6
$\begingroup$

In the paper "Łoś's theorem and the boolean prime ideal theorem imply axiom of choice" Howard has shown that Łoś's theorem and the boolean prime ideal theorem imply axiom of choice. At the end of the paper, author remarks that it was not known at the time whether BPIT can be removed from their argument, but he mentions that if there are no free ultrafilters on any set, then Łoś's theorem (pretty trivially) holds, while choice doesn't, so we might be unable to remove BPIT.

Indeed, two years later Blass has shown that it is consistent with ZF that there are no free ultrafilters on any set, and hence Łoś's theorem doesn't imply axiom of choice.

This argument shows that in order to deduce full axiom of choice from Łoś's theorem, we need some ultrafilters. Here comes my question:

Is it known whether Łoś's theorem implies full axiom of choice in conjuntion with:

There exists a free ultrafilter

or

There exists a free ultrafilter on every infinite set

Thanks in advance

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ That's an incredibly tough question to answer. If the answer is at all positive, this would require a very different proof (since just a simple reduction to Howard's proof is likely to be caught in the many years since that paper was published); if the answer is negative, this would require coming up with brand new models in which LT's holds and the axiom of choice fails (because I don't know if we even know of such models except Blass' model without ultrafilters). $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 19, 2015 at 1:10
  • $\begingroup$ @AsafKaragila Thanks for the reply. I hoped that there would have been some "simple reduction" over these years which I just couldn't have find be searching. Feel free to post this as an answer. $\endgroup$
    – Wojowu
    Commented Dec 19, 2015 at 7:47
  • $\begingroup$ In fact Howard applied only a special form of BPI, sometimes called the ultrafilter lemma, which claims that any filter over any set can be extended to an ultrafilter. $\endgroup$
    – Vladimir Kanovei
    Commented Mar 11, 2017 at 4:56

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .