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Apologies if this has been asked somewhere before, but I didn't see what I was looking for after several pages of Google results.

I was reading Jech's The Axiom of Choice and was introduced to the equivalence of the ultrafilter lemma and the compactness theorem for FOL. I gather that these are also equivalent to the completeness theorem. I am primarily interested in trying to understand the "ultrafilter lemma $\implies$ completeness/compactness" direction. Jech's own proof takes an odd route through "binary messes" and I don't really follow the last couple of steps. (It's on pp.17,18 of the 2008 Dover reprint, if one is interested).

I'm wondering if anyone knows a good reference for, or can provide, a readable proof that the ultrafilter lemma implies completeness or compactness, with a preference for completeness. Alternatively, if you have Jech's book on hand, helping me bridge the concluding paragraph of his $\textrm{(ii)}\implies\textrm{(iii)}$ step with the preceding steps would clear up much of my puzzlement. (Though in this case I can't offer any help in saying what, specifically, is missing in the way of a bridge).

Martin Sleziak
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Malice Vidrine
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  • For Compactness, a suitable ultraproduct of models of finite subsets is a model. – André Nicolas May 08 '14 at 02:43
  • @Andre - I've seen the ultraproduct proof of compactness before, but I wasn't sure I could show the product to be nonempty in the absence of Choice. Is this not as much of a problem as I think it is? – Malice Vidrine May 08 '14 at 02:58
  • You are right, it seems to be a problem. As someone who has used ultraproducts to prove things, I am insensitive to implicit uses of Choice. – André Nicolas May 08 '14 at 03:03
  • No worries; I have the dubious advantage of being oblivious to many standard theorems of $\mathsf{ZFC}$, due to excessive focus on $\mathsf{NF}$. – Malice Vidrine May 08 '14 at 03:09
  • @Andre: The conjunction of Compactness+Los Theorem is equivalent the axiom of choice over $\sf ZF$. – Asaf Karagila May 08 '14 at 07:09
  • Related: http://math.stackexchange.com/questions/402543/is-the-compactness-theorem-from-mathematical-logic-equivalent-to-the-axiom-of Let me see if I can find another reference to the one suggested by Mike. – Asaf Karagila May 08 '14 at 07:10
  • Incidentally, I got over my hump in Jech's proof. Nonetheless, I will try to remember to accept Asaf or Mike's answer when I can get my hands on the publications mentioned in their answer. – Malice Vidrine May 17 '14 at 02:20

2 Answers2

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I think you'll find what you're looking for in J.L. Bell and A.B. Slomson, "Models and Ultraproducts: an introduction" (Dover, 1969) in Chapter 5 Section 5: "The completeness theorem, the axiom of choice, and the ultrafilter theorem."

MikeC
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I didn't find many actual proofs of this equivalence. Henkin's original proof was announced on page 389 of the following paper, but doesn't seem to appear in press (interestingly enough the meeting itself had more than a handful of submissions related to the Prime Boolean Ideal theorems).

J. W. Green, The May meeting in Yosemite, Bull. Amer. Math. Soc. 60 (1954), no. 4, 386--399.

I have found in Halbeisen's Combinatorial Set Theory the same proof that you would find in Jech (in terms of equivalence, and going through the term "binary mess"). However Halbeisen writes in a different manner from Jech, and you might find it more readable there.

Lorenz J. Halbeisen, Combinatorial Set Theory: With a Gentle Introduction to Forcing, Springer Monographs in Mathematics, Springer 2011.

More specifically, this is Theorem 5.15 which appears on page 122. (You can also find a free copy on Halbeisen's homepage, although page numbering might differ there.)

Asaf Karagila
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