A Bounded Finite-One Degree Whose One-One Degrees Form Exactly a Dense Linear Order

Abstract

We construct a set $U\leq_T\emptyset''$ whose bounded finite-one degree, ordered internally by one-one reducibility, consists exactly of a countable dense linear order without endpoints. More precisely, \[ \left( \{[B]_1:B\equiv_{\mathrm{bfo}}U\},\leq_1 \right) \cong (\mathbb Q,\leq). \] This exact realization contrasts with two previous results. In earlier work, a bounded finite-one degree was constructed that contains a copy of $(\mathbb Q,\leq)$, but the same degree also contains an infinite antichain of one-one degrees and, more generally, embedded copies of all countable partial orders; thus the dense chain does not exhaust the degree. In a different direction, $m$-rigidity yields an almost-sure and comeager obstruction: for a measure-$1$ and comeager class of sets, the corresponding bounded finite-one degree contains an infinite antichain of one-one degrees and hence is not linearly ordered. The present construction shows that, despite this typical negative behaviour, exact dense linear order can occur. In particular, it gives an affirmative answer to Open Question~3 of Richter, Stephan, and Zhang. The proof has two main parts. A block homogenization construction produces a noncylindrical set $U$, a weak dyadic tower $(U_e)_{e\in\mathbb N}$, computable reservoirs of both colours, and a base absorption property. An abstract absorption-to-exhaustivity theorem then shows that every member of the bounded finite-one degree of $U$ is one-one equivalent to some finite autojoin $mU_e$, thereby yielding the exhaustive classification above.