w23-logic-2/inputs/lecture_20.tex

\lecture{20}{2024-01-15}{}

\begin{idea}
    We want to add a new object that satisfies certain condition.
    The elements of the forcing notion correspond to approximations
    of this object.

    A filter picks some information which we want to be true.
    Being a filter ensures that this information does not contradict itself.
\end{idea}

\begin{definition}
    Assume that $M$ is a transitive model of $\ZFC$,
    and $\bP \in M$ a poset.
    $G \subseteq \bP$ is said to be \vocab{$M$-generic for $\bP$} 
    if whenever $D \subseteq  \bP$ is dense
    and in $M$, then $G \cap D \neq \emptyset$.
\end{definition}
\begin{remark}
    That is the same as being $\{D \subseteq \bP \text{ dense } :  D \in M\}$-generic
    with generic defined as in \yaref{def:forcingwords}.
\end{remark}

% \begin{remark}
%     We can not prove that such a transitive model actually exists (we need inaccessible cardinals for that).
%     However there are transitive models of finite fragments of $\ZFC$.
%     % cf. Kuhnen, Chapter 7
%     So proofs using tansitive models can be made more general by
%     talking only about finite fragments, i.e.~a finite subset of the axioms of $\ZFC$ (since every proof can only
%     use finitely many axioms).
% \end{remark}

\begin{definition}[Cohen Forcing]
    \yalabel{Cohen Forcing}{Cohen Forcing}{def:cohenf}
    Let $\bP$ be the set of finite partial function $p$ from $\omega$ to $2$,
    i.e.~$\cP = 2^{<\omega}$.

    The order on $\bP$ is described by $q \le  p \colon\iff q \supseteq p$.
    $\bP$ is called the \vocab{Cohen forcing}.
\end{definition}
\begin{fact}
    Assume $X \subseteq 2^\omega$ is countable,
    Then there is $x \in 2^\omega \setminus X$.
\end{fact}
Of course we already know that, but let's use it
to test our machinery:
\begin{proof}
    Assume that $X = \{x_n \colon  n \in \omega\}$
    is an enumeration of $X$.
    Let $D_n = \{ p \in \bP : \exists i \in \dom(\bP).~x_n(i) \neq p(i)\}$.
    This makes sure that we get a ``new'' element
    not belonging to $X$.

    \begin{claim}
        $D_n$ is dense in $\bP$.
    \end{claim}
    \begin{subproof}
        Assume $q \in \bP$.
        Let $i = 1 + \max(\dom(q))$.
        Note that $i \not\in \dom(q)$.
        Let $p = q \cup \{(i, 1-x_n(i))\}$.
        Then $p \in D_n$.
    \end{subproof}
    Let $E_i = \{p \in \bP : i \in \dom(\bP)\}$.
    This makes sure that our ``new'' element
    is defined everywhere.
    \begin{claim}
        $\forall  i < \omega.~E_i \subseteq \bP \text{ is dense}$.
    \end{claim}
    \begin{subproof}
        Assume $q \in \bP$.
        If $ i \in \dom(q)$ pick $p = q \in E_i$.
        If $i \not\in \dom(q)$,
        let $p = q \cup \{(i,0)\} \in E_i$.
    \end{subproof}
    Let $\cD = \{D_n : n < \omega\} \cup \{E_i : i < \omega\}$.
    This is a countable subset of dense sets.
    By \yaref{lem:egenfilter}
    there is a $\cD$-generic filter $G$.
    Let $y = \bigcup G$.
    Note that $y$ is a function, since any two elements of $G$ are compatible.
\end{proof}
Note that the ``new'' element did already exists, so we used forcing language
to find it but didn't actually do anything.

\begin{lemma}
    Let $M$ be a transitive model of $\ZFC$
    and let $\bP = (P, \le ) \in M$.

    Let $D \subseteq  \bP$, $D \in M$, $p \in \bP$.
    Then
    \begin{enumerate}[(1)]
        \item $\bP$ is a partial order iff $M \models \text{``$\bP$ is a partial order''}$.
        \item $D$ is dense in  $\bP$ iff  $M \models \text{``$D$ is dense in $\bP$''}$.
        \item $D$ is dense below  $p$ iff  $M \models \text{``$D$ is dense below $p$''}$
            (this only makes sense if $p \in M$).
    \end{enumerate}
\end{lemma}
\begin{proof}
    All the definitions are $\Delta_0$, so we can apply
    \yaref{lem:d0absolute}.
\end{proof}

\begin{definition}
    Assume that $M$ is a transitive model of $\ZFC$
    and $\bP \in M$ is a poset.
    $G \subseteq \bP$ is called a
    \vocab{$\bP$-generic filter over $M$}
    or  \vocab{$M$-generic filter for $\bP$} 
    if
    \[
    \forall D \in M.~((D \subseteq \bP \text{ is dense}) \implies G \cap D \neq \emptyset).
    \]
\end{definition}
\begin{corollary}
    If $M$ is a countable transitive model of $\ZFC$,
    $\bP \in M$ is a poset
    and $p \in \bP$,
    then there is an $M$-generic filter $G \subseteq \bP$
    with $p \in G$.
\end{corollary}
\begin{remark}
    The filter usually exists outside of $M$.
    $M$ itself does not think that $M$ is countable,
    since $M \models \ZFC$.
    But from the outside, we see that $M$ is countable,
    so we can find a filter.
\end{remark}

\begin{definition}
    Assume that $\bP$ is a poset.
    $\bP$ is said to be \vocab{atomless} 
    if for all $p \in \bP$ there are $q,r \in \bP$
    such that
    \begin{enumerate}[(1)]
        \item $q \le p$, $r \le  p$,
        \item $q \bot r$.
    \end{enumerate}
\end{definition}
\begin{example}
    The \yaref{def:cohenf} is atomless.
\end{example}
Usually we are only interested in atomless partial orders.
\begin{lemma}
    Assume that $M$ is a transitive model of $\ZFC$,
    $\bP \in M$ an atomless poset
    and let $G \subseteq  \bP$ be $M$-generic for $\bP$.
    Then $G \not\in M$.
\end{lemma}
\begin{proof}
    Towards a contradiction assume $G \in M$.
    Define $D\coloneqq  \bP \setminus G$.
    We'll show that $D \subseteq \bP$ is dense,
    which is a contradiction, since $G$ was assumed to be $M$-generic.
    Let $q \in \bP$ and
    let $r,s$ be two extensions of $q$
    such that  $r \bot s$.
    These exist because $\bP$ is atomless.
    Since $G$ is a filter,
    it can contain at most one of $\{r,s\}$,
    wlog.~$s \not\in G$.
    In particular, $s \in D$ and $s \le q$.
    Hence $D$ is dense in $\bP$.
\end{proof}

\begin{lemma}
    Assume that $M$ is a transitive model of $\ZFC$,
    $\bP \in M$ a poset, $G \subseteq \bP$ an $M$-generic filter
    and $p \in G$.

    If $D$ is dense below $p$,
    then $G \cap D \neq  \emptyset$.
\end{lemma}
\begin{proof}
    Let $E = D \cup \{ q \in  \bP : q \bot p\}$.
    $E \subseteq  \bP$ is dense:
    Let $r \in  \bP$.
    \begin{itemize}
        \item If $r || p$ let $s \le r,p$.
            Since $D$ is dense below $p$,
            there exists $\overline{s} \in D$ such that $\overline{s} \le s$.
            Since $D \subseteq E$, $\overline{s} \in E$.
        \item If $r \bot p$, then it is obvious that $r \in E$.
    \end{itemize}

    Since $E \in M$, $G \cap E \neq \emptyset$.
    \begin{IEEEeqnarray*}{rCl}
        G \cap (D \cup \{q \in \bP : q \bot p\}) &\neq & \emptyset\\
        \implies (G\cap D) \cup \underbrace{(G \cap \{q \in \bP : q \bot p\})}_{\emptyset} & \neq & \emptyset\\
    \end{IEEEeqnarray*}
\end{proof}

\begin{definition}
    Assume that $\bP$ is a poset.
    \begin{enumerate}[(1)]
        \item $A \subseteq \bP$ is said to be an \vocab{antichain} 
            iff for all $p \neq q$ in $A$,
            $p \bot q$.
        \item  An antichain $A \subseteq \bP$ is a
            \vocab{maximal antichain}
            iff $\forall r \in \bP$, there exists $a \in \bP$
            such that $p || r$.
        \item $X \subseteq  \bP$ is said to be \vocab{open} 
            if $\forall  p \in X.~\forall q \le p.~ q \in X$.
    \end{enumerate}
\end{definition}
\begin{remark}
    Note that if $A$ is a maximal antichain
    in $\bP$, then it is maximal in $(\{A \subseteq \bP : A \text{ is an antichain}\}, \subseteq )$.
    Using $\AxC$, every antichain can be extend to a maximal antichain.

    The statement ``$A$ is an antichain'' is  $\Delta_0$.
    
    Note that ``every antichain of $\bP$ is countable'' is not necessarily
    absolute between transitive models of $\ZFC$.
    
\end{remark}
\begin{lemma}
    Assume that $M$ is a transitive model of $\ZFC$,
    $\bP \in M$ a poset and $G \subseteq \bP$ a filter.
    Then the following are equivalent:
    \begin{enumerate}[(1)]
        \item $G$ is $\bP$-generic over $M$.
        \item $G \cap A \neq \emptyset$ for every
            maximal antichain $A \in M$.
        \item $G \cap D \neq \emptyset$ for every dense
            open $D \in M$ with $D \subseteq \bP$
    \end{enumerate}
\end{lemma}
We'll prove this next time.