A closed subset of Baire space not Medvedev equivalent to any closed set of Cantor space

TL;DR

The paper demonstrates that there exists a closed subset of Baire space that is not Medvedev equivalent to any closed subset of Cantor space, answering a question about the relationship between these spaces.

Contribution

It provides a counterexample showing not all closed problems in Baire space are Medvedev equivalent to closed problems in Cantor space.

Findings

Existence of a closed subset of Baire space not Medvedev equivalent to any closed subset of Cantor space.

Counterexample to Shafer's question about Medvedev equivalence between closed sets in Baire and Cantor spaces.

Clarifies the limitations of Medvedev reducibility in relating closed sets across different spaces.

Abstract

For mass problems $P,Q\subseteq {\mathbb{N}^\mathbb{N}}$ (Baire space), $P$ is Medvedev reducible to $Q$ ($P\leq_sQ$) if for some Turing funcional $\Phi$, $\Phi(Q)\subseteq P$, and Medvedev equivalent to $Q$ if also $Q\leq_sP$. Shafer asked if every closed problem $P$ is Medvedev equivalent to a closed problem $Q$ with $Q\subseteq 2^\mathbb{N}$ (Cantor space). We show that this is not the case.