# If X is a subset of a complete boolean algebra, must there be a finite $Y \subset X$ s.t. sup(Y) = sup(X)?

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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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 6 '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 7 '12 at 5:06