Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

enter image description here

I do not see how the partition is well-defined.

By definition $A\neq\varnothing\mbox{ mod }D\iff A\notin I_{D}$.

Since D is a maximal filter $A\notin I_{D}\iff A\in D$ .

So $A\neq\varnothing\mbox{ mod }D\iff A\in D$ .

Thus, the first bullet says $\mathcal{A}\subseteq D$ . So in particular if $A,A'\in\mathcal{A}$ , $A\cap A'\in D$ . But the second bullet says for distinct $A,A'$ , we have $A\cap A'\in I_{D}\implies[A\cap A']^{C}\in D$ , which is impossible.

EDIT: $\mathcal G$ is independent modulo $D$ iff for all $l\in\omega$, forall {$g_1,...,g_l$}$\subseteq \mathcal G$ and {$j_1,...,j_l$}$\subseteq \omega$, we have {$n\in\omega:g_k(n)=j_k(n),1\leq k\leq l$}$\neq\varnothing$ mod $D$.

Kramer, Linus; Shelah, Saharon; Tent, Katrin; Thomas, Simon "Asymptotic cones of finitely presented groups". Adv. Math. 193 (2005), no. 1, 142–173.

The paper can be found here - Go to section 4. The "partition" definition is after Lemma 4.4.

share|cite|improve this question
\emptyset and \varnothing are symbols for the empty set. \phi is not. – Asaf Karagila Jul 11 '13 at 22:57
Could you explain what it means for $\mathcal{G} \subseteq \omega^{\omega}$ to be independent modulo a filter $D$ over $\omega$? – hot_queen Jul 11 '13 at 23:12
Maybe it does because $D$ is not unconditionally maximal. – hot_queen Jul 11 '13 at 23:16
What is $j_k(n)$ for $j_k \in \omega$? In any case, I do not see why $D$ should be an ultrafilter. – hot_queen Jul 11 '13 at 23:33
Please do not change the questions in such a drastic way. – Andrés E. Caicedo Apr 5 '14 at 5:58
up vote 1 down vote accepted

As was guessed in some of the comments, $D$ need not be an ultrafilter. The maximality statement in the quoted passage is intended to mean that $D$ is maximal among those filters modulo which $\mathcal G$ is independent. Unless $\mathcal G$ is empty, it cannot be independent modulo an ultrafilter.

To understand the general notion of partition modulo a filter (which makes sense for any filter, whether or not maximal with respect to some $\mathcal G$), it may be best to begin with the special case where $D$ is just $\{\omega\}$. Then the three bullet items say that (1) the partition consists of nonempty sets, (2) distinct sets in the partition are pairwise disjoint, and (3) the sets in the partition cover all of $\omega$ (that's easily equivalent to the literal reading: every nonempty subset of $\omega$ meets some set in the partition). So this is just the familiar notion of a partition of $\omega$.

When $D$ is not just $\{\omega\}$, one more ingredient goes into the definition: You are allowed to restrict attention to any set in $D$; equivalently, you are allowed to ignore anything that happens in a set in $I_D$. So the first bullet item says that the sets in the partition (I'll call them "blocks" for short) are not merely nonempty but remain so even when intersected with any element of $D$. So the effect of $D$ is to make this requirement more stringent than it was in the preceding paragraph. The second bullet item says that any two distinct blocks, though perhaps not actually disjoint, become disjoint when intersected with a suitable set in $D$ (which may depend on the particular blocks in question). So this second requirement has become less stringent. Finally, the third bullet item says that all the blocks together, though they might not cover all of $\omega$, must cover a set in $D$; thus, by restricting attention to that set in $D$, we no longer see any failure of covering. So the third requirement has also become less stringent. Altogether, the definition says that, if you focus on "almost all" of $\omega$, in the sense of $D$, then $\mathcal A$ will look like a partition.

This informal description can be made formal by abstracting a bit to deal with Boolean algebras. Specifically, the Boolean algebra $\mathcal P(\omega)$ of all subsets of $\omega$ has a quotient Boolean algebra $\mathcal P(\omega)/D$, obtained by identifying any two subsets of $\omega$ that agree almost everywhere with respect to $D$. That is, identify $a$ with $b$ if there is $d\in D$ with $a\cap d=b\cap d$. Then a partition $\mathcal A$ modulo $D$ is almost the same as a "partition of $1$" in this quotient algebra, i.e., a family of non-zero, pairwise disjoint elements of $\mathcal P(\omega)/D$ whose join (= supremum) is the top element $1$ of the algebra. The "almost" in the preceding sentence refers to the fact that a partition if $1$ is a family of elements of the quotient algebra $\mathcal P(\omega)/D$, i.e., a family of equivalence classes, while $\mathcal A$ is a set of representatives for such a family, one element from each of the equivalence classes in the family.

share|cite|improve this answer
Once again Boolean algebras save the day! – Asaf Karagila Jul 12 '13 at 2:01
Ok this is a very good answer. A couple of remarks: (1) I guess I should verify that there exists a maximal filter modulo which $\mathcal G$ is independent (probably by a standard Zorn's Lemma application). (2) In the last sentence of paragraph 3, do you mean "in the sense of $a$ $set$ $in$ $D$"? – Forever Mozart Jul 12 '13 at 2:25
@David (1) Yes, the existence of maximal filters modulo which $\mathcal G$ is independent is a straightforward application of Zorn's Lemma; the independence property is preserved when you form unions of nonempty chains of filters. To get started, you need that there is some filter modulo which $\mathcal G$ is independent. That's a condition on $\mathcal G$, equivalent to independence modulo $\{\omega\}$; presumably the paper verifies it for the $\mathcal G$'s that are actually used. – Andreas Blass Jul 12 '13 at 3:19
@David (2) The phrase "almost all of $\omega$ in the sense of $D$" means "all of some set in $D$". – Andreas Blass Jul 12 '13 at 3:21

Your Answer


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.