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Over any model M of $PA^-+I\Sigma_1^0$. Suppose $A\in [T]$ where $T$ is a $\Delta_2^0$-tree and $A$ is one isolated path. Further, $A$ is regular, i.e. $\forall n A\upharpoonright n$ has a code in $M$. A is easily seen to be $\Delta_2^0$. My question is does there exist a nice recursive approximation $\langle A_s\rangle$ of A such that $\forall n \exists t \forall s>t A_s\upharpoonright n =A\upharpoonright n$? This property could be expected for any $\Delta_2^0$ set in any $PA^-+B\Sigma_2^0$ model. However, I'm wondering if being an isolated path is enough restriction to force this nice property in $PA^-+I\Sigma_1^0$.

Edit: There might be some further restriction on the tree that I left out. Otherwise as François pointed out the restriction is vacuous. Let's say $T=\{\tau\in 2^{<M}: \forall u< |\tau| K(\tau\upharpoonright u)<K(u)+b\}$ where $K$ is the prefix-free Kolmogorov Complexity (it could be defined in models of fragments of arithmetic) and b is a constant. Actually, it corresponds to the K-trivial set via constant $b$ as in the standard case.

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  • $\begingroup$ Isn't every regular $\Delta^0_2$ set the unique branch through a $\Delta^0_2$-tree? $\endgroup$ Commented Jan 26, 2014 at 1:57
  • $\begingroup$ @FrançoisG.Dorais: You are right. Obviously I left out some details thinking that some restriction could be relaxed. See my edit. $\endgroup$
    – Jing Zhang
    Commented Jan 26, 2014 at 5:28

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