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Suppose Q is an atomless countable boolean algebra, and B is an arbitrary atomless boolean algebra. Q is unique modulo isomorphisms. There is a subalgebra in B that is isomorphic to Q. There is probably a mapping from B to Q that preserves all boolean operations, but I need something different. Let f be an epimorphism (cover) from B to Q with the following properties:

1) If b1≤b2 then f(b1)≤f(b2). (previously there was requirement that ∧ should be preserved, but actually I don't need it)

2) f(b1∨b2)=f(b1)∨f(b2)

3) f(b)=0 iff b=0

4) f(1)=1

The negation operation does not need to be preserved.

Does f exist for any B?

Example:

B = {periodic sequences of non-negative reals with integer period}⊆ 2{r≥0}

Q = {periodic sequences of non-negative integers}⊆ 2{k≥0}

f(α)={i : [i,i+1)∩α≠∅}

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First sentence: you should add that Q is also atomless (this was confusing me for a little while, and explains the second sentence). –  John Goodrick Dec 18 '09 at 21:05
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3 Answers

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The answer to the revised version of the question is Yes. In fact, there is no need to assume that B is atomless, but rather, only that it is infinite.

Suppose that B is any infinite Boolean algebra. It follows that there is a countable maximal antichain A subset B. The idea of the proof is to map A arbitrarily into your countable atomless Boolean algebra Q, and then extend to B in a way I will describe. Enumerate the maximal antichain A = { an | n in ω } and the nonzero elements of Q as { qn | n in ω}. We will associate an with qn. In order to define f, suppose that b is any element of B. Let Ab = { qn | b ∧ an not = 0 } be the associated set in Q. Define the function f:B to Q by f(b) = ∨ Ab, if Ab is finite, and otherwise f(b)=1.

We now make several observations about this function f. First, the function is clearly onto, since f(an)=qn. Also, f(1)=1, since 1 meets every element of A, and f(0)=0 since 0 meets no elements of A. Moreover, f(b)=0 iff b=0, since no nonzero element of B has zero meet with every element of A, as A was a maximal antichain.

Because (b ∨ c) ∧ a = (b ∧ a) ∨ (c ∧ a), it follows that Ab ∨ c = Ab union Ac. From this, it follows that f(b ∨ c) = f(b) ∨ f(c), since if either set is infinite, then the answer is 1, and if they are finite, we are taking the join of two finite joins. Thus, f is join-preserving.

It follows that f is an order-homomorphism, since b <= c implies b ∨ c = c implies f(b) ∨ f(c) = f(b ∨ c) = f(c) implies f(b) <= f(c).

So f has all the desired properties.

Note that f definitely does not respect negation, since f(neg an) = 1 for every n. And f definitely does not respect meet, since any two elements of A have meet 0, but the corresponding qn must sometimes be nonzero.

This construction has some affinity with your example. Namely, if you take the various half-open unit characteristic functions as the elements of the maximal antichain (and use the corresponding qn's), then your f and my construction are the same.

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Yeah, this seems to satisfy all requirements, though still not as structure-preserving as I need. Good starting point though, thanks. –  Grue Dec 23 '09 at 12:03
    
Thanks for accepting the answer. What other properties on f did you want? If you really want to handle every Boolean algebra, then it might not be possible to have much more. I say this because I found this answer in part by looking at particular Boolean algebras that I thought might be a counterexample to your property, and I proved for them that every f with your conditions is locally constant. That is, for these B every f has a maximal antichain A below whose elements f is constant. This suggested the solution I gave above. –  Joel David Hamkins Dec 23 '09 at 19:22
    
These problematic Boolean algebras are those corresponding for the forcing to add a Cohen subset to omega_1. This is the canonical method of forcing the Continuum Hypothesis. Such a B has a dense set that is countably closed (every countable descending sequence in the dense set has a nonzero lower bound). –  Joel David Hamkins Dec 23 '09 at 19:24
    
Joel, not every infinite Boolean algebra has a countably infinite maximal antichain! An example is $\mathcal P(\omega)/fin$ since no countably infinite family of almost disjoint subsets of $\omega$ is maximal almost disjoint. –  Stefan Geschke Aug 29 '10 at 10:34
    
Stefan, yes, I was thinking of complete Boolean algebras, which do always have countable maximal antichains. –  Joel David Hamkins Aug 29 '10 at 11:56
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Edit: This is the answer to the original question. See my other answer for the answer to the revised question.


The answer is that every such f is an isomorphism. Thus, there is such an f only when B is isomorphic to Q.

The main reason is that your conditions imply that f must preserve negation. That is, for any b in B, then

b ^ -b = 0, so you have 0 = f(0) = f(b ^ -b) = f(b) ^ f(-b),

and

b v -b = 1, so you have 1 = f(1) = f(b v -b) = f(b) v f(-b).

So f(b) and f(-b) are complements in Q, and thus f(-b) = -f(b) in Q.

Thus, f preserves all the Boolean algebra structure. Next, since you have f(b)=0 IFF b=0, it means that f must be an isomorphism.

In particular, there is no such f for your example B and Q, since B is uncountable.

If you relax that IFF, then it just means that Q is a quotient of B by the filter F=f^inverse(1).

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Ok, looks like I made a mistake. Actually I only need the order preserved, not &and; (I thought those were equivalent). The example must work, that's the whole reason for this question. The "iff" part is also important. I changed the requirement 1), how about this revised question? –  Grue Dec 18 '09 at 21:23
    
Preserving join implies preserving the order, since a is less than or equal to b iff a v b =b, so this would give f(a) v f(b) = f(b), which means f(a) is less than or equal to f(b). –  Joel David Hamkins Dec 18 '09 at 22:10
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Long comment (and I understand that the question was asked a while ago):

Your guess in the first paragraph is wrong: Not every infinite Boolean algebra admits a Boolean homomorphism onto the unique countable atomless Boolean algebra. An example is the power set algebra of the natural numbers.

Why? By Stone-duality, a Boolean homomorphism from $B$ to $Q$ corresponds to a continuous embedding from the Cantor space $2^{\mathbb N}$ into the Stone space $X$ of $B$. The Cantor space is infinite and metric and hence has a non-trivial convergent sequence. If $B$ is the powerset of $\mathbb N$, then $X$ is the Stone-Cech compactification of the discrete space $\mathbb N$, which (the compactification) does not have a non-trivial convergent sequence (this fact is quite shocking when you first hear about it, but it is well-known). In this case, there cannot be an embedding of the Cantor space into $X$, since such an embedding would have to preserve the nontrivial convergent sequence.

The argument actually can be extended to show the following: No infinite complete BA maps (by a homomorphism) onto any countably infinite BA.

Also note my comment saying that Joel's answer has a problem: Not every infinite BA has a countably infinite maximal antichain. I don't know at the moment whether this can be fixed. I will think about it. As Joel points out in his comment, everything works fine, i.e., there are maximal countably infinite antichains in every infinite complete Boolean algebra.

My argument above does not rule out a positive solution to your question in general since your are looking for something weaker than a homomorphism.
And by Joel's comment and the original answer, the maps that you are interested in actually exist for infinite complete Boolean algebras $B$.

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Stefan, note that my argument is fixed by using the completion of $B$, and then restricting back to $B$. –  Joel David Hamkins Aug 29 '10 at 12:12
    
I agree. Thanks for clearing this up. –  Stefan Geschke Aug 30 '10 at 5:48
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