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The question came up when I was trying to prove the compactness of the stone space S(B) of a complete boolean algebra B. Using only the basic facts regarding ultrafilters and boolean algebras, I cannot seem to find an answer.

Thank you very much, in advance.

John Toh
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The property in title seems require compactness. However, it seems not every complete boolean algebra is compact. there is a counterexample. Let $B=\wp(\omega)$, $X=\omega$. Note that $n=\{0,1,...,n-1\}$.

Then $\sup(X)=\bigcup(X)=\omega$, but for every finite $Y\subseteq X$, $\sup(Y)$ is definitely a finite number.

Popopo
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  • Hi, thanks for the reply. Yep I guess my approach was wrong. Would it be correct if I said that, given a subset X of B s.t. X is pairwise compatible (every pair has a non-zero meet), then there exists a filter of B containing X? – John Toh Nov 06 '12 at 17:17
  • @JohnToh Sure, $B$ itself is. Moreover ${z \in B|\bigwedge Y \le z \mbox{ for some finite collection } Y \mbox{ of } X}$ is the smallest one and which is also proper. – Popopo Nov 07 '12 at 05:06