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Let $S\subseteq B$ be a dense subset of a complete Boolean algebra $B$. Is is true that $\sum S = 1$? Jech seems to use this fact several times in his book (e.g. the proof of 7.15) but I have been unable to prove it, if it is true.

Is it true if we tighten the condition to open dense, i.e. whenever non-zero $u \le v$ for $v \in S$, then $ u \in S$ ?

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Suppose $\bigvee_{a \in S} a = b < 1$. Then in particular every element of $S$ is $\leq b$, so no nonzero element of $S$ is $\leq b^c$, because $b \wedge b^c = 0$. So $S$ is not dense.

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Thanks, was easier than I thought! –  Paul Slevin Feb 11 '12 at 1:47
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Yes, the join of any dense subset of a Boolean algebra is $1$, because otherwise it would be bounded by some $b$ less than $1$, and so $\neg b$ would have no elements of the dense set below it, contradicting density.

There is no need to assume that the Boolean algebra is complete.

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Well if B is not complete, is it possible that the dense subset does not have a join? I guess not because 1 is definitely an upper bound and there can be nothing less than it. –  Paul Slevin Feb 11 '12 at 1:49
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My argument did not assume that $\sup(S)$ exists, only that $1$ was not the supremum. This implies $S$ has a lower upper bound than $1$, which I called $b$, and so forth. So you do not need to assume that $\sup(S)$ exists to make the argument. –  JDH Feb 11 '12 at 1:54
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