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.
Maximal set of pairwise disjoint elements of a dense subset.
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set-theory
boolean-algebra
1 Answers
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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.
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0Thanks. R$e$vising for my $e$xam and hav$e$ forgotten a $fe$w things :) – 2012-05-01