\lecture{21}{2024-01-18}{}

\begin{goal}
    We want to show that certain statements are consistent with $\ZFC$
    (or  $\ZF$), for instance  $\CH$.

    We start with a model $M$ of $\ZFC$.
    Usually we want $M$ to be transitive.

    We want to enlarge $M$ to get a bigger model,
    where our desired statement holds,
    i.e.~add more reals to violate $\CH$.

    However we need to do this in a somewhat controlled way,
    so we can't just do it the way one builds field extensions.
    In particular, when trying to violate $\CH$ we need to make sure that
    we don't collapse cardinals.
\end{goal}
\begin{remark}
    The idea behind forcing is clever.
    Unfortunately an easy ``how could I have come up with this myself''-approach
    does not seem to exist.
\end{remark}
\begin{remark}
    How can a countable transitive model $M$ even exist?

    $M$ believes some statements that are wrong from the outside perspective.
    For example there exists $\aleph_1^M \in M$
    such that $M \models x = \aleph_1$.
    $\aleph_1^M$ is indeed an ordinal (since  being an ordinal is a  $\Sigma_0$-statement).
    However $\aleph_1^M$ is countable, since $M$ is countable
    and transitive.
    This is fine.
    (Note that ``$\aleph_1^M$ is uncountable'' is a $\Pi_1$-statement.)
\end{remark}

\begin{idea}[The method of \vocab{forcing}]
    Start with $M$, a countable transitive model of $\ZFC$
    and let $\mathbb{P} \in M$ be a partial order,
    where $p \le q$ means that $p$ has ``more information''
    than $q$.

    A filter $g \subseteq \mathbb{P}$ is $\mathbb{P}$-generic over $M$
    iff $g \cap D \neq  \emptyset$ for all dense $D \subseteq \mathbb{P}$,
    $D \in M$.

    Next steps:
    \begin{enumerate}[(1)]
        \item Define the \vocab{forcing extension} $M[g]$.
        \item Show that $M[g] \models \ZFC$.
        \item Determine other facts about (the theory of) $M[g]$.
            This depends on the partial order  $\mathbb{P}$ we chose
            in the beginning (and maybe $M$).
    \end{enumerate}
\end{idea}

\begin{example}[Prototypical example]
    Let $\bP = 2^{< \omega}, p \le q \mathop{:\iff} p \supseteq q$ be Cohen forcing,
    often denoted $\bC$.

    Let $M$ be a countable transitive model of $\ZFC$.
    Since the definition of $\bC$ is simple enough,
    $\bC \in M$.
    Let $g$ be $\bC$-generic over $M$.
    
    \begin{claim}
        \label{ex:cohen:c1}
        For each $n \in \omega$,
        the set $D_n \coloneqq  \{ p \in \bC : n \in  \dom(p)\}$
        is dense.
    \end{claim}
    \begin{subproof}
        This is trivial.
    \end{subproof}
    \begin{claim}
        \label{ex:cohen:c2}
        $D_n \in M$.
    \end{claim}
    \begin{subproof}
        The definition of $D_n$ is absolute.
    \end{subproof}
    \begin{claim}
        \label{ex:cohen:c3}
        If $p,q \in g \cap D_n$,
        then $p(n) = q(n)$.
    \end{claim}
    \begin{subproof}
        $g$ is a filter, so $p$ and $q$ are compatible.
        $p,q \in D_n$ makes sure that $p(n)$ and $q(n)$ are defined.
    \end{subproof}
    Let $x = \bigcup g$.
    By \yaref{ex:cohen:c3}, $x \in 2^{\le \omega}$.
    By \yaref{ex:cohen:c1} and \yaref{ex:cohen:c2},
    we have $g \cap D_n \neq \emptyset$ for all $n < \omega$,
    hence $n \in \dom(x)$ for all $n < \omega$.
    So $x \in 2^{\omega}$.

    \begin{claim}
        Let $z \in 2^{\omega}$, $z \in M$.
        Then $D^z = \{p \in \bC : \exists n \in \dom(p) .~p(n) \neq z(n)\} $
        is dense.
    \end{claim}
    \begin{subproof}
        Trivial.
    \end{subproof}
    \begin{claim}
        $D^z \in M$ for all $z \in 2^{< \omega}$ with $z \in M$.
        Therefore, $g \cap D^z \neq \emptyset$ for all $z \in M$,
        $z\colon  2^{<\omega}$.
        Hence $x \neq z$ for all $z \in M$, $z \in 2^{< \omega}$.
        In other words $x \not\in M$.
    \end{claim}

    The new real $x$ does not do too much damage to $M$
    when adding it.\footnote{We still need to make this precise.}
    (Some reals would completely kill the model.)

    Now let $\alpha$ be an ordinal in $M$.
    Let
    \begin{IEEEeqnarray*}{rCll}
        \bC(\alpha) &\coloneqq& \{p \colon &\text{$p$ is a function with domain $\alpha$,}\\
                    &&&\text{$p(\xi) \in \bC$ for all $\xi < \alpha$,}\\
                    &&&\text{$\{\xi < \alpha : p(\xi) \neq \emptyset\}$ is finite}\}
    \end{IEEEeqnarray*}
    ($\alpha$ many copies of $\bC$ with \vocab{finite support}).
    
    For $p, q \in \bC(\alpha)$ define $p \le q :\iff \forall \xi < \alpha .~p(\xi) \supseteq q(\xi)$.
    We have $\bC(\alpha) \in M$
    
    Let $g$ be $\bC(\alpha)$-generic over $M$.
    Let $x_\xi = \bigcup \{p(\xi) : p \in g\}$
    for $\xi < \alpha$.
    $x_\xi \in 2^{ \omega}$:
    For each $n < \omega$ and $\xi < \alpha$,
    \[
    D_{n,\xi} \coloneqq  \{ p \in \bC(\alpha) : n \in \dom(p(\xi))\}  \in M
    \]
    and $D_{n,\xi}$ is dense.

    \begin{claim}
        For all $\xi, \eta < \alpha$, $\xi \neq \eta$,
        \[
        D^{\xi, \eta} \coloneqq  \{ p \in \bC(\alpha) : \exists  n \in \dom(p(\xi)) \cap \dom(p(\eta)) .~
        p(\xi)(n) \neq  p(\eta)(n)\}
        \]
        we have that $D^{\xi, \eta} \in M$
        and is $D^{\xi, \eta}$ dense.
    \end{claim}
    Therefore if $\xi \neq \eta$, $x_\xi \neq x_\eta$.

    Currently this is not very exciting,
    since we only showed that for a countable transitive model $M$,
    there is a countable set of reals not contained in $M$.
    The interesting point will be, that we can actually add these reals
    to $M$.
\end{example}

Next steps:
\begin{itemize}
    \item Make sense of $M[g]$.
\end{itemize}