1. Jul 26, 2007

### CRGreathouse

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. Jul 29, 2007

### phoenixthoth

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...

3. Jul 29, 2007

### CRGreathouse

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.

4. Jul 31, 2007