\lecture{22}{2024-01-22}{More Forcing}

\begin{warning}+
    Forcing will not be relevant for the exam.
    Because of a lack of time, this is more of an outlook
    than a thorough presentation of the material.
\end{warning}

For the rest of the section, let us fix
a transitive model $M$ of $\ZFC$
a partial order $\mathbb{P}$
and an  $M$-generic filter $g$.

\begin{definition}[$\mathbb{P}$-names]
    For an ordinal $\alpha \in M$%
    \footnote{Recall that $\Ord_M = \Ord \cap M$.},
    let $M^{\mathbb{P}}_\alpha$,
    the \vocab{$\mathbb{P}$-names} in $M$ of rank $\le \alpha$,
    be defined as follows:

    \[
        \tau \in M_{\alpha}^{\mathbb{P}} :\iff
        \tau \in M \land
        \tau \subseteq  \mathbb{P} \times  \bigcup \{M_\beta^{\mathbb{P}}: \beta < \alpha\},
    \]
    i.e.~the elements of $\tau \in  M_\alpha^{\mathbb{P}}$
    are of the form $(p, \sigma)$,
    where $p \in \cP$ and $\sigma \in M^{\mathbb{P}}_\beta$
    for some $\beta < \alpha$.

    Finally $M^{\mathbb{P}} = \bigcup \{M_\alpha^{\mathbb{P}} : \alpha \in M\}$.
\end{definition}

Let $R$ be the relation on $M^{\mathbb{P}}$ defined by
$\sigma R \tau$ iff $\exists  p \in \mathbb{P}.~(p,\sigma) \in \tau$.
If $\tau \in M^\mathbb{P}$
and $(p, \sigma) \in \tau$,
then $\sigma \in \{p, \sigma\} \in (p,\sigma) \in \tau$,
so the relation $R$
is well founded.
\begin{definition}
    Let $\tau \in M_\alpha^{\mathbb{P}}$.
    Then $\tau^g$,
    the \vocab{$g$-interpretation of $\tau$},
    is defined to be
    \[
    \{\sigma^g : \exists  p \in g .~(p,\sigma) \in \tau\}.
    \]
\end{definition}

\begin{definition}
    $M[g]$,
    the forcing extension of $M$
    given by  $g$,
    is
    \[
    \{ \tau^g : \tau \in M^{\mathbb{P}}\} .
    \]
\end{definition}
\begin{lemma}
    $M[g]$ is transitive.
\end{lemma}
\begin{proof}
    Trivial!
\end{proof}
\begin{lemma}
    $M \cup \{g\} \subseteq  M[g]$.
\end{lemma}
\begin{proof}
    For all $x \in M$ we need to find a name \vocab{$\check{x}$}
    such that $\check{x}^g = x$.

    We can recursively (along $\in$) define
    \[
    \check{x} = \{(p, \check{y}) : p \in \mathbb{P} \land y \in x\}.
    \]
    By induction, $\check{x} \in M$ for all $x \in M$.
    \begin{claim}
        $\check{x}^g = x$.
    \end{claim}
    \begin{subproof}
        Recall that $\mathbb{P} \neq \emptyset$.
        Inductively, we get
        \begin{IEEEeqnarray*}{rCl}
            \check{x}^g &=& \{\check{y}^g : \exists  p \in  g.~(p,\check{y}) \in \check{x}\}\\
                        &\overset{\text{induction}}{=}& \{y : \exists  p \in  g.~(p,\check{y}) \in \check{x}\}\\
                        &\overset{\text{definition of $\check{x}$}}{=}& \{y : y \in x\} = x.
        \end{IEEEeqnarray*}
    \end{subproof}

    So $M \subseteq M[g]$.


    We also need a name for $g$.
    Let \vocab{$\dot{g}$}$ \coloneqq  \{(p, \check{p}) : p \in \mathbb{P}\}$.

    Indeed
    \begin{IEEEeqnarray*}{rCl}
        \dot{g}^g &=& \{\check{p}^g : \exists p \in g.~(p, \check{p}) \in \dot{g}\}\\
                  &=& \{p : p \in g\} = g.
    \end{IEEEeqnarray*}
\end{proof}


\begin{lemma}
    \label{lem:mgmodelexfundinfpairunion}
    $M[g] \models \AxExt, \AxFund, \AxInf, \AxPair, \AxUnion$.
\end{lemma}
\begin{proof}
    \begin{itemize}
        \item \AxExt:

            The formula $\forall x.~\forall y .~((\forall  z \in x.~z \in y \land \forall z \in y.~z \in x) \to  x = y)$
            is $\Pi_1$, hence it is true in $M[g]$
            by %TODO REF downward absolutenes.
        \item \AxFund:
            Again,
             \[
            \forall x.~(\exists  y \in x .~y = y \to  \exists  y \in x.~\forall z \in y.~z \not\in x)
            \]
            is $\Pi_1$.
        \item \AxInf
            can be written as
            \[
            \exists x .~(\underbrace{\neq  \in x \land \forall  y \in x.y \cup \{y\}  \in x}_{\Sigma_0}).
            \]
            We have $ \omega \in M \subseteq M[g]$,
            so $M[g] \models \AxInf$.
        \item \AxPair:
            Let us assume $x,y \in M[g]$,
            say $x = \tau^g$ and $y = \sigma^g$.
            Let  $\pi = \{(p,\tau) : p \in \mathbb{P} \} \cup \{(p,\sigma) : p \in \mathbb{P}\} \in M^{\mathbb{P}}$.
            Then $\pi^g = \{\tau^g, \sigma^g\} = \{x,y\}$,
            so $\{x,y\} \in M[g]$.
            As a $\cL_\in$-statement,
            $z = \{x,y\}$ is $\Sigma_0$,
            so $M[g] \models \text{``$\{x,y\}$ is the pair of $x$ and $y$''}$.
            Hence $M[g] \models \AxPair$.
        \item \AxUnion:
            Similar to \AxPair.\gist{\todo{Exercise}}{}
    \end{itemize}
\end{proof}

Still missing are
\begin{itemize}
    \item \AxPow,
    \item \AxAus,
    \item \AxRep,
    \item \AxC.
\end{itemize}

\begin{definition}[\vocab{Forcing relation}]
    Let $M$ be a countable transitive model of $\ZFC$
    and let $\mathbb{P} \in M$ be a partial order.
    Let $p \in \mathbb{P}$
    and let $\phi$ be a $\cL_{\in }$-formula.

    Let $\tau_1,\ldots, \tau_k \in M^{\mathbb{P}}$ be names.

    We say that
    $p$ \vocab[force]{forces}  $\phi(\tau_1,\ldots, \tau_k)$,
    \[
        p  \Vdash^{\mathbb{\cP}}_{M} \phi(\tau_1, \ldots, \tau_k),
    \]
    if for all $h \subseteq \mathbb{P}$
    which are $\mathbb{P}$-generic over $M$
    with $p \in h$,
    \[
    M[h] \models\phi(\tau_1^h, \ldots, \tau_k^h).
    \]
\end{definition}
\begin{theorem}
    Fix an $\cL_\in$-formula $\phi$.
    Then the relation
    \[
        R = \{(p,\tau_1,\ldots,\tau_k : p  \Vdash ^{\mathbb{P}}_M \phi(\tau_1,\ldots,\tau_k)\}
    \]
    is definable over $M$
    (in the parameter $\mathbb{P}$).
\end{theorem}
\begin{proof}
    Omitted.
\end{proof}

\begin{theorem}[\vocab{Forcing Theorem}]
    Let $M$,  $\mathbb{P}$, $g$,
    be as above,
    let $\phi$ be a formula,
    and let $\tau_1, \ldots, \tau_k \in M^{\mathbb{P}}$.
    Then the following are equivalent:
    \begin{enumerate}[(1)]
        \item $M[g] \models \phi(\tau_1^g, \ldots, \tau_k^g)$.
        \item There is some $p \in g$ with
            \[
            p  \Vdash^{\mathbb{P}}_M \phi(\tau_1,\ldots, \tau_k).
            \]
            
    \end{enumerate}
\end{theorem}
\begin{proof}
    Omitted.
\end{proof}

\begin{theorem}
    $M[g] \models \ZFC$.
\end{theorem}
\begin{proof}
    We have already shown a part of this in
    \yaref{lem:mgmodelexfundinfpairunion}.

    Let us show that $M[g] \models \AxAus$,
    the rest is similar and left as an exercise.%
    \footnote{or done next semester in Logic IV!}

    Let $\phi$ be a formula,
    let $a, x_1,\ldots,x_k \in M[g]$.
    We need to see
    \[M[g] \models \exists y.~y = \{z  \in a : \phi(z, x_1,\ldots,x_k)\}.\]

    If suffices to show that there is some $y \in M[g]$
    with $y = \{ z \in a : M[g] \models \phi(z, x_1,\ldots,x_k)\}$.

    For this, let us construct a name for  $y$.
    Let $a = \tau^g$, $x_i = \sigma_i^g$.

    Let
    \[
    \pi = \{(p,\rho) : \exists \overline{p} > p.~(\overline{p}, \rho) \in \tau \land
    p  \Vdash^{\mathbb{P}}_M \phi(\rho, \sigma_1, \ldots, \sigma_k)\}.
    \]

    We have $\pi \in M$, since the relation
    $ \Vdash^{\mathbb{P}}_M$ can be defined in $M$.

    Let $z \in a$ such that $M[g] \models \phi(z, x_1,\ldots,x_n)$.
    We have $z = \rho^g$
    for some  $\rho$
    and there is  $\overline{p} \in g$ with $(\overline{p}, \rho) \in \pi$.
    Now $M[g] = \phi(\rho^g, \sigma_1^g,\ldots\sigma_k^g)$.

    Let $p'  \Vdash^{\mathbb{P}}_M \phi(\rho, \sigma_1,\ldots, \sigma_k)$,
    where $p' \in g$.
    We have $p', \overline{p} \in g$,
    so there is some $p \le  p', \overline{p}$ with $p \in g$.
    Then $(p,\rho) \in \pi$,
    so $\rho^g \in \pi^g$.

    This shows that
    \[
    \{z \in a : M[g] \models \phi(z,x_1,\ldots,x_k)\}  \subseteq \pi^g.
    \]
    The other inclusion is easy.

\end{proof}