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

Let $B$ be a complete Boolean algebra, and suppose $D \subseteq B$ is a dense subset. How is it possible to construct a maximal set of pairwise disjoint elements of $D$? Is it true that $\sum S = \sum D$ ? I want to do this since it is used frequently to construct partitions of $B$. Any help would be appreciated.

share|cite|improve this question
up vote 1 down vote accepted

You know that $\bigvee D=1$. If $\bigvee S=s<1$, there is a $d\le\lnot s$ in $D$, and $S\cup\{d\}$ is a pairwise disjoint subset of $D$ strictly larger than $S$.

To get a maximal pairwise disjoint subset of $D$, I'd simply appeal to the axiom of choice.

share|cite|improve this answer
Thanks. Revising for my exam and have forgotten a few things :) – Paul Slevin May 1 '12 at 1:23

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.