\lecture{19}{2024-01-11}{Forcing}

Beginning with this lecture,
the material is no longer relevant for the exam.

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}
    \label{lem:pi1downardsabsolute}
    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 and 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}