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The question I have is more of a curiosity, and that is why I decided to post here instead of Mathoverflow.

Before posing the question, let me set up some background.

Background: Let $\Omega$ be a Boolean algebra. For the purposes of this post a measure $\nu$ on $\Omega$ is a finitely additive map $\Omega\to A$ with values in a linear space $A$. If $A$ is a commutative, unital algebra, $\nu$ is multiplicative or spectral if $\nu(E\cap F)= \nu(E)\nu(F)$ and $\nu(1)= 1$. I will denote by $\mathbf{A}(\Omega, V)$ the linear space of $V$-measures on $\Omega$ and by $\mathbf{AS}(\Omega, A)$ the set of spectral measures.

note: the field of scalars is irrelevant; take the real field for the sake of determinateness.

Theorem 1: The functor $V\mapsto \mathbf{A}(\Omega, V)$ is representable, that is, there is a linear space $S(\Omega)$ and a natural isomorphism $\mathbf{A}(\Omega, V)\cong \mathbf{Vect}(S(\Omega), V)$.

$S(\Omega)$ is the space of simple elements. Denote by $\chi$ the universal measure $\Omega\to S(\Omega)$ and call elements of the form $\chi(E)$ characteristic.

note: By Stone duality, $S(\Omega)$ is isomorphic to the linear space of linear combinations of characteristic functions of clopen sets of the Stone space of $\Omega$. Thus the monikers "simple" and "characteristic elements". Stone duality needs the Boolean prime ideal theorem.

We can introduce a commutative unital algebra structure on $S(\Omega)$ by $\chi(E)\chi(F)= \chi(E\cap F)$ and then extending linearly (or juggling around the universal property). $\chi$ is now spectral and in fact the universal spectral measure.

Theorem 2: The functor $A\mapsto \mathbf{AS}(\Omega, A)$ is naturally isomorphic to $\mathbf{CAlg}(S(\Omega), A)$.

In fact, we have a functor $\mathbf{CAlg}\to \mathbf{Bool}$ that sends a commutative, unital algebra to the Boolean algebra of its idempotents. It is now easy to see that this functor is right adjoint to $S$ and, in particular, $S$ is cocontinuous.

Problem: An easy application of the Boolean prime ideal theorem (BPI for short) implies that if $E\in \Omega$ is non-zero then $\chi(E)$ is likewise non-zero. In particular, $S(\Omega)$ is non-trivial ($0\neq 1$). Conversely, it is not difficult to see that if $1\neq 0$ then if $E\in \Omega$ is non-zero then $\chi(E)$ is non-zero. My question is can the non-triviality of $S(\Omega)$ be proved without recourse to the BPI? Say, in ZF alone? ZF + P with some P weaker than BPI?

note: BPI is weaker than AC by a well-known result of Halpern and Levy.

Here is what I managed to do until now. Theorem 2 guarantees that as long as we have a non-trivial spectral measure on $\Omega$, non-triviality of $S(\Omega)$ drops out. $S$ of the initial Boolean algebra (the non-trivial one constituted solely by the bottom and top elements) is the scalar field, thus, if we have an ultrafilter on $\Omega$ we can prove that $S(\Omega)$ is non-trivial. In particular, if $\Omega$ has even just one atom we are done. But I am at a loss as how to proceed in the case of arbitrary atomless algebras. If we can factor some morphism into a Boolean algebra with non-trivial $S$ then the problem would also be solved, but this is really not an advancement over constructing ultrafilters directly.

At this point, and after not managing to get non-triviality of $S(\Omega)$ without BPI I started wondering if non-triviality of $S(\Omega)$ is actually equivalent to BPI, but the only thing I accomplished is to narrow down to the problem of constructing one algebra morphism from $S(\Omega)$ to the real field -- not much, that is.

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Now when you say simple, in what sort of context do you mean that? Also, when you say $\Omega$ is a Boolean algebra, and $\nu(E\cap F)$, do you mean $E\cdot F$? –  Asaf Karagila Oct 13 '11 at 0:08
    
Also, BPI is equivalent to Stone's Representation Theorem, so it's more than just "needs BPI", it is in fact equivalent to it. So it seems to me that you're asking whether or not there is a known characterization of this families of Boolean algebras, such that the restriction of BPI to this family is equivalent to some nicely put choice principle? –  Asaf Karagila Oct 13 '11 at 0:10
    
@Asaf Karagila: For your first comment, $\cap$ is binary infimum in the Boolean algebra and concatenation is multiplication in the target algebra. The "simple" adjective is just borrowing the measure theoretic terminology of simple functions = linear combinations of characteristic functions -- the note justifies the borrowing. For your second comment, yes you are correct BPI is equivalent to Stone's representation theorem. I am unable to parse your reformulation of my question though; is the question not clear? I have added a couple of sentences to, hopefully, make it clearer. –  G. Rodrigues Oct 13 '11 at 1:01
    
I see. I am used to multiplicative notations. Either way, I am not too comfortable with the categorical feeling I get from the question. I'm not sure I can answer it, but I'll give it some thought. –  Asaf Karagila Oct 13 '11 at 9:48
    
Regarding the first sentence of the question, I don't believe there is any incompatibility between curiosity and mathoverflow, and nor should there be. In particular, I would find this question to be on-topic for MO. Questions about the sufficiency for a purpose of strictly intermediate weak forms of the axiom of choice certainly are the right level for MO. –  JDH Dec 17 '11 at 23:59

1 Answer 1

up vote 3 down vote accepted

The existence of non-trivial finitely additive measures on arbitrary non-degenerate Boolean algebras (and therefore the non-triviality of $S(\Omega)$) is known to be strictly weaker than the Boolean prime ideal theorem but not provable in ZF alone. If I remember correctly, it's also known to be equivalent to the Hahn-Banach theorem. (The standard reference for such questions is the book "Consequences of the Axiom of Choice" by Paul Howard and Jean Rubin, along with its associated web site.)

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Thanks, you are indeed correct. The existence of non-trivial finitely additive measures is equivalent to Hahn-Banach. This is a result of Luxemburg in a paper of 1969. The proof can be found in E. Shecter's "Handbook of Functional Analysis". And Hahn-Banach is weaker than BPI (Pincus?). –  G. Rodrigues Dec 27 '12 at 13:07
    
One correction and one clarification on my previous comment. The name of Schecter's book is "Handbook of analysis and its foundations". That BPI implies HB is a result of Luxemburg. That the converse does not hold is a result of Pincus in "Independence of the prime ideal theorem from the Hahn Banach theorem". The paper contains the references to the cited Luxemburg papers. It is available online but I cannot seem to insert a link to it, so just google or go to projecteuclid and search there. –  G. Rodrigues Dec 29 '12 at 13:49

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