Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Background/motivation: I'm investigating the construction of models for a first-order modal system (S5) as products of classical models. Since ultraproducts are all classical models and I need non-classical ones as well, I need to look at reduced products where the filter is not an ultrafilter. This leads me to ask about filters in general:

J.L. Bell & A.B. Slomson, in Models and Ultraproducts (p. 116), state and prove:

Lemma 1.17. Let I be a countable set. Then the collections of non-principal, $\omega$-incomplete, uniform, and regular ultrafilters on I all coincide.

Suppose I alter their definitions slightly so the above properties are all defined for filters in general, then modify the lemma to assert that it holds for filters in general. Would that be true? Can anyone supply a reference to a proof or disproof? Thanks.

share|improve this question
1  
In the statement of Lemma 1.17, "complete" should be "incomplete. Also, when you "alter the definitions slightly" do you just mean to change "ultrafilter" to "filter" in Bell and Slomson's definitions, or do you envision other slight changes? For example, many people define "non-principal" to mean "contains the complement of every finite set." That's equivalent to the definition in Bell and Slomson for ultrafilters but not for filters in general. –  Andreas Blass Nov 21 '10 at 22:47
    
Andreas - I changed "complete" to "incomplete"; thanks. -- [B&S] already define a "principal" filter as one generated by a fixed non-empty subset of I; let's say a filter is "non-principal" if it's not principal. Let's say a filter F is "uniform" if all its members are of the same cardinality as I. It's "$\omega$-complete" if it contains the intersection of any countable collection of its members, and "$\omega$-incomplete" otherwise. "Regular" is a bit complicated to define in a comment. –  MikeC Nov 21 '10 at 23:45
add comment

1 Answer 1

up vote 3 down vote accepted

If you use literally the definitions in Bell and Slomson, only changing "ultrafilter" to "filter," and if, as in the lemma you cited, you're interested only in flters on a countable set, then I believe non-principal is equivalent to $\omega$-incomplete, while "regular" is strictly stronger and "uniform" is strictly weaker. Unfortunately, I don't have time right now to check this carefully, so I hope someone will object loudly if I've messed it up.

Now that I have a bit more time, let me add the counterexamples that justify "strictly". Partition $\omega$ into two infinite pieces $A$ and $B$. The principal filter $F_0$ generated by $A$ is uniform; that establishes the second "strictly" above. For the first, let $U$ be a nonprincipal ultrafilter that contains $B$, and let $F_1=U\cap F_0$. Then $F_1$ is nonprincipal (the intersection of all the sets in it is $A$, which isn't in it), but it is not regular. (A function $f$ as in Bell and Slomson's definition of "regular" on page 114 would have to send each $a\in A$ to a finite set $f(a)$ that contains all elements $j$ of $\omega$, a contradiction.)

share|improve this answer
    
Thanks, that's very interesting. I expected the answer to be simpler. –  MikeC Nov 21 '10 at 23:43
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.