# Axioms for atomless Boolean algebras

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"

• 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. – t.b. Jul 22 '11 at 16:47
• Do the primitives in your axioms for Boolean algebra include the "<" relation? If not you may have to re-phrase your new axiom. – GEdgar Jul 22 '11 at 16:53
• Theo -- hi again -- Jech's book isn't part of my library, I'll see if I can get a copy. Thanks. – MikeC Jul 22 '11 at 17:14
• 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. – MikeC Jul 22 '11 at 17:16
• 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). – t.b. Jul 22 '11 at 17:39

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).
• @Michael Carroll: You certainly knew what was needed. The theory is a nice early example, because a back-and-forth argument proves $\omega$-categoricity. – André Nicolas Jul 22 '11 at 17:38