Re: Axiom of Multiple Choice

Arturo Magidin <magidin@xxxxxxxxxxxxxxxxx> wrote:
>
> In searching for a proof that "Every vector space has a basis" implies
> the Axiom of Choice, Jose Carlos Santos steered me towards a paper of
> J.D. Alpern that proves that "In every vector space every spanning set
> contains a basis" implies the Axiom of Choice; and a paper of
> Andreas Blass that notes that Halpern is using a weaker axiom system
> than ZF, and where he proves that "Every vector space has a basis"
> implies what he terms the "Axiom of Multiple Choice":
>
> AoMC: For every nonempty family of nonempty sets, there is a
> function assigning to each set in the family a finite
> nonempty subset of itself.
>
> Blass then mentions that ZF |- (AoMC --> AoC).
>
> Why does this assertion not imply that in ZF every family of finite
> nonempty sets has a choice function?
>
> Can someone sketch or direct me to a proof of AoMC => AoC?

You might find the answer here (I don't have time to check)

Levy, A. Axioms of multiple choice. Fund. Math. 50, 1962, 475-483
http://matwbn.icm.edu.pl/ksiazki/fm/fm50/fm5040.pdf

--Bill Dubuque
.

Relevant Pages
Quantcast