Does $\mathsf{DC}$ imply $\mathsf{AC}_ω$, uniformly?
Alessandro Andretta, Lorenzo Notaro
TL;DR
The paper investigates whether dependent choice DC can uniformly imply countable choice AC_ω across all sets X. It demonstrates that AC_ω(ℝ) suffices to obtain AC_ω(X) from DC(X) in many cases, but it is consistent with ZF that there exists A ⊆ ℝ with DC(A) while AC_ω(A) fails, established via an ω-long iteration of symmetric extensions starting from a Cohen model. The construction shows A is not separable and contains no nonempty perfect subset, yet DC(A) holds, highlighting a separability/topology aspect of the counterexample. The work further analyzes the behavior of DC under finite unions, contrasts with the Feferman-Levy model where such a counterexample cannot occur, and discusses definability concerns and higher descriptive-set-theoretic implications for the counterexample.
Abstract
The Axiom of Dependent Choice $\mathsf{DC}$ and the Axiom of Countable Choice $\mathsf{AC}_ω$ are two weak forms of the Axiom of Choice that can be stated for a specific set: $\mathsf{DC}(X)$ asserts that any total binary relation on $X$ has an infinite chain, while $\mathsf{AC}_ω(X)$ asserts that any countable collection of nonempty subsets of $X$ has a choice function. It is well-known that $\mathsf{DC} \Rightarrow \mathsf{AC}_ω$. We study for which sets and under which hypotheses $\mathsf{DC}(X) \Rightarrow \mathsf{AC}_ω(X)$, and then we show it is consistent with $\mathsf{ZF}$ that there is a set $A \subseteq \mathbb{R}$ for which $\mathsf{DC} (A)$ holds, but $\mathsf{AC}_ω(A)$ fails.