%! TEX root = FC.tex
% vim: tw=50
% 15/03/2025 12PM

TODO

\subsubsection*{What about $2^{\aleph_1}$?}

More on \glspl{nn}:

\begin{fcdefnstar}[$lambda$-nice names]
  \glsnoundefn{lnn}{$\lambda$-nice
  name}{$\lambda$-nice names}%
  Generalise ``\glspl{nn}'' to \emph{$\lambda$-nice
  names}:

  Let
  \[
    \OL
    = \{A_\alpha : \alpha < \lambda\}
  \]
  be a family of $\lambda$ many maximal
  $\Pbb$-\glspl{antic}:
  \[
    \tau_{\OL}
    \defeq \{(\check{\alpha}, p) : p \in A_\alpha\}
  \]
  for $\alpha \in \lambda$.

  These are names for subsets of $\lambda$.
\end{fcdefnstar}

Observe that our theorem ``every $A \subseteq
\lambda$ in $M[G]$ has a \gls{lnn}'' still goes
through.

If $\kappa$ is an $M$-cardinal such that every
\gls{antic} of $\Pbb$ has size $\le \kappa$. [On
\es{3}, this is called the $\kappa^+$-chain
condition.]

Let $\mu \defeq |\Pbb|$ (in $M$). Then
\[
  (\mu^\kappa)^\lambda
  = \mu^{\kappa \cdot \lambda}
\]
is an upper bound on the number of \glspl{lnn}.

Thus
\[
  M[G] \models 2^\lambda \le (\mu^{\kappa \cdot
  \lambda})^M
.\]

\textbf{Question:} Forcing with $\Pbb \defeq
\Fn(\omega \times \aleph_2, 2)$ and calculate
$2^{\aleph_1}$. Assume $M \models \gch$. Then:
\begin{align*}
  \mu = |\Pbb|
  &= \aleph_2 \\
  \lambda
  &= \aleph_1 \\
  \lambda
  &= \aleph_0
\end{align*}
(since $\Pbb$ has \gls{ccc}).
So
\[
  M[G] \models 2^\lambda \le \aleph_2^{\aleph_1
  \cdot \aleph_0} = (\aleph_2^{\aleph_1})^M
.\]
Calculate $(\aleph_2^{\aleph_1})^M$:
\[
  \aleph_2^{\aleph_1}
  \stackrel{\text{Hausdorff's formula}}{=}
  \aleph_2 \cdot 2^{\aleph_1}
  \stackrel{\gch}{=} \aleph_2 \cdot \aleph_2
  = \aleph_2
.\]
Together:
\[
  M[G] \models 2^{\aleph_1} = \aleph_2 =
  2^{\aleph_0}
.\]
\textbf{Question:} Is it possible to get PSA, i.e.
\[
  \forall \kappa, \lambda, 
  ~~\kappa < \lambda
  \to 2^\kappa < 2^\lambda
\]
without $\ch$.

In particular, can we get
\begin{align*}
  2^{\aleph_0}
  &= \aleph_2 \\
  2^{\aleph_1}
  &= \aleph_3
\end{align*}

\textbf{First idea:} Force with $\Fn(\aleph_1
\times \aleph_3, 2)$.
\begin{enumerate}[(1)]
  \item Yields $\aleph_3$ many subsets of
    $\aleph_1$.
  \item Still has \gls{ccc}, so all cardinals are
    preserved.
  \item How many \glsref[lnn]{$\aleph_1$-nice
    names} are there:
    \[
      \aleph_3^{\aleph_1 \cdot \aleph_0}
      = \aleph_3^{\aleph_1}
      \stackrel{\text{Hausdorff}}{=} \aleph_3
      \cdot 2^{\aleph_1}
      \stackrel{\gch}{=} \aleph_3
    .\]
    Get: $2^{\aleph_1} = \aleph_3$ in $M[G]$.
\end{enumerate}
Unfortunately,
\[
  M[G] \models 2^{\aleph_0} = \aleph_3
.\]
Interpret the generic object $G$ as $f_\alpha \to
\aleph_1 \to 2$ for $\alpha < \aleph_3$.

Define $g_\alpha \defeq f_\alpha|_\omega$.

\textbf{Claim:} For $\alpha \neq \alpha'$,
$g_\alpha \neq g_\alpha'$. This is since
\[
  \mathcal{D}_{\alpha, \alpha'}
  \defeq \{p : \exists u \in \omega, g_\alpha(u)
  \neq g_{\alpha'}(u)\}
\]
is still \gls{dense}.

So, forcing with $\Fn(\aleph_1 \times \aleph_3,
2)$ gives the same situation as forcing with
$\Fn(\omega \times \aleph_3, 2)$ for
$2^{\aleph_0}$, $2^{\aleph_1}$.

\textbf{Idea:}
\[
  \Fn(X, Y, \kappa)
  \defeq \{p : \dom(p) \subseteq X, \range(p)
  \subseteq Y, |p| \le \kappa\}
.\]
Thus $\Fn(X, Y) = \Fn(X, Y, \aleph_0)$.

Consider
\[
  \Pbb
  \defeq \Fn(\aleph_1 \times \aleph_3, 2, \aleph_1)
.\]

\textbf{Properties:}
\begin{enumerate}[(1)]
  \item We still have $M[G] \models 2^{\aleph_1}
    \ge \aleph_3^M$.
  \item Not clear that this forcing is preserving
    cardinals!
\end{enumerate}

First goal: \emph{What about preserving
cardinals?}

Clearly, $\Fn(\aleph_1 \times \aleph_3, 2,
\aleph_1)$ does not have the \gls{ccc} anymore.

With example (38) (on \es{3}), we need to figure
out the chain condition of $\Fn(\aleph_1 \times
\aleph_3, 2, \aleph_1)$.

We need a $\Delta$-system lemma for this: If
$\lambda$ is regular ($\cf \lambda = \lambda$) and
$\OL$ is a family of sets of size $< \lambda$ of
size $\lambda^+$. Then there is a $\Delta$-system
$\mathcal{D} \subseteq \OL$ of size $\lambda^+$.

This $\Delta$-system lemma gives with the same
proof as before:

$\Fn(\aleph_1 \times \aleph_3, 2, \aleph_1)$ has
the $\aleph_2^M$-chain condition.

So: $\Fn(\aleph_1 \times \aleph_3, 2, \aleph_1)$
preserves cardinals $\ge \aleph_2^M$.

\subsubsection*{Closure}

\begin{fcdefnstar}[$lambda$-closed]
  \glsadjdefn{lclosed}{$\lambda$-closed}{forcing}%
  A forcing $\Pbb$ is called
  \emph{$\lambda$-closed} if any family
  $\{p_\alpha : \alpha < \gamma\}$ for $\gamma <
  \lambda$ that is a descending chain:
  \[
    \alpha < \beta
    \implies p_\beta < p_\alpha
  \]
  there is $q$ such that $q \le p_\alpha$ for all
  $\alpha < \gamma$.
\end{fcdefnstar}

\begin{example*}
  $\Fn(\aleph_1 \times \aleph_3, 2, \aleph_1)$ is
  \glsref[lclosed]{$\aleph_1$-cloesd}.

  [If $\{p_\alpha\}$ is a descending chain, then
  $\bigcup_{\alpha < \gamma} p_\alpha$ is a
  condition in $\Fn(\aleph_1 \times \aleph_3, 2,
  \aleph_1)$.]
\end{example*}

\begin{fcthm}[]
  Assuming:
  - $\Pbb$ is \gls{lclosed} and $\kappa <
  \lambda$
  Then:
  \[
    \mathcal{P}(\kappa) \cap M
    = \mathcal{P}(\kappa) \cap M[G]
  .\]
\end{fcthm}

\begin{fccoro}[]
  Forcing with \cloze{$\Fn(\aleph_1 \times \aleph_3, 2,
  \aleph_1)$}
  \divcloze{
  \begin{cenum}[(a)]
    \item Does not change $\mathcal{P}(\omega)$.
    \item Therefore preserves $\aleph_1$ (see
      \es{1} and the relation between codes for
      countable well-orders and preserving
      $\aleph_1$).
  \end{cenum}
  }
\end{fccoro}

\textbf{Summary:} Forcing with $\Fn(\aleph_1
\times \aleph_3, 2, \aleph_1)$ over a model of
$\gch$ gives $M[G]$ with:
\begin{enumerate}[(1)]
  \item The same cardinals (cardinals $\ge
    \aleph_2$ preserved by $\aleph_2$-chain
    condition; $\aleph_1$ preserved by
    $\aleph_1$-closure).
  \item $2^{\aleph_1} \ge \aleph_3$ (standard).
  \item $2^{\aleph_0} = \aleph_1$ (by corollary 1 to
    the closure theorem).
  \item Calculate number of \glspl{nn}:
  \[
    \aleph_3^{\aleph_1 \cdot \aleph_1}
    = \aleph_3^{\aleph_1}
    = \aleph_3 \cdot 2^{\aleph_1}
    = \aleph_3 \cdot \aleph_2
    = \aleph_3
  .\]
\end{enumerate}
Hence $2^{\aleph_1} = \aleph_3$.
