How much choice for free ultrafilters?

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In ZF set theory, the ultrafilter theorem is known to imply the Axiom of Choice (AC), but it remains uncertain if the converse is false, specifically if there exists a model of ZF where the ultrafilter theorem holds without AC. The discussion explores whether weaker forms of AC, such as countable choice, could still imply the ultrafilter theorem. Additionally, participants consider the implications of the ultrafilter theorem on countable or dependent choice and its relevance to constructing nonstandard analysis (NSA). It is suggested that countable choice might suffice for NSA, particularly due to the compactness theorem's role in logic. The conversation emphasizes the need for further investigation into these relationships.
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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?
 
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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.
 
phoenixthoth said:
Sorry this wasn't more helpful.

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.
 
CRGreathouse said:
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?

The following link offers some discussion related to your questions.

http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist_2006;task=show_msg;msg=2629

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".
 
fopc said:
The following link offers some discussion related to your questions.

http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist_2006;task=show_msg;msg=2629

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".

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).
 
The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive, can we find an alternative operator, _A' , so that ( H_A)_A' makes sense? Isn't Pearson Neyman related to this? Hope I'm making sense. Edit: I was motivated by a superficial similarity of the idea with double transposition of matrices M, with ## (M^{T})^{T}=M##, and just wanted to see if it made sense to talk...

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