Recursively Pointed Sacks Forcing and Preserving $\omega_1$

Let $\mathbb{P}$ denote recursively pointed Sacks forcing. This is forcing with recursively pointed perfect trees ordered by inclusion. A tree $T \subseteq {}^{<\omega}2$ is recursively pointed if and only if for all $f \in [T]$, $T \leq_T f$.

The question is whether $\mathbb{P}$ preserves $\aleph_1$.

From the paper The Derived Model Theorem II by Yizhen Zhu which is based on some notes of Woodin (this can be found at http://graduate.math.nus.edu.sg/~g0700513/der.pdf), near the bottom of page 5, it writes "It is a basic property of recursively pointed Sacks forcing that every countable set of ordinals in the generic extension is covered by a countable set in the ground model" This implies that $\aleph_1$ is preserved.

However, I believe that a fundamental property of recursively pointed Sacks forcing is that the generic real $x_G$ has the property that for all $x \in ({}^\omega 2)^V$, $x \leq_T x_G$. Since there are only countably many reals that are Turing reducible to any give real, the ground model reals are countable in the generic extension. Sacks proves a version of this in Countable Admissible Ordinals and Hyperdegrees for hyperarithmetic reducibility. Is this fact still true with recursively pointed Sacks forcing?

The following Mathoverflow question Does forcing with recursively pointed perfect trees add a Turing degree that is minimal over $V$? also suggest that the generic real for recursively pointed Sacks forcing computes all ground model reals.

Am I misunderstanding something here. It appears that the generic real computing all ground model reals implies $\aleph_1$ is collapse, but the claim from the above paper implies $\aleph_1$ is preserved. Thanks for any clarification.

• If $T$ is any recursively pointed tree, then can't we refine it to a smaller tree $T'\leq T$ consisting of (the initial segments of) just one branch? And surely in this case $T'$ is recursively pointed, since $T'$ in this case would be computable from its only branch. So it seems that this forcing, the way you have described it, is atomic? If so, this interpretation of the forcing would make it trivial. Could you clarify? – Joel David Hamkins Oct 13 '14 at 22:44
• I suspect it should be "recursively pointed perfect trees." – Noah Schweber Oct 13 '14 at 22:52
• That would solve my issue, and I expect that you are correct. – Joel David Hamkins Oct 13 '14 at 22:53
• A question about the cited claim: is it, "given any set of ordinals $X\in V[G]$ such that $V[G]\models$ "$X$ is countable," there is a set of ordinals $Y\in V$ such that $Y\supseteq X$ and $Y$ is countable in $V$?" or ". . . in V[G]?" If the latter, I don't see how that implies that $\aleph_1$ is preserved. – Noah Schweber Oct 13 '14 at 22:54
• @NoahS I think it is countable in $V$. After the quote, the author writes "Hence in $M_\delta$ there exists $q \leq p$ and a countable set a such that $q \Vdash ran(\dot w) \subseteq \hat{a}$ ... since $a$ is countable in $M_\delta$!". It appears that the author is forcing in $M_\delta$, i.e. $M_\delta$ is a the ground model and the author writes $a$ is countable in $M_\delta$. – William Oct 13 '14 at 23:18