arXiv · 2602.19960
Rigid many-one degrees contain infinite antichains of $1$-degrees
Abstract
Odifreddi asked whether every non-irreducible many-one degree must contain an infinite antichain of one-one degrees. Positive answers are known for computably enumerable many-one degrees (Degtev) and, more recently, for many-one degrees admitting a $\Delta^0_2$ representative (Batyrshin). In this note we isolate a rigidity principle behind these phenomena. Call a set $A\subseteq\omega$ \emph{$m$-rigid} if every total computable $m$-autoreduction of $A$ is eventually the identity. We prove that if $A$ is $m$-rigid, then its many-one degree $\deg_m(A)$ contains an infinite antichain of $1$-degrees. The proof uses a uniform duplication construction: for each computable parameter $S$ we define $B_S\equiv_m A$ so that any injective reduction $B_S\le_1 B_T$ induces an $m$-autoreduction of $A$ and therefore forces $S\subseteq^{*}T$. Choosing an almost-inclusion infinite antichain of computable sets yields the desired infinite $1$-antichain inside $\deg_m(A)$. As applications, Jockusch's rigidity theorem implies that every $1$-generic set is $m$-rigid, giving a comeager family of positive instances. Moreover, $m$-rigidity holds with Lebesgue measure $1$ (indeed, every Martin-L\"of random real is $m$-rigid). Consequently, Odifreddi's Question~5 has a positive answer \emph{with probability $1$} for a fair-coin random $A\in 2^\omega$; any counterexample (if it exists) is confined to a null set (and, by genericity, also to a meager set).
Explore related subjects
Keep this discovery
Explore connections, maps & timelines
Patrizio Cintioli. 2026-02-23. Rigid many-one degrees contain infinite antichains of $1$-degrees. https://arxiv.org/abs/2602.19960
Cite the original work for its findings. Save a collection to share your selection of sources.