Is it possible for countably closed forcing to collapse $\aleph_2$ to $\aleph_1$ without collapsing the continuum?
I got the answer from Stevo Todorcevic last weekend at the MAMLS conference in honor of Richard Laver in Boulder, CO. He told me that it is an unpublished result of his that any semi-proper forcing which collapses $\aleph_2$ collapses the continuum. As countably closed forcing is semi-proper, the answer to my question is no. Stevo sketched a proof for me, but I do not remember it well enough to reproduce it here.
Here is for the benefit of the curious (and it answers the original question): By a result of Baumgartner-Taylor (Saturation properties of ideals in generic extensions. I) we know $[\omega_2]^\omega$ can be splitted into $2^\omega$ many disjoint stationary subsets. For any proper forcing $\mathbb{P}$ that collapses $\omega_2$, we know in the generic extension there exists a club subset of $[\omega_2^V]^\omega$ of size $\omega_1$. Since stationary sets are preserved when forcing with proper $\mathbb{P}$, we can define an injection from $(2^\omega)^V$ into $\omega_1$ in $V[G]$.
(For completeness) For the case of semi-proper forcings, the following will give the result:
- We can partition $[\omega_2]^\omega$ into $2^\omega$ disjoint stationary subsets $S_r$ such that for each club $C\subset [\omega_2]^\omega$, $|S_r\cap C|\geq 2^\omega$ (just greedily union up the partition obtained from Baumgartner-Taylor)
- (Todorcevic) For each stationary $S\subset [\omega_2]^\omega$, we can find its stationary core $S^*\subset S$ such that $|S-S^*|\leq \aleph_2$ and if $S^*$ is stationary then it remains so after any semiproper forcing. (the proof is pretty clever but I believe something written is coming out)
- Assuming $2^\omega > \omega_2$, and apply 2 to each $S_r$ we see by the consideration of cardinality that the hypothesis is satisfied. But then we proceed as in the proper case.
Here is, I think, a partial answer. I believe I can show that as long as a countably closed forcing adds a new $\omega_1$-sequence, the continuum is collapsed below the size of the poset. I am not sure if you can do better.
Prop. Let $\mathbb{P}$ be a countably closed notion of forcing such that $\Vdash\dot{f}:\omega_1\rightarrow ON,\dot{f}\not\in V$. Then, if $G$ is $\mathbb{P}$-generic over $V$ we will have $V[G]\vDash 2^\omega\leq |\mathbb{P}|$.
Pf: It's enough to show that in $V[G]$, $|\mathcal{P}(\omega)\cap V|\leq |\mathbb{P}|$ (because $\mathbb{P}$ is countably closed). Note that for each $p\in\mathbb{P}$ there is some $\alpha<\omega_1$ such that $p$ doesn't decide $\dot{f}(\alpha)$ (otherwise $f$ could be defined in $V$); let $\alpha(p)$ denote the least such $\alpha$. Let $\beta_0(p)<\beta_1(p)$ be the least ordinals $\beta$ such that there's $q\leq p$ for which $q\Vdash\dot{f}(\alpha(p))=\beta$.
Fix in $V$ a well-ordering $\prec$ of $\mathbb{P}$. Now, working in $V[G]$, we associate to each $q\in\mathbb{P}$ an $x_q\subseteq\omega$ as follows. Inductively define a descending sequence of conditions $q_0\geq q_1\ldots \geq q_n\ldots $ by $q_0=q$, $q_{n+1}$ is the $\prec$-least member of $G$ below $q_n$ which decides $\dot{f}(\alpha(q_n))$. Let $x_q= \{n\in \omega|f(\alpha(q_n))=\beta_0(q_n)\}$ .
To finish we just have to show that for each $x\in\mathcal{P}(\omega)$ that the set $D_x=\{r\in\mathbb{P}|r\Vdash(\exists q\in \dot{G})x=\dot{x_q}\}$ is dense. Let $p\in\mathbb{P}$ be a fixed condition. Inductively define $p_0\geq p_1\geq \ldots p_n\geq $. Set $p_0=p$. If $n\in x$ set $p_{n+1}$ to be the $\prec$-least member of $\mathbb{P}$ with $p_{n+1}\leq p_n$ and $p_{n+1}\Vdash\dot{f}(\alpha(p_n))=\beta_0(p_n)$; if $n\not\in x$ then do the same thing but have $p_{n+1}\Vdash\dot{f}(\alpha(p_n))=\beta_1(p_n)$. Then let $r$ be below all the $p_n$. Then $r\in D_x$, with $p$ as our witnessing $q$.