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How much choice for free ultrafilters?

  1. Jul 26, 2007 #1


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    In ZF, AC implies the ultrafilter theorem (every Boolean algebra has a free ultrafilter).

    • Is the converse known to be false? That is, is there a model of ZF where the ultrafilter theorem is true and AC is false?
    • Does some weaker version of AC (countable choice, for example) imply the ultrafilter theorem?
    • Does the ultrafilter theorem imply countable or dependent choice?
    • In particular, how much choice is needed to build nonstandard analysis? Can it be done with less than AC?
  2. jcsd
  3. Jul 29, 2007 #2
    My guess is a countable choice would be enough for NSA, though I'm not sure. I'm guessing that because the compactness theorem in logic gives NSA and it's only needed in a countable case. (If [x]P(x) means there exists an x such that P(x) is true, and
    S={[x]x<1, [x]x<1/2, [x]x<1/3, [x]x<1/4,...} has a model for every finite subset of S and so S has a model.) I really don't know if that has anything to do with what you're asking...

    Sorry this wasn't more helpful.
  4. Jul 29, 2007 #3


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    Are you kidding? That's great! It gives me an angle to start from, and it looks like it solves the biggest part of my question off the bat.

    I'm going to look into this in more detail; I'll post again if I find something.
  5. Jul 31, 2007 #4
    The following link offers some discussion related to your questions.


    There's a link provided in the above to a page by Eric Schechter that's worth a read.
    Also, you might check out his book on "...analysis and its foundations".
  6. Jul 31, 2007 #5


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    Thanks for the link! That really seems to answer the question. Now I only need the weaker ultrafilter theorem, but I think the two may be ZF-equivalent. I'll look some more into that.

    I've been Schechter's site, but somehow I missed that page (or that part).
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