TITLE: Can noncomputable sets be distinguishable in $RCA_0$?
QUESTION [12 upvotes]: Say that a set $X\subseteq\omega$ is distinguishable if there is some Turing machine $\Phi_e$ which, when given two sets exactly one of which is $X$, can determine which set is $X$. Formally, $X$ is distinguishable if there is some Turing machine $\Phi_e$ such that for all $Y\not=X$, $$\Phi_e^{X\oplus Y}(0)=0,\quad \Phi_e^{Y\oplus X}(0)=1.$$ (Think of "$0$" and "$1$" as meaning "Left" and "Right.")
Clearly every computable set is distinguishable; it is not hard to show (see my answer to https://math.stackexchange.com/questions/1189370/is-there-a-turing-machine-that-can-distinguish-the-halting-problem-among-others) that the converse also holds. My question is about the reverse mathematics of the converse. Specifically, my proof used Weak Konig's Lemma, and I don't immediately see a way to do without it. So my question is:

Is it consistent with $RCA_0$ that there is a non-computable, distinguishable set?


Motivation: Over the last year or so I've developed an interest in "alternate computability theories" - e.g. see Visser's delightfully-named paper "Oracle bites theory" at http://www.phil.uu.nl/preprints/lgps/authors/visser/oracle-bites-theory/pdf. I'm especially interested in "almost computable" sets in such theories, and distinguishable sets might provide such an example.

REPLY [4 votes]: Extending D. Turetsky's answer:  If $\mathcal{M}$ is a Turing ideal containing no PA degree, then every $\Delta_2$ degree in $\mathcal{M}$ contains a set that is computably distinguishable within $\mathcal{M}$.
Lemma. If $\mathcal{M}$ contains no PA degree, then $\mathcal{M}$'s version of the Cantor space $2^\omega$ is computably homeomorphic to $\mathcal{M}$'s version of the Baire space $\omega^\omega$.
Proof. Fix a computable binary tree with no path, and name its leaf nodes $\tau_1,\tau_2,\ldots$.  Every $f \in 2^\omega$ is a concatenation of infinitely many $\tau_{n_1}\tau_{n_2}\cdots$ and through this maps to the string $n_1n_2\cdots \in \omega^\omega$.
Proof of main claim. Fix a $\Delta_2$ set $D$ in $\mathcal{M}$.  Construct a computable tree $T \subseteq \omega^{<\omega}$ whose unique path $P$ is $D$'s computation function, i.e., the map taking $n$ to the least $s$ for which $D_s\cap[0,n] = D \cap [0,n]$.  Transform $T$ using the Lemma into a computable $T^* \subseteq 2^{<\omega}$.  Then $T^*$ has a unique path $P^* \equiv_T P \equiv_T D$, and this $P^*$ is distinguishable by the procedure which takes $X,Y$ and checks to see if their initial segments are all in $T^*$. $\square$
I'd be curious to know whether your proof of distinguishable-equals-computable really needs $WKL$ in the form of a $PA(X)$ degree, or whether it can be finessed down to a $PA$.  It would narrow the gap if you're looking for a reversal.