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I'm embarrassed to be asking, but: "Write down a set of axioms for the theory of atomless Boolean algebras."

This is Exercise 1.14 in Chapter 9 of "Models and Ultraproducts" by Bell and Slomson. I'm trying to read this on my own. I have no math community except this one.

Clearly we need the axioms for Boolean algebras, and then at least one more. The additional one might say something like: For any non-zero element x, there's a non-zero element y such that 0 < y < x.

But I don't know if that's anywhere close to correct. My further problem is that even if it were, I wouldn't know how to prove I had an axiom system for atomless Boolean algebras. Thanks for any help.

"Clueless in Tucson"

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    $\begingroup$ Two things: 1. No need to apologize, that's what this site is for. 2. For what it's worth, your additional axiom is the one that I would write down as well, and Jech in his 3rd edition of Set Theory, page 79 seems to agree. $\endgroup$ – t.b. Jul 22 '11 at 16:47
  • $\begingroup$ Do the primitives in your axioms for Boolean algebra include the "<" relation? If not you may have to re-phrase your new axiom. $\endgroup$ – GEdgar Jul 22 '11 at 16:53
  • $\begingroup$ Theo -- hi again -- Jech's book isn't part of my library, I'll see if I can get a copy. Thanks. $\endgroup$ – MikeC Jul 22 '11 at 17:14
  • $\begingroup$ GEdgar -- yes, i was aware i was fudging a bit by using stricly less than. i assume there's away around that but i'll have to check to be sure. $\endgroup$ – MikeC Jul 22 '11 at 17:16
  • $\begingroup$ Maybe this link to the second edition on Google Books works for you (the paragraph I'm referring to -- the very first paragraph, beginning on the previous page -- is the one I meant). if you want someone to be notified by your mentioning the name, add an "@" in front of the user name, e.g. @GEdgar (only one person besides the owner of the thread can be notified per comment). $\endgroup$ – t.b. Jul 22 '11 at 17:39
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An atom is a minimal nonzero element. Your axiom says no such minimal element exists. Any model that satisfies the axioms is clearly a Boolean algebra. Could a model that satisfies your axiom have a minimal nonzero element? I would try a proof by contradiction.

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  • $\begingroup$ Jay -- thanks, that makes sense. my background in formal theories bit me. i was thinking of a proof in the sense that every sentence valid in all models is provable from the axioms, but that's not what we're looking for here. $\endgroup$ – MikeC Jul 22 '11 at 17:12
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If we want to axiomatize Boolean algebras in terms of partial order, we want to specify that the order is dense in the usual sense. So a possible additional axiom is $$\forall x \forall y(x<y\implies \exists z(x \lt z \land z \lt y)).$$

But this can be derived from the weaker-seeming axiom $$\forall y(\lnot(y=0) \implies \exists z(0 \lt z \land z \lt y))$$ that you suggested. Thus your axiomatization is perfectly correct and complete (pun).

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  • $\begingroup$ Andre -- Thanks. I guess I'm not as clueless as I thought (maybe). $\endgroup$ – MikeC Jul 22 '11 at 17:17
  • $\begingroup$ @Michael Carroll: You certainly knew what was needed. The theory is a nice early example, because a back-and-forth argument proves $\omega$-categoricity. $\endgroup$ – André Nicolas Jul 22 '11 at 17:38

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