Turing independent refinement

Question

Suppose $\kappa< 2^{\aleph_0}$ and $\langle P_i : i < \kappa\rangle$ is a sequence of perfect subsets of $2^{\omega}$. Can we find $Q_i \subseteq P_i$ for $i < \kappa$ such that each $Q_i$ is perfect and for every $x_i \in Q_i$ (for $i < \kappa$), the set $\{x_i: i < \kappa\}$ is Turing independent?

Answer 1

If Martin's axiom holds, then the answer is yes. In particular, under the continuum hypothesis, the answer is yes. In any case, the answer is yes when $\kappa$ is countable. More generally, the answer is yes if $\text{MA}_\kappa$ holds. (See update below for improvements.)

Let's consider the case first that $\kappa$ is countable. Suppose that we have countably many perfect sets $\langle P_n\mid n\in\omega\rangle$, each of which is the set of branches through a perfect tree $T_n$, so that $P_n$ consists of the branches through $T_n$.

I propose to refine these trees using the construction method of this answer, in order to construct perfect subtrees $S_n\subseteq T_n$ whose sets of branches $Q_n$ will be the desired perfect refinements with pairwise Turing incompatible branches.

Specifically, we build the subtrees $S_n$ in a sequence of stages. At each stage, we are committed to only finitely much information altogether about which nods are in $S_n$, and at each stage, we end-extend the current approximation to $S_n$. At a given stage, we consider whether a given program $e$ might compute a branch through $S_m$ using an oracle that is a branch through $S_n$. Call this requirement $R_{e,n,m}$.

We can meet this requirement by extending the branches of $S_n$ sufficiently so that program $e$ determines a branch higher than the current branches we have promised about $S_m$, in such a way that we can extend our promise of $S_m$ to the next stage so as to avoid it. In this way, we are fulfilling requirement $R_{e,n,m}$.

I am not saying that this process is computable, since perhaps no extension of the current promise to $S_n$ will enable $e$ to halt sufficiently; but the point is that in this case, we needn't worry about this program, since it will not be giving us a branch through $S_m$. So I am computing the subtrees using the jumps of the oracles.

We can also fold in stages of the construction to ensure that the trees $S_n$ are all branching, so that each $Q_n$ will be a perfect set.

In this way, in $\omega$ many stages, we construct the perfect refinements $Q_n\subseteq P_n$ so that no real in $Q_n$ computes any real in $Q_m$ for $n\neq m$, as desired. We can even arrange the construction so that no branch through any $S_n$ computes any other distinct branch through any $S_m$, including $n=m$.

Now, consider the case of general $\kappa$, under the assumption that $\text{MA}_\kappa$ holds. In this case, we can consider the forcing to add the subtrees $S_n$ with finite conditions, each specifying a finite piece of $S_n$ inside $T_n$, with finite support, ordered by end-extension. This forcing is isomorphic to adding $\kappa$ many Cohen reals, and is therefore c.c.c. Thus, by Martin's axiom, there is a way of choosing the subtrees so as to meet all the requirements $R_{e,\alpha,\beta}$, for Turing programs $e$ and distinct $\alpha,\beta<\kappa$. Each requirement corresponds to a dense set in the forcing, and there are only $\kappa$ many requirements.

Update. I've now realized several improvements.

Let us call your principle the perfect set refinement property ($\text{PSR}_\kappa$).

Theorem. The perfect set refinement property $\text{PSR}_\kappa$ follows from $\text{MA}_\kappa(\text{Cohen})$.

Proof. The principle $\text{MA}_\kappa(\text{Cohen})$ is the very weak version of Martin's axiom, which applies only to the forcing $\text{Add}(\omega,1)$ that adds a single Cohen real. Fix any family of perfect sets $P_\alpha$ for $\alpha<\kappa$. Consider the forcing to add a single Cohen real. Such forcing adds a size-continuum family of pairwise mutually generic Cohen reals $c_\alpha$ for $\alpha<\mathfrak{c}$. Use these reals to pick out perfect subtrees $S_\alpha\subset T_\alpha$. By the argument above, paths through these trees will be Turing incomparable. And so we will have realized our desired family in the extension by adding a Cohen real. The actual properties needed in this argument are only the $\kappa$ many dense sets corresponding to the pairwise independence. So we don't need an actual Cohen real, but only $\text{MA}_\kappa(\text{Cohen})$. $\Box$

Theorem. $\text{PSR}_\kappa$ holds after forcing to add $\theta$ many Cohen reals, for any $\theta\geq\kappa^+$.

Proof. Suppose $G$ is $V$-generic for the forcing to add $\theta$ many Cohen reals, where $\theta\geq\kappa^+$. If $\langle P_\alpha\mid\alpha<\kappa\rangle$ is a family of perfect sets, then this sequence is added by the restriction of $G$ to a set of size $\kappa$. On one of the remaining coordinates, we have added a Cohen real generically over that part of $G$, and so we have created the desired perfect refinement already in $V[G]$. $\Box$

Corollary. One can force the full perfect set refinement property, $\text{PSR}_\kappa$ for all $\kappa<\mathfrak{c}$, simply by adding sufficiently many Cohen reals.

Meanwhile, one might ask why the OP has insisted that $\kappa<\mathfrak{c}$. The reason is that allowing $\kappa$ to be the continuum itself (or larger) makes the principle false.

Theorem. $\text{PSR}_\kappa$ is false for $\kappa\geq\mathfrak{c}$.

Proof. For each real $x$, there is a perfect set $P_x$ of reals $y$ such that $x\leq_T y$. Consider any perfect refinement $Q_x\subseteq P_x$. Let $y\in Q_x$ and consider any $z\in Q_y$. It follows that $y\leq_T z$, and so these are not Turing independent. So there is no independent perfect refinement of this continuum-sized family. $\Box$