4
$\begingroup$

Blass (1984) shows that the existence of Hamel basis for arbitrary vector space over any field implies the axiom of choice. However such implication needs the axiom of regularity. As in Blass' article, existence of basis just implies axiom of multiple choice, strictly weaker than AC when without assuming regularity.

In the article he says whether the existence of basis implies the axiom of choice in $\mathsf{ZF - regularity}$ remains open. However it has been 31 years since the paper published and I wonder there is a progress about it. I would appreciate your answer.

$\endgroup$
  • 2
    $\begingroup$ I'm not aware of any result removing the need for regularity here. $\endgroup$ – Andreas Blass Dec 13 '15 at 10:31
  • $\begingroup$ So, @Asaf, have you found every question with "Hamel basis" in it now? $\endgroup$ – Gerry Myerson Aug 12 '16 at 23:14
  • 1
    $\begingroup$ @Gerry: I found every question tagged with the tag [hamel-basis] now. And since I already removed the tag from all the questions earlier today, it only made sense to do that again today (and if you'll check, you'll see that all these questions had the tag added about the same time, so none of these were my bumps). $\endgroup$ – Asaf Karagila Aug 12 '16 at 23:19
  • $\begingroup$ @Asaf, indeed. Gives tag-team wrestling a whole different meaning. $\endgroup$ – Gerry Myerson Aug 13 '16 at 4:55
8
$\begingroup$

As far as I know, no real progress has been made.

Two papers worth mentioning, however, are these:

  1. Paul Howard, If vector spaces are projective modules then multiple choice holds, MLQ Math. Log. Q. 51 (2005), no. 2, 187--190.

  2. Marianne Morillon, Linear forms and the axiom of choice, Comment. Math. Univ. Carolin. 50 (2009), no. 3, 421--431.

Neither proves anything significant regarding the question whether or not the existence of bases implies choice in $\sf ZFA$ (equivalently, $\sf ZF-Reg$). But both make something which can be considered progress towards such answer, even if that progress is not a big step forward.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.