set theory - Is it possible for countably closed forcing to collapse $aleph_2$ to $aleph_1$ without collapsing the continuum?

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 $Vdashdot{f}:omega_1rightarrow ON,dot{f}notin V$. Then, if $G$ is $mathbb{P}$-generic over $V$ we will have $V[G]vDash 2^omegaleq |mathbb{P}|$.

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 $pinmathbb{P}$ 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 $qleq p$ for which $qVdashdot{f}(alpha(p))=beta$.

Fix in $V$ a well-ordering $prec$ of $mathbb{P}$. Now, working in $V[G]$, we associate to each $qinmathbb{P}$ an $x_qsubseteqomega$ as follows. Inductively define a descending sequence of conditions $q_0geq q_1ldots geq q_nldots$ 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 $x_q= {nin omega|f(alpha(q_n))=beta_0(q_n)}$ .

To finish we just have to show that for each $xinmathcal{P}(omega)$ that the set $D_x={rinmathbb{P}|rVdash(exists qin dot{G})x=dot{x_q}}$ is dense. Let $pinmathbb{P}$ be a fixed condition. Inductively define $p_0geq p_1geq ldots p_ngeq$. Set $p_0=p$. If $nin x$ set $p_{n+1}$ to be the $prec$-least member of $mathbb{P}$ with $p_{n+1}leq p_n$ and $p_{n+1}Vdashdot{f}(alpha(p_n))=beta_0(p_n)$; if $nnotin x$ then do the same thing but have $p_{n+1}Vdashdot{f}(alpha(p_n))=beta_1(p_n)$. Then let $r$ be below all the $p_n$. Then $rin D_x$, with $p$ as our witnessing $q$.

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