%! 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 $2^{\aleph_1} = \aleph_3$.