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\begin{center}
  \begin{tabular}{|m{8em}|m{8em}|m{8em}|m{8em}|}
    \hline
    Forcing
    & Property
    & Preservation
    & Arithmetic \\
    \hline
    $\Fn(\omega \times \aleph_2, 2)$
    & \gls{ccc}
    & all cardinals
    & $2^{\aleph_0} = \aleph_2$, $2^{\aleph_1} =
    \aleph_2$ \\
    \hline
    $\Fn(\omega \times \aleph_3, 2)$
    & \gls{ccc}
    & all cardinals
    & $2^{\aleph_0} = \aleph_3$, $2^{\aleph_1} =
    \aleph_3$, $2^{\aleph_2} = \aleph_3$ \\
    \hline
    $\Fn(\aleph_1 \times \aleph_3, 2, \aleph_1)$
    & $\aleph_1$-closed.

    If $M \models \ch$,
    then $\aleph_2$-chain condition
    & $\aleph_1$ preserved. (Closure lemma [not
    yet proved])

    $\kappa \ge \aleph_2$ preserved
    & If $M \models \ch$, then $2^{\aleph_0} =
    \aleph_1$, $2^{\aleph_1} = \aleph_3$. \\
    \hline
  \end{tabular}
\end{center}

Note $\gch \to \psa$, but our model of $\neg \ch$
fails $\psa$.

Question: Can we have $\aleph_1 < 2^{\aleph_0} <
2^{\aleph_1}$?

Closure lemma:

\begin{theorem*}
  If $\Pbb$ is $\lambda$-closed and $\kappa <
  \lambda$, then $\mathcal{P}(\kappa) \cap M =
  \mathcal{P}(\kappa) \cap M[G]$.
\end{theorem*}

$\lambda$-closed: every descending sequence of
length $< \lambda$ has a lower bound.

\begin{proof}
  Let $f \in M[G]$, $f : \kappa \to 2$ and assume
  towards contradiction that $f \notin M$. $\to$ $f
  \notin B$.
  \[
    B
    \defeq \{f \in M \st f : M \to 2\}
  .\]
  Let $\tau$ be a name for $f$.

  By Forcing Theorem, there is $p \in G$ such that
  $p \forces \tau : \check{\kappa} \to \check{2}
  \wedge \tau \notin \check{B}$.

  Construct a $\kappa$-sequence of conditions
  $p_\alpha$ TODO

  TODO

  The sequence $\{p_\alpha : \alpha \le \kappa\}$
  is defined in $M$ (by Definability Theorem), so
  we can define
  \[
    g(\alpha) = 1:
    \iff p_{\alpha + 1} \forces
    \tau(\check{\alpha}) = \check{1}
  .\]
  Then $g \in M$.

  But now $p_\kappa \forces \tau(\check{\alpha}) =
  \check{1}$ or $p_\kappa \forces
  \tau(\check{\alpha}) = \check{0}$ for all
  $\alpha$.

  uso $p_k \forces \tau = \check{g}$. Hence
  $p_\kappa \forces \tau \in \check{B}$.

  But $p_\kappa \le p \forces \tau \notin
  \check{B}$. Contradiction.
\end{proof}

Note that while $\Fn(X, Y, \aleph_1)$ is
\emph{always} $\aleph_1$-closed, the chain
condition depended on the value of
$\aleph_1^{\aleph_0}$.

The partial order $\Fn(\aleph_1 \times \aleph_3,
2, \aleph_1)$ has in general the
$(2^{\aleph_0})^+$-chain condition.

$\ch$ implies $(2^{\aleph_0})^+ = \aleph_2$, so
\emph{all} cardinals are preserved.

However, if $2^{\aleph_0} > \aleph_1$, then there
is a gap and we do not know whether TODO.

If $\mathcal{M} \models 2^{\aleph_0} = \aleph_2$,
does
\[
  \Fn(\aleph_1^M \times \aleph_3^M, 2, \aleph_1^M)
\]
preserve $\aleph_2^M$?

Answer: $\Fn(\lambda^+ \times \kappa, 2,
\lambda^+)$ \emph{always adds} a surjection from
$\lambda^+$ to $2^\lambda \cap M$. ($*$)

Application: If $\lambda = \aleph_2$ and $M =
2^{\aleph_0} = \aleph_2$, then $\Fn(\aleph_1
\times \kappa, 2, \aleph_1^M$ adds a surjection
from $\aleph_1^M$ onto $\mathcal{P}(\aleph_0) \cap
M$, i.e. $\aleph_2^M$. So $|\aleph_2^M| =
\aleph_1^{M[G]}$.

\begin{proof}[[Proof of ($*$)]]
  A generic for $\Pbb$ ``is'' a map $f : \lambda^+
  \times \kappa \to 2$.

  Define $h : \lambda^+ \to 2^\lambda$ by
  \[
    h(\alpha)(\beta) = 1:
    \iff f(\alpha, \beta) = 1
  .\]
  Claim: $h$ is a surjection onto $2^\lambda \cap
  M$.

  If $g \in M$, $g : \lambda \to 2$, consider
  \[
    \mathcal{D}_g
    \defeq \{p \st \exists \alpha < \lambda^+,
    \forall \beta < \lambda, p(\alpha, \beta) = 1
    \iff g(\beta) = 1\}
  .\]
  This is \gls{dense}, and thus $g \in \range(h)$.
\end{proof}

Back to our question: Can we get $\aleph_1 <
2^{\aleph_0} < 2^{\aleph_1}$?

Start with $M \models \gch$. Consider
\[
  \Pbb
  \defeq \Fn(\aleph_1 \times \aleph_3, 2,
  \aleph_1) \cap M
\]
and let $G$ be \dgen[\Pbb] over $M$. Consider
\[
  \Qbb
  \defeq \Fn(\omega \times \aleph_2, 2) \cap M[G]
,\]
and let $H$ be \dgen[\Qbb] over $M[G]$.

Claim: $M[G][H] \models \aleph_1 < 2^{\aleph_0} <
2^{\aleph_1}$.

($M[G][H]$ is a \emph{forcing iteration}).

\begin{enumerate}[(1)]
  \item $\Pbb$ is cardinal preserving over $M$
    since $M \models \gch$. So $\aleph_n^M =
    \aleph_n^{M[G]}$.
  \item $\Qbb$ has \gls{ccc}, so is cardinal
    preserving: $\implies$
    \[
      \aleph_n^{M[G][H]}
      = \aleph_n^{M[G]}
      = \aleph_n^M
    .\]
  \item $M[G] \models 2^{\aleph_0} = \aleph_1
    \wedge 2^{\aleph_1} = \aleph_3$ (lecture 15;
    since $M \models \ch$).
  \item Then $M[G][H] \models 2^{\aleph_0} =
    \aleph_2 \wedge 2^{\aleph_1} \ge \aleph_3$
    (using a \gls{nn} analysis, we could calculate
    $2^{\aleph_1} = \aleph_3$).
\end{enumerate}
This proves the claim.

Important: The order of forcings matters!

Suppose $M \models \gch$.
\[
  \Qbb'
  \defeq \Fn(\omega \times \aleph_2, 2) \cap M
.\]
$H$ \dgen[\Qbb'] over $M$.
\[
  \Pbb'
  \defeq \Fn(\aleph_1 \times \aleph_3, 2,
  \aleph_1) \cap M[H]
.\]
$G$ \dgen[\Pbb'] over $M[H]$.

Consider $M[H][G]$. Then
\[
  M[H] \models 2^{\aleph_0} = \aleph_2 =
  2^{\aleph_1}
.\]
But that means that forcing with $\Pbb'$
\emph{will} collapse $\aleph_2^M =
\aleph_2^{M[H]}$.

Since $\Pbb'$ is $\aleph_1$-closed,
\[
  \mathcal{P}(\omega) \cap M[H]
  = \mathcal{P}(\omega) \cap M[H][G]
.\]
In particular, $M[H][G] \models \ch$. So, this
order does not achieve what we want.

\subsubsection*{Final Remark on Forcing CH}

Assume $M \models 2^{\aleph_0} = \aleph_2$.

Question: Can you obtain $M[G] \models \ch$?

The natural forcing would be
\[
  \Pbb
  \defeq \Fn(\omega, \aleph_1^M)
.\]
This collapses $\aleph_1^M$; it does not have the
\gls{ccc}, but since it has size $\aleph_1^M$, it
has the $\aleph_2^M$-chain condition, so all
cardinals $\ge \aleph_2^M$ are preserved.

Clearly therefore:
\[
  M[G]
  \models |\mathcal{P}(\omega) \cap M| =
  \aleph_2^M = \aleph_1^{M[G]}
.\]
But: is $|\mathcal{P}(\omega) \cap M| =
|\mathcal{P}(\omega) \cap M[G]|$?

\glsref[nn]{Nice names}: gives upper bound of
\[
  (\aleph_1^M)^{\aleph_1^M \cdot \aleph_0}
  = (2^{\aleph_1^M})^M
.\]
That's not surprising, since any $A \subseteq
\aleph_1$ in $M$ becomes a new subset of $\omega$
in $M[G]$ via the new bijection between $\omega$
and $\aleph_1^M$.
