%! TEX root = FC.tex % vim: tw=50 % 18/03/2025 12PM \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 $\aleph_1^M$.