Definable Hamel bases and $AC_ω(R)$

Abstract

There is a model of ZF with a ∆ 1 3 definable Hamel bases in which ACω(R) fails.Answering a question from [13, p. 433] it was shown in [2] that there is a Hamel basis in the Cohen-Halpern-Lévy model. In this paper we show that in a variant of this model, there is a projective, in fact ∆ 1 3 , Hamel basis. Throughout this paper, by a Hamel basis we always mean a basis for R, construed as a vector space over Q. We denote by E 0 the Vitali equivalence relation, xE 0 y iff x − y ∈ Q for x, y ∈ R. We also write [x] E0 = {y : yE 0 x} for the E 0 -equivalence class of x. A transversal for the set of all E 0 -equivalence classes picks exactly one member from each [x] E0 . The range of any such transversal is also called a Vitali set. If we identify R with the Cantor space ω 2, then xE 0 y iff {n :at most countable for every null set N ⊂ R ("null" in the sense of Lebesgue measure). A set B ⊂ R is a Bernstein set iff B ∩ P = ∅ = P \ B for every perfect set P ⊂ R. A Burstin basis is a Hamel basis which is also a Bernstein set. It is easy to see that B ⊂ R is a Burstin basis iff B is a Hamel basis and B ∩ P = ∅ for every perfectBy AC ω (R) we mean the statement that for all sequences (A n : n < ω) such that ∅ = A n ⊂ R for all n < ω there is some choice function f : ω → R, i.e., f (n) ∈ A n for all n < ω.D. Pincus and K. Prikry study the Cohen-Halpern-Lévy model H in [13]. The model H is obtained by adding a countable set of Cohen reals (say over L) without adding their enumeration; H does not satisfy AC ω (R). It is shown in [13] that there is a Luzin set in H, so that in ZF, the existence of a Luzin set does not even imply AC ω (R). [2, Theorems 1.7 and 2.1] show that in H there is a Bernstein set as well as a Hamel basis. As in ZF the existence of a Hamel basis implies the existence of a Vitali set, the latter also reproves Feferman's result (see [13]) according to which there is a Vitali set in H.