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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.

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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.

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    @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.2012-11-07