## Is it possible for countably closed forcing to collapse $\aleph_2$ to $\aleph_1$ without collapsing the continuum?

Suppose the continuum is larger than $\aleph_2$. Does there exist a countably closed notion of forcing that collapses $\aleph_2$ to $\aleph_1$, but does not collapse the continuum to $\aleph_1$? Moreover, does there exist such a forcing notion that is separative and has size continuum? It is known (see below) that the canonical collapse Coll$(\aleph_1, \aleph_2)$ collapses the continuum. Trying something like the canonical collapse relativized to some inner model will fail to answer the question, because this forcing will not be countably closed in V.

Background information:

This question came up as a result of my studies of the following theorem.

Let $\kappa < \theta$ be cardinals, with $\kappa$ regular and $\theta^{<\kappa} = \theta$. Then any forcing of size $\theta^{<\kappa}$ which is separative and $<\kappa$ closed and which collapses $\theta$ to $\kappa$ is forcing equivalent to the canonical collapse forcing Coll$(\kappa, \theta)$.

I want to know whether this theorem still holds in the case where $\theta^{<\kappa} = \theta$ fails. The question above is the simplest possible such case.

The reason why Coll$(\aleph_1, \aleph_2)$ collapses the continuum (when CH fails) is that we can think of $\aleph_1$ as $\aleph_1$ many $\aleph_0$-blocks. Consider only the elements of Coll$(\aleph_1, \aleph_2)$ such that on each $\aleph_0$ block, they are either fully defined or fully undefined. This is a dense set in Coll$(\aleph_1, \aleph_2)$, and it's isomorphic to Coll$(\aleph_1, \aleph_2^{\aleph_0})$ = Coll$(\aleph_1, \bf{c})$.

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## 2 Answers

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 He told me that it is an unpublished result of his that any semi-proper forcing which collapses $\aleph_2$ collapses the continuum.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.

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 There is a closely related result for proper forcings (and with a very nice proof), by Shelah and Roslanowski, in "More forcing notions imply diamond." (A proper forcing of size $\le{\mathfrak c}$ that collapses $\omega_2$ forces diamond.) Stevo's result is proved by a completely different argument, though. – Andres Caicedo May 16 2010 at 6:12 Has Stevo's argument come out yet? Please update us when it does. – François G. Dorais♦ Aug 20 2011 at 1:26

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 $q$.

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Justin, if P is countably closed and nontrivial, then P must have size at least continuum. This is because one can build a copy of 2^{<omega} inside P by splitting on finite levels, and then using countable closure to extend it to 2^omega. Thus, P must have antichains of size continuum. I think this trivializes your stated limitation that the continuum is collapsed below the size of P. (But perhaps your idea can be improved..?) – Joel David Hamkins Jan 15 2010 at 0:45