What the Vitali set, Bernstein sets and free ultrafilters cost in choice

Choice enters this page in three genuinely different ways.

First, Assuming choice on the cosets of $\mathbb{Q}$ in $\mathbb{R}$, a Vitali set in $[0,1]$ exists uses a selector on the family of rational-equivalence classes meeting $[0,1]$. The later theorem that a Vitali set is nonmeasurable uses only countably many translates of an already chosen selector, so the cost is concentrated in the existence step, not in the measure argument.

Second, Assuming the real line can be well ordered, a Bernstein set exists uses a well-order of the real line and a transfinite construction through the perfect subsets. That is a different cost from the Vitali selector: the page isolates it because a Bernstein set is built by repeatedly choosing fresh points from a well-ordered development, not by one choice function on one fixed family.

Third, A free ultrafilter on $\mathbb{N}$, viewed as a subset of $\{0,1{\}}^{\mathbb{N}}$ and hence of $[0,1]$, is not Lebesgue measurable is intentionally one-directional. It proves what follows from being given a free ultrafilter, namely nonmeasurability; it does not produce a free ultrafilter. The existence cost is recorded elsewhere in What the ultrafilter lemma costs: a choice principle strictly weaker than AC.

The published remarks Solovay's model: ZF + DC with every set of reals measurable ‡ and Shelah 1984: the inaccessible is needed for measurability, not for the Baire property ‡ explain why none of these pathologies can be read as consequences of ZF + DC alone: relative to the stated consistency hypotheses, ZF + DC can coexist with all sets of reals being measurable, and with all sets of reals having the Baire property.