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Previous lecture: Suppose $M$ a \gls{ctm} of
$\zfc$.

TODO

\textbf{Question:} ($*$) What are the possible values for
$2^{\aleph_0}$?

Mentioned last lecture: not all values are
possible. In particular, $2^{\aleph_0} \neq
\aleph_\omega$.

\begin{fcdefnstar}[Cofinal]
  \glsnoundefn{cofinal}{cofinal}{subset}%
  $C \subseteq \kappa$ is \emph{cofinal} ($=$
  unbounded) if $\forall \lambda < \kappa, \exists
  \gamma \in C, \gamma \ge \lambda$. We can then
  define:
  \[
    \cfinternal \kappa
    \defeq \{|C| : \text{$C$ is cofinal}\}
  .\]
\end{fcdefnstar}

\begin{example}
  $\cf \aleph_1 = \aleph_1$, $\cf \aleph_\omega =
  \aleph_0$.
\end{example}

\begin{fclemma}[Kőnig's Lemma]
  \label{lemma:konnig}
  $\kappa^{\cf \kappa} > \kappa$.
\end{fclemma}

Then \courseref{LST} \es{4} Q10 is the special
case $\kappa = \aleph_\omega$, $\cf \kappa =
\aleph_0$ of \nameref{lemma:konnig}.

Consequence for $2^{\cf \kappa}$: $(2^{\cf
\kappa})^{\cf \kappa} = 2^{\cf \kappa}$, thus
$2^{\cf \kappa} \neq \kappa$.

Preview of answer to ($*$): Every value not
prohibited by \nameref{lemma:konnig} is possible.

Now: $\aleph_2$!

\begin{fcdefnstar}[]
  \glssymboldefn{ol}%
  \glsnoundefn{nn}{nice name}{nice names}%
  If $\Pbb$ is any \gls{forcing}, let
  \[
    \olinternal
    \defeq \{A_n : n \in \omega\}
  \]
  be any $\omega$-sequence of \glspl{antic} in
  $\Pbb$.

  Let
  \[
    \tau_{\olinternal}
    \defeq \{(\check{n}, p) : p \in A_n\}
  .\]
  We call these \emph{nice names}.
\end{fcdefnstar}

If $|\Pbb| = \kappa$ and $\Pbb$ has \gls{ccc},
then there are at most $\kappa^{\aleph_0}$ many
\glspl{antic} and thus at most
$(\kappa^{\aleph_0})^{\aleph_0} = \kappa^{\aleph_0
\cdot \aleph_0} = \kappa^{\aleph_0}$ many
$\omega$-sequences of \glspl{antic} and thus
\glspl{nn}.

\begin{fcthm}[]
  Assuming:
  - $M[G] \models x \subseteq \omega$
  Then:
  there is a \gls{nn} $\tau$ such that $\val(\tau,
  G) = x$.
\end{fcthm}

\begin{note*}
  The Theorem does not need any assumptions about
  $\Pbb$.
\end{note*}

\begin{proof}
  Start with $M$ such that $\val(\mu, G) = x$
  (possibly not \glsref[nn]{nice}). Fix $n \in
  \omega$. Fix either a well-ordering of $\Pbb$
  (possibly using $\ac$ to get one) and build a
  maximal \gls{antic} $A_n$ such that $\forall p
  \in A_n$, $p \forces \check{n} \in \mu$.

  TODO.

  Claim: $\val(\mu, G) = \val(\tau_{\OL}, G)$.

  $\supseteq$: If $n \in \val(\tau_{OL}, G)$, then
  there is $p \in G$ such that $(\check{n}, p) \in
  \tau_{\OL}$, and then $p \in A_n$, so $p \forces
  \check{n} \in \mu$. So $n \in \val(\mu, G)$.

  $\subseteq$: If $n \in \val(\mu, G)$, then by
  Forcing Theorem, get $q \in G$ such that $q
  \forces \check{n} \in \mu$.

  Subclaim: $A_n \cap G \neq \emptyset$. Indeed,
  by our lemma on compatibility (\es{3}), get $q'
  \in G$ such that $q' \perp p$ for all $p \in
  A_n$. Find $r \le q, q'$. Then $r \forces
  \check{n} \in \mu$. But that is in contradiction
  to $A_n$ being maximal.

  So find $p \in G \cap A_n$. By definition,
  $(\check{n}, p) \in \tau_{\OL}$ and $p \in G$.
  So $n \in \val(\tau_{\OL}, G)$.
\end{proof}

\begin{fccoro}[]
  Assuming:
  - $\Pbb$ has \gls{ccc}
  - $M \models |\Pbb| = \kappa \wedge \lambda =
  \kappa^{\aleph_0}$
  Then:
  $M[G] \models 2^{\aleph_0} \le \lambda$.
\end{fccoro}

\begin{proof}
  Follows directly from:
  \begin{enumerate}[(a)]
    \item Theorem.
    \item Calculation of the number of \glspl{nn}.
  \end{enumerate}
\end{proof}

\subsubsection*{Main Application}

If $\Pbb = \Fn(\omega \times \aleph_2^M, 2)$, then
$|\Pbb| = \aleph_2^M$.

Calculate in $M$, $\aleph_2^{\aleph_0}$.

Hausdorff's Formula:
\[
  \aleph_{\alpha + 1}^{\aleph_\beta}
  = \aleph_{\alpha + 1} \cdot
  \aleph_\alpha^{\aleph_\beta}
.\]
So in particular:
\begin{align*}
  \aleph_1^{\aleph_0}
  &= \aleph_1 \cdot \aleph_0^{\aleph_0} \\
  &= 2^{\aleph_0} \\
  \aleph_2^{\aleph_0}
  &= \aleph_2 \cdot \aleph_1^{\aleph_0} \\
  &= \aleph_2 \cdot 2^{}
\end{align*}
So $\aleph_2^{\aleph_0} = \max(\aleph_2,
2^{\aleph_0})$.

By this calculation, if $M \models 2^{\aleph_0}
\le \aleph_2$, then $M[G] \models 2^{\aleph_0} \le
\aleph_2$.

\begin{corollary}[]
  If $M \models \ch$, then $M[G] \models
  2^{\aleph_0} = \aleph_2$.
\end{corollary}

\begin{remark*}
  This proof also shows that if
  \[
    M \models 2^{\aleph_0} \le \aleph_n
  \]
  and $G$ is \dgen[\Pbb] over $M$ where $\Pbb =
  \Fn(\omega \times \aleph_2^M, 2)$, then
  \[
    M[G] \models 2^{\aleph_0}
    \le \aleph_n
  .\]
  
  Corollary: If $M \models \ch$, then $M[G]
  \models 2^{\aleph_0} = \aleph_n$.
\end{remark*}

\begin{remark*}
  What happens at $\aleph_\omega$?
  \[
    \Pbb \defeq \Fn(\omega \times \aleph_\omega^M,
    2)
  .\]
  By general theory, $M[G] \models 2^{\aleph_0}
  \ge \aleph_\omega$, but \nameref{lemma:konnig}
  gives $2^{\aleph_0} \ge \aleph_{\omega + 1}$.

  What about the lower bound?

  Our theorem and counting of \glspl{nn} yields
  \[
    M[G] \models 2^{\aleph_0} \le
    \ub{\aleph_\omega^{\aleph_0}}_{> \aleph_\omega}
  .\]
  If $M \models \gch$, then
  \[
    \aleph_\omega^{\aleph_0}
    \le \aleph_\omega^{\aleph_\omega}
    = \aleph_{\omega + 1}
  .\]
  So by \nameref{lemma:konnig},
  $\aleph_\omega^{\aleph_0} = \aleph_{\omega +
  1}$.

  Therefore, if $M \models \gch$, then $M[G]
  \models 2^{\aleph_0} = \aleph_{\omega + 1}$.
\end{remark*}

TODO

First limit cardinal that is a possible value of
$2^{\aleph_0}$ is $\aleph_{\omega_1}$.

Clearly, $\Pbb = \Fn(\omega \times
\aleph_{\omega_1}^M, 2)$ adds injection from
$\aleph_{\omega_1}^M$ into $\mathcal{P}(\omega)$.
So $M[G] \models 2^{\aleph_0} \ge
\aleph_{\omega_1}$.

Count \glspl{nn}: $|\Pbb|^{\aleph_0} =
\aleph_{\omega_1}^{\aleph_0}$.

If for all $\alpha < \omega_1$,
$\aleph_\alpha^{\aleph_0} \le \aleph_{\omega_1}$
($*$),
then $\aleph_{\omega_1}^{\aleph_0} =
\aleph_{\omega_1}$.
\begin{align*}
  \aleph_{\omega_1}^{\aleph_0}
  &= \aleph_{\omega_1} \\
  &= \ub{\{f \st f : \omega \to
  \aleph_{\omega_1}\}}_{\eqdef X} \\
  &= \bigcup_{\alpha < \omega_1} \{f \st f :
  \omega \to \aleph_\alpha\}
\end{align*}
(since $\omega_1$ has no \gls{cofinal}
$\omega$-sequence).

Thus $|X| \le \aleph_1 \cdot \aleph_{\omega_1} =
\aleph_{\omega_1}$ (by assumption ($*$)).

Thus $M[G] \models 2^{\aleph_0} =
\aleph_{\omega_1}$.