Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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$ ?

share|cite|improve this question
up vote 4 down vote accepted

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.

share|cite|improve this answer
Thanks, was easier than I thought! – Paul Slevin Feb 11 '12 at 1:47

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.

share|cite|improve this answer
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
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.