DMC, Multiple Choice, and AC qualifications

DC implies DMC. DMC together with finite-selection countable choice implies DC. In $\mathrm{Z}\mathrm{F}$, Multiple Choice is equivalent to AC, but that equivalence must not be imported into $\mathrm{Z}\mathrm{F}\mathrm{A}$; the cited permutation-model strictness results are $\mathrm{Z}\mathrm{F}\mathrm{A}$ qualifications only. No strict DMC-versus-DC claim is made over $\mathrm{Z}\mathrm{F}$.

• The implication and the equivalence. DC and finite multiple selections proves over $\mathrm{Z}\mathrm{F}$ that $\mathrm{D}\mathrm{C}\leftrightarrow (\mathrm{D}\mathrm{M}\mathrm{C}\wedge {\mathrm{A}\mathrm{C}}_{\omega ,\mathrm{f}\mathrm{i}\mathrm{n}})$, which gives both the implication DC to DMC and the converse implication from DMC plus finite-selection countable choice; the principles are those of The axiom of dependent choice: a relation in which every element is related to something admits an $\mathbb{N}$-indexed chain and Dependent multiple choice in finite-level tree form.

• Multiple choice in ZF. Multiple choice is equivalent to AC in ZF proves $\mathrm{M}\mathrm{C}\leftrightarrow \mathrm{A}\mathrm{C}$ in $\mathrm{Z}\mathrm{F}$ (Multiple choice and dependent multiple choice, The Axiom of Choice). That is a theorem about $\mathrm{Z}\mathrm{F}$; it says nothing about $\mathrm{Z}\mathrm{F}\mathrm{A}$, where the same sentence is not available at this point in the library.

• The ZFA qualification. The separations known for DMC are obtained in permutation models with atoms; they are therefore theorems about $\mathrm{Z}\mathrm{F}\mathrm{A}$ and are not transferred to $\mathrm{Z}\mathrm{F}$ by this item. In particular the strictness of DMC below DC over $\mathrm{Z}\mathrm{F}$ is not asserted, and the only ZF-side nonprovability recorded here is If ZF is consistent, DMC is not provable in ZF, conditional on $\mathrm{Con}(\mathrm{Z}\mathrm{F})$.

• Consumers. Any use of "DMC is weaker than DC" must name the theory: the $\mathrm{Z}\mathrm{F}\mathrm{A}$ model supplies the separation there, while over $\mathrm{Z}\mathrm{F}$ the question is open, as recorded by the dated status item of this page.