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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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    @Michael: Your notification arrived! Congratulations for the first ping! The [FAQ is here](http://math.stackexchange.com/faq) (link top right of the screen), but I don't think it mentions the notifications. I don't really use Google books for reading, but for looking things up quickly, its very useful. However, Google likes to hide the things I *really* want to look at, which makes it of somewhat limited use... I prefer printed books as well!2011-07-22

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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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    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.2011-07-22
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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

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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    @Michael Carroll: You certainly knew what was needed. The theory is a nice early example, because a back-and-forth argument proves $\omega$-categoricity.2011-07-22