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\textbf{Summary:} If $G$ is $\Fn(\omega \times
\aleph_2^M, 2)$-generic, then $M[G] \models \zfc +
$there is an injection from $\aleph_2^M$ into
$\mathcal{P}(\omega)$.

This implies $M[G] \models \zfc + 2^{\aleph_0} \ge
\aleph_2$ if we have $\aleph_1^M = \aleph^{M[G]}$,
$\aleph_2^M = \aleph_2^{M[G]}$. This is the goal
for today's lecture.

Note that our proof above is \emph{not good
enough} for the statement ($*$) from Lecture 8:

\begin{quote}
  For all $T \subseteq \zfc$ finite, there exists
  $T^* \subseteq \zfc$ finite such that if $M$ is
  a \gls{ctm} of $T^*$, then there is $N \supseteq
  M$ \gls{ctm} of $T^* + \neg \ch$.
\end{quote}

For this, we need to look more carefully at the
proof of the GMT:
\[
  M \models \zfc
  \implies M[G] \models \zfc
.\]

The proof proceeds AXIOM BY AXIOM and thus for
each $\varphi \in \zfc$, we find finite
$S_\varphi$ such that $M \models S_\varphi$
implies $M[G] \models \varphi$.

Let $S \subseteq \zfc$ finite such that $S$ proves
that all relevant notions (name, value, \ldots)
are well-defined and absolute. Then for $T
\subseteq \zfc$ finite, define
\[
  T^* \defeq S \cup \bigcup_{\varphi \in T}
  S_\varphi
.\]
Then, the proof shows ($*$).

Let's prove $\neg\ch$ in $M[G]$.

\begin{fcdefnstar}[Preserves cardinals]
  \glsadjdefn{prescard}{preserves
  cardinals}{forcing poset}%
  We say $\Pbb$ \emph{preserves cardinals} if
  $\forall G$ \dgen[\Pbb] over $M$, ``$\kappa$ is
  a cardinal'' is absolute between $M$ and $M[G]$.
\end{fcdefnstar}

\begin{fcdefn}[Countable chain condition]
  \glsadjdefn{ccc}{countable chain
  condition}{forcing poset}%
  We say $\Pbb$ has the \emph{countable chain
  condition} (c.c.c.) if every antichain (note the
  anti!) in $\Pbb$ is countable.
\end{fcdefn}

\begin{fcthm}[]
  \label{thm:ccc2prescard}
  Assuming:
  - $\Pbb$ has \gls{ccc}
  Then:
  $\Pbb$ \gls{prescard}.
\end{fcthm}

\begin{fcthm}[]
  \label{thm:hasccc}
  $\Fn(\omega \times \aleph_2^M, 2)$ has \cloze{the
  \gls{ccc}.}
\end{fcthm}

\begin{fccoro}[]
  Assuming:
  - $G$ is \dgen[\Fn(\omega \times \aleph_2^M,
  2)] over $M$, then
  \[
    M[G] \models \zfc + 2^{\aleph_0} \ge \aleph_2
  .\]
\end{fccoro}

\begin{fclemma}[]
  Assuming:
  - $M \models \Pbb$ has \gls{ccc}
  - $X, Y \in M$
  - $G$ is \dgen[\Pbb] over $M$
  - $f : X \to Y$, $f \in M[G]$
  Then:
  there is $F \in M$ such that $\forall x \in X,
  F(x) \subseteq  Y$, $\forall x \in X, f(x) \in
  F(x)$ and $M \models \forall x \in X, F(x)
  \text{ is countable}$.
\end{fclemma}

\begin{proof}[Proof of \cref{thm:ccc2prescard}]
  Suppose $M \models \kappa \text{is a cardinal}$,
  $M[G] \models \text{$\kappa$ is not a
  cardinal}$, so there is $\lambda < \kappa$ and
  $f \in M[G]$, $f : \lambda \to \kappa$, $f$ is a
  surjection. Apply the lemma to get $F$.

  Define $R \defeq \bigcup_{\alpha < \lambda}
  F(\alpha)$.

  Since $\forall x, f(x) \in F(x)$, $R = \kappa$.

  But $M \models |R| = \aleph_0 \cdot \lambda =
  \lambda < \kappa$. But then $M$ thinks that
  $\kappa$ is not a cardinal, contradiction.
\end{proof}

Now we prove the lemma:

\begin{proof}
  Let $F(x) \defeq \{y \in Y : \exists p \in \Pbb,
  p \forces \tau(\check{x}) = \check{y}\}$.

  Fix $\tau$ a name for $f$.

  By Definability Theorem , $F \in M$.

  Then $\forall x \in X, F(x) \subseteq Y$ follows
  from definition.

  $\forall x \in X, f(x) \in F(x)$ follows from
  Forcing Theorem.

  Let's look at $M \models F(x)$ is countable: If
  $y \in F(x)$, let $p_y$ be such that $p_y
  \forces \tau(\check{x}) = \check{y}$.

  If $y \neq y'$, then $p_y \perp p_{y'}$. Thus
  \[
    \{p_y : y \in F(x)\}
  \]
  is an antichain.

  By \gls{ccc}, it's countalbe. So $F(x)$ was
  countable.
\end{proof}

TODO

\begin{proof}[Proof of \cref{thm:hasccc}]
  Actually, we prove $\Fn(X, Y)$ has \gls{ccc}
  whenever $Y$ is countable.

  $\Delta$-systems form \es{3} Q33 are also called
  \emph{quasi-disjoint families}.

  (A family of finite sets $\mathcal{D}$ is called
  a $\Delta$-system if there is a finite set $R$
  (called the root of the $\Delta$-system) such
  that for all $D, D' \in \mathcal{D}$, if $D \neq
  D'$ then $D \cap D' = R$).

  $\Delta$-system lemma: Any countable family of
  finite sets contains an uncountable
  $\Delta$-system.

  Take any $A \subseteq \Pbb$ uncountable and
  prove that it's not an antichain.

  If $p \in A$, then $\dom(p) \subseteq X$ finite.
  Consider
  \[
    S \defeq \{\dom(p) : p \in A\}
  .\]
  That's an uncountable family of finite sets, so
  by $\Delta$-system lemma, find $\Delta$-system
  $\mathcal{D} \subseteq S$ uncountable.

  Let $r \subseteq X$ finite be the root of
  $\mathcal{D}$.

  Since $Y$ is countable, there are only countably
  many functions $q : r \to Y$. Since $D$ is
  uncountable, by pigeonhole principle, there are
  $p, q$ such that $\dom(p), \dom(q) \in D$ and
  $p|_r = q|_r$. But since $\dom(p) \cap \dom(q) =
  r$, $p$ and $q$ are compatible.

  So $A$ is not an antichain.
\end{proof}

We got $M[G] \models 2^{\aleph_0} \ge \aleph_2$.

Next time: What is the size of $2^{\aleph_0}$ in
$M[G]$?

Remember: \courseref{LST} \es{4}: $\zfc \syn
2^{\aleph_0} \neq \aleph_\omega$.

What if $\aleph_2$ was one of the forbidden
values?

\begin{remark*}
  Obtaining $M[G] \models 2^{\aleph_0} =
  \aleph_2$ cannot be quite as general as this
  proof: if $M \models 2^{\aleph_0} > \aleph_2$,
  then this will remain true in $M[G]$.
\end{remark*}

\begin{remark*}
  If $G$ is \dgen[\Fn(\omega \times \aleph_\alpha^M,
  2)] over $M$, then
  \[
    M[G] \models 2^{\aleph_0} \ge \aleph_\alpha
  .\]
\end{remark*}
