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A tree $T$ on $\omega$ is recursively pointed if it is recursive in each of its branches. We can consider a variant of Sacks forcing where the conditions are recursively pointed perfect trees ordered by inclusion. Given a $V$-generic filter $G$ for this forcing, we can define the real $x_G = \bigcap_{T\in G} [T]$. For every real $x \in V$ we have $x <_\text{T} x_G$. Given a real $x \in V[G]$ with $x <_\text{T} x_G$, must we have $x \in V$?

Probably the answer to this question is well-known, but I'm afraid I don't see how the proof of the minimality property for Sacks forcing (as given by Jech) might adapt to recursively pointed trees.

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If you two haven't worked this out already, it seems to me that $x_G$ is not minimal as a Turing degree above $V$, but is a minimal $V$ degree. For the first, since $x_G$ computes $0'$, it follows that $x_G$ is the join of two mutually 1-generic reals, neither of which computes $x_G$ and at least one of which is not in $V$. However, if $x_G$ computes $x$ and $x$ is not in $V$, then there is an $a$ in $V$ such that $x\oplus a$ computes $x_G$. This is by the usual minimality argument: $a$ is the degree of the splitting tree for the functional used to compute $x$ from $x_G$.

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Let $\Phi$ be such that $ x = \Phi^{x_G}$. Since $x\not\ge_T x_G$, it must be that $x_G$ belongs to a tree $T\in V$ with no $\Phi$-splittings. Now given $n$ we can search through $T$ for a string $\sigma$ making $\Phi^\sigma(n)$ converge, and then it must be that $\Phi^\sigma(n)=x(n)$. Thus $x\le_T T$ and so $x\in V$.

(Edited)

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    $\begingroup$ Could you explain a bit more? I can guess the definition of $\Phi$-splitting, but I don't see how to prove what you said. $\endgroup$
    – Trevor Wilson
    Jul 7, 2014 at 3:48
  • $\begingroup$ Thanks. Now can you say why the second sentence is true? $\endgroup$
    – Trevor Wilson
    Jul 7, 2014 at 5:32
  • $\begingroup$ I guess the issue is that $x\not\ge_T x_G$ needs to imply $x\oplus T\not\ge_T x_G$. If all the trees are $A$-recursive for a fixed $A$ then that should be okay, but if $A$ is allowed to vary then you maybe want to have a counting of all the reals in $V$... $\endgroup$
    – Bjørn Kjos-Hanssen
    Jul 7, 2014 at 6:01
  • $\begingroup$ I see, that does seem to be the issue. The trees have unbounded Turing degrees because I really do want to consider all trees in $V$, and I would like to show that the Sacks real added by this particular forcing notion is minimal over $V$, not just that we can get some real that is minimal over $V$. But maybe there is some way to show that the case "$x_G \le_\text{T} x \oplus T$ and $x_G \not\le_\text{T} x$" cannot occur in this situation? $\endgroup$
    – Trevor Wilson
    Jul 7, 2014 at 16:44
  • $\begingroup$ As for "minimal over $V$", would you be okay with weakening the assumption "$x<_T x_G$" to "$x\le_T x_G$ and $x_G\not\in L(x)$"? $\endgroup$
    – Bjørn Kjos-Hanssen
    Jul 7, 2014 at 19:43

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