Countably closed forcing that collapses $\mathfrak{c} $ to$\aleph_1$

Question

A paper I'm reading claims that we can find a countably closed forcing notion that collapses $\mathfrak{c}$ to $\aleph_1$, but I can't think of one. I know of the Lévy Collapse, but I don't think that is countably closed, since the conditions are $\textit{finite}$ partial functions.

Answer 1

Just tweak the Levy collapse: force with countable partial injections from $\mathfrak{c}$ to $\aleph_1$.

(Note that it's crucial here that the codomain be uncountable: the poset of countable partial injections from $\mathfrak{c}$ to $\aleph_0$ is trivial as far as forcing is concerned since the set of minimal elements is dense, and indeed it's easy to prove that no countably closed forcing can collapse $\mathfrak{c}$ to $\aleph_0$.)

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EDIT: in light of Andreas Blass' answer, let me add a comparison of the two forcing notions involved, $\mathbb{P}$ = countable partial injections $\mathfrak{c}\rightarrow\omega_1$ and $\mathbb{Q}$ = countable partial functions $\omega_1\rightarrow 2$.

• $\mathbb{Q}$-generics yield $\mathbb{P}$-generics: given $g:\omega_1\rightarrow 2$ coming from a sufficiently generic $\mathbb{Q}$-filter, let $g^\circ:2^\omega\rightarrow\omega_1$ be the map given by $$g^\circ(r)=\min\{\alpha: \forall n\in\omega(r(n)=g(\omega\alpha+n))\}.$$ Now let $$\hat{g}:2^\omega\rightarrow\omega_1: r\mapsto otp(ran(g^\circ\upharpoonright\alpha))$$ be the "transitive codomain collapse" of $g^\circ$ (think about how $ran(g^\circ)\subsetneq \omega_1$). It's easy to check that this $\hat{g}$ is correspondingly $\mathbb{P}$-generic. Precisely, if $g$ comes from a $\mathbb{Q}$-generic-over-$M$ filter with $M$ a c.t.m., then $\hat{g}$ is $\mathbb{P}$-generic over that same $M$.

• $\mathbb{P}$-generics yield $\mathbb{Q}$-generics: given $g:\mathfrak{c}\rightarrow\omega_1$ coming from a sufficiently generic $\mathbb{P}$-filter, let $g':\omega_1\rightarrow 2$ be the map given by $$g'(\alpha)=\begin{cases} 0 & \mbox{ if }g(\alpha)\mbox{ is a limit ordinal},\\ 1 & \mbox{ if }g(\alpha)\mbox{ is a successor ordinal}\\ \end{cases}$$ (keeping in mind that $\omega_1\le\mathfrak{c}$ so this makes sense). As above, $g'$ is "as generic as" $g$.

So in a precise sense, Andreas' answer and my answer are the same.

Answer 2

Noah Schweber's answer is the most natural one, but another very simple forcing that does the job is the set of countable partial functions from $\omega_1$ to $2$. That is, add a "Cohen-generic" map $G:\omega_1\to2$ with countable conditions. The reason this works is that, by a density argument, every function $\omega\to2$ (i.e., every real) occurs as $n\mapsto G(\alpha+n)$ for some $\alpha\in\omega_1$.