$.
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\end{notation}
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\begin{lemma}
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\label{lem:d0absolute}
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Let $M$ be transitive, $\phi \in \Delta_0$
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and $a_0,\ldots, a_n \in M$.
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Then $M \models\phi(a_0,\ldots,a_n)$
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iff $V \models\phi(a_0,\ldots,a_n)$.
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\end{lemma}
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\begin{proof}
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Clearly $M \models a_i \in a_j \iff V \models a_i \in a_j$
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and $M \models a_i = a_j \iff V \models a_i = a_j$,
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i.e.~the lemma holds for atomic $\phi$.
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It is clear that if $M \models \phi_i \iff V \models \phi_i, i = 1,2$,
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then also $M \models \lnot \phi_i \iff V \models \lnot \phi_i$
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and $M \models \phi_1 \land \phi_2 \iff V \models \phi_1 \land \phi_2$.
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Assume that the lemma holds for $\phi$.
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Then it also holds for $\exists a_i \in a_j.~\phi$:
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We have that $a_i \in a_j$ is atomic and by the assumption that the lemma holds for $\phi$
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so since $M$ is transitive,
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a witness can be transferred from $V$ to $M$ and vice versa.
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The case of $\forall a_i \in a_j.~\phi$ can be treated similarly.
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\end{proof}
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A similar arguments yields \vocab{upwards absoluteness} for $\Sigma_1$-formulas
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and \vocab{downwards absoluteness} for $\Pi_1$-formulas:
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\begin{lemma}
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\label{lem:pi1downardsabsolute}
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Let $M$ be transitive.
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Let $\phi(x_0,\ldots,x_n) \in \cL_\in$ and $a_0,\ldots,a_n \in M$.
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Then
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\begin{itemize}
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\item If $\phi$ is $\Sigma_1$, then
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\[
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M \models\phi(a_0,\ldots,a_n) \implies V \models \phi(a_0,\ldots,a_n).
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\]
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\item If $\phi$ is $\Pi_1$, then
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\[
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V \models\phi(a_0,\ldots,a_n) \implies M \models \phi(a_0,\ldots,a_n).
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\]
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\end{itemize}
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\end{lemma}
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\begin{definition}
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Assume that $T$ is a theory and $\phi \in \cL_\in $ a formula
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We say that $\phi$ is \vocab{$\Delta_1^T$}
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iff there are formulas $\psi$, $\tau$
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such that $\psi \in \Sigma_1$, $\tau \in \Pi_1$
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and
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\[
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T \vdash\phi \leftrightarrow \psi \leftrightarrow\tau.
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\]
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\end{definition}
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Again by a similar argument we get:
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\begin{lemma}
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Let $M$ be a transitive model of a theory $T$.
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Let $\phi$ be a $\Delta_1^T$ formula
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and $a_0,\ldots,a_n \in M$.
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Then $M \models \phi(a_0,\ldots,a_n) \iff V \models \phi(a_0,\ldots,a_n)$.
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\end{lemma}
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\begin{lemma}
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Let $\phi$ denote the statement ``$R$ is a well-founded relation''.
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Then $\phi \in \Delta_1^{\ZFC^-}$.
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\end{lemma}
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\begin{proof}
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$\phi$ is equivalent to
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\begin{itemize}
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\item $R$ is a relation ($\Delta_0$) and
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\item $\forall b.~b \cap \ran(R) = \emptyset \lor \exists x \in b.~\text{``$x$ is $R$-minimal''}$.
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\end{itemize}
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We only need to care about the second point.
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This is equivalent (using \AxC!) to
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the statement that there is no
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\[
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f\colon \omega \to \dom(R) \cup \ran(R) \text{ such that } \forall n < \omega.~f(n+1)Rf(n),
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\]
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which can be written as a $\Pi_1$-formula.
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With the help of ranks, we can also write it as a $\Sigma_1$-formula:
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\[
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\exists r\colon \OR \to \dom(R) \cup \ran(R).~
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\forall x \in \dom(R) \cup \ran(R).~ r(x) = \{\sup(r(y) +1) : y R x\}.
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\]
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So $\phi \in \Delta_1^{\ZFC^-}$.
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\end{proof}
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\begin{lemma}
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Assume that $M$ is transitive. Then
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\begin{enumerate}[(1)]
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\item $M \models \AxExt$.
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\item $M \models \AxFund$.
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\item If $\omega \in M$, then $M \models \AxInf$.
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\item If $M$ is closed under $(x,y) \mapsto \{x,y\}$,
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then $M \models \AxPair$.
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\item If $M$ is closed under $x \mapsto \bigcup x$,
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then $M \models \AxUnion$.
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\end{enumerate}
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\end{lemma}
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\begin{proof}
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\begin{enumerate}[(1)]
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\item Let $x, y \in M$
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such that $M \models \forall t .~t \in x \iff t \in y$.
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Since $M$ is transitive $V \models \forall t.~t \in x \iff t \in y$.
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Since $V \models \Ext$, we can apply
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$V \models x = y \iff M \models x = y$.
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\item We need to show $M \models \forall y \neq \emptyset.~\exists x \in y.~x \cap y = \emptyset$.
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Let $y \in M$. Since $V \models \AxFund$,
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$V \models \exists x \in y .~x \cap y = \emptyset$.
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Note that this is a $\Delta_0$-formula,
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hence $M \models \exists x \in y.~x \cap y = \emptyset$.
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\item By assumption $\omega \in M$.
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Since $M$ is transitive, we get $\omega \subseteq M$.
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Hence $\omega$ is a witness for $\AxInf$.
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\item Trivial.
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\item Trivial.
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\end{enumerate}
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\end{proof}
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\section{Forcing}
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Recall that a structure $\bP = (P, \le )$
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is a partially ordered set (\vocab{poset})
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if $\le $ is reflexive, symmetric and transitive.
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\begin{definition}
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\label{def:forcingwords}
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A non-empty poset $\bP = (P, \le )$ is called a \vocab{forcing notion}.
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The elements of $P$ are called \vocab{conditions}.
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If $q \le p$ we say that $q$ is \vocab{stronger} than $p$.%
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\footnote{i.e.~it carries more information.}
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$D \subseteq P$ is called \vocab{dense}
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iff $\forall p \in P.~\exists q \in D.~q \le p$.
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Let $p \in P, D \subseteq P$. Then $D$ is \vocab{dense below $p$}
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iff $\forall P \ni q \le p.~\exists r \in D.~r \le q$.
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$G \subseteq P$ is called a \vocab{filter}
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iff
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\begin{enumerate}[(1)]
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\item $\forall p,q \in G.~\exists r \in G.~r \le p \land r \le q$.
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\item $(p \in G \land p \le q) \implies q \in G$.
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\end{enumerate}
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For $p,q \in P$ we say that $p$ and $q$ are \vocab{compatible},
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$p || q$,
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iff $\exists r \in P .~r \le p \land r \le q$.
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Otherwise they are \vocab{incompatible}, $p \perp q$.
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Let $\cD$ be a family of dense subsets of $P$
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and $G$ a filter.
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We say that $G$ is \vocab{$\cD$}-generic
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iff $\forall D \in \cD.~G \cap D \neq \emptyset$.
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\end{definition}
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\begin{lemma}
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\label{lem:egenfilter}
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Let $\cP = (P, \le )$ be a poset,
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$\cD$ a countable family of dense subsets of $P$
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and $p \in P$.
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Then there exists a $\cD$-generic filter $G \subseteq P$
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such that $p \in G$.
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\end{lemma}
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\begin{proof}
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Fix $p$ as above.
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Let $\langle D_n : n < \omega \rangle$ be an enumeration of $\cD$.
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Let $p_0 \le p$ be such that $p_0 \in D_0$.
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If $p_n$ is given, let $p_{n+1} \le p_n$ be such that
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$p_{n+1} \in D_{n+1}$.
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This is possible since $\cD$ is a collection of dense sets.
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Define $ G \coloneqq \{ q \in P : \exists n.~p_n \le q\}$.
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$G$ is a filter:
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Let $r,q \in G$. Let $n_r, n_q < \omega$ such that $p_{n_r} \le r$
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and $p_{n_q} \le q$.
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Let $m = \max \{n_r, n_q\}$.
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Then $p_m$ is a common extension.
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Clearly $G$ is $\cD$-generic.
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\end{proof}
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