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$(\aleph_1, \bf{c})$.
Best Answer
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 He told me that it is an unpublished result of his that any semi-proper forcing which collapses $\aleph_2$ collapses the continuum.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.