\lecture{23}{2024-01-25}{Negation of CH is consistent to ZFC}

\begin{goal}
    We want to construct a model of $\ZFC$
    such that  $2^{\aleph_0} \ge  \aleph_2$.
    % = ?
\end{goal}

Let $M$ be a countable transitive model of $\ZFC$.
Suppose that  $M \models \CH$
(otherwise we are done).

Let $\alpha = \omega_2^M$.

Let $\bC(\alpha) \coloneqq  \{p : p\colon  \alpha \to  \bC \text{ is a function such that } \{\xi < \alpha : p(\xi) \neq \emptyset\} \text{ is finite} \}$,
ordered by $p \le_{\bC(\alpha)} q$ iff
$p(\xi) \le_{\bC} q(\xi)$ for all $\xi < \alpha$.

Recall that $\bC$ is the set of finite sequences of natural numbers
ordered by  $p \le_{\bC} q$ iff $p \supseteq q$.

Let $g $ be $\bC(\alpha)$-generic over $M$.
For $\xi < \alpha$ let $x_\xi = \bigcup \{ p(\xi) : p \in g\}$.
We have already seen that $x_\xi\colon \omega \to  \omega$
is a function
and $x_\xi \neq  x_\eta$ for $\xi \neq \eta$.

We have $M[g] \models\ZFC$.\footnote{We only handwaved this step.}
As $g \in M[g]$,
we have $\langle x_\xi : \xi < \alpha\rangle \in M[g]$.
Therefore $M[g] \models \text{``$2^{\aleph_0} \ge \alpha$''}$.
Also $\alpha = \omega_2^M$.
However the proof is not finished yet,
since we need to make sure,
that $M[g]$ does not collapse cardinals.

We only have $M[g] \models 2^{\aleph_0} \ge \aleph_2^M$,
i.e.~we need to see $\aleph_2^{M[g]} = \aleph_2^M$.

\begin{claim}
    Every cardinal of $M$ is still
    a cardinal of $M[g]$.
\end{claim}
This suffices, because
then $\aleph_0^M = \aleph_0^{M[g]}$,
$\aleph_1^M = \aleph_1^{M[g]}$,
$\aleph_2^{M}= \aleph_2^{M[g]}$, $\ldots$

\begin{definition}
    Let $(\bP, \le )$ be a partial order.
    We say that $\bP$ has the
    \vocab{countable chain condition} 
    (\vocab{c.c.c.})%
    \footnote{it should really be the ``countable antichain condition''}
    iff there is no uncountable
    antichain,
    i.e.~every uncountable $V \subseteq \bP$
    contains compatible $p \neq q$.
\end{definition}

We shall prove:
\begin{claim}
    \label{l23:c:1}
    For all $\beta$, $\bC(\beta)$ has the c.c.c.
\end{claim}
\begin{claim}
    \label{l23:c:2}
    If $\bP \in M$ and
        $M \models \text{`` $\bP$ has the c.c.c.''}$
        and $h$ is generic over $M$,
        then all $M$-cardinals are still
        $M[h]$ cardinals.
        \footnote{Being a cardinal is  $\Pi_1$,
            so $M[h]$ cardinals are always $M$ cardinals.}
\end{claim}
\begin{yarefproof}{l23:c:2}
    Suppose not.
    Let $\kappa$ be minimal such that $M \models \text{``$\kappa$ is a cardinal''}$,
    but $M[h] \models \text{``$\kappa$ is not a cardinal''}$.
    Then $\kappa = (\lambda^+)^M$ for some unique $M$-cardinal $\lambda < \kappa$.
    By minimality, $\lambda$ is also an $M[h]$-cardinal.

    Let $f \in M[h]$ be such that $M[h] \models \text{``$f$ is a surjection from $\lambda$ onto $\kappa$''}$.
    There is a name $\tau \in M^{\bP}$ with $\tau^h = f$.

    We then have some $p \in h$
    with $p  \Vdash_M^\bP \text{``$\tau$ is a surjection from $\check{\lambda}$ onto $\check{\kappa}$''}$.

    Let $\xi < \lambda$.
    Consider $X_\xi \coloneqq  \{\eta < \kappa: \exists  q \le_{\bP} p .~q  \Vdash \tau(\check{\xi}) = \check{\eta}\} \in M$.

    $X_\xi$ is countable in $M$ by the following argument (in $M$):
    For every $\eta \in X_\xi$,
    let  $q_\eta \le p$ be such that $q_\xi  \Vdash^{\bP}_M \tau(\check{\xi}) = \check{\eta}$.
    The set $\{q_\eta : \eta \in X_\xi\}$ is an antichain
    as for $\eta_1 \neq \eta_2$ we have that $q_{\eta_i}  \Vdash \tau(\check{\xi}) = \check{\eta_i}$,
    so they are not compatible.
    So $\{q_\eta : \eta \in X_\xi\}$ is countable by the c.c.c.
    Thus $X_\xi$ is countable.



    Therefore we may define a function in $M$
    \begin{IEEEeqnarray*}{rCl}
        F\colon \lambda \times \omega &\longrightarrow & \kappa
    \end{IEEEeqnarray*}
    such that for all $\xi < \lambda$
    \[
    \{F(\xi,n) : n < \omega\}  = X_\xi.
    \]
    
    $F$ is surjective since $f$ is surjective:
    For $\eta < \kappa$,
    there is some $\xi < \lambda$ such that $M[h] \models \text{``$f(\xi) = \eta$''}$,
    there is some $\overline{q} \in h$ with $\overline{q}  \Vdash ^{\bP}_M \tau(\check{\xi}) = \check{\eta}$.
    Pick $q \le \overline{q},p$.
    This shows $\eta \in X_\xi$
    hence $\eta = F(\xi, n)$ for some $n$.
    But $|\lambda \times \omega| = |\lambda| = \lambda$,
    so in $M$ there is a surjection $F' \colon  \lambda \to  \kappa$,
    but $\kappa$ is a cardinal in $M$ $\lightning$.
\end{yarefproof}

\begin{yarefproof}{l23:c:1}
    Omitted.
    % TODO combinatorial argument
\end{yarefproof}



